chapter 4 three major classes of chemical · pdf file · 2014-10-26chapter 4 three...

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CHAPTER 4 THREE MAJOR CLASSES OF CHEMICAL REACTIONS FOLLOW–UP PROBLEMS 4.1A Plan: Examine each compound to see what ions, and how many of each, result when the compound is dissolved

in water and match one compound’s ions to those in the beaker. Use the total moles of particles and the molar ratio in the compound’s formula to find moles and then mass of compound. Solution: a) LiBr(s) 2H O→ Li+(aq) + Br–(aq)

Cs2CO3(s) 2H O→ 2Cs+(aq) + CO32–(aq)

BaCl2(s) 2H O→ Ba2+(aq) + 2Cl–(aq) Since the beaker contains +2 ions and twice as many –1 ions, the electrolyte is BaCl2.

b) Mass (g) of BaCl2 = ( )2

2 2 22 2

2

1 mol BaCl 208.2 g BaCl0.05 mol Ba ions3 Ba particles1 mol BaCl1 Ba particle 1 mol Ba ions

++

+ +

= 31.2 g BaCl2 4.1B Plan: Write the formula for sodium phosphate and then write a balanced equation showing the ions that result

when sodium phosphate is placed in water. Use the balanced equation to determine the number of ions that result when 2 formula units of sodium phosphate are placed in water. The molar ratio from the balanced equation gives the relationship between the moles of sodium phosphate and the moles of ions produced. Use this molar ratio to calculate the moles of ions produced when 0.40 mol of sodium phosphate is placed in water. Solution: a) The formula for sodium phosphate is Na3PO4. When the compound is placed in water, 4 ions are produced for each formula unit of sodium phosphate: three sodium ions, Na+, and 1 phosphate ion, PO4

3–. Na3PO4(s) 2H O→ 3Na+(aq) + PO4

3–(aq) If two formula units of sodium phosphate are placed in water, twice as many ions should be produced: 2Na3PO4(s) 2H O→ 6Na+(aq) + 2PO4

3–(aq) Any drawing should include 2 phosphate ions (each with a 3– charge) and 6 sodium ions (each with a 1+ charge).

b) Moles of ions = 0.40 mol Na3PO4 � 4 mol ions1 mol Na3PO4

� = 1.6 moles of ions

4.2A Plan: Write an equation showing the dissociation of one mole of compound into its ions. Use the given

information to find the moles of compound; use the molar ratio between moles of compound and moles of ions in the dissociation equation to find moles of ions.

Solution: a) One mole of KClO4 dissociates to form one mole of potassium ions and one mole of perchlorate ions.

KClO4(s) 2H O→ K+(aq) + ClO4–(aq)

Therefore, 2 moles of solid KClO4 produce 2 mol of K+ ions and 2 mol of ClO4– ions.

b) Mg(C2H3O2)2(s) 2H O→ Mg2+(aq) + 2C2H3O2–(aq)

First convert grams of Mg(C2H3O2)2 to moles of Mg(C2H3O2)2 and then use molar ratios to determine the moles of each ion produced.

Moles of Mg(C2H3O2)2 = ( ) 2 3 2 22 3 2 2

2 3 2 2

1 mol Mg(C H O )354 g Mg(C H O )

142.40 g Mg(C H O )

= 2.48596 mol

The dissolution of 2.48596 mol Mg(C2H3O2)2(s) produces 2.49 mol Mg2+ and (2 x 2.48596) = 4.97 mol C2H3O2

–

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c) (NH4)2CrO4(s) 2H O→ 2NH4+(aq) + CrO4

2–(aq) First convert formula units to moles.

Moles of (NH4)2CrO4 = ( )24 4 2 423

1 mol (NH ) CrO1.88 x 10 FU

6.022 x 10 FU

= 3.121886 mol

The dissolution of 3.121886 mol (NH4)2CrO4(s) produces (2 x 3.121886) = 6.24 mol NH4+ and 3.12 mol CrO4

2–

. 4.2B Plan: Write an equation showing the dissociation of one mole of compound into its ions. Use the given

information to find the moles of compound; use the molar ratio between moles of compound and moles of ions in the dissociation equation to find moles of ions.

Solution: a) One mole of Li2CO3 dissociates to form two moles of lithium ions and one mole of carbonate ions.

Li2CO3(s) 2H O→ 2Li+(aq) + CO32–(aq)

Therefore, 4 moles of solid Li2CO3 produce 8 mol of Li+ ions and 4 mol of CO32– ions.

b) Fe2(SO4)3(s) 2H O→ 2Fe3+(aq) + 3SO42–(aq)

First convert grams of Fe2(SO4)3 to moles of Fe2(SO4)3 and then use molar ratios to determine the moles of each ion produced.

Moles of Fe2(SO4)3= (112 g Fe2(SO4)3) � 1 mol Fe2(SO4)3399.88 g Fe2(SO4)3

� = 0.2801 = 0.280 mol Fe2(SO4)3

The dissolution of 0.280 mol Fe2(SO4)3(s) produces 0.560 mol Fe3+ (2 x 0.2801) and 0.840 mol SO42–

(3 x 0.2801) c) Al(NO3)3(s) 2H O→ Al3+(aq) + 3NO3

–(aq) First convert formula units of Al(NO3)3 to moles of Al(NO3)3 and then use molar ratios to determine the moles of each ion produced.

Moles of Al(NO3)3= (8.09 x 1022 formula units Al(NO3)3) � 1 mol Al(NO3)36.022 x 1023formula units Al(NO3)3

�

= 0.1343 mol Al(NO3)3 The dissolution of 0.134 mol Al(NO3)3 (s) produces 0.134 mol Al3+ and 0.403 mol NO3

– (3 x 0.1343) 4.3A Plan: Convert the volume from mL to liters. Convert the mass to moles by dividing by the molar mass of KI.

Divide the moles by the volume in liters to calculate molarity. Solution:

Amount (moles) of KI = 6.97 g KI � 1 mol KI166.0 g KI

� = 0.0420 mol KI

M = �0.0420 mol KI100. mL

� �1000 mL1 L

� = 0.420 M 4.3B Plan: Convert the volume from mL to liters. Convert the mass to moles by first converting mg to g and then

dividing by the molar mass of NaNO3. Divide the moles by the volume in liters to calculate molarity. Solution:

Amount (moles) of NaNO3 = 175 mg NaNO3 � 1 g1000 mg

� � 1 mol NaNO3

85.00 g NaNO3� = 0.00206 mol NaNO3

M = �0.00206 mol NaNO3

15.0 mL� �1000 mL

1 L � = 0.137 M

4.4A Plan: Divide the mass of sucrose by its molar mass to change the grams to moles. Divide the moles of sucrose by

the molarity to obtain the volume of solution.

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Solution: Volume (L) of solution =

( ) 12 22 1112 22 11

12 22 11 12 22 11

1 mol C H O 1 L135 g C H O342.30 g C H O 3.30 mol C H O

= 0.11951 = 0.120 L

Road map:

Divide by M (g/mol) Divide by M (mol/L) 4.4B Plan: Convert the volume from mL to L; then multiply by the molarity of the solution to obtain the moles of

H2SO4. Solution:

Amount (mol) of H2SO4 = 40.5 mL H2SO4 � 1 L1000 mL

� �0.128 mol H2SO4

1 L� = 0.00518 mol H2SO4

Road map:

1000 mL = 1 L Multiply by M

(1 L soln = 0.128 mol H2SO4) 4.5A Plan: Multiply the volume and molarity to calculate the number of moles of sodium phosphate in the solution.

Write the formula of sodium phosphate. Determine the number of each type of ion that is included in each formula unit. Use this information to determine the amount of each type of ion in the described solution.

Solution:

Amount (mol) of Na3PO4 = 1.32 L �0.55 mol Na3PO4

1 L� = 0.7260 = 0.73 mol Na3PO4

In each formula unit of Na3PO4, there are 3 Na+ ions and 1 PO43– ion.

Amount (mol) of Na+ = 0.73 mol Na3PO4 � 3 mol Na+

1 mol Na3PO4� = 2.2 mol Na+

Amount (mol) of PO43– = 0.73 mol Na3PO4 � 1 mol PO4

3-

1 mol Na3PO4� = 0.73 mol PO4

3–

Mass (g) of sucrose

Amount (moles) of sucrose

Volume (L) of solution

Volume (mL) of soln

Volume (L) of soln

Amount (mol) H2SO4

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4.5B Plan: Write the formula of aluminum sulfate. Determine the number of aluminum ions in each formula unit. Calculate the number of aluminum ions in the sample of aluminum sulfate. Convert the volume from mL to L. Divide the number of moles of aluminum ion by the volume in L to calculate the molarity of the solution.

Solution: In each formula unit of Al2(SO4)3, there are 2 Al3+ ions.

Amount (mol) of Al3+ = 1.25 mol Al2(SO4)3 � 2 mol Al3+

1 mol Al2(S04)3� = 2.50 mol Al3+

M = �2.50 mol Al3+

875 mL� �1000 mL

1 L � = 2.86 M

4.6A Plan: Determine the new volume from the dilution equation (Mconc)(Vconc) = (Mdil)(Vdil). Solution:

Vdil = MconcVconc

Mdil =

(4.50 M)(60.0 mL)(1.25 M)

= 216 mL

4.6B Plan: Determine the new concentration from the dilution equation (Mconc)(Vconc) = (Mdil)(Vdil). Convert the

molarity (mol/L) to g/mL in two steps (one step is moles to grams, and the other step is L to mL). Solution:

( )( )3

conc concdil 3

dil

7.50 M 25.0 m

500.m= =

M VM

V = 0.375 M

Concentration (g/mL) =3

2 4 2 4

2 4

0.375 mol H SO 98.08 g H SO 10 L1 L 1 mol H SO 1 mL

−

= 0.036780 = 3.68x10–2 g/mL solution 4.7A Plan: Count the number of particles in each solution per unit volume.

Solution: Solution A has 6 particles per unit volume while Solution B has 12 particles per unit volume. Solution B is more concentrated than Solution A. To obtain Solution B, the total volume of Solution A was reduced by half:

( )( )( )

dil dilconc

conc

6 particles 1.0 mL12 particles

= =N V

VN

= 0.50 mL

Solution C has 4 particles and is thus more dilute than Solution A. To obtain Solution C, ½ the volume of solvent must be added for every volume of Solution A:

( )( )( )

conc concdil

dil

6 particles 1.0 mL4 particles

= =N V

VN

= 1.5 mL

4.7B Plan: Count the number of particles in each solution per unit volume. Determine the final volume of the solution.

Use the dilution equation, (Nconc)(Vconc) = (Ndil)(Vdil), to determine the number of particles that will be present when 300. mL of solvent is added to the 100. mL of solution represented in circle A. (Because M is directly proportional to the number of particles in a given solution, we can replace the molarity terms in the dilution equation with terms representing the number of particles.) Solution: There are 12 particles in circle A, 3 particles in circle B, 4 particles in circle C, and 6 particles in circle D. The concentrated solution (circle A) has a volume of 100. mL. 300. mL of solvent are added, so the volume of the diluted solution is 400. mL.

Ndil = NconcVconc

Vdil =

(12 particles)(100.0 mL)(400. mL)

= 3 particles

There are 3 particles in circle B, so circle B represents the diluted solution.

4.8A Plan: Determine the ions present in each substance on the reactant side and write new cation-anion combinations. Use Table 4.1 to determine if either of the combinations of ions is not soluble. If a precipitate forms there will be

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a reaction and chemical equations may be written. The molecular equation simply includes the formulas of the substances and balancing. In the total ionic equation, all soluble substances are written as separate ions. The net ionic equation comes from the total ionic equation by eliminating all substances appearing in identical form (spectator ions) on each side of the reaction arrow.

Solution: a) The resulting ion combinations that are possible are iron(III) phosphate and cesium chloride. According to Table 4.1, iron(III) phosphate is insoluble, so a reaction occurs. We see that cesium chloride is soluble.

Total ionic equation: Fe3+(aq) + 3Cl–(aq) + 3Cs+(aq) + PO4

3–(aq) → FePO4(s) + 3Cl–(aq) + 3Cs+(aq) Net ionic equation: Fe3+(aq) + PO4

3–(aq) → FePO4(s) b) The resulting ion combinations that are possible are sodium nitrate (soluble) and cadmium hydroxide (insoluble). A reaction occurs.

Total ionic equation: 2Na+(aq) + 2OH–(aq) + Cd2+(aq) + 2NO3

–(aq) → Cd(OH)2(s) + 2Na+(aq) + 2NO3–(aq)

Note: The coefficients for Na+ and OH– are necessary to balance the reaction and must be included. Net ionic equation: Cd2+(aq) + 2OH–(aq) → Cd(OH)2(s)

c) The resulting ion combinations that are possible are magnesium acetate (soluble) and potassium bromide (soluble). No reaction occurs.

4.8B Plan: Determine the ions present in each substance on the reactant side and write new cation-anion combinations.

Use Table 4.1 to determine if either of the combinations of ions is not soluble. If a precipitate forms there will be a reaction and chemical equations may be written. The molecular equation simply includes the formulas of the substances and balancing. In the total ionic equation, all soluble substances are written as separate ions. The net ionic equation comes from the total ionic equation by eliminating all substances appearing in identical form (spectator ions) on each side of the reaction arrow.

Solution: a) The resulting ion combinations that are possible are silver chloride (insoluble, an exception) and barium nitrate (soluble). A reaction occurs.

Total ionic equation: 2Ag+(aq) + 2NO3

–(aq) + Ba2+(aq) + 2Cl–(aq) → 2AgCl(s) + Ba2+(aq) + 2NO3–(aq)

Net ionic equation: Ag+(aq) + Cl–(aq) → AgCl(s)

b) The resulting ion combinations that are possible are ammonium sulfide (soluble) and potassium carbonate (soluble). No reaction occurs. c) The resulting ion combinations that are possible are lead(II) sulfate (insoluble, an exception) and nickel(II) nitrate (soluble). A reaction occurs.

Total ionic equation: Ni2+(aq) + SO4

2–(aq) + Pb2+(aq) + 2NO3–(aq) → PbSO4(s) + Ni2+(aq) + 2NO3

–(aq) Net ionic equation: Pb2+(aq) + SO4

2–(aq) → PbSO4 (s)

4.9A Plan: Look at the ions (number and charge) produced when each of the given compounds dissolves in water and find the match to the ions shown in the beaker. Once the ions in each beaker are known, write new cation-anion combinations and use Table 4.1 to determine if any of the combination of ions is not soluble. If a precipitate forms there will be a reaction and chemical equations may be written. The molecular equation simply includes the formulas of the substances and must be balanced. In the total ionic equation, all soluble substances are written as separate ions. The net ionic equation comes from the total ionic equation by eliminating all substances appearing in identical form (spectator ions) on each side of the reaction arrow. Solution: a) Beaker A has four ions with a +2 charge and eight ions with a –1 charge. The beaker contains dissolved Zn(NO3)2 which dissolves to produce Zn2+ and NO3

– ions in a 1:2 ratio. The compound PbCl2 also has a +2 ion and –1 ion in a 1:2 ratio but PbCl2 is insoluble so ions would not result from this compound.

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b) Beaker B has three ions with a +2 charge and six ions with a –1 charge. The beaker contains dissolved Ba(OH)2 which dissolves to produce Ba2+ and OH– ions in a 1:2 ratio. Cd(OH)2 also has a +2 ion and a –1 ion in a 1:2 ratio but Cd(OH)2 is insoluble so ions would not result from this compound. c) The resulting ion combinations that are possible are zinc hydroxide (insoluble) and barium nitrate (soluble). The precipitate formed is Zn(OH)2. The spectator ions are Ba2+ and NO3

–. Balanced molecular equation: Zn(NO3)2(aq) + Ba(OH)2(aq) → Zn(OH)2(s) + Ba(NO3)2(aq) Total ionic equation: Zn2+(aq) + 2NO3

–(aq) + Ba2+(aq) + 2OH– (aq) → Zn(OH)2(s) + Ba2+(aq) + 2NO3– (aq)

Net ionic equation: Zn2+(aq) + 2OH–(aq) → Zn(OH)2(s) d) Since there are only six OH– ions and four Zn2+ ions, the OH– is the limiting reactant.

Mass of Zn(OH)2= (6 OH– ions) �0.050 mol OH−

1 OH− particle � �1 mol Zn(OH)2

2 mol OH− � �99.40 g Zn(OH)2 1 mol Zn(OH)2

�

= 14.9100 = 15 g Zn(OH)2 4.9B Plan: Look at the ions (number and charge) produced when each of the given compounds dissolves in water and

find the match to the ions shown in the beaker. Once the ions in each beaker are known, write new cation-anion combinations and use Table 4.1 to determine if any of the combination of ions is not soluble. If a precipitate forms there will be a reaction and chemical equations may be written. The molecular equation simply includes the formulas of the substances and must be balanced. In the total ionic equation, all soluble substances are written as separate ions. The net ionic equation comes from the total ionic equation by eliminating all substances appearing in identical form (spectator ions) on each side of the reaction arrow. Solution: a) Beaker A has eight ions with a +1 charge and four ions with a –2 charge. The beaker contains dissolved Li2CO3 which dissolves to produce Li+ and CO3

2–ions in a 2:1 ratio. The compound Ag2SO4 also has a +1 ion and –2 ion in a 2:1 ratio but Ag2SO4 is insoluble so ions would not result from this compound. b) Beaker B has three ions with a +2 charge and six ions with a –1 charge. The beaker contains dissolved CaCl2 which dissolves to produce Ca2+ and Cl– ions in a 1:2 ratio. Ni(OH)2 also has a +2 ion and a –1 ion in a 1:2 ratio but Cd(OH)2 is insoluble so ions would not result from this compound. c) The resulting ion combinations that are possible are calcium carbonate (insoluble) and lithium chloride (soluble). The precipitate formed is CaCO3. The spectator ions are Li+ and Cl–. Balanced molecular equation: Li2CO3(aq) + CaCl2(aq) → CaCO3(s) + 2LiCl(aq) Total ionic equation:

2Li+(aq) + CO32– (aq) + Ca2+(aq) + 2Cl– (aq) → CaCO3(s) + 2Li+(aq) + 2Cl– (aq)

Net ionic equation: Ca2+(aq) + CO3

2–(aq) → CaCO3 (s) d) Since there are only four CO3

2– ions and three Ca2+ ions, the Ca2+ is the limiting reactant.

Mass of CaCO3= (3 Ca2+ ions) � 0.20 mol Ca2+

1 Ca2+ particle � �1 mol CaCO3

1 mol Ca2+ � �100.09 g CaCO3

1 mol CaCO3� = 60.0540 = 60. g CaCO3

4.10A Plan: We are given the molarity and volume of calcium chloride solution, and we must find the volume of sodium

phosphate solution that will react with this amount of calcium chloride. After writing the balanced equation, we find the amount (mol) of calcium chloride from its molarity and volume and use the molar ratio to find the amount (mol) of sodium phosphate required to react with the calcium chloride. Finally, we use the molarity of the sodium phosphate solution to convert the amount (mol) of sodium phosphate to volume (L).

Solution: The balanced equation is: 3CaCl2(aq) + 2Na3PO4(aq) Ca3(PO4)2(s) + 6NaCl(aq) Finding the volume (L) of Na3PO4 needed to react with the CaCl2:

Volume (L) of Na3PO4 = 0.300 L CaCl2 �0.175 moles CaCl21 L

� �2 mol Na3PO4

3 mol CaCl2� � 1 L

0.260 mol Na3PO4�

= 0.1346 = 0.135 L Na3PO4 4.10B Plan: We are given the mass of silver chloride produced in the reaction of silver nitrate and sodium chloride, and

we must find the molarity of the silver nitrate solution. After writing the balanced equation, we find the amount

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(mol) of silver nitrate that produces the precipitate by dividing the mass of silver chloride produced by its molar mass and then using the molar ratios from the balanced equation. Then we calculate the molarity by dividing the moles of silver nitrate by the volume of the solution (in L). The balanced equation is: AgNO3(aq) + NaCl(aq) AgCl(s) + NaNO3(aq)

Amount (mol) of AgNO3 = 0.148 g AgCl � 1 mol AgCl143.4 g AgCl

� �1 mol AgNO3

1 mol AgCl� = 0.001032 = 1.03x10–3 mol AgNO3

Molarity (M) of AgNO3 = �1.03 x 10–3mol AgNO3 45.0 mL

� �1000 mL1 L

� = 2.29 x 10–2 M

4.11A Plan: Multiply the volume in liters of each solution by its molarity to obtain the moles of each reactant. Write a

balanced equation. Use molar ratios from the balanced equation to determine the moles of lead(II) chloride that may be produced from each reactant. The reactant that generates the smaller number of moles is limiting. Change the moles of lead(II) chloride from the limiting reactant to the grams of product using the molar mass of lead(II) chloride.

Solution: (a)The balanced equation is: Pb(C2H3O2)2(aq) + 2NaCl(aq) → PbCl2(s) + 2NaC2H3O2(aq)

Moles of Pb(C2H3O2)2 = ( )3

2 3 2 21.50 mol Pb(C H O )10 L268 mL1 mL 1 L

−

= 0.402 mol Pb(C2H3O2)2

Moles of NaCl = ( )310 L 3.40 mol NaCl130 mL

1 mL 1 L

−

= 0.442 mol NaCl

Moles of PbCl2 from Pb(C2H3O2)2 = ( ) 22 3 2 2

2 3 2 2

1 mol PbCl0.402 mol Pb(C H O )

1 mol Pb(C H O )

= 0.402 mol PbCl2

Moles of PbCl2 from NaCl = ( ) 21 mol PbCl0.442 mol NaCl

2 mol NaCl

= 0.221 mol PbCl2

The NaCl is limiting. The mass of PbCl2 may now be determined using the molar mass.

Mass (g) of PbCl2 = ( ) 22

2

278.1 g PbCl0.221 mol PbCl

1 mol PbCl

= 61.4601 = 61.5 g PbCl2

(b) Ac is used to represent C2H3O2: Amount (mol) Pb(Ac)2 + 2NaCl → PbCl2 + 2NaAc Initial 0.402 0.442 0 0 Change – 0.221 – 0.442 +0.221 +0.442 Final 0.181 0 0.221 0.442 4.11B Plan: Write a balanced equation. Multiply the volume in liters of each solution by its molarity to obtain the

moles of each reactant. Use molar ratios from the balanced equation and the molar mass of iron(III) hydroxide to determine the mass (g) of iron(III) hydroxide that may be produced from each reactant. The smaller mass is the amount of product actually made.

Solution: (a)The balanced equation is: FeCl3(aq) + 3NaOH(aq) → 3NaCl(aq) + Fe(OH)3(s) Mass(g) of Fe(OH)3 produced from FeCl3 =

155 mL FeCl3 � 1 L1000 mL

� �0.250 mol FeCl31 L

� �1 mol mol Fe(OH)31 mol FeCl3

� � 106.87 g Fe(OH)31 mol mol Fe(OH)3

� = 4.14 g Fe(OH)3

Mass(g) of Fe(OH)3 produced from NaOH =

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215 mL NaOH � 1 L1000 mL

� �0.300 mol NaOH1 L

� �1 mol mol Fe(OH)33 mol NaOH

� � 106.87 g Fe(OH)31 mol mol Fe(OH)3

� = 2.30 g Fe(OH)3

NaOH produces the smaller amount of product, so it is the limiting reagent and 2.30 g of Fe(OH)3 are produced. (b) To complete the reaction table, we need the moles of FeCl3 and NaOH that react. Remember that NaOH is the

limiting reagent and will be consumed in this reaction. Amount (mol) of FeCl3 =

155 mL FeCl3 � 1 L1000 mL

� �0.250 mol FeCl31 L

� = 0.0388 mol FeCl3

Amount (mol) of NaOH =

215 mL NaOH � 1 L1000 mL


� = 0.0645 mol NaOH

Amount (mol) FeCl3 + 3NaOH → 3NaCl + Fe(OH)3 Initial 0.0388 0.0645 0 0 Change – 0.0215 –0.0645 +0.0645 +0.0215 Final 0.0173 0 0.0645 0.0215 4.12A Plan: Convert the given volume from mL to L and multiply by the molarity (mol/L) to find moles of KOH. Each

mole of the strong base KOH will produce one mole of hydroxide ions. Finally, multiply the amount (mol) of hydroxide ions by Avogadro’s number to calculate the number of hydroxide ions produced.

Solution: KOH(s) 2H O→ K+(aq) + OH–(aq)

No. of OH– ions produced = 451 mL NaOH � 1 L1000 mL

� �1.20 mole KOH1 L

� � 1 mol OH-

1 mole KOH� �6.022 x 10 23 OH- ions

1 mole OH- � = 3.2591 x 1023 = 3.26 x 1023 OH– ions

4.12B Plan: Convert the given volume from mL to L and multiply by the molarity (mol/L) to find moles of HCl. Each

mole of the strong acid HCl will produce one mole of hydrogen ions. Finally, multiply the amount (mol) of hydrogen ions by Avogadro’s number to calculate the number of hydrogen ions produced.

Solution: HCl(g) 2H O→ H+(aq) + Cl–(aq)

No. of H+ ions produced = 65.5 mL HCl � 1 L1000 mL

� �0.722 mole HCl1 L

� � 1 mol H+

1 mole HCl� �6.022 x 10 23 H+ ions

1 mole H+ � = 2.8479 x 1022 = 2.85 x 1022 H+ ions

4.13A Plan: According to Table 4.2, both reactants are strong and therefore completely dissociate in water. Thus, the key

reaction is the formation of water. The other product of the reaction is soluble. Solution: Molecular equation: 2HNO3(aq) + Ca(OH)2(aq) → Ca(NO3)2(aq) + 2H2O(l) Total ionic equation: 2H+(aq) + 2NO3

–(aq) + Ca2+(aq) + 2OH–(aq) → Ca2+(aq) + 2NO3–(aq) + 2H2O(l)

Net ionic equation: 2H+(aq) + 2OH–(aq) → 2H2O(l) which simplifies to H+(aq) + OH–(aq) → H2O(l) 4.13B Plan: According to Table 4.2, both reactants are strong and therefore completely dissociate in water. Thus, the key

reaction is the formation of water. The other product of the reaction is soluble. Solution: Molecular equation: HI(aq) + LiOH(aq) → LiI(aq) + H2O(l) Total ionic equation: H+(aq) + I–(aq) + Li+(aq) + OH–(aq) → Li+(aq) + I–(aq) + H2O(l) Net ionic equation: H+(aq) + OH–(aq) → H2O(l)

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4.14A Plan: The reactants are a weak acid and a strong base. The acidic species is H+ and a proton is transferred to OH– from the acid. The only spectator ion is the cation of the base. Solution: Molecular equation: 2HNO2(aq) + Sr(OH)2(aq) → Sr(NO2)2(aq) + 2H2O(l) Ionic equation: 2HNO2(aq) + Sr2+(aq) + 2OH–(aq) → Sr2+(aq) + 2NO2

– (aq) + 2H2O(l) Net ionic equation: 2HNO2(aq) + 2OH–(aq) → 2NO2

– (aq) + 2H2O(l) or HNO2(aq) + OH–(aq) → NO2

– (aq) + H2O(l)

HNO2(aq) + OH-(aq) NO2-(aq) + H2O(l)

The salt is Sr(NO2)2, strontium nitrite, and the spectator ion is Sr2+. 4.14B Plan: The reactants are a strong acid and the salt of a weak base. The acidic species is H3O+ and a proton is

transferred to the weak base HCO3– to form H2CO3, which then decomposes to form CO2 and water.

Solution: Molecular equation: 2HBr(aq) + Ca(HCO3)2(aq) → CaBr2(aq) + 2H2CO3(aq) Ionic equation: 2H3O+(aq) + 2Br–(aq) + Ca2+(aq) + 2HCO3

–(aq) → 2CO2(g) + 4H2O(l) + 2Br–(aq) + Ca2+(aq) Net ionic equation: 2H3O+(aq) + 2HCO3

–(aq) → 2CO2(g) + 4H2O(l) or H3O+(aq) + HCO3–(aq) → CO2(g) + 2H2O(l)

H3O+(aq) + HCO3-(aq) CO2(g) + 2H2O(l)

The salt is CaBr2, calcium bromide. 4.15A Plan: Write a balanced equation. Determine the moles of HCl by multiplying its molarity by its volume, and,

through the balanced chemical equation and the molar mass of aluminum hydroxide, determine the mass of aluminum hydroxide required for the reaction.

Solution: The balanced equation is: Al(OH)3(s) + 3HCl(aq) → AlCl3(aq) + 3H2O(l)

Mass(g) of Al(OH)3 = 3.4 x 10–2 L HCl �0.10 mol HCl1 L

� �1 mol Al(OH)33 mol HCl

� �78.00 g Al(OH)3

1 mol Al(OH)3�

= 0.08840 = 0.088 g Al(OH)3 4.15B Plan: Write a balanced equation. Determine the moles of NaOH by multiplying its molarity by its volume, and,

through the balanced chemical equation and the molar mass of acetylsalicylic acid, determine the mass of acetylsalicylic acid in the tablet.

Solution: The balanced equation is: HC9H7O4(aq) + NaOH(aq) → NaC9H7O4(aq) + H2O(l)

Mass(g) of HC9H7O4= 14.10 mL NaOH � 1 L1000 mL


� �1 mol HC9H7O4

1 mol NaOH� �180.15 g HC9H7O4

1 mol HC9H7O4�

= 0.3251 = 0.325 g HC9H7O4

4.16A Plan: Write a balanced chemical equation. Determine the moles of HCl by multiplying its molarity by its volume, and, through the balanced chemical equation, determine the moles of Ba(OH)2 required for the reaction. The amount of base in moles is divided by its molarity to find the volume.

Solution: The molarity of the HCl solution is 0.1016 M. However, the molar ratio is not 1:1 as in the example problem. According to the balanced equation, the ratio is 2 moles of acid per 1 mole of base:

2HCl(aq) + Ba(OH)2(aq) → BaCl2(aq) + 2H2O(l)

4-10

Volume (L) of Ba(OH)2 = ( )3

2

2

1 mol Ba(OH)10 L 0.1016 mol HCl L50.00 mL1 mL L 2 mol HCl 0.1292 mol Ba(OH)

−

= 0.0196594 = 0.01966 L Ba(OH)2 solution 4.16B Plan: Write a balanced chemical equation. Determine the moles of H2SO4 by multiplying its molarity by its

volume, and, through the balanced chemical equation, determine the moles of KOH required for the reaction. The amount of base in moles is divided by its volume (in L) to find the molarity.

Solution: The balanced equation is: 2KOH(aq) + H2SO4(aq) → K2SO4(aq) + 2H2O(l)

Amount (mol) of KOH = 20.00 mL H2SO4 � 1 L1000 mL

� �0.2452 mol H2SO4

1 L� � 2 mol KOH

1 mol H2SO4� = 0.009808 mol

KOH

Molarity = �0.009808 mol KOH18.15 mL

� �1000 mL1 L

� = 0.5404 M 4.17A Plan: Apply Table 4.4 to the compounds. Do not forget that the sum of the O.N.’s (oxidation numbers) for a

compound must sum to zero, and for a polyatomic ion, the sum must equal the charge on the ion. Solution:

a) Sc = +3 O = –2 In most compounds, oxygen has a –2 O.N., so oxygen is often a good starting point. If each oxygen atom has a –2 O.N., then each scandium must have a +3 oxidation state so that the sum of O.N.’s equals zero: 2(+3) + 3(–2) = 0. b) Ga = +3 Cl = –1 In most compounds, chlorine has a –1 O.N., so chlorine is a good starting point. If each chlorine atom has a –1 O.N., then the gallium must have a +3 oxidation state so that the sum of O.N.’s equals zero: 1(+3) + 3(–1) = 0. c) H = +1 P = +5 O = –2 The hydrogen phosphate ion is HPO4

2–. Again, oxygen has a –2 O.N. Hydrogen has a +1 O.N. because it is combined with nonmetals. The sum of the O.N.’s must equal the ionic charge, so the following algebraic equation can be solved for P: 1(+1) + 1(P) + 4(–2) = –2; O.N. for P = +5. d) I = +3 F = –1 The formula of iodine trifluoride is IF3. In all compounds, fluorine has a –1 O.N., so fluorine is often a good starting point. If each fluorine atom has a –1 O.N., then the iodine must have a +3 oxidation state so that the sum of O.N.’s equal zero: 1(+3) + 3(–1) = 0.

4.17B Plan: Apply Table 4.4 to the compounds. Do not forget that the sum of the O.N.’s (oxidation numbers) for a compound must sum to zero, and for a polyatomic ion, the sum must equal the charge on the ion.

Solution: a) K = +1 C = +4 O = –2 In all compounds, potassium has a +1 O.N, and in most compounds, oxygen has a –2 O.N. If each potassium has a +1 O.N. and each oxygen has a –2 O.N., carbon must have a +4 oxidation state so that the sum of the O.N.’s equals zero: 2(+1) + 1(+4) + 3(–2) = 0. b) N = –3 H = +1 When it is combined with a nonmetal, like N, hydrogen has a +1 O.N. If hydrogen has a +1 O.N., nitrogen must have a –3 O.N. so the sum of the O.N.’s equals +1, the charge on the polyatomic ion: 1(–3) + 4(+1) = +1. c) Ca = +2 P = –3 Calcium, a group 2A metal, has a +2 O.N. in all compounds. If calcium has a +2 O.N., the phosphorus must have a –3 O.N. so the sum of the O.N.’s equals zero: 3(+2) + 2(–3) = 0. d) S = +4 Cl = –1 Chlorine, a group 7A nonmetal, has a –1 O.N. when it is in combination with any nonmetal except O or other halogens lower in the group. If chlorine has an O.N. of –1, the sulfur must have an O.N. of +4 so that the sume of the O.N.’s equals zero: 1(+4) + 4(–1) = 0.

4.18A Plan: Apply Table 4.4 to determine the oxidation numbers for all the compounds in the reaction. Do not forget

that the sum of the O.N.’s (oxidation numbers) for a compound must sum to zero, and for a polyatomic ion, the sum must equal the charge on the ion. After determining oxidation numbers for all atoms in the reaction, identify the atoms for which the oxidation numbers have changed from the left hand side of the equation to the right hand side of the equation. If the oxidation number of a particular atom increases, that atom has been oxidized, and the compound, element, or ion containing that atom on the reactant side of the equation is the reducing agent. If the oxidation number of a particular atom decreases, that atom has been reduced, and the compound, element, or ion containing that atom on the reactant side of the equation is the oxidizing agent.

4-11

Solution: a) Oxidation numbers in NCl3: N = +3, Cl = –1

Oxidation numbers in H2O: H = +1, O = –2 Oxidation numbers in NH3: N = –3, H = +1 Oxidation numbers in HOCl: H = +1, O = –2, Cl = +1 The oxidation number of nitrogen decreases from +3 to –3, so N is reduced and NCl3 is the oxidizing agent. The oxidation number of chlorine increases from –1 to +1, so Cl is oxidized and NCl3 is also the reducing agent. b) Oxidation numbers in AgNO3: N = +1, N = +5, O = –2 Oxidation numbers in NH4I: N = –3, H = +1, I = –1 Oxidation numbers in AgI: Ag = +1, I = –1 Oxidation numbers in NH4NO3: N (in NH4

+) = –3, H = +1, N (in NO3–) = +5, O = –2

None of the oxidation numbers change, so this is not an oxidation-reduction reaction. c) Oxidation numbers in H2S: H = +1, S = –2 Oxidation numbers in O2: O = 0 Oxidation numbers in SO2: S = +4, O = –2 Oxidation numbers in H2O: H = +1, O = –2 The oxidation number of oxygen decreases from 0 to –2, so O is reduced and O2 is the oxidizing agent. The oxidation number of sulfur increases from –2 to +4, so S is oxidized and H2S is the reducing agent.

4.18B Plan: Apply Table 4.4 to determine the oxidation numbers for all the compounds in the reaction. Do not forget

that the sum of the O.N.’s (oxidation numbers) for a compound must sum to zero, and for a polyatomic ion, the sum must equal the charge on the ion. After determining oxidation numbers for all atoms in the reaction, identify the atoms for which the oxidation numbers have changed from the left hand side of the equation to the right hand side of the equation. If the oxidation number of a particular atom increases, that atom has been oxidized, and the compound, element, or ion containing that atom on the reactant side of the equation is the reducing agent. If the oxidation number of a particular atom decreases, that atom has been reduced, and the compound, element, or ion containing that atom on the reactant side of the equation is the oxidizing agent.

Solution: a) Oxidation numbers in SiO2: Si = +4, O = –2

Oxidation numbers in HF: H = +1, F = –1 Oxidation numbers in SiF4: Si = +4, F = –1 Oxidation numbers in H2O: H = +1, O = –2 None of the oxidation numbers change, so this is not an oxidation-reduction reaction. b) Oxidation numbers in C2H6: C = –3, H = +1 Oxidation numbers in O2: O = 0 Oxidation numbers in CO2: C = +4, O = –2 Oxidation numbers in H2O: H = +1, O = –2 The oxidation number of oxygen decreases from 0 to –2, so O is reduced and O2 is the oxidizing agent. The oxidation number of carbon increases from –3 to +4, so C is oxidized and C2H6 is the reducing agent. c) Oxidation numbers in CO: C = +2, O = –2 Oxidation numbers in I2O5: I = +5, O = –2 Oxidation numbers in I2: I = 0 Oxidation numbers in CO2: C = +4, O = –2 The oxidation number of iodine decreases from +5 to 0, so I is reduced and I2O5 is the oxidizing agent. The oxidation number of carbon increases from +2 to +4, so C is oxidized and CO is the reducing agent.

4.19A Plan: Write a balanced reaction. Find moles of KMnO4 by multiplying its volume and molarity and determine

the moles of Ca2+ in the sample using the molar ratio from the reaction. Divide moles of Ca2+ by the volume of the milk sample to obtain molarity. In part (b), moles of Ca2+ will be converted to grams. Solution: 2KMnO4(aq) + 5CaC2O4(s) + H2SO4(aq) → 2MnSO4(aq) + K2SO4(aq) + 5CaSO4(s) + 10CO2(g) + 8H2O(l)

Moles of KMnO4 = ( )33

44.56 x 10 mol KMnO10 L6.53 mL1 mL 1 L

−−

4-12

= 2.97768 x 10–5 mol KMnO4

Molarity of Ca2+ = ( )2+

5 2 44 3

4 2 4

5 mol CaC O 1 mol Ca 1 1 mL2.97768 x 10 mol KMnO2 mol KMnO 1 mol CaC O 2.50 mL 10 L

−−

= 2.97768 x 10–2 = 2.98 x 10–2 mol/L = 2.98 x 10–2 M Ca2+

b) Concentration (g/L) = 2 2+ 2+

2+2.97768 x 10 mol Ca 40.08 g Ca

L 1 mol Ca

−

= 1.193454 = 1.19 g/L

This concentration is consistent with the typical value in milk.

4.19B Plan: Find moles of K2Cr2O7 by multiplying its volume and molarity, and determine the moles of Fe2+ in the sample using the molar ratio from the reaction provided in the problem statement (remember that the Fe2+ is part of the FeSO4 compound and that there is one Fe2+ ion for every formula unit of FeSO4). Divide moles of Fe2+ by the volume of the FeSO4 solution to obtain molarity. In part (b), convert the moles of Fe2+ to grams. Then divide the mass of iron ion by the total mass of the ore and multiply by 100% to find the mass percent of iron in the ore. Solution: a) The balanced equation (provided in the problem statement) is: 6FeSO4(aq) + K2Cr2O7(aq) + 7H2SO4(aq) → 3Fe2(SO4)3(aq) + Cr2(SO4)3(aq) + 7H2O(l) + K2SO4(aq)

Amount (mol) of Fe2+ = 21.85 mL K2Cr2O7 � 1 L1000 mL

� �0.250 mol K2Cr2O7

1 L� � 6 mole FeSO4

1 mol K2Cr2O7� � 1 mol Fe2+

1 mol FeSO4�

= 0.0328 mol Fe2+

Molarity of Fe2+ = �0.0328 mol Fe2+

30.0 mL� �1000 mL

1 L� = 1.0933 = 1.09 M

b) Mass (g) of iron = 0.0328 mol Fe2+ � 1 mol Fe1 mol Fe2+� �

55.85 g Fe1 mol Fe

� = 1.8319 = 1.83 g Fe

Mass % of iron in sample = �1.83 g Fe2.58 g ore

�(100%) = 70.9302 = 70.9%

4.20A Plan: To classify a reaction, compare the number of reactants used versus the number of products formed. Also

examine the changes, if any, in the oxidation numbers. Recall the definitions of each type of reaction: Combination: X + Y → Z; Decomposition: Z → X + Y Single displacement: X + YZ → XZ + Y Solution:

a) Combination; S8(s) + 16F2(g) → 8SF4(g) O.N.: S = 0 F = 0 S = +4 F = –1

Sulfur changes from 0 to +4 oxidation state; it is oxidized and S8 is the reducing agent. Fluorine changes from 0 to –1 oxidation state; it is reduced and F2 is the oxidizing agent.

b) Displacement; 2CsI(aq) + Cl2(aq) → 2CsCl(aq) + I2(aq) O.N.: Cs = +1 Cl = 0 Cs = +1 I = 0 I = –1 Cl = –1

Total ionic eqn: 2Cs+(aq) + 2I–(aq) + Cl2(aq) → 2Cs+(aq) + 2Cl–(aq) + I2(aq) Net ionic eqn: 2I–(aq) + Cl2(aq) → 2Cl–(aq) + I2(aq)

Iodine changes from –1 to 0 oxidation state; it is oxidized and CsI is the reducing agent. Chlorine changes from 0 to –1 oxidation state; it is reduced and Cl2 is the oxidizing agent.

c) Displacement; 3Ni(NO3)2(aq) + 2Cr(s) → 2Cr(NO3)3(aq) + 3Ni(s) O.N.: Ni = +2 Cr = 0 Cr = +3 Ni = 0 N = +5 N = +5 O = –2 O = –2

Total ionic eqn: 3Ni+2(aq) + 6NO3–(aq) + 2Cr(s) → 2Cr+3(aq) + 6NO3

–(aq) + 3Ni(s) Net ionic eqn: 3Ni+2(aq) + 2Cr(s) → 2Cr+3(aq) + 3Ni(s)

Nickel changes from +2 to 0 oxidation state; it is reduced and Ni(NO3)2 is the oxidizing agent. Chromium changes from 0 to +3 oxidation state; it is oxidized and Cr is the reducing agent.

4-13

4.20B Plan: To classify a reaction, compare the number of reactants used versus the number of products formed. Also examine the changes, if any, in the oxidation numbers. Recall the definitions of each type of reaction:

Combination: X + Y → Z; Decomposition: Z → X + Y Single displacement: X + YZ → XZ + Y Solution:

a) Displacement; Co(s) + 2HCl(aq) → CoCl2(aq) + H2(g) (molecular equation) O.N.: Co = 0 H = +1, Cl = –1 Co = +2, Cl = –1 H = 0 Co(s) + 2H+(aq) + 2Cl–(aq) → Co2+(aq) + 2Cl–(aq) + H2(g) (total ionic equation) Co(s) + 2H+(aq) → Co2+(aq) + H2(g) (net ionic equation)

Cobalt changes from 0 to +2 oxidation state; it is oxidized and Co is the reducing agent. Hydrogen changes from +1 to 0 oxidation state; it is reduced and HCl is the oxidizing agent.

b) Combination; 2CO (g) + O2(g) → 2CO2(g) O.N.: C = +2, O = –2 O = 0 C = +4, O = –2

Carbon changes from +2 to +4 oxidation state; it is oxidized and CO is the reducing agent. Oxygen changes from 0 to –2 oxidation state; it is reduced and O2 is the oxidizing agent.

c) Decomposition; 2N2O5(s) → 4NO2(g) + O2(g) O.N.: N = +5 N = +4 O = 0 O = –2 O = –2

Nitrogen changes from +5 to +4 oxidation state; it is reduced and N2O5 is the oxidizing agent. Oxygen changes from –2 to 0 oxidation state; it is oxidized and N2O5 is also the reducing agent.

END–OF–CHAPTER PROBLEMS 4.1 Plan: Review the discussion on the polar nature of water.

Solution: Water is polar because the distribution of its bonding electrons is unequal, resulting in polar bonds, and the shape of the molecule (bent) is unsymmetrical.

4.2 Plan: Review the discussion on water soluble compounds.

Solution: Ionic and polar covalent compounds are most likely to be soluble in water. Because water is polar, the partial charges in its molecules are able to interact with the charges, either ionic or dipole-induced, in other substances.

4.3 Plan: Solutions that conduct an electric current contain electrolytes.

Solution: Ions must be present in an aqueous solution for it to conduct an electric current. Ions come from ionic compounds

or from other electrolytes such as acids and bases. 4.4 Plan: Review the discussion on ionic compounds in water. Solution:

The ions on the surface of the solid attract the water molecules (cations attract the “negative” ends and anions attract the “positive” ends of the water molecules). The interaction of the solvent with the ions overcomes the attraction of the oppositely charged ions for one another, and they are released into the solution. 4.5 Plan: Recall that ionic compounds dissociate into their ions when dissolved in water. Examine the charges of the

ions in each scene and the ratio of cations to anions. Solution: a) CaCl2 dissociates to produce one Ca2+ ion for every two Cl– ions. Scene B contains four 2+ ions and twice that number of 1– ions.

b) Li2SO4 dissociates to produce two Li+ ions for every one SO42– ion. Scene C contains eight 1+ ions

and half as many 2– ions. c) NH4Br dissociates to produce one NH4

+ ion for every one Br– ion. Scene A contains equal numbers of 1+ and 1– ions.

4-14

4.6 Plan: Write the formula for magnesium nitrate and note the ratio of magnesium ions to nitrate ions. Solution: Upon dissolving the salt in water, magnesium nitrate, Mg(NO3)2, would dissociate to form one Mg2+ ion for every two NO3

– ions, thus forming twice as many nitrate ions. Scene B best represents a volume of magnesium nitrate solution. Only Scene B has twice as many nitrate ions (red circles) as magnesium ions (blue circles).

4.7 Plan: Review the discussion of ionic compounds in water. Solution:

In some ionic compounds, the force of the attraction between the ions is so strong that it cannot be overcome by the interaction of the ions with the water molecules. These compounds will be insoluble in water.

4.8 Plan: Review the discussion of covalent compounds in water.

Solution: The interaction with water depends on the structure of the molecule. If the interaction is strong, the substance will

be soluble; otherwise, the substance will not be very soluble. Covalent compounds that contain polar groups interact well with the polar solvent water and therefore dissolve in water. Covalent compounds that do

not contain polar bonds are not soluble in water.

4.9 Plan: Review the discussion of covalent compounds in water. Solution:

Some covalent compounds that contain the hydrogen atom dissociate into ions when dissolved in water. These compounds form acidic solutions in water; three examples are HCl, HNO3, and HBr. 4.10 Plan: Count the total number of spheres in each box. The number in box A divided by the volume change in each part will give the number we are looking for and allow us to match boxes. Solution: The number in each box is: A = 12, B = 6, C = 4, and D = 3. a) When the volume is tripled, there should be 12/3 = 4 spheres in a box. This is box C. b) When the volume is doubled, there should be 12/2 = 6 spheres in a box. This is box B. c) When the volume is quadrupled, there should be 12/4 = 3 spheres in a box. This is box D. 4.11 Plan: Recall that molarity = moles of solute/volume (L) of solution.

Solution: a) Mdil = molarity of the diluted solution Mconc = molarity of the concentrated solution

Vdil = volume of the diluted solution Vconc = volume of the concentrated solution The equation works because the quantity (moles) of solute remains the same when a solution is diluted; only the amount of solvent changes. M x V = moles; Mdil x Vdil = Mconc x Vconc molesdil = molesconc

b) Molarity = moles soluteliters solution

Moles CaCl2 = molarity x liters of solution; Mass CaCl2 = molarity x liters of solution x molar mass of CaCl2. 4.12 Plan: Recall that molarity = moles of solute/volume (L) of solution. Here you can use the number of particles in

place of moles of solute. Solution: a) Solution B has the highest molarity as it has the largest number of particles, 12, in a volume of 50 mL. b) Solutions A and F both have 8 particles in a volume of 50 mL and thus the same molarity. Solutions C, D, and E all have 4 particles in a volume of 50 mL and thus have the same molarity. c) Mixing Solutions A and C results in 8 + 4 = 12 particles in a volume of 100 mL. That is a lower molarity than that of Solution B which has 12 particles in a volume of 50 mL or 24 particles in a volume of 100 mL. d) Adding 50 mL to Solution D would result in 4 particles in a total volume of 100 mL; adding 75 mL to Solution F would result in 4 particles in a volume of 100 mL. The molarity of each solution would be the same.

4-15

e) Solution A has 8 particles in a volume of 50 mL while Solution E has the equivalent of 4 particles in a volume of 50 mL. The molarity of Solution E is half that of Solution A. Therefore half of the volume, 12.5 mL, of Solution E must be evaporated. When 12.5 mL of solvent is evaporated from Solution E, the result will be 2 particles in 12.5 mL or 8 particles in 50 mL as in Solution A.

4.13 Plan: Remember that molarity is moles of solute/volume of solution. Solution: Volumes may not be additive when two different solutions are mixed, so the final volume may be slightly different from 1000.0 mL. The correct method would state, “Take 100.0 mL of the 10.0 M solution and add water until the total volume is 1000 mL.”

4.14 Plan: Compounds that are soluble in water tend to be ionic compounds or covalent compounds that have polar

bonds. Many ionic compounds are soluble in water because the attractive force between the oppositely charged ions in an ionic compound are replaced with an attractive force between the polar water molecule and the ions when the compound is dissolved in water. Covalent compounds with polar bonds are often soluble in water since the polar bonds of the covalent compound interact with those in water. Solution: a) Benzene, a covalent compound, is likely to be insoluble in water because it is nonpolar and water is polar. b) Sodium hydroxide (NaOH) is an ionic compound and is therefore likely to be soluble in water.

c) Ethanol (CH3CH2OH) will likely be soluble in water because it contains a polar –OH bond like water. d) Potassium acetate (KC2H3O2) is an ionic compound and will likely be soluble in water.

4.15 Plan: Compounds that are soluble in water tend to be ionic compounds or covalent compounds that have polar

bonds. Many ionic compounds are soluble in water because the attractive force between the oppositely charged ions in an ionic compound are replaced with an attractive force between the polar water molecule and the ions when the compound is dissolved in water. Covalent compounds with polar bonds are often soluble in water since the polar bonds of the covalent compound interact with those in water. Solution: a) Lithium nitrate is an ionic compound and is expected to be soluble in water.

b) Glycine (H2NCH2COOH) is a covalent compound, but it contains polar N–H and O–H bonds. This would make the molecule interact well with polar water molecules, and make it likely that it would be soluble.

c) Pentane (C5H12) has no bonds of significant polarity, so it would be expected to be insoluble in the polar solvent water. d) Ethylene glycol (HOCH2CH2OH) molecules contain polar O–H bonds, similar to water, so it would be expected to be soluble.

4.16 Plan: Substances whose aqueous solutions conduct an electric current are electrolytes such as ionic compounds,

acids, and bases. Solution: a) Cesium bromide, CsBr, is a soluble ionic compound, and a solution of this salt in water contains Cs+ and Br– ions. Its solution conducts an electric current.

b) HI is a strong acid that dissociates completely in water. Its aqueous solution contains H+ and I– ions, so it conducts an electric current. 4.17 Plan: Substances whose aqueous solutions conduct an electric current are electrolytes such as ionic compounds,

acids, and bases. Solution: a) Potassium sulfate, K2SO4, is an ionic compound that is soluble in water, producing K+ and SO4

2– ions. Its solution conducts an electric current. b) Sucrose is neither an ionic compound, an acid, nor a base, so it would be a nonelectrolyte (even though it’s soluble in water). Its solution does not conduct an electric current.

4.18 Plan: To determine the total moles of ions released, write an equation that shows the compound dissociating into

ions with the correct molar ratios. Convert mass and formula units to moles of compound and use the molar ratio to convert moles of compound to moles of ions.

Solution:

4-16

a) Each mole of NH4Cl dissolves in water to form 1 mole of NH4+ ions and 1 mole of Cl– ions, or a total of 2

moles of ions: NH4Cl(s) → NH4+(aq) + Cl–(aq).

Moles of ions = ( )44

2 mol ions0.32 mol NH Cl1 mol NH Cl

= 0.64 mol of ions

b) Each mole of Ba(OH)2•8H2O forms 1 mole of Ba2+ ions and 2 moles of OH– ions, or a total of 3 moles of ions: Ba(OH)2•8H2O(s) → Ba2+(aq) + 2OH–(aq). The waters of hydration become part of the larger bulk of water. Convert mass to moles using the molar mass.

Moles of ions = ( ) 2 22 2

2 2 2 2

1 mol Ba(OH) •8H O 3 mol ions25.4 g Ba(OH) •8H O315.4 g Ba(OH) •8H O 1 mol Ba(OH) •8H O

= 0.2415980 = 0.242 mol of ions c) Each mole of LiCl produces 2 moles of ions (1 mole of Li+ ions and 1 mole of Cl– ions):

LiCl(s) → Li+(aq) + Cl–(aq). Recall that a mole contains 6.022x1023 entities, so a mole of LiCl contains 6.022x1023 units of LiCl, more easily expressed as formula units.

Moles of ions = ( )1923

1 mol LiCl 2 mol ions3.55x10 FU LiCl1 mol LiCl6.022 x10 FU LiCl

= 1.17901x10–4 = 1.18x10–4 mol of ions 4.19 Plan: To determine the total moles of ions released, write an equation that shows the compound dissociating into


Solution: a) Each mole of Rb2SO4 dissolves in water to form 2 moles of Rb+ ions and 1 mole of SO4

2– ions, or a total of 3 moles of ions: Rb2SO4(s) → 2Rb+(aq) + SO4

2–(aq).

Moles of ions = ( )2 42 4

3 mol ions0.805 mol Rb SO1 mol Rb SO

= 2.415 = 2.42 mol of ions

b) Each mole of Ca(NO3)2 forms 1 mole of Ca2+ ions and 2 moles of NO3– ions, or a total of

3 moles of ions: Ca(NO3)2(s) → Ca2+(aq) + 2NO3–(aq). Convert mass to moles using molar mass.

Moles of ions = ( )3 3 23 2

3 2 3 2

1 mol Ca(NO ) 3 mol ions3.85x10 g Ca(NO )164.10 g Ca(NO ) 1 mol Ca(NO )

−

= 7.03839x10–5 = 7.04x10–5 mol of ions c) Each mole of Sr(HCO3)2 produces 3 moles of ions (1 mole of Sr2+ ions and 2 moles of HCO3

– ions): Sr(HCO3)2(s) → Sr2+(aq) + 2HCO3

–(aq). Recall that a mole contains 6.022x1023 entities, so a mole of Sr(HCO3)2 contains 6.022x1023 units of Sr(HCO3)2, more easily expressed as formula units.

Moles of ions = ( )19 3 23 2 23

3 23 2

1 mol Sr(HCO ) 3 mol ions4.03x10 FU Sr(HCO )1 mol Sr(HCO )6.022 x10 FUSr(HCO )

= 2.0076x10–4 = 2.01x10–4 mol of ions 4.20 Plan: To determine the total moles of ions released, write an equation that shows the compound dissociating into


Solution: a) Each mole of K3PO4 forms 3 moles of K+ ions and 1 mole of PO4

3– ions, or a total of 4 moles of ions: K3PO4(s) → 3K+(aq) + PO4

3–(aq)


4 mol ions0.75 mol K PO1 mol K PO

= 3.0 mol of ions.

b) Each mole of NiBr2•3H2O forms 1mole of Ni2+ ions and 2 moles of Br– ions, or a total of 3 moles of ions: NiBr2•3H2O(s) → Ni2+(aq) + 2Br –(aq). The waters of hydration become part of the larger bulk of water.

4-17

Convert mass to moles using the molar mass.

Moles of ions = ( )3 2 22 2

2 2 2 2

1 mol NiBr •3H O 3 mol ions6.88 x 10 g NiBr •3H O272.54 g NiBr •3H O 1 mol NiBr •3H O

−

= 7.5732x10–5 = 7.57x10–5 mol of ions c) Each mole of FeCl3 forms 1mole of Fe3+ ions and 3 moles of Cl– ions, or a total of 4 moles of ions:

FeCl3(s) → Fe3+(aq) + 3Cl–(aq). Recall that a mole contains 6.022x1023 entities, so a mole of FeCl3 contains 6.022x1023 units of FeCl3, more easily expressed as formula units.


33

1 mol FeCl 4 mol ions2.23x10 FU FeCl1 mol FeCl6.022x10 FU FeCl

= 0.148124 = 0.148 mol of ions 4.21 Plan: To determine the total moles of ions released, write an equation that shows the compound dissociating into


Solution: a) Each mole of Na2HPO4 forms 2 moles of Na+ ions and 1 mole of HPO4

2– ions, or a total of 3 moles of ions: Na2HPO4(s) → 2Na+(aq) + HPO4

2–(aq).


3 mol ions0.734 mol Na HPO1 mol Na HPO

= 2.202 = 2.20 mol of ions

b) Each mole of CuSO4•5H2O forms 1 mole of Cu2+ ions and 1 mole of SO42– ions, or a total of 2 moles of ions:

CuSO4•5H2O(s) → Cu+2(aq) + SO42–(aq). The waters of hydration become part of the larger bulk of water.

Convert mass to moles using the molar mass.

Moles of ions = ( ) 4 24 2

4 2 4 2

1 mol CuSO •5H O 2 mol ions3.86 g CuSO •5H O 249.69 g CuSO •5H O 1 mol CuSO •5H O

= 3.0918x10–2 = 3.09x10–2 mol of ions c) Each mole of NiCl2 forms 1mole of Ni2+ ions and 2 moles of Cl– ions, or a total of 3 moles of ions: NiCl2(s) → Ni2+(aq) + 2Cl–(aq). Recall that a mole contains 6.022x1023 entities, so a mole of NiCl2 contains 6.022x1023 units of NiCl2, more easily expressed as formula units.


22

1 mol NiCl 3 mol ions8.66x10 FU NiCl1 mol NiCl6.022x10 FU NiCl

= 4.31418x10–3 = 4.31x10–3 mol of ions

4.22 Plan: In all cases, use the known quantities and the definition of molarity moles solute L of solution

=

M to find the

unknown quantity. Volume must be expressed in liters. The molar mass is used to convert moles to grams. The chemical formulas must be written to determine the molar mass. (a) You will need to convert milliliters to liters, multiply by the molarity to find moles, and convert moles to mass in grams. (b) Convert mass of solute to moles and volume from mL to liters. Divide the moles by the volume. (c) Multiply the molarity by the volume.

Solution: a) Calculating moles of solute in solution:

Moles of Ca(C2H3O2)2 = ( )3

2 3 2 20.267 mol Ca(C H O )10 L185.8 mL1 mL 1 L

−

= 0.0496086 mol Ca(C2H3O2)2

Converting from moles of solute to grams:

Mass (g) of Ca(C2H3O2)2 = ( ) 2 3 2 22 3 2 2

2 3 2 2

158.17 g Ca(C H O )0.0496086 mol Ca(C H O )

1 mol Ca(C H O )

= 7.84659 = 7.85 g Ca(C2H3O2)2

4-18

b) Converting grams of solute to moles:

Moles of KI = ( ) 1 mol KI21.1 g KI166.0 g KI

= 0.127108 moles KI

Volume (L) = ( )310 L500. mL

1 mL

−

= 0.500 L

Molarity of KI = L 0.500

KI mol 127108.0 = 0.254216 = 0.254 M KI

c) Moles of NaCN = ( ) 0.850 mol NaCN145.6 L1 L

= 123.76 = 124 mol NaCN


=

M to find the

unknown quantity. Volume must be expressed in liters. The molar mass is used to convert moles to grams. The chemical formulas must be written to determine the molar mass. (a) You will need to convert mass of solute to moles and divide by the molarity to obtain volume in liters, which is then converted to milliliters. (b) Multiply the volume by the molarity to obtain moles of solute. Use Avogadro’s number to determine the number of ions present. (c) Divide mmoles by milliliters; molarity may not only be expressed as moles/L, but also as mmoles/mL.

Solution: a) Converting mass of solute to moles:

Moles of KOH = ( ) 1 mol KOH8.42 g KOH56.11 g KOH

= 0.15006 mol KOH

Volume (L) of KOH solution = ( ) 1 L0.15006 mol KOH2.26 mol

= 0.066398 L KOH solution

Volume (mL) of KOH solution = ( ) 31 L0.066398 L KOH

10 mL−

= 66.39823 = 66.4 mL KOH solution

b) Moles of CuCl2 = ( ) 22.3 mol CuCl52 L

L

= 119.6 mol CuCl2

Moles of Cu2+ ions = ( )2

22

1 mol Cu119.6 mol CuCl1 mol CuCl

+

= 119.6 mol Cu2+ ions

Converting moles of ions to number of ions:

Number of Cu2+ ions = ( )23 2

22

6.022 x10 Cu ions119.6 mol Cu ions1 mol Cu ions

++

+

= 7.2023x1025 = 7.2x1025 Cu2+ ions

c) M glucose = 135 mmol glucose

275 mL

= 0.490909 = 0.491 M glucose

Note: Since 1 mmol is 10–3 mol and 1 mL is 10–3 L, we can use these units instead of converting to mol and L since molarity is a ratio of mol/L. Molarity may not only be expressed as moles/L, but also as mmoles/mL.


=

M to find the

unknown quantity. Volume must be expressed in liters. The molar mass is used to convert moles to grams. The chemical formulas must be written to determine the molar mass. (a) Convert volume in milliliters to liters, multiply the volume by the molarity to obtain moles of solute, and convert moles to mass in grams. (b) The simplest way will be to convert the milligrams to millimoles. Molarity may not only be expressed as moles/L, but

4-19

also as mmoles/mL. (c) Convert the milliliters to liters and find the moles of solute and moles of ions by multiplying the volume and molarity. Use Avogadro’s number to determine the number of ions present.

Solution: a) Calculating moles of solute in solution:

Moles of K2SO4 = ( )3 2

2 410 L 5.62 x10 mol K SO475 mL1 mL L

− −

= 0.026695 mol K2SO4

Converting moles of solute to mass:

Mass (g) of K2SO4 = ( ) 2 42 4

2 4

174.26 g K SO0.026695 mol K SO

1 mol K SO

= 4.6519 = 4.65 g K2SO4

b) Calculating mmoles of solute:

Mmoles of CaCl2 = 2 2

2

7.25 mg CaCl 1 mmol CaCl1 mL 110.98 mg CaCl

= 0.065327 mmoles CaCl2

Calculating molarity:

Molarity of CaCl2 = 20.065327 mmol CaCl1 mL

= 0.065327 = 0.0653 M CaCl2

If you believe that molarity must be moles/liters then the calculation becomes:

Molarity of CaCl2 = 3

2 23

2

7.25 mg CaCl 1 mol CaCl10 g 1 mL1 mL 1 mg 110.98 g CaCl10 L

−

−

= 0.065327 = 0.0653 M CaCl2

Notice that the two central terms cancel each other. c) Converting volume in L to mL:

Volume (L) = ( )310 L1 mL

1 mL

−

= 0.001 L

Calculating moles of solute and moles of ions:

Moles of MgBr2 = ( ) 20.184 mol MgBr0.001 L

1 L

= 1.84x10–4 mol MgBr2

Moles of Mg2+ ions = ( )2

42

2

1 mol Mg1.84 x 10 mol MgBr1 mol MgBr

+−

= 1.84x10–4 mol Mg2+ ions

Number of Mg2+ ions = ( )23 2

4 22

6.022 x10 Mg ions1.84x10 Mg ions1 mol Mg ions

+− +

+

= 1.1080x1020 = 1.11x1020 Mg2+ ions


=

M to find the

unknown quantity. Volume must be expressed in liters. The molar mass is used to convert moles to grams. The chemical formulas must be written to determine the molar mass. (a) Convert mass of solute to moles and volume from mL to liters. Divide the moles by the volume. (b) You will need to convert mass of solute to moles and divide by the molarity to obtain volume in liters. (c) Divide the mmoles of solute by the molarity to obtain volume in mL. Solution: a) Calculating moles of solute:

Moles of AgNO3 = ( ) 33

3

1 mol AgNO46.0 g AgNO

169.9 g AgNO

= 0.2707475 mol AgNO3

Volume (L) = ( )310 L335 mL

1 mL

−

= 0.335 L

4-20

Calculating the molarity:

Molarity of AgNO3 = 30.2707574 mol AgNO0.335 L

= 0.80823 = 0.808 M AgNO3

b) Calculating moles of solute:

Moles of MnSO4 = ( ) 44

4

1 mol MnSO63.0 g MnSO

151.00 g MnSO

= 0.417218 mol MnSO4

Calculating volume of solution:

Volume (L) of solution = ( )44

1 L0.417218 mol MnSO0.385 mol MnSO

= 1.08368 = 1.08 L MnSO4 solution

c) Volume (mL) of ATP solution = ( ) 21mL1.68 mmol ATP

6.44 x10 mmol ATP−

= 26.087 = 26.1 mL ATP solution 4.26 Plan: To determine the total moles of ions released, write an equation that shows the compound dissociating into

ions with the correct molar ratios. Convert the information given to moles of compound and use the molar ratio to convert moles of compound to moles of ions. Avogadro’s number is used to convert moles of ions to numbers of ions.

Solution: a) Each mole of AlCl3 forms 1mole of Al3+ ions and 3 moles of Cl– ions: AlCl3(s) → Al3+(aq) + 3Cl–(aq). Molarity and volume must be converted to moles of AlCl3.

Moles of AlCl3 = ( )3

30.45 mol AlCl10 L130. mL1 mL L

−

= 0.0585 mol AlCl3

Moles of Al3+ = ( )3

33

1 mol Al0.0585 mol AlCl1 mol AlCl

+

= 0.0585 = 0.058 mol Al3+

Number of Al3+ ions = ( )23 3

33

6.022x10 Al0.0585 mol Al1 mol Al

++

+

= 3.52287x 1022 = 3.5x1022 Al3+ ions

Moles of Cl– = ( )33

3 mol Cl0.0585 mol AlCl1 mol AlCl

−

= 0.1755 = 0.18 mol Cl–

Number of Cl– ions = ( )236.022x10 Cl0.1755 mol Cl

1 mol Cl

−−

−

= 1.05686x1023 = 1.1x1023 Cl– ions

b) Each mole of Li2SO4 forms 2 moles of Li+ ions and 1 mole of SO4

2– ions: Li2SO4(s) → 2Li+(aq) + SO42–(aq).

Moles of Li2SO4 = ( )3

2 4 2 4

2 4

2.59 g Li SO 1 mol Li SO10 L9.80 mL1 mL 1 L 109.94 g Li SO

−

= 2.3087x10–4 mol Li2SO4

Moles of Li+ = ( )42 4

2 4

2 mol Li2.3087x10 mol Li SO1 mol Li SO

+−

= 4.6174x10–4 = 4.62x10–4 mol Li+

Number of Li+ ions = ( )23

4 6.022 x 10 Li4.6174x10 mol Li1 mol Li

+− +

+

= 2.7806x1020 = 2.78x1020 Li+ ions

Moles of SO42– = ( )

24 4

2 42 4

1 mol SO2.3087x10 mol Li SO

1 mol Li SO

−−

= 2.3087x10–4 = 2.31x10–4 mol SO42–

4-21

Number of SO42– ions = ( )

23 24 2 4

4 24

6.022 x 10 SO2.3087x10 mol SO

1 mol SO

−− −

−

= 1.39030x1020 = 1.39x1020 SO42– ions

c) Each mole of KBr forms 1 mole of K+ ions and 1 mole of Br– ions: KBr(s) → K+(aq) + Br–(aq).

Moles of KBr = ( )3 22

2310 L 3.68x10 FU KBr 1 mol KBr245 mL1 mL L 6.022x10 FU KBr

−

= 0.01497 mol KBr

Moles of K+ = ( ) 1 mol K0.01497 mol KBr1 mol KBr

+

= 0.01497 = 1.50x10–2 mol K+

Number of K+ ions = ( )236.022 x10 K0.01497 mol K

1 mol K

++

+

= 9.016x1021 = 9.02x1021 K+ ions

Moles of Br– = ( ) 1 mol Br0.01497 mol KBr1 mol KBr

−

= 0.01497 = 1.50x10–2 mol Br–

Number of Br– ions = ( )236.022 x10 Br0.01497 mol Br

1 mol Br

−−

−

= 9.016x1021 = 9.02x1021 Br– ions

4.27 Plan: To determine the total moles of ions released, write an equation that shows the compound dissociating into

ions with the correct molar ratios. Convert the information given to moles of compound and use the molar ratio to convert moles of compound to moles of ions. Avogadro’s number is used to convert moles of ions to numbers of ions.

Solution: a) Each mole of MgCl2 forms 1 mole of Mg2+ ions and 2 moles of Cl– ions: MgCl2(s) → Mg2+(aq) + 2Cl–(aq).

Moles of MgCl2 = ( )3

21.75 mol MgCl10 L88.mL1 mL L

−

= 0.154 mol MgCl2

Moles of Mg2+ = ( )2

22

1 mol Mg0.154 mol MgCl1 mol MgCl

+

= 0.154 = 0.15 mol Mg2+

Number of Mg2+ ions = ( )23 2

22

6.022x10 Mg0.154 mol Mg1 mol Mg

++

+

= 9.27388x1022 = 9.3x1022 Mg2+ ions


2 mol Cl0.154 mol MgCl1 mol MgCl

−

= 0.308 = 0.31 mol Cl–

Number of Cl– ions = ( )236.022x10 Cl0.308 mol Cl

1 mol Cl

−−

−

= 1.854776x1023 = 1.9x1023 Cl– ions

b) Each mole of Al2(SO4)3 forms 2 moles of Al3+ ions and 3 moles of SO42– ions:

Al2(SO4)3(s) → 2Al3+(aq) + 3SO42–(aq).

Moles of Al2(SO4)3 = ( )3

2 4 3 2 4 3

2 4 3

0.22 g Al (SO ) 1 mol Al (SO )10 L321 mL1 mL 1 L 342.14 g Al (SO )

−

= 2.06406x10–4 mol Al2(SO4)3

Moles of Al3+ = ( )3

42 4 3

2 4 3

2 mol Al2.06406x10 mol Al (SO )1 mol Al (SO )

+−

= 4.12812x10–4 = 4.1x10–4 mol Al3+

Number of Al3+ ions = ( )23 3

4 33

6.022x10 Al4.12812x10 mol Al1 mol Al

+− +

+

= 2.4860x1020 = 2.5x1020 Al3+ ions

4-22


24 4

2 4 32 4 3

3 mol SO2.06406x10 mol Al (SO )

1 mol Al (SO )

−−

= 6.19218x10–4 = 6.2x10–4 mol SO42

Number of SO42– ions = ( )

23 24 2 4

4 24

6.022 x10 SO6.19218x10 mol SO

1 mol SO

−− −

−

= 3.7289x1020 = 3.7x1020 SO42– ions

c) Each mole of CsNO3 forms 1 mole of Cs+ ions and 1 mole of NO3– ions: CsNO3(s) → Cs+(aq) + NO3

–(aq)

Moles of CsNO3 = ( )21

3 323

3

8.83x10 FU CsNO 1 mol CsNO1.65 LL 6.022x10 FU CsNO

= 0.024194 mol CsNO3

Moles of Cs+ = ( )33

1 mol Cs0.024194 molCsNO1 mol CsNO

+

= 0.024194 = 0.0242 mol Cs+

Number of Cs+ ions = ( )236.022x10 Cs0.024194 mol Cs

1 mol Cs

++

+

= 1.45695x1022 = 1.46x1022 Cs+ ions

Moles of NO3– = ( ) 3

33

1 mol NO0.024194 molCsNO

1 mol CsNO

−

= 0.024194 = 0.0242 mol NO3–

Number of NO3– ions = ( )

233

33

6.022x10 NO0.024194 mol NO1 mol NO

−−

−

= 1.45695x1022 = 1.46x1022 NO3– ions

4.28 Plan: These are dilution problems. Dilution problems can be solved by converting to moles and using the new

volume; however, it is much easier to use M1V1 = M2V2. The dilution equation does not require a volume in liters; it only requires that the volume units match. In part c), it is necessary to find the moles of sodium ions in each separate solution, add these two mole amounts, and divide by the total volume of the two solutions. Solution:

a) M1 = 0.250 M KCl V1 = 37.00 mL M2 = ? V2 = 150.00 mL M1V1= M2V2

1 12

2

x =

M VMV

= ( )( )0.250 37.00 mL

150.0 mLM

= 0.061667 = 0.0617 M KCl

b) M1 = 0.0706 M (NH4)2SO4 V1 = 25.71 mL M2 = ? V2 = 500.00 mL M1V1= M2V2

1 12

2

x =

M VMV

= ( )( )0.0706 25.71 mL

500.0 mLM

= 0.003630 = 0.00363 M (NH4)2SO4

c) Moles of Na+ from NaCl solution = ( )310 L 0.348 mol NaCl 1 mol Na3.58 mL

1 mL 1 L 1 mol NaCl

− +

= 0.00124584 mol Na+

Moles of Na+ from Na2SO4 solution = ( )3 2

2 4

2 4

10 L 6.81x10 mol Na SO 2 mol Na500. mL1 mL 1 L 1 mol Na SO

− − +

= 0.0681 mol Na+ Total moles of Na+ ions = 0.00124584 mol Na+ ions + 0.0681 mol Na+ ions = 0.06934584 mol Na+ ions Total volume = 3.58 mL + 500. mL = 503.58 mL = 0.50358 L

Molarity of Na+ = total moles Na ions 0.06934584 mol Na ions = total volume 0.50358 L

+ +

= 0.1377057 = 0.138 M Na+ ions

4.29 Plan: These are dilution problems. Dilution problems can be solved by converting to moles and using the new

volume; however, it is much easier to use M1V1 = M2V2. The dilution equation does not require a volume in liters; it only requires that the volume units match. Solution:

4-23

a) M1 = 2.050 M Cu(NO3)2 V1 = ? M2 = 0.8543 M Cu(NO3)2 V2 = 750.0 mL M1V1= M2V2

2 21

1

x =

M VVM

= ( )( )0.8543 750.0 mL

2.050 M

M = 312.5488 = 312.5 mL

b) M1 = 1.63 M CaCl2 M1 Cl– = 2

2

1.63 mol CaCl 2 mol Cl1 L 1 mol CaCl

−

= 3.26 M Cl– ions

M1 = 3.26 M Cl– V1 = ? M2 = 2.86x10–2 M Cl– ions V2 = 350. mL M1V1= M2V2

2 21

1

x =

M VVM

= ( ) ( )22.86x10 350. mL

3.26

M

M

−

= 3.07055= 3.07 mL

c) M1 = 0.155 M Li2CO3 V1 = 18.0 mL M2 = 0.0700 M Li2CO3 V2 = ? M1V1= M2V2

1 12

2

x =

M VVM

= ( )( )

( )0.155 18.0 mL

0.0700 M

M = 39.8571 = 39.9 mL

4.30 Plan: Use the density of the solution to find the mass of 1 L of solution. Volume in liters must be converted to

volume in mL. The 70.0% by mass translates to 70.0 g solute/100 g solution and is used to find the mass of HNO3 in 1 L of solution. Convert mass of HNO3 to moles to obtain moles/L, molarity.

Solution:

a) Mass (g) of 1 L of solution = ( ) 31 mL 1.41 g solution1 L solution

1 mL10 L−

= 1410 g solution

Mass (g) of HNO3 in 1 L of solution = ( ) 370.0 g HNO1410 g solution

100 g solution

= 987 g HNO3/L

b) Moles of HNO3 = ( ) 33

3

1 mol HNO987 g HNO

63.02 g HNO

= 15.6617 mol HNO3

Molarity of HNO3 = 315.6617 mol HNO1 L solution

= 15.6617 = 15.7 M HNO3

4.31 Plan: Use the molarity of the solution to find the moles of H2SO4 in 1 mL. Convert moles of H2SO4 to mass of

H2SO4, divide that mass by the mass of 1 mL of solution, and multiply by 100 for mass percent. Use the density of the solution to find the mass of 1 mL of solution.

Solution:

a) Moles of H2SO4 in 1 mL = 3

2 418.3 mol H SO 10 L1 L 1 mL

−

= 1.83x10–2 mol H2SO4/mL

b) Mass of H2SO4 in 1 mL = ( )2 2 42 4

2 4

98.08 g H SO1.83x10 mol H SO

1 mol H SO−

= 1.79486 g H2SO4

Mass of 1 mL of solution = ( ) 1.84 g1 mL1 mL

= 1.84 g solution

Mass percent = ( )2 4mass of H SO100

mass of solution = ( )2 41.79486 g H SO

1001.84 g solution

= 97.5467 = 97.5% H2SO4 by mass

4.32 Plan: Recall the definition of molarity moles solute L of solution

=

M . Convert mass of solute to moles and volume from

mL to liters. Divide the moles by the volume. Solution:

4-24

Moles of NaClO = ( ) 1 mol NaClO20.5 g NaClO74.44 g NaClO

= 0.2785896 mol NaClO

Volume (L) = ( )310 L375 mL

1 mL

−

= 0.375 L

Molarity of NaClO = 0.2753896 mol NaClO

0.375 L

= 0.73437 = 0.734 M NaClO

4.33 Plan: The first part of the problem is a simple dilution problem (M1V1 = M2V2). The volume in units of gallons

can be used. In part b), convert mass of HCl to moles and use the molarity to find the volume that contains that number of moles.

Solution: a) M1 = 11.7 M V1 = ? M2 = 3.5 M V2 = 3.0 gal

2 21

1

x =

M VVM

= ( )( )3.5 3.0 gal

11.7 M

M = 0.897436 gal

Instructions: Be sure to wear goggles to protect your eyes! Pour approximately 2.0 gal of water into the container. Add slowly and with mixing 0.90 gal of 11.7 M HCl into the water. Dilute to 3.0 gal with water. b) Converting from mass of HCl to moles of HCl:

Moles of HCl = ( ) 1 mol HCl9.66 g HCl36.46 g HCl

= 0.264948 mol HCl

Converting from moles of HCl to volume:

Volume (mL) of solution = ( ) 31 L 1 mL0.264948 mol HCl

11.7 mol HCl 10 L−

= 22.64513 = 22.6 mL muriatic acid solution 4.34 Plan: Convert the mass of the seawater in kg to g and use the density to convert the mass of the seawater to

volume in L. Convert mass of each compound to moles of compound and then use the molar ratio in the dissociation of the compound to find the moles of each ion. The molarity of each ion is the moles of ion divided by the volume of the seawater. To find the total molarity of the alkali metal ions [Group 1A(1)], add the moles of the alkali metal ions and divide by the volume of the seawater. Perform the same calculation to find the total

molarity of the alkaline earth metal ions [Group 2A(2)] and the anions (the negatively charged ions). Solution: a) The volume of the seawater is needed.

Volume (L) of seawater = ( )3 3 3

310 g cm 1 mL 10 L1.00 kg1 kg 1.025 g 1 mL1 cm

−

= 0.97560976 L

The moles of each ion are needed. If an ion comes from more than one source, the total moles are needed. NaCl: Each mole of NaCl forms 1 mole of Na+ ions and 1 mole of Cl– ions: NaCl(s) → Na+(aq) + Cl–(aq)

Moles of NaCl = ( ) 1 mol NaCl26.5 g NaCl58.44 g NaCl

= 0.4534565 mol NaCl

Moles of Na+ = ( ) 1 mol Na0.4534565 mol NaCl1 mol NaCl

+

= 0.4534565 mol Na+

Moles of Cl– = ( ) 1 mol Cl0.4534565 mol NaCl1 mol NaCl

−

= 0.4534565 mol Cl–

MgCl2: Each mole of MgCl2 forms 1 mole of Mg2+ ions and 2 moles of Cl– ions: MgCl2(s) → Mg2+(aq) + 2Cl–(aq)

4-25

Moles of MgCl2 = ( ) 22

2

1 mol MgCl2.40 g MgCl

95.21 g MgCl

= 0.025207 mol MgCl2


22

1 mol Mg0.025207 mol MgCl1 mol MgCl

+

= 0.025207 mol Mg2+


2 mol Cl0.025207 mol MgCl1 mol MgCl

−

= 0.050415 mol Cl–

MgSO4: Each mole of MgSO4 forms 1 mole of Mg2+ ions and 1 mole of SO4

2– ions: MgSO4(s) → Mg2+(aq) + SO42–(aq)

Moles of MgSO4 = ( ) 44

4

1 mol MgSO3.35 g MgSO

120.37 g MgSO

= 0.0278308 mol MgSO4


44

1 mol Mg0.0278308 mol MgSO1 mol MgSO

+

= 0.0278308 mol Mg2+


24

44

1 mol SO0.0278308 mol MgSO

1 mol MgSO

−

= 0.0278308 mol SO42–

CaCl2: Each mole of CaCl2 forms 1 mole of Ca2+ ions and 2 moles of Cl– ions: CaCl2(s) → Ca2+(aq) + 2Cl–(aq)

Moles of CaCl2 = ( )2

22

2 2

1 mol CaCl 1 mol Ca1.20 g CaCl110.98 g CaCl 1 mol CaCl

+

= 0.0108128 mol CaCl2

Moles of Ca2+ = ( )2

22

1 mol Ca0.0108128 mol CaCl1 mol CaCl

+

= 0.0108128 mol Ca2+


2 mol Cl0.0108128 mol CaCl1 mol CaCl

−

= 0.0216255 mol Cl–

KCl: Each mole of KCl forms 1 mole of K+ ions and 1 mole of Cl– ions: KCl(s) → K+(aq) + Cl–(aq)

Moles of KCl = ( ) 1 mol KCl1.05 g KCl74.55 g KCl

= 0.0140845 mol KCl

Moles of K+ = ( ) 1 mol K0.0140845 mol KCl1 mol KCl

+

= 0.0140845 mol K+

Moles of Cl– = ( ) 1 mol Cl0.0140845 mol KCl1 mol KCl

−

= 0.0140845 mol Cl–

NaHCO3: Each mole of NaHCO3 forms 1 mole of Na+ ions and 1 mole of HCO3

– ions: NaHCO3(s) → Na+(aq) + HCO3–(aq)

Moles of NaHCO3 = ( ) 33

3

1 mol NaHCO0.315 g NaHCO

84.01 g NaHCO

= 0.00374955 mol NaHCO3

Moles of Na+ = ( )33

1 mol Na0.00374955 mol NaHCO1 mol NaHCO

+

= 0.00374955 mol Na+

Moles of HCO3– = ( ) 3

33

1 mol HCO0.00374955 mol NaHCO

1 mol NaHCO

−

= 0.00374955 mol HCO3–

4-26

NaBr Each mole of NaBr forms 1 mole of Na+ ions and 1 mole of Br– ions: NaBr(s) → Na+(aq) + Br–(aq)

Moles of NaBr = ( ) 1 mol NaBr0.098 g NaBr102.89 g NaBr

= 0.0009524735 mol NaBr

Moles of Na+ = ( ) 1 mol Na0.0009524735 mol NaBr1 mol NaBr

+

= 0.0009524735 mol Na+

Moles of Br– = ( ) 1 mol Br0.0009524735 mol NaBr1 mol NaBr

−

= 0.0009524735 mol Br–

Total moles of each ion: Cl–: 0.4534565 + 0.050415 + 0.0216255 + 0.0140845 = 0.5395815 mol Cl– Na+: 0.4534565 + 0.00374955 + 0.0009524735 = 0.458158523 mol Na+ Mg2+: 0.025207 + 0.0278285 = 0.0530355 mol Mg2+ SO4

2–: 0.0278308 mol SO42–

Ca2+: 0.0108128 mol Ca2+ K+: 0.0140845 mol K+ HCO3

–: 0.00374955 mol HCO3–

Br–: 0.0009524735 mol Br– Dividing each of the numbers of moles by the volume (0.97560976 L) and rounding to the proper number of significant figures gives the molarities.

M = molL

M Cl– = 0.5395815 mol Cl0.97560976 L

−

= 0.55307 = 0.553 M Cl–

M Na+ = 0.45815823 mol Na0.97560976 L

+

= 0.469612 = 0.470 M Na+

M Mg2+ = 20.0530355 mol Mg

0.97560976 L

+

= 0.054361 = 0.0544 M Mg2+

M SO42– =

240.0278308 mol SO

0.97560976 L

−

= 0.028526 = 0.0285 M SO42–

M Ca2+ = 20.0108128 mol Ca

0.97560976 L

+

= 0.011083 = 0.0111 M Ca2+

M K+ = 0.0140845 mol K0.97560976 L

+

= 0.014437 = 0.0144 M K+

M HCO3– = 30.00374955 mol HCO

0.97560976 L

−

= 0.003843 = 0.00384 M HCO3–

M Br– = 0.0009524735 mol Br0.97560976 L

−

= 0.0009763 = 0.00098 M Br–

b) The alkali metal cations are Na+ and K+. Add the molarities of the individual ions. 0.469612 M Na+ + 0.014437 M K+ = 0.484049 = 0.484 M total for alkali metal cations c) The alkaline earth metal cations are Mg2+ and Ca2+. Add the molarities of the individual ions. 0.054361 M Mg2+ + 0.011083 M Ca2+ = 0.065444 = 0.0654 M total for alkaline earth cations d) The anions are Cl–, SO4

2–, HCO3–, and Br–. Add the molarities of the individual ions.

0.55307 M Cl– + 0.028526 M SO42– + 0.003843 M HCO3

– + 0.0009763 M Br– = 0.5864153 = 0.586 M total for anions

4-27

4.35 Plan: Use the molarity and volume of the ions to find the moles of each ion. Multiply the moles of each ion by that ion’s charge to find the total moles of charge. Since sodium ions have a +1 charge, the total moles of charge equals the moles of sodium ions. Solution:


3 0.015 mol Ca1.0 x10 LL

+

= 15 mol Ca2+

Moles of charge from Ca2+ = ( )22

2 mol charge15 mol Ca1 mol Ca

++

= 30. mol charge from Ca2+

Moles of Fe3+ = ( )3

3 0.0010 mol Fe1.0x10 LL

+

= 1.0 mol Fe3+

Moles of charge from Fe3+ = ( )33

3 mol charge1.0 mol Fe1 mol Fe

++

= 3.0 mol charge from Fe3+

Total moles of charge = 30. mol + 3.0 mol = 33 mol charge

Moles Na+ = ( ) 1 mol Na33 mol charge1 mol charge

+

= 33 mol Na+

4.36 Plan: Review the definition of spectator ions.

Solution: Ions in solution that do not participate in the reaction do not appear in a net ionic equation. These spectator ions

remain as dissolved ions throughout the reaction. These ions are only present to balance charge. 4.37 Plan: Write the total ionic and net ionic equations for the reaction given. The total ionic equation shows all

soluble ionic substances dissociated into ions. The net ionic equation eliminates the spectator ions. New equations may be written by replacing the spectator ions in the given equation by other spectator ions.

Solution: The reaction given has the following total ionic and net ionic equations:

Total ionic equation: Ba2+(aq) + 2NO3–(aq) + 2Na+(aq) + CO3

2–(aq) → BaCO3(s) + 2Na+(aq) + 2NO3–(aq)

The spectator ions are underlined and are omitted: Net ionic equation: Ba2+(aq) + CO3

2–(aq) → BaCO3(s) New equations will contain a soluble barium compound and a soluble carbonate compound. The “new” equations are: Molecular: BaCl2(aq) + K2CO3(aq) → BaCO3(s) + 2KCl(aq) Total ionic: Ba2+(aq) + 2Cl–(aq) + 2K+(aq) + CO3

2–(aq) → BaCO3(s) + 2K+(aq) + 2Cl–(aq) Molecular: BaBr2(aq) + (NH4)2CO3(aq) → BaCO3(s) + 2NH4Br(aq) Total ionic: Ba2+(aq) + 2Br–(aq) + 2NH4

+(aq) + CO32–(aq) → BaCO3(s) + 2NH4

+(aq) + 2Br–(aq) 4.38 If the electrostatic attraction between the ions is greater than the attraction of the ions for water molecules, the ions will form a precipitate. This is the basis for the solubility rules. 4.39 Plan: Write the new cation-anion combinations as the products of the reaction and use the solubility rules to determine if any of the new combinations are insoluble. The spectator ions are the ions that are present in the soluble ionic compound. Solution:

a) Ca(NO3)2(aq) + 2NaCl(aq) → CaCl2(aq) + 2NaNO3(aq) Since the possible products (CaCl2 and NaNO3) are both soluble, no reaction would take place.

b) 2KCl(aq) + Pb(NO3)2(aq) → 2KNO3(aq) + PbCl2(s) According to the solubility rules, KNO3 is soluble but PbCl2 is insoluble so a precipitation reaction takes place. The K+ and NO3

– would be spectator ions, because their salt is soluble. 4.40 Plan: Use the solubility rules to predict the products of this reaction. Ions not involved in the precipitate are

spectator ions and are not included in the net ionic equation.

4-28

Solution: Assuming that the left beaker is AgNO3 (because it has gray Ag+ ions) and the right must be NaCl, then the NO3

– is blue, the Na+ is brown, and the Cl– is green. (Cl– must be green since it is present with Ag+ in the precipitate in the beaker on the right.) Molecular equation: AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)

Total ionic equation: Ag+(aq) + NO3–(aq) + Na+(aq) + Cl–(aq) → AgCl(s) + Na+(aq) + NO3

–(aq) Net ionic equation: Ag+(aq) + Cl–(aq) → AgCl(s) 4.41 Plan: Write the new cation-anion combinations as the products of the reaction and use the solubility rules to

determine if any of the new combinations are insoluble. The total ionic equation shows all soluble ionic substances dissociated into ions. The spectator ions are the ions that are present in

the soluble ionic compound. The spectator ions are omitted from the net ionic equation. Solution: a) Molecular: Hg2(NO3)2(aq) + 2KI(aq) → Hg2I2(s) + 2KNO3(aq) Total ionic: Hg2

2+(aq) + 2NO3–(aq) + 2K+(aq) + 2I–(aq) → Hg2I2(s) + 2K+(aq) + 2NO3

–(aq) Net ionic: Hg2

2+(aq) + 2I–(aq) → Hg2I2(s) Spectator ions are K+ and NO3

–. b) Molecular: FeSO4(aq) + Sr(OH)2(aq) → Fe(OH)2(s) + SrSO4(s) Total ionic: Fe2+(aq) + SO4

2–(aq) + Sr2+(aq) + 2OH–(aq) → Fe(OH)2(s) + SrSO4(s) Net ionic: This is the same as the total ionic equation because there are no spectator ions. 4.42 Plan: Write the new cation-anion combinations as the products of the reaction and use the solubility rules to

determine if any of the new combinations are insoluble. The total ionic equation shows all soluble ionic substances dissociated into ions. The spectator ions are the ions that are present in

the soluble ionic compound. The spectator ions are omitted from the net ionic equation. Solution: a) Molecular: 3CaCl2(aq) + 2Cs3PO4(aq) → Ca3(PO4)2(s) + 6CsCl(aq) Total ionic: 3Ca2+(aq) + 6Cl–(aq) + 6Cs+(aq) + 2PO4

3–(aq) → Ca3(PO4)2(s) + 6Cs+(aq) + 6Cl–(aq) Net ionic: 3Ca2+(aq) + 2PO4

3–(aq) → Ca3(PO4)2(s) Spectator ions are Cs+ and Cl–. b) Molecular: Na2S(aq) + ZnSO4(aq) → ZnS(s) + Na2SO4(aq) Total ionic: 2Na+(aq) + S2–(aq) + Zn2+(aq) + SO4

2–(aq) → ZnS(s) + 2Na+(aq) + SO42–(aq)

Net ionic: Zn2+(aq) + S2–(aq) → ZnS(s) Spectator ions are Na+ and SO4

2–. 4.43 Plan: A precipitate forms if reactant ions can form combinations that are insoluble, as determined by the solubility rules in Table 4.1. Create cation-anion combinations other than the original reactants and determine if they are insoluble. Any ions not involved in a precipitate are spectator ions and are omitted from the net ionic equation. Solution: a) NaNO3(aq) + CuSO4(aq) → Na2SO4(aq) + Cu(NO3)2(aq)

No precipitate will form. The ions Na+ and SO42– will not form an insoluble salt according to the first solubility

rule which states that all common compounds of Group 1A ions are soluble. The ions Cu2+ and NO3– will not

form an insoluble salt according to the solubility rule #2: All common nitrates are soluble. There is no reaction. b) A precipitate will form because silver ions, Ag+, and bromide ions, Br–, will combine to form a solid salt, silver

bromide, AgBr. The ammonium and nitrate ions do not form a precipitate. Molecular: NH4Br(aq) + AgNO3(aq) → AgBr(s) + NH4NO3(aq) Total ionic: NH4

+(aq) + Br–(aq) + Ag+(aq) + NO3–(aq) → AgBr(s) + NH4

+(aq) + NO3–(aq)

Net ionic: Ag+(aq) + Br–(aq) → AgBr(s) 4.44 Plan: A precipitate forms if reactant ions can form combinations that are insoluble, as determined by the solubility rules in Table 4.1. Create cation-anion combinations other than the original reactants and determine if they are insoluble. Any ions not involved in a precipitate are spectator ions and are omitted from the net ionic equation. Solution:

a) Barium carbonate (BaCO3) precipitates since the solubility rules state that all common carbonates are insoluble.

4-29

Molecular: K2CO3(aq) + Ba(OH)2(aq) → BaCO3(s) + 2KOH(aq) Total ionic: 2K+(aq) + CO3

2–(aq) + Ba2+(aq) + 2OH–(aq) → BaCO3(s) + 2K+(aq) + 2OH–(aq) Net ionic: Ba2+(aq) + CO3

2–(aq) → BaCO3(s) b) Aluminum phosphate (AlPO4) precipitates since most common phosphates are insoluble; the sodium nitrate is soluble.

Molecular: Al(NO3)3(aq) + Na3PO4(aq) → AlPO4(s) + 3NaNO3(aq) Total ionic: Al3+(aq) + 3NO3

–(aq) + 3Na+(aq) + PO43–(aq) → AlPO4(s) + 3Na+(aq) + 3NO3

–(aq) Net ionic: Al3+(aq) + PO4

3–(aq) → AlPO4(s) 4.45 Plan: A precipitate forms if reactant ions can form combinations that are insoluble, as determined by the solubility rules in Table 4.1. Create cation-anion combinations other than the original reactants and determine if they are insoluble. Any ions not involved in a precipitate are spectator ions and are omitted from the net ionic equation. Solution:

a) New cation-anion combinations are potassium nitrate (KNO3) and iron(III) chloride (FeCl3). The solubility rules state that all common nitrates and chlorides (with some exceptions) are soluble, so no precipitate forms.

3KCl(aq) + Fe(NO3)3(aq) → 3KNO3(aq) + FeCl3(aq) b) New cation-anion combinations are ammonium chloride and barium sulfate. The solubility rules state that most chlorides are soluble; however, another rule states that sulfate compounds containing barium are insoluble. Barium sulfate is a precipitate and its formula is BaSO4.

Molecular: (NH4)2SO4(aq) + BaCl2(aq) → BaSO4(s) + 2NH4Cl(aq) Total ionic: 2NH4

+(aq) + SO42–(aq) + Ba2+(aq) + 2Cl–(aq) → BaSO4(s) + 2NH4

+(aq) + 2Cl–(aq) Net ionic: Ba2+(aq) + SO4

2–(aq) → BaSO4(s) 4.46 Plan: A precipitate forms if reactant ions can form combinations that are insoluble, as determined by the solubility rules in Table 4.1. Create cation-anion combinations other than the original reactants and determine if they are insoluble. Any ions not involved in a precipitate are spectator ions and are omitted from the net ionic equation. Solution:

a) New cation-anion combinations are nickel(II) sulfide (NiS) and sodium sulfate (Na2SO4). The solubility rules state that all compounds of Group 1A(1) like sodium are soluble; another rule states that common sulfide compounds are insoluble. Nickel(II) sulfide is a precipitate.

Molecular: Na2S(aq) + NiSO4(aq) → NiS(s) + Na2SO4(aq) Total ionic: 2Na+(aq) + S–2(aq) + Ni2+(aq) + SO4

2–(aq) → NiS(s) + 2Na+(aq) + SO42–(aq)

Net ionic: Ni2+(aq) + S2–(aq) → NiS(s) b) New cation-anion combinations are lead(II) bromide (PbBr2) and potassium nitrate (KNO3). The solubility rules state that all common nitrate compounds are soluble; another rule states that while most common bromide compounds are soluble, lead(II) bromide is insoluble. Lead(II) bromide is a precipitate.

Molecular: Pb(NO3)2(aq) + 2KBr(aq) → PbBr2(s) + 2KNO3(aq) Total ionic: Pb2+(aq) + 2NO3

–(aq) + 2K+(aq) + 2Br–(aq) → PbBr2(s) + 2K+(aq) + 2NO3–(aq)

Net ionic: Pb2+(aq) + 2Br–(aq) → PbBr2(s) 4.47 Plan: Write a balanced equation for the chemical reaction described in the problem. By applying the solubility rules to the two possible products (NaNO3 and PbI2), determine that PbI2 is the precipitate. By using molar relationships, determine how many moles of Pb(NO3)2 are required to produce 0.628 g of PbI2. The molarity is calculated by dividing moles of Pb(NO3)2 by its volume in liters. Solution: The reaction is: Pb(NO3)2(aq) + 2NaI(aq) → PbI2(s) + 2NaNO3(aq).

Moles of Pb(NO3)2 = ( ) 3 222

2 2

1 mol Pb(NO )1 mol PbI0.628 g PbI

461.0 g PbI 1 mol PbI

= 0.001362256 mol Pb(NO3)2

Moles of Pb2+ = moles of Pb(NO3)2 = 0.001362256 mol Pb2+

Molarity of Pb2+ = 2

2 3moles Pb 0.001362256 mol 1 mL =

38.5 mLvolume of Pb 10 L

+

+ −

= 0.035383 = 0.0354 M Pb2+

4-30

4.48 Plan: Write a balanced equation for the chemical reaction described in the problem. By applying the solubility rules to the two possible products (KNO3 and AgCl), determine that AgCl is the precipitate. By using molar relationships, determine how many moles of AgNO3 are required to produce 0.842 g of AgCl. The molarity is calculated by dividing moles of AgNO3 by its volume in liters. Solution:

The reaction is AgNO3(aq) + KCl(aq) → AgCl(s) + KNO3(aq).

Moles of AgNO3 = ( ) 31 mol AgNO1 mol AgCl0.842 g AgCl143.4 g AgCl 1 mol AgCl

= 0.0058717 mol AgNO3

Moles of Ag+ = moles of AgNO3 = 0.0058717 mol Ag+

Molarity of Ag+ = 3moles Ag 0.0058717 mol 1 mL =

25.0 mLvolume of Ag 10 L

+

+ −

= 0.2348675 = 0.235 M Ag+

4.49 Plan: The first step is to write and balance the chemical equation for the reaction. Multiply the molarity and

volume of each of the reactants to determine the moles of each. To determine which reactant is limiting, calculate the amount of barium sulfate formed from each reactant, assuming an excess of the other reactant. The reactant that produces less product is the limiting reagent. Use the limiting reagent and the mole ratio from the balanced chemical equation to determine the mass of barium sulfate formed.

Solution: The balanced chemical equation is: BaCl2(aq) + Na2SO4(aq) → BaSO4(s) + 2NaCl(aq)

Moles of BaCl2 = ( )3

20.160 mol BaCl10 L35.0 mL1 mL 1 L

−

= 0.00560 mol BaCl2

Finding the moles of BaSO4 from the moles of BaCl2 (if Na2SO4 is in excess):

Moles of BaSO4 from BaCl2 = ( ) 42

2

1 mol BaSO0.00560 moL BaCl

1 mol BaCl

= 0.00560 mol BaSO4

Moles of Na2SO4 = ( )3

2 40.065 mol Na SO10 L58.0 mL1 mL 1 L

−

= 0.00377 mol Na2SO4

Finding the moles of BaSO4 from the moles of Na2SO4 (if BaCl2 is in excess):

Moles BaSO4 from Na2SO4 = ( ) 42 4

2 4

1 mol BaSO0.00377 moL Na SO

1 mol Na SO

= 0.00377 mol BaSO4

Sodium sulfate is the limiting reactant. Converting from moles of BaSO4 to mass:

Mass (g) of BaSO4 = ( ) 44

4

233.4 g BaSO0.0377 moL BaSO

1 mol BaSO

= 0.879918 = 0.88 g BaSO4

4.50 Plan: The first step is to write and balance the chemical equation for the reaction. To determine which reactant is

limiting, calculate the amount of iron(III) sulfide formed from each reactant, assuming an excess of the other reactant. The reactant that produces less product is the limiting reagent. Multiply the molarity and volume (in L) of each of the reactants to determine the moles of each. Use the mole ratio from the balanced chemical equation and the molar mass of iron(III) sulfide to determine the mass of iron(III) sulfide formed.

Solution: The balanced chemical equation is: 2FeCl3(aq) + 3CaS(aq) → Fe2S3(s) + 3CaCl2(aq) Finding the mass of Fe2S3 from the molarity and volume of FeCl3 (if CaS is in excess): Mass of Fe2S3 from FeCl3 =

62.0 mL FeCl3�1 L

103 mL� �0.135 mol FeCl3

1 L� �1 mol Fe2S3

2 mol FeCl3� �207.88 g Fe2S3

1 mol Fe2S3� = 0.870 g Fe2S3

Finding the mass of Fe2S3 from the molarity and volume of CaS (if FeCl3 is in excess): Mass of Fe2S3 from CaS =

4-31

45.0 mL CaS� 1 L 103 mL

� �0.285 mol CaS1 L

� �1 mol Fe2S3

3 mol CaS� �207.88 g Fe2S3

1 mol Fe2S3� = 0.889 g Fe2S3

FeCl3 is the limiting reactant.

0.870 g of Fe2S3 are formed. 4.51 Plan: A precipitate forms if reactant ions can form combinations that are insoluble, as determined by the solubility rules in Table 4.1. Create cation-anion combinations other than the original reactants and determine if they are insoluble. Any ions not involved in a precipitate are spectator ions and are omitted from the net ionic equation. Use the molar ratio in the balanced net ionic equation to calculate the mass of product. Solution:

a) The yellow spheres cannot be ClO4– or NO3

– as these ions form only soluble compounds. So the yellow sphere must be SO4

2–. The only sulfate compounds possible that would be insoluble are Ag2SO4 and PbSO4. The precipitate has a 1:1 ratio between its ions. Ag2SO4 has a 2:1 ratio between its ions. Therefore the blue spheres are Pb2+ and the yellow spheres are SO4

2–. The precipitate is thus PbSO4. b) The net ionic equation is Pb2+(aq) + SO4

2– (aq) → PbSO4(s).

c) Mass (g) of PbSO4 = ( )4 2+

2+ 4 42+ 2+

4

5.0x10 mol Pb 1 mol PbSO 303.3 g PbSO10 Pb spheres1 mol PbSO1 Pb sphere 1 mol Pb

−

= 1.5165 = 1.5 g PbSO4 4.52 Plan: A precipitate forms if reactant ions can form combinations that are insoluble, as determined by the solubility rules in Table 4.1. Create cation-anion combinations other than the original reactants and determine if they are insoluble. Any ions not involved in a precipitate are spectator ions and are omitted from the net ionic equation. Use the molar ratio in the balanced net ionic equation to calculate the mass of product. Solution:

a) There are 9 purple spheres representing cations and 7 green spheres representing anions. In the precipitate, there are 8 purple spheres (cations) and 4 green spheres (anions), indicating a 2:1 ratio between cation and anion in the compound. Only Reaction 3 produces a precipitate (Ag2SO4) fitting this description: Li2SO4(aq) + 2AgNO3(aq) → 2LiNO3(aq) + Ag2SO4(s) Reaction 1 does not produce a precipitate since all common nitrate and chloride compounds are soluble. Reaction 2 does not produce a precipitate since all common perchlorate and chloride compounds are soluble. Reaction 4 produces a precipitate, PbBr2, but it has a cation:anion ratio of 1:2, instead of 2:1. Total ionic equation for Reaction 3 = 2Li+(aq) + SO4

2–(aq) + 2Ag+(aq) + 2NO3–(aq) → 2Li+(aq) + 2NO3

–(aq) + Ag2SO4(s) Net ionic equation = 2Ag+(aq) + SO4

2–(aq) → Ag2SO4(s) b) There are 4 unreacted spheres of ions.

Number of ions = ( )3 232.5x10 mol ions 6.022x10 ions4 spheres

1 sphere 1 mol ions

−

= 6.022x1021 = 6.0x1021 ions

c) Mass (g) of solid =

( )3 2

2 4 2 4 2 44 2

2 44

2.5x10 mol SO ions 1 mol Ag SO 311.9 g Ag SO4 spheres of SO ions1 sphere 1 mol Ag SO1 mol SO

− −−

−

= 3.119 = 3.1 g solid 4.53 Plan: Multiply the molarity of the CaCl2 solution by its volume in liters to determine the number of moles of

CaCl2 reacted. Write the precipitation reaction using M to represent the alkali metal. Use the moles of CaCl2 reacted and the molar ratio in the balanced equation to find the moles of alkali metal carbonate. Divide the mass of that carbonate by the moles to obtain the molar mass of the carbonate. Subtract the mass of CO3 from the molar mass to obtain the molar mass of the alkali metal, which can be used to identify the alkali metal and the formula and name of the compound. Solution: The reaction is: M2CO3(aq) + CaCl2(aq) → CaCO3(s) + 2MCl(aq). Since the alkali metal M forms a +1 ion and carbonate is a –2 ion, the formula is M2CO3.

4-32

Moles of CaCl2 = ( )3

20.350 mol CaCl10 L31.10 mL1 mL 1 L

−

= 0.010885 mol CaCl2

Moles of M2CO3 = ( ) 2 32

2

1 mol M CO0.010885 mol CaCl

1 mol CaCl

= 0.010885 mol M2CO3

Molar mass (g/mol) of M2CO3 = 2 3

2 3

mass of M CO 1.50 g = moles of M CO 0.010885 mol

= 137.804 g/mol

Molar mass (g/mol) of M2 = molar mass of M2CO3 – molar mass of CO3 = 137.804 g/mol – 60.01 g/mol = 77.79 g/mol Molar mass (g/mol) of M = 77.79 g/mol/2 = 38.895 g/mol = 38.9 g/mol This molar mass is closest to that of potassium; the formula of the compound is K2CO3, potassium carbonate. 4.54 Plan: Multiply the molarity of the AgNO3 solution by its volume in liters to determine the number of moles of

AgNO3 reacted. Write the precipitation reaction using M to represent the metal in the chloride compound. Use the moles of AgNO3 reacted and the molar ratio in the balanced equation to find the moles of metal chloride. Divide the mass of the chloride by the moles to obtain the molar mass of the chloride. Subtract the mass of Cl2 from the molar mass to obtain the molar mass of the metal, which can be used to identify the metal and the formula and name of the compound. Solution: The reaction is: MCl2(aq) + 2AgNO3(aq) → 2AgCl(s) + M(NO3)2(aq).

Since the metal M is a +2 ion and chloride is a –1 ion, the formula of the metal chloride is MCl2.

Moles of AgNO3 = ( )3

30.515 mol AgNO10 L22.40 mL1 mL 1 L

−

= 0.011536 mol AgNO3

Moles of MCl2 = ( ) 23

3

1 mol MCl0.011536 mol AgNO

2 mol AgNO

= 0.005768 mol MCl2

Molar mass (g/mol) of MCl2 = 2

2

mass of MCl 0.750 g = moles of MCl 0.005768 mol

= 130.0277 g/mol

Molar mass (g/mol) of M = molar mass of MCl2 – molar mass of Cl2 = 130.0277 g/mol – 70.90 g/mol = 59.1277 = 59.1 g/mol This molar mass is closest to that of nickel; the compound is NiCl2, nickel(II) chloride. 4.55 Plan: Write a balanced equation for the reaction. Find the moles of AgNO3 by multiplying the molarity and

volume of the AgNO3 solution; use the molar ratio in the balanced equation to find the moles of Cl– present in the 25.00 mL sample. Then, convert moles of Cl– into grams, and convert the sample volume into grams using the given density. The mass percent of Cl– is found by dividing the mass of Cl– by the mass of the sample volume and multiplying by 100.

Solution: The balanced equation is AgNO3(aq) + Cl–(aq) → AgCl(s) + NO3

–(aq).

Moles of AgNO3 = ( )3

30.2970 mol AgNO10 L53.63 mL1 mL L

−

= 0.01592811 mol AgNO3

Mass (g) of Cl– = ( )33

1 mol Cl 35.45 g Cl0.01592811 mol AgNO1 mol AgNO 1 mol Cl

−

−

= 0.56465 g Cl–

Mass (g) of seawater sample = ( ) 1.024 g25.00 mLmL

= 25.60 g sample

Mass % Cl– = mass Cl x 100%mass sample

− = 0.56465 g Cl x 100%

25.60 g sample

− = 2.20566 = 2.206% Cl–

4-33

4.56 Plan: Write the reaction between aluminum sulfate and sodium hydroxide and check the solubility rules to determine the precipitate. Spectator ions are omitted from the net ionic equation. Find the moles of sodium hydroxide by multiplying its molarity by its volume in liters; find the moles of aluminum sulfate by converting grams per liter to moles per liter and multiplying by the volume of that solution. To determine which reactant is limiting, calculate the amount of precipitate formed from each reactant, assuming an excess of the other reactant, using the molar ratio from the balanced equation. The smaller amount of precipitate is the answer. Solution: a) According to the solubility rules, most common sulfate compounds are soluble, but most common hydroxides are insoluble. Aluminum hydroxide is the precipitate. Total ionic equation: Al2(SO4)3(aq) + 6NaOH(aq) → 3Na2SO4(aq) + 2Al(OH)3(s)

Net ionic equation: Al3+(aq) + 3OH–(aq) → Al(OH)3(s)

b) Moles of Al2(SO4)3 = ( )3

2 4 3 2 4 3

2 4 3

15.8 g Al (SO ) 1 mol Al (SO )10 L627 mL1 mL L 342.14 g Al (SO )

−

= 0.028955 mol Al2(SO4)3

Mass (g) of Al(OH)3 from Al2(SO4)3 = ( ) 3 32 4 3

2 4 3 3

2 mol Al(OH) 78.00 g Al(OH)0.028955 mol Al (SO )

1 mol Al (SO ) 1 mol Al(OH)

= 4.5170 g Al(OH)3

Moles of NaOH = ( )310 L 0.533 mol NaOH185.5 mL

1 mL L

−

= 0.0988715 mol NaOH

Mass (g) of Al(OH)3 from NaOH = ( ) 3 3

3

2 mol Al(OH) 78.00 g Al(OH)0.0988715 mol NaOH

6 mol NaOH 1 mol Al(OH)

= 2.570659 = 2.57 g Al(OH)3 NaOH is the limiting reagent. 4.57 Plan: Write the chemical reaction between the two reactants. Then write the total ionic equation in which all soluble ionic substances are dissociated into ions. Omit spectator ions in the net ionic equation. Solution: The molecular equation is: H2SO4(aq) + Sr(OH)2(aq) → SrSO4(s) + 2H2O(l) The total ionic equation is: 2H+(aq) + SO4

2–(aq) + Sr2+(aq) + 2OH–(aq) → SrSO4(s) + 2H2O(l) According to the solubility rules, SrSO4 is insoluble and therefore does not dissociate into ions. Since there are no spectator ions, the total and net ionic equations are the same. 4.58 Plan: H+ is used to represent an acid while OH– is used to represent the base. In a neutralization

reaction, stoichiometric amounts of an acid and a base react to form water. Solution: The general equation for a neutralization reaction is H+(aq) + OH–(aq) → H2O(l).

4.59 Plan: Review the section on acid-base reactions.

Solution: a) Any three of HCl, HBr, HI, HNO3, H2SO4, or HClO4

b) Any three of NaOH, KOH, Ca(OH)2, Sr(OH)2, Ba(OH)2 c) Strong acids and bases dissociate 100% into ions in aqueous solution. 4.60 Plan: Review the section on acid-base reactions.

Solution: a) There are many possibilities including: acetic acid (HC2H3O2), chlorous acid (HClO2), and nitrous acid (HNO2). All acids are weak except for the six strong acids listed in the text.

b) NH3 c) Strong acids and bases dissociate 100% into ions and are therefore strong electrolytes; weak acids and bases dissociate much less than this (typically less than 10%) in aqueous solution and are therefore weak electrolytes. The

4-34

electrical conductivity of a solution of a strong acid or base would be much higher than that of a weak acid or base of equal concentration.

4.61 Plan: Formation of a gaseous product or a precipitate drives a reaction to completion.

Solution: a) The formation of a gas, SO2(g), and formation of water drive this reaction to completion, because both products

remove reactants from solution. b) The formation of a precipitate, Ba3(PO4)2(s), will cause this reaction to go to completion. This reaction is one between an acid and a base, so the formation of water molecules through the combination of H+ and OH– ions also drives the reaction. 4.62 Plan: Since strong acids and bases dissociate completely in water, these substances can be written as ions in a

total ionic equation; since weak acids and bases dissociate into ions only to a small extent, these substances appear undissociated in total ionic equations. Solution: a) Acetic acid is a weak acid and sodium hydroxide is a strong base: Molecular equation: CH3COOH(aq) + NaOH(aq) → CH3COONa(aq) + H2O(l) Total ionic equation: CH3COOH (aq) + Na+(aq) + OH–(aq) → Na+(aq) + CH3COO–(aq) + H2O(l)

Net ionic equation (remove the spectator ion Na+): CH3COOH(aq) + OH–(aq) → CH3COO–(aq) + H2O(l) Hydrochloric acid is a strong acid: Molecular equation: HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) Total ionic equation: H+(aq) + Cl–(aq) + Na+(aq) + OH–(aq) → Na+(aq) + Cl–(aq) + H2O(l) Net ionic equation (remove the spectator ions Na+ and Cl–): H+(aq) + OH–(aq) → H2O(l) The difference in the net ionic equation is due to the fact that CH3COOH is a weak acid and dissociates very little

while HCl is a strong acid and dissociates completely. b) When acetic acid dissociates in water, most of the species in the solution is un-ionized acid, CH3COOH(aq); the amounts of its ions, H+ and CH3COO– , are equal but very small: [CH3COOH] >> [H+] = [CH3COO–].

4.63 Plan: The acids in this problem are all strong acids, so you can assume that all acid molecules dissociate completely to yield H+ ions and associated anions. One mole of HClO4, HNO3, and HCl each produce one mole of H+ upon dissociation, so moles H+ = moles acid. Calculate the moles of acid by multiplying the molarity (moles/L) by the volume in liters. Solution: a) HClO4(aq) → H+(aq) + ClO4

–(aq)

Moles H+ = mol HClO4 = ( )

L 1mol 0.25 L 40.1 = 0.35 mol H+

b) HNO3(aq) → H+(aq) + NO3–(aq)

Moles H+ = mol HNO3 = ( )310 L 0.92 mol6.8 mL

1 mL 1 L

−

= 6.256x10–3 = 6.3x10–3 mol H+

c) HCl(aq) → H+(aq) + Cl–(aq)

Moles H+ = mol HCl = ( ) 0.085 mol2.6 L1 L

= 0.221 = 0.22 mol H+

4.64 Plan: The acids in this problem are all strong acids, so you can assume that all acid molecules dissociate completely to yield H+ ions and associated anions. One mole of HBr, HI, and HNO3 each produce one mole of H+ upon dissociation, so moles H+ = moles acid. Calculate the moles of acid by multiplying the molarity (moles/L) by the volume in liters. Solution: a) HBr(aq) → H+(aq) + Br–(aq)

Moles H+ = mol HBr = ( )310 L 0.75 mol1.4 mL

1 mL 1 L

−

= 1.05x10–3 = 1.0x10–3 mol H+

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b) HI(aq) → H+(aq) + I–(aq)

Moles H+ = mol HI = ( )310 L 1.98 mol2.47 mL

1 mL 1 L

−

= 4.8906x10–3 = 4.89x10–3 mol H+

c) HNO3(aq) → H+(aq) + NO3–(aq)

Moles H+ = mol HNO3 = ( )310 L 0.270 mol395 mL

1 mL 1 L

−

= 0.10665 = 0.107 mol H+

4.65 Plan: Remember that strong acids and bases can be written as ions in the total ionic equation but weak acids and

bases cannot be written as ions. Omit spectator ions from the net ionic equation. Solution: a) KOH is a strong base and HBr is a strong acid; both may be written in dissociated form. KBr is a soluble compound since all Group 1A(1) compounds are soluble. Molecular equation: KOH(aq) + HBr(aq) → KBr(aq) + H2O(l)

Total ionic equation: K+(aq) + OH–(aq) + H+(aq) + Br–(aq) → K+(aq) + Br–(aq) + H2O(l) Net ionic equation: OH–(aq) + H+(aq) → H2O(l) The spectator ions are K+(aq) and Br–(aq). b) NH3 is a weak base and is written in the molecular form. HCl is a strong acid and is written in the dissociated form (as ions). NH4Cl is a soluble compound, because all ammonium compounds are soluble.

Molecular equation: NH3(aq) + HCl(aq) → NH4Cl(aq) Total ionic equation: NH3(aq) + H+(aq) + Cl–(aq) → NH4

+(aq) + Cl–(aq) Net ionic equation: NH3(aq) + H+(aq) → NH4

+(aq) Cl– is the only spectator ion. 4.66 Plan: Remember that strong acids and bases can be written as ions in the total ionic equation but weak acids and

bases cannot be written as ions. Omit spectator ions from the net ionic equation. Solution: a) CsOH is a strong base and HNO3 is a strong acid; both may be written in dissociated form. CsNO3 is a soluble compound since all nitrate compounds are soluble. Molecular equation: CsOH(aq) + HNO3(aq) → CsNO3(aq) + H2O(l)

Total ionic equation: Cs+(aq) + OH–(aq) + H+(aq) + NO3–(aq) → Cs+(aq) + NO3

–(aq) + H2O(l) Net ionic equation: OH–(aq) + H+(aq) → H2O(l) Spectator ions are Cs+ and NO3

–. b) HC2H3O2 is a weak acid and is written in the molecular form. Ca(OH)2 is a strong base and is written in the dissociated form (as ions). Ca(C2H3O2)2 is a soluble compound, because all acetate compounds are soluble. Molecular equation: Ca(OH)2(aq) + 2HC2H3O2(aq) → Ca(C2H3O2)2(aq) + 2H2O(l)

Total ionic equation: Ca2+(aq) + 2OH–(aq) + 2HC2H3O2 (aq) → Ca2+(aq) + 2C2H3O2–(aq) + 2H2O(l)

Net ionic equation: OH–(aq) + HC2H3O2(aq) → C2H3O2–(aq) + H2O(l)

Spectator ion is Ca2+. 4.67 Plan: Write an acid-base reaction between CaCO3 and HCl. Remember that HCl is a strong acid.

Solution: Calcium carbonate dissolves in HCl(aq) because the carbonate ion, a base, reacts with the acid to form H2CO3 which decomposes into CO2(g) and H2O(l).

CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2CO3(aq) Total ionic equation: CaCO3(s) + 2H+(aq) + 2Cl–(aq) → Ca2+(aq) + 2Cl–(aq) + H2O(l) + CO2(g) Net ionic equation: CaCO3(s) + 2H+(aq) → Ca2+(aq) + H2O(l) + CO2(g)

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4.68 Plan: Write an acid-base reaction between Zn(OH)2 and HNO3. Remember that HNO3 is a strong acid. Solution: Zinc hydroxide dissolves in HCl(aq) because the hydroxide ion, a base, reacts with the acid to form soluble zinc nitrate and water. Zn(OH)2(s) + 2HNO3(aq) → Zn(NO3)2(aq) + 2H2O(aq)

Total ionic equation: Zn(OH)2(s) + 2H+(aq) + 2NO3

–(aq) → Zn2+(aq) + 2NO3–(aq) + 2H2O(l)

Net ionic equation: Zn(OH)2(s) + 2H+(aq) → Zn2+(aq) + 2H2O(l) 4.69 Plan: Convert the mass of calcium carbonate to moles, and use the mole ratio in the balanced chemical equation

to find the moles of hydrochloric acid required to react with these moles of calcium carbonate. Use the molarity of HCl to find the volume that contains this number of moles. Solution:

2HCl(aq) + CaCO3(s) → CaCl2(aq) + CO2(g) + H2O(l) Converting from grams of CaCO3 to moles:

Moles of CaCO3 = ( ) 33

3

1 mol CaCO16.2 g CaCO

100.09 g CaCO

= 0.161854 mol CaCO3

Converting from moles of CaCO3 to moles of HCl:

Moles of HCl = ( )33

2 mol HCl0.161854 mol CaCO1 mol CaCO

= 0.323708 mol HCl

Converting from moles of HCl to volume:

Volume (mL) of HCl = ( ) 31 L 1 mL0.323708 mol HCl

0.383 mol HCl 10 L−

= 845.1906 = 845 mL HCl solution

4.70 Plan: Convert the volume of NaOH solution to liters and multiply by the molarity of the solution to obtain

moles of NaOH. Use the mole ratio in the balanced chemical equation to find the moles of NaH2PO4 required to react with these moles of NaOH. Finally, convert moles of NaH2PO4 to moles. Solution:

NaH2PO4(s) + 2NaOH(aq) → Na3PO4(aq) + 2H2O(l)

Volume (L) = ( )310 L43.74 mL

1 mL

−

= 0.04374 mL

Finding moles of NaOH:

Moles of NaOH = ( ) 0.285 mol NaOH0.04374 L1 L

= 0.0124659 mol NaOH

Converting from moles of NaOH to moles of NaH2PO4:

Moles of NaH2PO4 = ( ) 2 41 mol NaH PO0.0124659 mol NaOH

2 mol NaOH

= 0.00623295 mol NaH2PO4

Converting from moles of NaH2PO4 to mass:

Mass (g) of NaH2PO4 = ( ) 2 42 4

2 4

119.98 g NaH PO0.00623295 mol NaH PO

1 mol NaH PO

= 0.747829 = 0.748 g NaH2PO4

4.71 Plan: Write a balanced equation. Find the moles of KOH from the molarity and volume information and use the

molar ratio in the balanced equation to find the moles of acid present. Divide the moles of acid by its volume to determine the molarity. Solution: The reaction is: KOH(aq) + CH3COOH(aq) → CH3COOK(aq) + H2O(l)

4-37

Moles of KOH = ( )310 L 0.1180 mol KOH25.98 mL

1 mL L

−

= 0.00306564 mol KOH

Moles of CH3COOH = ( ) 31mol CH COOH0.00306564 mol KOH

1 mol KOH

= 0.00306564 mol CH3COOH

Molarity of CH3COOH = 33

0.00306564 mol CH COOH 1 mL52.50 mL 10 L−

= 0.05839314 = 0.05839 M CH3COOH

4.72 Plan: Write a balanced equation. Find the moles of NaOH from the molarity and volume information and use the

molar ratio in the balanced equation to find the moles of acid present. Divide the moles of acid by its volume to determine the molarity. Solution: The reaction is: 2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2H2O(l)


1 mL L

−

= 0.00485625 mol NaOH

Moles of H2SO4 = ( ) 2 41mol H SO0.00485625 mol NaOH

2 mol NaOH

= 0.002428125 mol H2SO4

Molarity of H2SO4 = 2 43

0.002428125 mol H SO 1 mL25.00 mL 10 L−

= 0.097125 = 0.09712 M H2SO4

4.73 Plan: Write a balanced equation. Find the moles of H2SO4 from the molarity and volume information and use the

molar ratio in the balanced equation to find the moles of NaHCO3 required to react with that amount of H2SO4. Divide the moles of NaHCO3 by its molarity to find the volume. Solution: The reaction is: 2NaHCO3(aq) + H2SO4(aq) → Na2SO4(aq) + 2 H2O(l) + 2CO2(g)

Moles of H2SO4 = ( )3

2 42.6 mol H SO10 L88 mL1 mL L

−

= 0.2288 mol H2SO4

Moles of NaHCO3 = ( ) 32 4

2 4

2mol NaHCO0.2288 mol H SO

1 mol H SO

= 0.4576 mol NaHCO3

Volume (mL) of NaHCO3 = ( )3 33

1 L 1 mL0.4576 mol NaHCO1.6 mol NaHCO 10 L−

= 286 = 2.9 x 102 mL NaHCO3 4.74 Plan: Write a balanced equation. Find the moles of HCl added from the molarity and volume information and use

the molar ratio in the balanced equation to find the moles of NaOH present. Find the volume of NaOH from the difference in buret readings. Divide the moles of NaOH by its volume to determine the molarity. Solution: The reaction is: NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)

Moles of HCl = ( )310 L 0.1528 mol HCl25.00 mL

1 mL L

−

= 0.00382 mol HCl

Moles of NaOH = ( ) 1mol NaOH0.00382 mol HCl1 mol HCl

= 0.00382 mol NaOH

Volume (mL) of NaOH = final buret reading – initial buret reading = 39.21 mL – 2.24 mL = 36.97 mL NaOH

Molarity of NaOH =3

0.00382 mol NaOH 1 mL36.97 mL 10 L−

= 0.103327 = 0.1033 M NaOH

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4.75 Plan: Write balanced equations for the reaction of NaOH with oxalic acid, benzoic acid, and HCl. Find the moles of added NaOH from the molarity and volume information; then use the molarity and volume information for HCl to find the moles of HCl required to react with the excess NaOH. Use the molar ratio in the NaOH/HCl reaction to find the moles of excess NaOH. The moles of NaOH required to titrate the acid samples is the difference of the added NaOH and the excess NaOH. Let x = mass of benzoic acid and 0.3471 – x = mass of oxalic acid. Convert the mass of each acid to moles using the molar mass and use the molar ratios in the balanced reactions to find the amounts of each acid. Mass percent is calculated by dividing the mass of benzoic acid by the mass of the sample and multiplying by 100. Solution: Oxalic acid is H2C2O4 and benzoic acid is HC7H5O2. The reactions are: NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)

2NaOH(aq) + H2C2O4(aq) → Na2C2O4(aq) + 2H2O(l) NaOH(aq) + HC7H5O2(aq) → NaC7H5O2(aq) + H2O(l)

Moles of NaOH added = ( )310 L 0.1000 mol NaOH100.0 mL

1 mL 1 L

−

= 0.01000 mol NaOH

Moles of added HCl = ( )310 L 0.2000 mol HCl20.00 mL

1 mL 1 L

−

= 0.004000 mol HCl

Moles of excess NaOH = ( ) 1 mol NaOH0.004000 mol HCl1 mol HCl

= 0.004000 mol NaOH

Moles of NaOH required to titrate sample = moles NaOH added – moles excess NaOH = 0.01000 mol – 0.004000 mol = 0.006000 mol NaOH

Let x = mass of HC7H5O2 and 0.3471 – x = mass of H2C2O4 Moles of NaOH required to titrate HC7H5O2 =

( ) 7 5 27 5 2

7 5 2 7 5 2

1 mol HC H O 1 mol NaOHx g HC H O122.12 g HC H O 1 mol HC H O

= 0.008189x

Moles of NaOH required to titrate H2C2O4 = ( ) 2 2 42 2 4

2 2 4 2 2 4

1 mol H C O 2 mol NaOH(0.3471 x) g H C O90.04 g H C O 1 mol H C O

−

= 0.007710 – 0.02221x Moles of NaOH required to titrate sample = 0.006000 mol = 0.008189x + (0.007710 – 0.02221x) 0.006000 = 0.007710 – 0.014021x –0.001710 = –0.014021x 0.12196 = x = mass of HC7H5O2

Mass % of HC7H5O2 = ( ) ( )7 5 2mass of HC H O 0.12196 g100 = 100mass of sample 0.3471 g

= 35.1368 = 35.14%

4.76 Plan: Balance the reaction. Convert the amount of UO2 from kg to g to moles; use the molar ratio in the balanced reaction to find the moles of HF required to react with the moles of UO2. Divide moles of HF by its molarity to calculate the volume. Solution:

The reaction is: UO2(s) + 4HF(aq) → UF4(s) + 2H2O(l)

Moles of UO2 = ( )3

22

2

1 mol UO10 g2.15 kg UO1 kg 270.0 g UO

= 7.96296 mol UO2

Moles of HF = ( )22

4 mol HF7.96296 mol UO1 mol UO

= 31.85184 mol HF

Volume (L) of HF = ( ) 1 L31.85184 mol HF2.40 mol HF

= 13.2716 = 13.3 L HF

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4.77 Plan: Write balanced reactions between HNO3 and each of the bases. Find the moles of HNO3 from its molarity and volume. Let x = mass of Al(OH)3 and 0.4826 – x = mass of Mg(OH)2. Convert the mass of each base to moles using the molar mass and use the molar ratios in the balanced reactions to find the amounts of each base. Mass percent is calculated by dividing the mass of Al(OH)3 by the mass of the sample and multiplying by 100. Solution: The reactions are: 3HNO3(aq) + Al(OH)3(aq) → Al(NO3)3(aq) + 3H2O(l)

2HNO3(aq) + Mg(OH)2(aq) → Mg(NO3)2(aq) + 2H2O(l)

Moles of HNO3 = ( )3

31.000 mol HNO10 L17.30 mL1 mL 1 L

−

= 0.0173 mol HNO3

Let x = mass of Al(OH)3 and 0.4826 – x = mass of Mg(OH)2 Moles of HNO3 required to titrate Al(OH)3 =

( ) 3 33

3 3

1 mol Al(OH) 3 mol HNOx g Al(OH)

78.00 g Al(OH) 1 mol Al(OH)

= 0.038462x

Moles of HNO3 required to titrate Mg(OH)2 =

( ) 322

2 2

2 mol HNO 1 mol Mg(OH)(0.4826 x) g Mg(OH)

58.33 g Mg(OH) 1 mol Mg(OH)

−

= 0.01655 – 0.03429x Moles of HNO3 required to titrate sample = 0.0173 mol = 0.038462x + (0.01655 – 0.03429x)

0.0173 = 0.004172x + 0.01655 0.17977 g = x = mass of Al(OH)3

Mass % of Al(OH)3 = ( ) ( )3mass of Al(OH) 0.17977 g100 = 100mass of sample 0.4826 g

= 37.2503 = 37.25%

4.78 Plan: Consult the Rules for Assigning an Oxidation Number. The sum of the O.N. values for the atoms in a molecule equals zero, while the sum of the O.N. values for the atoms in an ion equals the ion’s charge. Solution:

a) Since H is bonded to a nonmetal, the O.N. of each hydrogen is +1, for a total of +2. The O.N.s must add to zero, so the O.N. of S is –2.

b) The O.N. of each oxygen is –2 for a total of –6. Since the O.N.s must add to –2, the charge of the ion, the O.N. of S is +4. 4.79 Plan: Use the Rules for Assigning an Oxidation Number to assign each atom an O.N. If the O.N.s of atoms change as the reactants become products, the reaction is a redox reaction. Solution:

The O.N. of hydrogen is +1 in NH3, for a total of +3. The O.N. of N is then –3. In HCl, the O.N. of H is +1 and that of Cl is –1. The oxidation numbers of N, H, and Cl are the same in the products. No, since no element changes oxidation number, this cannot be a redox reaction.

4.80 Plan: Recall that oxidation is the loss of electrons and reduction is the gain of electrons. Solution:

An oxidizing agent causes something else to be oxidized; ie., to lose electrons. The oxidizing agent accepts these electrons and is reduced. 4.81 Plan: Recall that oxidation is the loss of electrons and reduction is the gain of electrons. Solution: The electrons that a substance gains during reduction must come from somewhere. So there must be an oxidation in which electrons are lost, to provide the electrons gained during reduction. 4.82 Plan: An oxidizing agent gains electrons and therefore has an atom whose oxidation number decreases during the

reaction. Use the Rules for Assigning an Oxidation Number to assign S in H2SO4 an O.N. and see if this oxidation number changes during the reaction. An acid transfers a proton during reaction.

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Solution: a) In H2SO4, hydrogen has an O.N. of +1, for a total of +2; oxygen has an O.N. of –2 for a total of –8. The S has an O.N. of +6. In SO2, the O.N. of oxygen is –2 for a total of –4 and S has an O.N. of +4. So the S has been reduced from +6 to +4 and is an oxidizing agent. Iodine is oxidized during the reaction. b) The oxidation number of S is +6 in H2SO4; in BaSO4, Ba has an O.N. of +2, the four oxygen atoms have a total O.N. of –8, and S is again +6. Since the oxidation number of S (or any of the other atoms) did not change, this is not a redox reaction. H2SO4 transfers a proton to F– to produce HF, so it acts as an acid.

4.83 Plan: Use the Rules for Assigning an Oxidation Number to assign each atom an oxidation number.

An oxidizing agent gains electrons and therefore has an atom whose oxidation number decreases during the reaction. A reducing agent loses electrons and therefore has an atom whose oxidation number increases during the reaction. Solution: +3 –4 +2 –3 +1 +4 –2 0 +1 –2 8NH3(g) + 6NO2(g) → 7N2(g) + 12H2O(l) Oxidizing agent = NO2; Reducing agent = NH3; the NO2 is the oxidizing agent (O.N. of N decreases from +4 in NO2 to 0 in N2), and the NH3 is the reducing agent (O.N. of N increases from –3 in NH3 to 0 in N2).

4.84 Plan: Consult the Rules for Assigning an Oxidation Number. The sum of the O.N. values for the atoms in a molecule equals zero, while the sum of the O.N. values for the atoms in an ion equals the ion’s charge. Solution: a) CF2Cl2. The rules dictate that F and Cl each have an O.N. of –1; two F and two Cl yield a sum of –4, so the

O.N. of C must be +4. C = +4 b) Na2C2O4. The rule dictates that Na [Group 1A(1)] has a +1 O.N.; rule 5 dictates that O has a –2 O.N.; two Na and four O yield a sum of –6 [2(+1) + 4(–2)]. Therefore, the total of the O.N.s on the two C atoms is +6 and each C is +3. C = +3 c) HCO3

–. H is combined with nonmetals and has an O.N. of +1; O has an O.N. of –2 and three O have a sum of –6. To have an overall oxidation state equal to –1, C must be +4 because (+1) + (+4) + (–6) = –1. C = +4

d) C2H6. Each H has an O.N. of +1; six H gives +6. The sum of O.N.s for the two C atoms must be –6, so each C is –3. C = –3 4.85 Plan: Consult the Rules for Assigning an Oxidation Number. The sum of the O.N. values for the atoms in a molecule equals zero. Solution:

a) KBr. Potassium [Group 1A(1)] has an O.N. of +1. Therefore, the O.N. of Br must be –1. Br = –1 b) BrF3. According to Rule 6, a halogen has an O.N. of –1 when it is in combination with a halogen lower in the group. So the O.N. of F is –1 and the total for the three F atoms is –3; the O.N. of Br must be +3. Br = +3 c) HBrO3. Hydrogen has an O.N. of +1; the O.N. of oxygen is –2 and the total of three oxygen atoms is –6. To have an overall sum of O.N.s = 0, Br must be +5 since [(+1) + (+5) + (–6)] = 0. Br = +5

d) CBr4. The O.N. of halogens in combination with a nonmetal like carbon is –1; the total of four Br atoms is –4. The O.N. of C must be +4. Br = –1 4.86 Plan: Consult the Rules for Assigning an Oxidation Number. The sum of the O.N. values for the atoms in a molecule equals zero, while the sum of the O.N. values for the atoms in an ion equals the ion’s charge. Solution:

a) NH2OH. Hydrogen has an O.N. of +1, for a total of +3 for the three hydrogen atoms. Oxygen has an O.N. of –2. The O.N. of N must be –1 since [(–1) + (+3) + (–2)] = 0. N = –1 b) N2F4. The O.N. of each fluorine is –1 for a total of –4; the sum of the O.N.s for the two N atoms must be

+4, so each N has an O.N. of +2. N = +2 c) NH4

+. The O.N. of each hydrogen is +1 for a total of +4; the O.N. of nitrogen must be –3 since the overall sum of the O.N.s must be +1: [(–3) + (+4)] = +1 N = –3

d) HNO2. The O.N. of hydrogen is +1 and that of each oxygen is –2 for a total of –4 from the oxygens. The O.N. of nitrogen must be +3 since [(+1) + (+3) + (–4)] = 0. N = +3

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4.87 Plan: Consult the Rules for Assigning an Oxidation Number. The sum of the O.N. values for the atoms in a molecule equals zero. Solution:

a) SOCl2. The O.N. of oxygen is –2 and that of each chlorine is –1 for a total of –2 for the two chlorine atoms. The O.N. of sulfur must be +4 since [(+4) + (–2) + (–2)] = 0. S = +4

b) H2S2. The O.N. of each hydrogen is +1, for a total of +2. The sum of the O.N.s of the two sulfur atoms must equal –2, so the O.N. of each S atom is –1. S = –1 c) H2SO3. The O.N. of each hydrogen atom is +1 for a total of +2; the O.N. of each oxygen atom is –2 for a total of –6. The O.N. of the sulfur must be +4 since [(+2) + (+4) + (–6)] = 0. S = +4 d) Na2S. The O.N. of each sodium [Group 1A(1)] is +1, for a total of +2. The O.N. of sulfur is –2. S = –2 4.88 Plan: Consult the Rules for Assigning an Oxidation Number. The sum of the O.N. values for the atoms in a molecule equals zero, while the sum of the O.N. values for the atoms in an ion equals the ion’s charge. Solution:

a) AsH3. H is combined with a nonmetal, so its O.N. is +1 (Rule 3). Three H atoms have a sum of +3. To have a sum of 0 for the molecule, As has an O.N. of –3. As = –3

b) H2AsO4–. The O.N. of H in this compound is +1, for a total of +2. The O.N. of each oxygen is –2, for a total

of –8. As has an O.N. of +5 since [(+2) + (+5) + (–8)] = –1, the charge of the ion. As = +5 c) AsCl3. Each chlorine has an O.N. of –1, for a total of –3. The O.N. of As is +3. As = +3 4.89 Plan: Consult the Rules for Assigning an Oxidation Number. The sum of the O.N. values for the atoms in a molecule equals zero, while the sum of the O.N. values for the atoms in an ion equals the ion’s charge. Solution: a) H2P2O7

2–. The O.N.of each hydrogen is +1, for a total of +2; the O.N. of each oxygen is –2, for a total of –14. The sum of the O.N.s of the two phosphorus atoms must be +10 since [(+2) + (+10) + (–14)] = –2, the charge of the ion. Each of the two phosphorus atoms has an O.N. of +5. P = +5 b) PH4

+. The O.N. of each hydrogen is +1, for a total of +4. The O.N. of P is –3 since [(–3) + (+4)] = +1, the charge of the ion. P = –3

c) PCl5. The O.N. of each Cl is –1, for a total of –5. The O.N. of P is therefore +5. P = +5 4.90 Plan: Consult the Rules for Assigning an Oxidation Number. The sum of the O.N. values for the atoms in a molecule equals zero, while the sum of the O.N. values for the atoms in an ion equals the ion’s charge. Solution:

a) MnO42–. The O.N. of each oxygen is –2, for a total of –8; the O.N. of Mn must be +6 since [(+6) + (–8)] = –2,

the charge of the ion. Mn = +6 b) Mn2O3. The O.N. of each oxygen is –2, for a total of –6; the sum of the O.N.s of the two Mn atoms must be +6. The O.N. of each manganese is +3. Mn = +3 c) KMnO4. The O.N. of potassium is +1 and the O.N. of each oxygen is –2, for a total of –8. The O.N. of Mn is +7 since [(+1) + (+7) + (–8)] = 0. Mn = +7 4.91 Plan: Consult the Rules for Assigning an Oxidation Number. The sum of the O.N. values for the atoms in a molecule equals zero, while the sum of the O.N. values for the atoms in an ion equals the ion’s charge. Solution: a) CrO3. The O.N. of each oxygen atom is –2, for a total of –6. The O.N. of chromium must be +6. Cr = +6

b) Cr2O72–. The O.N. of each oxygen is –2, for a total of –14. The sum of the O.N.s of the two chromium atoms

must be +12 since [(+12) + (–14)] = –2, the charge of the ion. Each of the two chromium atoms has an O.N. of +6. Cr = +6 c) Cr2(SO4)3. It is convenient to treat the polyatomic ion SO4

2– as a unit with a –2 charge, for a total of –6 for the three sulfate ions. The sum of the two chromium atoms must be +6 and the O.N. of each chromium atom is +3. Cr = +3

4.92 Plan: First, assign oxidation numbers to all atoms following the rules. The reactant that is the reducing

agent contains an atom that is oxidized (O.N. increases from the left side to the right side of the equation). The reactant that is the oxidizing agent contains an atom that is reduced (O.N. decreases from the left side to the right side of the equation). Recognize that the agent is the compound that contains the atom that is oxidized or reduced, not just the atom itself.

4-42

Solution: a) +2 +6 –8 –8 –4 +2

+1 +3 –2 +7 –2 +1 +2 +4 –2 +1 –2 5H2C2O4(aq) + 2MnO4

–(aq) + 6H+(aq) → 2Mn2+(aq) + 10CO2(g) + 8H2O(l) Mn in MnO4

– changes from +7 to +2 (reduction). Therefore, MnO4– is the oxidizing agent. C in H2C2O4

changes from +3 to +4 (oxidation), so H2C2O4 is the reducing agent. b) –6 +2 0 +1 +5 –2 +2 +2 –2 +1 –2

3Cu(s) + 8H+(aq) + 2NO3–(aq) → 3Cu2+(aq) + 2NO(g) + 4H2O(l)

Cu changes from 0 to +2 (is oxidized) and Cu is the reducing agent. N changes from +5 (in NO3–) to +2 (in

NO) and is reduced, so NO3– is the oxidizing agent.


agent contains an atom that is oxidized (O.N. increases from the left side to the right side of the equation). The reactant that is the oxidizing agent contains an atom that is reduced (O.N. decreases from the left side to the right side of the equation). Recognize that the agent is the compound that contains the atom that is oxidized or reduced, not just the atom itself. Solution: 0 +1 +2 0

a) Sn(s) + 2H+(aq) → Sn2+(aq) + H2(g) Sn changes from 0 to +2 (is oxidized) so Sn is the reducing agent. H changes from +1 to 0 (is reduced) so H+ is the oxidizing agent. b) +2 –2 +2 +1 +1 –1 +2 +3 +1 –2 2H+(aq) + H2O2(aq) + 2Fe2+(aq) → 2Fe3+(aq) + 2H2O(l) Oxygen changes from –1 in H2O2 to –2 in H2O (is reduced) so H2O2 is the oxidizing agent. Fe changes from +2 to +3 (is oxidized) so Fe2+ is the reducing agent.

4.94 Plan: First, assign oxidation numbers to all atoms following the rules. The reactant that is the reducing agent contains an atom that is oxidized (O.N. increases from the left side to the right side of the equation). The reactant that is the oxidizing agent contains an atom that is reduced (O.N. decreases from the left side to the right side of the equation). Recognize that the agent is the compound that contains the atom that is oxidized or reduced, not just the atom itself. Solution: a) –6 –6 –4 +2 +1 –1 0 +5 –2 +4 –1 +4 –2 +1 –2 8H+(aq) + 6Cl–(aq) + Sn(s) + 4NO3

–(aq) → SnCl62–(aq) + 4NO2(g) + 4H2O(l)

Nitrogen changes from an O.N. of +5 in NO3– to +4 in NO2 (is reduced) so NO3

– is the oxidizing agent. Sn changes from an O.N. of 0 to an O.N. of +4 in SnCl6

2– (is oxidized) so Sn is the reducing agent. b) –8 +2 +7 –2 –1 +1 0 +2 +1 –2 2MnO4

–(aq) + 10Cl–(aq) + 16H+(aq) → 5Cl2(g) + 2Mn2+(aq) + 8H2O(l) Manganese changes from an O.N. of +7 in MnO4

– to an O.N. of +2 in Mn2+ (is reduced) so MnO4– is the

oxidizing agent. Chlorine changes its O.N. from –1 in Cl– to 0 as the element Cl2 (is oxidized) so Cl– is the reducing agent.


agent contains an atom that is oxidized (O.N. increases from the left side to the right side of the equation). The reactant that is the oxidizing agent contains an atom that is reduced (O.N. decreases from the left side to the right side of the equation). Recognize that the agent is the compound that contains the atom that is oxidized or reduced, not just the atom itself. Solution:

a) +12 –14 –6 –8 +2 +1 +6 –2 +4 –2 +3 +6 –2 +1 –2

8H+(aq) + Cr2O72–(aq) + 3SO3

2–(aq) → 2Cr3+(aq) + 3SO42–(aq) + 2H2O(l)

4-43

Chromium changes from an O.N. of +6 in Cr2O7

2–to +3 in Cr3+ (is reduced) so Cr2O72– is the oxidizing agent.

Sulfur changes from an O.N. of +4 in SO32– to +6 in SO4

2– (is oxidized) so SO32– is the reducing agent.

b) –6 +2 –8 +4 +3 +5 –2 0 –2 +1 +1 –2 +2 –2 +1 –3 +1

NO3–(aq) + 4Zn(s) + 7OH–(aq) + 6H2O(l) → 4Zn(OH)4

2–(aq) + NH3(aq) Nitrogen changes from an O.N. of +5 in NO3

– to an O.N. of –3 in NH3 (is reduced) so NO3– is the oxidizing

agent. Zinc changes from an O.N. of 0 to an O.N. of +2 in Zn(OH)4

2– (is oxidized) so Zn is the reducing agent. 4.96 Plan: Find the moles of MnO4

– from the molarity and volume information. Use the molar ratio in the balanced equation to find the moles of H2O2. Multiply the moles of H2O2 by its molar mass to determine the mass of H2O2 present. Mass percent is calculated by dividing the mass of H2O2 by the mass of the sample and multiplying by 100. Assign oxidation numbers to all atoms following the rules. The reactant that is the reducing agent contains an atom that is oxidized (O.N. increases from the left side to the right side of the equation). Solution:

a) Moles of MnO4– = ( )

340.105 mol MnO10 L43.2 mL

1 mL L

−−

= 4.536x10–3 = 4.54x10–3 mol MnO4–

b) Moles of H2O2 = ( )3 2 24

4

5 mol H O4.536x10 MnO2 mol MnO

− −−

= 0.01134 = 0.0113 mol H2O2

c) Mass (g) of H2O2 = ( ) 2 22 2

2 2

34.02 g H O0.01134 mol H O

1 mol H O

= 0.3857868 = 0.386 g H2O2

d) Mass percent of H2O2 = ( ) ( )2 2 2 2mass of H O 0.3857868 g H O100 = 100

mass of sample 14.8 g sample = 2.606668 = 2.61% H2O2

e) –8 +2 –2 +2 +7 –2 +1 –1 +1 0 +2 +1 –2

2MnO4–(aq) + 5H2O2(aq) + 6H+(aq) → 5O2(g) + 2Mn2+(aq) + 8H2O(l)

The O.N. of oxygen increases from –1 in H2O2 to 0 in O2 and is therefore oxidized while the O.N. of Mn decreases from +7 in MnO4

– to +2 in Mn2+ and is reduced. H2O2 is the reducing agent. 4.97 Plan: Find the moles of Cr2O7

2– from the molarity and volume information. Use the molar ratio in the balanced equation to find the moles of C2H5OH and multiply the moles of C2H5OH by its molar mass to determine the mass of C2H5OH present. Mass percent is calculated by dividing the mass of C2H5OH by the mass of the sample and multiplying by 100. Solution:

Moles of Cr2O72– = ( )

232 70.05961 molCr O10 L35.46 mL

1 mL 1 L

−−

= 0.0021138 mol Cr2O72–

Moles of C2H5OH = ( )2 2 52 7 2

2 7

1 mol C H OH0.0021138 mol Cr O

2 mol Cr O−

−

= 0.0010569 mol C2H5OH

Mass (g) of C2H5OH = ( ) 2 52 5

2 5

46.07 g C H OH0.0010569 mol C H OH

1 mol C H OH

= 0.0486914 g C2H5OH

Mass percent of C2H5OH = ( ) ( )2 5 2 5mass of C H OH 0.0486914 g C H OH100 = 100

mass of sample 28.00 g sample

= 0.173898 = 0.1739% C2H5OH 4.98 Plan: The three types of redox reactions are combination, decomposition, and displacement. In a combination

reaction, two or more reactants form one product, so the number of substances decreases. In a decomposition reaction, one reactant forms two or more products, so the number of substances increases. In a displacement reaction, the number of substances is the same, but atoms exchange places.

4-44

Solution: a) decomposition b) combination c) displacement

4.99 Plan: Recall that a reactant breaks down into two or more products in a decomposition reaction, while

reactants combine to form a product in a combination reaction. Solution: By definition, elements cannot decompose into anything simpler, so they could not be reactants in a

decomposition reaction. 4.100 Plan: Review the types of redox reaction discussed in this section.

Solution: Combination, decomposition, and displacement reactions generally produce only one compound; combustion reactions, however, often produce both carbon dioxide and water.

4.101 Plan: Recall that combustion is a reaction in which a reactant is combined with elemental oxygen. Solution: Yes, all combustion reactions are redox reactions since the oxidation number of O2 will change from zero (in the element) to some negative value (typically –2) in the product(s) during the reaction.

4.102 Plan: Recall that two reactants are combined to form one product in a combination reaction. A reaction is a redox

reaction only if the oxidation numbers of some atoms change during the reaction. Solution: A common example of a combination/redox reaction is the combination of a metal and nonmetal to form an ionic salt, such as 2Mg(s) + O2(g) → 2MgO(s) in which magnesium is oxidized and oxygen is reduced. A common example of a combination/non-redox reaction is the combination of a metal oxide and water to form an acid, such as: CaO(s) + H2O(l) → Ca(OH)2(aq).

4.103 Plan: In a combination reaction, two or more reactants form one product. In a decomposition reaction, one

reactant forms two or more products. In a displacement reaction, atoms or ions exchange places. Balance the reactions by inspection. Solution: a) Ca(s) + 2H2O(l) → Ca(OH)2(aq) + H2(g)

Displacement: one Ca atom displaces 2 H atoms. b) 2NaNO3(s) → 2NaNO2(s) + O2(g)

Decomposition: one reactant breaks into two products. c) C2H2(g) + 2H2(g) → C2H6(g) Combination: two reactants combine to form one product. 4.104 Plan: In a combination reaction, two or more reactants form one product. In a decomposition reaction, one

reactant forms two or more products. In a displacement reaction, atoms or ions exchange places. Balance the reactions by inspection. Solution: a) 2HI(g) → H2(g) + I2(g) Decomposition: one reactant breaks into two products.

b) Zn(s) + 2AgNO3(aq) → Zn(NO3)2(aq) + 2Ag(s) Displacement: one Zn displaces 2 Ag atoms.

c) 2NO(g) + O2(g) → N2O4(l) Combination: two reactants combine to form one product. 4.105 Plan: In a combination reaction, two or more reactants form one product. In a decomposition reaction, one

reactant forms two or more products. In a displacement reaction, atoms or ions exchange places. Balance the reactions by inspection. Solution:

a) 2Sb(s) + 3Cl2(g) → 2SbCl3(s)

4-45

Combination: two reactants combine to form one product. b) 2AsH3(g) → 2As(s) + 3H2(g) Decomposition: one reactant breaks into two products. c) Zn(s) + Fe(NO3)2(aq) → Zn(NO3)2(aq) + Fe(s) Displacement: one Zn displaces one Fe atom. 4.106 Plan: In a combination reaction, two or more reactants form one product. In a decomposition reaction, one

reactant forms two or more products. In a displacement reaction, atoms or ions exchange places. Balance the reactions by inspection. Solution: a) Mg(s) + 2H2O(g) → Mg(OH)2(s) + H2(g) Displacement: one Mg displaces two H atoms.

b) Cr(NO3)3(aq) + Al(s) → Al(NO3)3(aq) + Cr(s) Displacement: one Al displaces one Cr atom.

c) PF3(g) + F2(g) → PF5(g) Combination: two reactants combine to form one product. 4.107 Plan: In a combination reaction, two or more reactants form one product. Two elements as reactants often results

in a combination reaction. In a decomposition reaction, one reactant forms two or more products; one reactant only often indicates a decomposition reaction. In a displacement reaction, atoms or ions exchange places. An element and a compound as reactants often indicate a displacement reaction. Balance the reactions by inspection. Solution: a) The combination between a metal and a nonmetal gives a binary ionic compound.

Sr(s) + Br2(l) → SrBr2(s) b) Many metal oxides release oxygen gas upon thermal decomposition.

2Ag2O(s) →∆ 4Ag(s) + O2(g) c) This is a displacement reaction. Mn is a more reactive metal and displaces Cu2+ from solution. Mn(s) + Cu(NO3)2(aq) → Mn(NO3)2(aq) + Cu(s) 4.108 Plan: In a combination reaction, two or more reactants form one product. Two elements as reactants often results

in a combination reaction. In a decomposition reaction, one reactant forms two or more products; one reactant only often indicates a decomposition reaction. In a displacement reaction, atoms or ions exchange places. An element and a compound as reactants often indicate a displacement reaction. Balance the reactions by inspection. Solution: a) This is a displacement reaction. Active metals like Mg can displace hydrogen from acid. Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g) b) Some compounds undergo decomposition to their elements during electrolysis in which electrical energy is absorbed.

2LiCl(l) electricity→ 2Li(l) + Cl2(g) c) This is a displacement reaction in which cobalt displaces tin.

SnCl2(aq) + Co(s) → CoCl2(aq) + Sn(s) 4.109 Plan: In a combination reaction, two or more reactants form one product. Two elements as reactants often results

in a combination reaction. In a decomposition reaction, one reactant forms two or more products; one reactant only often indicates a decomposition reaction. In a displacement reaction, atoms or ions exchange places. An element and a compound as reactants often indicate a displacement reaction. Balance the reactions by inspection. Solution: a) The combination of two nonmetals gives a covalent compound. N2(g) + 3H2(g) → 2NH3(g)

b) Some compounds undergo thermal decomposition to simpler substances.

2NaClO3(s) →∆ 2NaCl(s) + 3O2(g) c) This is a displacement reaction. Active metals like Ba can displace hydrogen from water.

4-46

Ba(s) + 2H2O(l) → Ba(OH)2(aq) + H2(g) 4.110 Plan: In a combination reaction, two or more reactants form one product. Two elements as reactants often results

in a combination reaction. In a decomposition reaction, one reactant forms two or more products; one reactant only often indicates a decomposition reaction. In a displacement reaction, atoms or ions exchange places. An element and a compound as reactants often indicate a displacement reaction. Balance the reactions by inspection. Solution: a) This is a displacement reaction in which iron displaces hydrogen. Fe(s) + 2HClO4(aq) → Fe(ClO4)2(aq) + H2(g) b) The combination of two nonmetals gives a covalent compound.

S8(s) + 8O2(g) →∆ 8SO2(g) c) Some compounds undergo decomposition to their elements during electrolysis in which electrical energy is absorbed.

BaCl2(l) electricity→ Ba(l) + Cl2(g) 4.111 Plan: In a combination reaction, two or more reactants form one product. Two elements as reactants often results

in a combination reaction. In a decomposition reaction, one reactant forms two or more products; one reactant only often indicates a decomposition reaction. In a displacement reaction, atoms or ions exchange places. An element and a compound as reactants often indicate a displacement reaction. Balance the reactions by inspection. Solution: a) Cs, a metal, and I2, a nonmetal, combine to form the binary ionic compound, CsI.

2Cs(s) + I2(s) → 2CsI(s) b) Al is a stronger reducing agent than Mn and is able to displace Mn from solution, i.e., cause the reduction from Mn2+(aq) to Mn0(s). 2Al(s) + 3MnSO4(aq) → Al2(SO4)3(aq) + 3Mn(s)

c) This is a combination reaction in which sulfur dioxide, SO2, a nonmetal oxide, combines with oxygen, O2, to form the higher oxide, SO3.

2SO2(g) + O2(g) →∆ 2SO3(g) It is not clear from the problem, but energy must be added to force this reaction to proceed.

d) Butane is a four carbon hydrocarbon with the formula C4H10. It burns in the presence of oxygen, O2, to form carbon dioxide gas and water vapor. Although this is a redox reaction that could be balanced using the oxidation number method, it is easier to balance by considering only atoms on either side of the equation. First, balance carbon and hydrogen (because they only appear in one species on each side of the equation), and then balance oxygen.

2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(g) e) Total ionic equation in which soluble species are shown dissociated into ions: 2Al(s) + 3Mn2+(aq) + 3SO4

2–(aq) → 2Al3+(aq) + 3SO42–(aq) + 3Mn(s)

Net ionic equation in which the spectator ions are omitted: 2Al(s) + 3Mn2+(aq) → 2Al3+(aq) + 3Mn(s) Note that the molar coefficients are not simplified because the number of electrons lost (6 e–) must equal the electrons gained (6 e–). 4.112 Plan: In a combination reaction, two or more reactants form one product. Two elements as reactants often results

in a combination reaction. In a decomposition reaction, one reactant forms two or more products; one reactant only often indicates a decomposition reaction. In a displacement reaction, atoms or ions exchange places. An element and a compound as reactants often indicate a displacement reaction. Balance the reactions by inspection. Solution: a) Pentane is a five carbon hydrocarbon with the formula C5H12. It burns in the presence of oxygen, O2, to form carbon dioxide gas and water vapor. Although this is a redox reaction that could be balanced using the oxidation number method, it is easier to balance by considering only atoms on either side of the equation. First, balance carbon and hydrogen (because they only appear in one species on each side of the equation), and then balance oxygen. C5H12(l) + 8O2(g) → 5CO2(g) + 6H2O(g)

4-47

b) Phosphorus trichloride, PCl3, is a nonmetal halide that combines with additional halogen, to form the higher halide, PCl5. PCl3(l) + Cl2(g) → PCl5(s) c) This is a displacement reaction. Active metals like Zn can displace hydrogen from acid.

Zn(s) + 2HBr(aq) → ZnBr2(aq) + H2(g) d) This is a displacement reaction in which bromine displaces iodine. A halogen higher in the periodic table can displace a halogen that is lower. 2KI(aq) + Br2(l) → 2KBr(aq) + I2(s) e) Total ionic equation in which soluble species are shown dissociated into ions: 2K+(aq) + 2I–(aq) + Br2(l) → 2K+(aq) + 2Br–(aq) + I2(s) Net ionic equation in which the spectator ions are omitted: 2I–(aq) + Br2(l) → I2(s) + 2Br–(aq) 4.113 Plan: Write a balanced equation that shows the decomposition of HgO to its elements. Convert the mass of HgO

to moles and use the molar ratio from the balanced equation to find the moles and then the mass of O2. Perform the same calculation to find the mass of the other product.

Solution:

a) The balanced chemical equation is 2HgO(s) →∆ 2Hg(l) + O2(g).

Moles of HgO = ( )310 g 1 mol HgO4.27 kg HgO

1 kg 216.6 g HgO

= 19.71376 mol HgO

Moles of O2 = ( ) 21 mol O19.71376 mol HgO

2 mol HgO

= 9.85688 mol O2

Mass (g) of O2 = ( ) 22

2

32.00 g O9.85688 mol O

1 mol O

= 315.420 = 315 g O2

b) The other product is mercury.

Moles of Hg = ( ) 2 mol Hg19.71376 mol HgO2 mol HgO

= 19.71376 mol Hg

Mass (kg) Hg = ( ) 3200.6 g Hg 1 kg19.71376 mol Hg1 mol Hg 10 g

= 3.95458 = 3.95 kg Hg

4.114 Plan: Write a balanced equation that shows the decomposition of calcium chloride to its elements. Convert the

mass of CaCl2 to moles and use the molar ratio from the balanced equation to find the moles and then the mass of Cl2. Perform the same calculation to find the mass of the other product.

Solution: The balanced chemical equation is CaCl2(l) →elect Ca(l) + Cl2(g). Note: The reaction cannot be done in the presence of water as elemental calcium would displace the hydrogen from the water.

Moles of CaCl2 = ( ) 22

2

1 mol CaCl874 g CaCl

110.98 g CaCl

= 7.87529 mol CaCl2

Moles of Cl2 = ( ) 22

2

1 mol Cl7.87529 mol CaCl

1 mol CaCl

= 7.87529 mol Cl2

Mass (g) of Cl2 = ( ) 22

2

70.90 g Cl7.87529mol Cl

1 mol Cl

= 558.358 = 558 g Cl2

The other product is calcium.

4-48

Moles of Ca = ( )22

1 mol Ca7.87529 mol CaCl1 mol CaCl

= 7.87529 mol Ca

Mass (g) of Ca = ( ) 40.08 g Ca7.87529 mol Ca1 mol Ca

= 315.64 = 316 g Ca

4.115 Plan: To determine the reactant in excess, write the balanced equation (metal + O2 → metal oxide), convert

reactant masses to moles, and use molar ratios to see which reactant makes the smaller (“limiting”) amount of product. Use the limiting reactant to calculate the amount of product formed. Use the molar ratio to find the amount of excess reactant required to react with the limiting reactant; the amount of excess reactant that remains is the initial amount of excess reactant minus the amount required for the reaction.

Solution: The balanced equation is 4Li(s) + O2(g) → 2Li2O(s).

a) Moles of Li2O if Li limiting = ( ) 22 mol Li O1 mol Li1.62 g Li6.941 g Li 4 mol Li

= 0.1166979 mol Li2O

Moles of Li2O if O2 limiting = ( ) 2 22

2 2

1 mol O 2 mol Li O6.50 g O

32.00 g O 1 mol O

= 0.40625 mol Li2O

Li is the limiting reactant since it produces the smaller amount of product; O2 is in excess. b) Using Li as the limiting reagent, 0.1166979 = 0.117 mol Li2O is formed. c) Li is limiting, thus there will be none remaining (0 g Li).

Mass (g) of Li2O = ( ) 22

2

29.88 g Li O0.1166979 mol Li O

1 mol Li O

= 3.4869 = 3.49 g Li2O

Mass (g) of O2 reacted = ( ) 2 2

2

1 mol O 32.00 g O1 mol Li1.62 g Li6.941 g Li 4 mol Li 1 mol O

= 1.867166 g O2

Remaining O2 = initial amount – amount reacted = 6.50 g O2 – 1.867166 g O2 = 4.632834 = 4.63 g O2 4.116 Plan: To determine the reactant in excess, write the balanced equation (metal + N2 → metal nitride), convert

reactant masses to moles, and use molar ratios to see which reactant makes the smaller (“limiting”) amount of product. Use the limiting reactant to calculate the amount of product formed. Use the molar ratio to find the amount of excess reactant required to react with the limiting reactant; the amount of excess reactant that remains is the initial amount of excess reactant minus the amount required for the reaction.

Solution:

The balanced equation is 3Mg(s) + N2(g) →∆ Mg3N2(s).

a) Moles of Mg3N2 if Mg is limiting = ( ) 3 21 mol Mg N1 mol Mg2.22 g Mg24.31 g Mg 3 mol Mg

= 0.030440 mol Mg3N2

Moles of Mg3N2 if N2 is limiting = ( ) 3 222

2 2

1 mol Mg N1 mol N3.75 g N

28.02 g N 1 mol N

= 0.13383 mol Mg3N2

Mg is the limiting reactant since it produces the smaller amount of product; N2 is present in excess. b) Using Mg as the limiting reactant, 0.030440 = 0.0304 mol Mg3N2 is formed. c) There will be 0 Mg remaining since it is the limiting reagent and will be completely consumed.

Mass (g) of Mg3N2 = ( ) 3 23 2

3 2

100.95 g Mg N0.030440 mol Mg N

1 mol Mg N

= 3.07292 = 3.07 g Mg3N2

Mass (g) of N2 reacted = ( ) 2 2

2

1 mol N 28.02 g N1 mol Mg2.22 g Mg24.31 g Mg 3 mol Mg 1 mol N

= 0.852933 g N2

Remaining N2 = initial amount – amount reacted = 3.75 g N2 – 0.852933 g N2 = 2.897067 = 2.90 g N2

4-49

4.117 Plan: Since mass must be conserved, the original amount of mixture – amount of residue = mass of oxygen produced. Write a balanced equation and use molar ratios to convert from the mass of oxygen produced to the amount of KClO3 reacted. Mass percent is calculated by dividing the mass of KClO3 by the mass of the sample and multiplying by 100. Solution:

2KClO3(s) →∆ 2KCl(s) + 3O2(g) Mass (g) of O2 produced = mass of mixture – mass of residue = 0.950 g – 0.700 g = 0.250 g O2

Mass (g) of KClO3 = ( ) 3 322

2 2 3

2 mol KClO 122.55 g KClO1 mol O0.250g O

32.00 g O 3 mol O 1 mol KClO

= 0.63828125 g KClO3

Mass % KClO3 = ( )3mass of KClO100%

mass of sample = ( )30.63828125 g KClO

100%0.950 g sample

= 67.1875 = 67.2% KClO3

4.118 Plan: Since mass must be conserved, the original amount of mixture – amount of remaining solid = mass of

carbon dioxide produced. Write a balanced equation and use molar ratios to convert from the mass of CO2 produced to the amount of CaCO3 reacted. Mass percent is calculated by dividing the mass of CaCO3 by the mass of the sample and multiplying by 100. Solution:

CaCO3(s) →∆ CaO(s) + CO2(g) Mass (g) of CO2 produced = mass of mixture – mass of remaining solid = 0.693 g – 0.508 g = 0.185 g CO2

Mass (g) of CaCO3 = ( ) 3 322

2 2 3

1 mol CaCO 100.09 g CaCO1 mol CO0.185g CO

44.01 g CO 1 mol CO 1 mol CaCO

= 0.420737 g CaCO3

Mass % CaCO3 = ( )3mass of CaCO100%

mass of sample = ( )30.420737 g CaCO

100%0.693 g sample

= 60.7124 = 60.7% CaCO3

4.119 Plan: Write the balanced equation for the displacement reaction, convert reactant masses to moles, and use molar

ratios to see which reactant makes the smaller (“limiting”) amount of product. Use the limiting reactant to calculate the amount of product formed.

Solution: The balanced reaction is 2Al(s) + Fe2O3(s) → 2Fe(l) + Al2O3(s).

Moles of Fe if Al is limiting = ( )310 g 1 mol Al 2 mol Fe1.50 kg Al

1 kg 26.98 g Al 2 mol Al

= 55.59674 mol Fe

Moles of Fe if Fe2O3 is limiting = ( )2 32 3

2 mol Fe25.0 mol Fe O1 mol Fe O

= 50.0 mol Fe

Fe2O3 is the limiting reactant since it produces the smaller amount of Fe; 50.0 moles of Fe forms.

Mass (g) of Fe = ( ) 55.85 g Fe50.0 mol Fe1 mol Fe

= 2792.5 = 2790 g Fe = 2.79 kg Fe

4.120 Plan: In ionic compounds, iron has two common oxidation states, +2 and +3. First, write the balanced equations

for the formation and decomposition of compound A. Then, determine which reactant is limiting by converting reactant masses to moles, and using molar ratios to see which reactant makes the smaller (“limiting”) amount of product. From the amount of the limiting reactant calculate how much compound B will form. Solution:

Compound A is iron chloride with iron in the higher, +3, oxidation state. Thus, the formula for compound A is FeCl3 and the correct name is iron(III) chloride. The balanced equation for formation of FeCl3 is: 2Fe(s) + 3Cl2(g) → 2FeCl3(s) To find out whether 50.6 g Fe or 83.8 g Cl2 limits the amount of product, calculate the number of moles of iron(III) chloride that could form based on each reactant.

4-50

Moles of FeCl3 if Fe is limiting = ( ) 32 mol FeCl1 mol Fe50.6 g Fe55.85 g Fe 2 mol Fe

= 0.905998 mol FeCl3

Moles of FeCl3 if Cl2 is limiting = ( ) 322

2 2

2 mol FeCl1 mol Cl83.8 g Cl

70.90 g Cl 3 mol Cl

= 0.787964 mol FeCl3

Since fewer moles of FeCl3 are produced from the available amount of chlorine, the chlorine is the limiting reactant, producing 0.787964 mol of FeCl3. The FeCl3 decomposes to FeCl2 with iron in the +2 oxidation state. FeCl2 is compound B. The balanced equation for the decomposition of FeCl3 is:

2FeCl3(s) → 2FeCl2(s) + Cl2(g)

Mass (g) of FeCl2 = ( ) 2 23

3 2

2 mol FeCl 126.75 g FeCl0.787964 mol FeCl

2 mol FeCl 1 mol FeCl

= 99.874 = 99.9 g FeCl2

4.121 Plan: Use the volume and molarity of HCl to find the moles of HCl added to the metal. That number of moles

minus the moles of HCl that remain after reaction gives the moles of HCl that actually reacted. Use the moles of reacted HCl and the mole ratio in the balanced chemical equation to find the moles of Mg that reacted. Convert moles of Mg to mass of Mg, divide that mass by the mass of impure metal sample, and multiply by 100 to find mass %. Solution:

Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)

Moles of HCl added = ( )0.750 mol HCl 0.100 L1 L

= 0.0750 mol HCl

Moles of HCl reacting with Mg = moles of added HCl – moles of HCl remaining = 0.0750 mol – 0.0125 mol HCl = 0.0625 mol HCl

Moles of Mg reacting = ( ) 1 mol Mg0.0625 mol HCl2 mol HCl

= 0.03125 mol Mg

Mass (g) of Mg = ( ) 24.31 g Mg0.03125 mol Mg1 mol Mg

= 0.7596875 g Mg

Mass percent Mg = ( ) ( )mass of Mg 0.7596875 g Mg100 = 100mass of sample 1.32 g sample

= 57.552 = 57.6% Mg

4.122 The equilibrium state is dynamic, because the forward and reverse processes continue even after apparent change has ceased. 4.123 The system must be closed so the gaseous product cannot escape. 4.124 Acetic acid molecules are constantly and randomly colliding with each other and with water molecules.

Occasionally, an H+ is transferred from an acetic acid molecule to a water molecule because of one of these collisions. At first, this process occurs at a larger rate than the reverse transfer of an H+ from a H3O+ to a CH3COO–

. As the concentration of H+ and CH3COO– build up, the rate of transfer of H+ between them equals the rate of transfer of H+ between CH3COOH and H2O and equilibrium is reached when about 2% of the CH3COOH molecules have ionized.

4.125 Plan: Review the concept of dynamic equilibrium.

Solution: 2NO(g) + Br2(g) ↔ 2NOBr(g) On a molecular scale, chemical reactions are dynamic. If NO and Br2 are placed in a container, molecules of NO and molecules of Br2 will react to form NOBr. Some of the NOBr molecules will decompose and the resulting NO and Br2 molecules will recombine with different NO and Br2 molecules to form more NOBr. In this sense, the reaction is dynamic because the original NO and Br2 pairings do not remain permanently attached to each

4-51

other. Eventually, the rate of the forward reaction (combination of NO and Br2) will equal the rate of the reverse reaction (decomposition of NOBr) at which point the reaction is said to have reached equilibrium. If you could take “snapshot” pictures of the molecules at equilibrium, you would see a constant number of reactant (NO, Br2) and product molecules (NOBr), but the pairings would not stay the same.

4.126 Plan: Write the balanced equation for the combination reaction to produce ammonia. Convert reactant masses

to moles, and use molar ratios to see which reactant makes the smaller (“limiting”) amount of product. Use the limiting reactant to calculate the amount of product formed. This is the theoretical yield of ammonia. Divide the actual yield of ammonia by the theoretical yield and multiply by 100 to obtain the percent yield. Use molar ratios to determine the amount of N2 and H2 that reacted to form the product. The difference between the initial moles of reactants and the amounts that actually reacted will give the moles of reactant remaining at equilibrium.

Solution: a) The reaction is N2(g) + 3H2(g) → 2NH3(g).

First, the limiting reactant must be found.

Moles of NH3 if H2 is limiting = ( ) 322

2 2

2 mol NH1 mol H10.0 g H

2.016 g H 3 mol H

= 3.306878 mol NH3

Moles of NH3 if N2 is limiting = ( ) 322

2 2

2 mol NH1 mol N20.0 g N

28.02 g N 1 mol N

= 1.42755 mol NH3

N2 is the limiting reactant since it produces the smaller amount of product.

Mass (g) of NH3 = ( ) 33

3

17.03 g NH1.42755 mol NH

1 mol NH

= 24.311 g NH3

% yield = ( )actual yield 100theoretical yield

= ( )15.0 g 10024.311 g

= 61.700 = 61.7%

b) Moles of H2 initially present = ( ) 22

2

1 mol H10.0 g H

2.016 g H

= 4.9603 mol H2

Moles of N2 initially present = ( ) 22

2

1 mol N20.0 g N

28.02 g N

= 0.71378 mol N2

Moles of H2 required to produce 15.0 g NH3 = ( ) 3 23

3 3

1 mol NH 3 mol H15.0 g NH

17.03 g NH 2 mol NH

= 1.3212 mol H2

Moles of N2 required to produce 15.0 g NH3 = ( ) 3 23

3 3

1 mol NH 1 mol N15.0 g NH

17.03 g NH 2 mol NH

= 0.4404 mol N2

Moles of H2 at equilibrium = initial moles – reacted moles = 4.9603 mol – 1.3212 mol = 3.6391 = 3.64 mol H2 Moles of N2 at equilibrium = initial moles – reacted moles = 0.71378 mol – 0.4404 mol = 0.27338 = 0.273 mol N2

4.127 Plan: Ferrous ion is Fe2+. Write a reaction to show the conversion of Fe to Fe2+. Convert the mass of Fe in a 125-g

serving to the mass of Fe in a 737-g sample. Use molar mass to convert mass of Fe to moles of Fe and use Avogadro’s number to convert moles of Fe to moles of ions. Solution: a) Fe oxidizes to Fe2+ with a loss of 2 electrons. The H+ in the acidic food is reduced to H2 with a gain of 2 electrons. The balanced reaction is:

Fe(s) + 2H+(aq) → Fe2+(aq) + H2(g) O.N.: 0 +1 +2 0

b) Mass (g) of Fe in the jar of tomato sauce = ( )349 mg Fe 10 g737 g sauce

125 g sauce 1 mg

−

= 0.288904 g Fe

4-52

Number of Fe2+ ions = ( )2 23 2

21 mol Fe 1 mol Fe 6.022 x10 Fe ions0.288904 g Fe

55.85 g Fe 1 mol Fe 1 mol Fe

+ +

+

= 3.11509x1021 = 3.1x1021 Fe2+ ions per jar of sauce 4.128 Plan: Write balanced equations for the conversion of CaCO3 to CaO, the reaction of CaO with SO2, and the

combustion of sulfur to produce SO2. Add these three reactions to obtain the overall reaction. Use the mass % of sulfur in the coal sample to find the mass and then moles of sulfur present. Use the molar ratio in the overall balanced reaction to find the mass of CaCO3 required to react with this amount of sulfur, assuming a 70% yield. Solution: The reactions are:

CaCO3(s) →∆ CaO(s) + CO2(g)

CaO(s) + SO2(g) →∆ CaSO3(s)

S(s) + O2(g) →∆ SO2(g)

Overall reaction: CaCO3(s) + S(s) + O2(g) → CaSO3(s) + CO2(g)

Moles of sulfur = ( )3

4 10 g 0.33% 1 mol S8.5x10 kg coal1 kg 100% 32.06 g S

= 8749.220 mol S

Mass (g) CaCO3 = ( ) 3 3

3

1 mol CaCO 100.09 g CaCO 100%8749.220 mol S1 mol S 1 mol CaCO 70%

= 1.25101x106 = 1.3x106 g CaCO3 needed 4.129 Plan: Convert the mass of glucose to moles and use the molar ratios from the balanced equation to find the

moles of ethanol and CO2. The amount of ethanol is converted from moles to grams using its molar mass. The amount of CO2 is converted from moles to volume in liters using the conversion factor given. Solution:

Moles of C2H5OH = ( ) 6 12 6 2 56 12 6

6 12 6 6 12 6

1 mol C H O 2 mol C H OH100. g C H O

180.16 g C H O 1 mol C H O

= 1.11012 mol C2H5OH

Mass (g) of C2H5OH = ( ) 2 52 5

2 5

46.07 g C H OH1.11012 molC H OH

1 mol C H OH

= 51.143 = 51.1 g C2H5OH

Moles of CO2 = ( ) 6 12 6 26 12 6

6 12 6 6 12 6

1 mol C H O 2 mol CO100. g C H O

180.16 g C H O 1 mol C H O

= 1.11012 mol CO2

Volume (L) of CO2 = ( ) 22

2

22.4 L CO1.11012 mol CO

1 mol CO

= 24.8667 = 24.9 L CO2

4.130 Plan: Find the moles of KMnO4 from the molarity and volume information. Use the molar ratio in the balanced

equation to find the moles and then mass of iron. Mass percent is calculated by dividing the mass of iron by the mass of the sample and multiplying by 100. Solution:

Moles of KMnO4 = ( )3

40.03190 mol MnO10 L39.32 mL1 mL L

−−

= 0.00125431 mol KMnO4

4-53

Mass (g) of Fe = ( )2

4 24

5 mol Fe 55.85 g Fe0.00125431 mol MnO1 mol MnO 1 mol Fe

+−

− +

= 0.350266 g Fe

Mass % of Fe = ( ) ( )mass of Fe 0.350266 g100 = 100mass of sample 1.1081 g

= 31.6096 = 31.61% Fe

4.131 Plan: Remember that spectator ions are omitted from net ionic equations. Assign oxidation numbers to all

atoms in the titration reaction, following the rules. The reactant that is the reducing agent contains an atom that is oxidized (O.N. increases from the left side to the right side of the equation). The reactant that is the oxidizing agent contains an atom that is reduced (O.N. decreases from the left side to the right side of the equation). Find the number of moles of KMnO4 from the molarity and volume and use molar ratios in the balanced equations to find the moles and then mass of CaCl2. Mass percent is calculated by dividing the mass of CaCl2 by the mass of the sample and multiplying by 100.

Solution: a) The reaction is: Na2C2O4(aq) + CaCl2(aq) → CaC2O4(s) + 2NaCl(aq) The total ionic equation in which soluble substances are dissociated into ions is: 2Na+(aq) + C2O4

2–(aq) + Ca2+(aq) + 2Cl–(aq) → CaC2O4(s) + 2Na+(aq) + 2Cl–(aq) Omitting Na+ and Cl– as spectator ions gives the net ionic equation:

Ca2+(aq) + C2O42–(aq) → CaC2O4(s)

b) You may recognize this reaction as a redox titration, because the permanganate ion, MnO4–, is a common

oxidizing agent. The MnO4– oxidizes the oxalate ion, C2O4

2– to CO2. Mn changes from +7 to +2 (reduction) and C changes from +3 to +4 (oxidation). The equation that describes this process is: H2C2O4(aq) + K+(aq) + MnO4

–(aq) → Mn2+(aq) + CO2(g) + K+(aq) H2C2O4 is a weak acid, so it cannot be written in a fully dissociated form. KMnO4 is a soluble salt, so it can be written in its dissociated form. K+(aq) is omitted in the net ionic equation because it is a spectator ion. We will balance the equation using the oxidation number method. First assign oxidation numbers to all elements in the reaction:

+2 +6 –8 –8 –4 +1 +3 –2 +7 –2 +2 +4 –2

H2C2O4(aq) + MnO4–(aq) → Mn2+(aq) + 2CO2(g)

Identify the oxidized and reduced species and multiply one or both species by the appropriate factors to make the electrons lost equal the electrons gained. Each C in H2C2O4 increases in O.N. from +3 to +4 for a total gain of 2 electrons. Mn in MnO4

– decreases in O.N. from +7 to +2 for a loss of 5 electrons. Multiply C by 5 and Mn by 2 to get an electron loss = electron gain of 10 electrons.

5H2C2O4(aq) + 2MnO4–(aq) → 2Mn2+(aq) + 10CO2(g)

Adding water and H+(aq) to finish balancing the equation is appropriate since the reaction takes place in acidic medium. Add 8H2O(l) to right side of equation to balance the oxygen and then add 6H+(aq) to the left to balance hydrogen. 5H2C2O4(aq) + 2MnO4

–(aq) + 6H+(aq) → 10CO2(g) + 2Mn2+(aq) + 8H2O(l) c) Mn in MnO4

– decreases in O.N. from +7 to +2 and is reduced. KMnO4 is the oxidizing agent. d) C in H2C2O4 increases in O.N. from +3 to +4 and is oxidized. H2C2O4 is the reducing agent. e) The balanced equations provide the accurate molar ratios between species.

Moles of MnO4– = ( )

340.1019 mol KMnO10 L37.68 mL

1 mL 1 L

−

= 0.0038396 mol MnO4–

Moles of H2C2O4 = ( ) 2 2 44

4

5 mol H C O0.0038396 mol MnO

2 mol MnO−

−

= 0.009599 mol H2C2O4

Mass (g) CaCl2 = ( ) 2 22 2 4

2 2 4 2

1 mol CaCl 110.98 g CaCl0.009599 mol H C O

1 mol H C O 1 mol CaCl

= 1.06530 g CaCl2

Mass percent CaCl2 = ( )2mass of CaCl 100

mass of sample= ( )21.06530 g CaCl

1001.9348 g mixture

= 55.05995 = 55.06% CaCl2

4-54

4.132 Plan: Write balanced equations for the two acid-base reactions. Find the moles of H2SO4 from the molarity and volume information and use the molar ratio in the balanced equation for the reaction of H2SO4 and NaOH to find the moles of NaOH used in the titration. Divide the moles of NaOH by its volume to determine its molarity.

Then find the moles of NaOH used in the titration of HCl by multiplying the NaOH molarity by its volume; use the molar ratio in this reaction to find moles of HCl. Dividing moles of HCl by its volume gives its molarity. Solution: Write the balanced chemical equations:

NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l) 2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2H2O(l) Determine the NaOH concentration from the reaction of NaOH with H2SO4:

Moles of H2SO4 = ( )3

2 40.0782 mol H SO10 L50.0 mL1 mL L

−

= 0.00391 mol H2SO4

Moles of NaOH = ( )2 42 4

2 mol NaOH0.00391 mol H SO1 mol H SO

= 0.00782 mol NaOH

Molarity of NaOH = 30.00782 1 mL18.4 mL 10 L−

= 0.425 M NaOH

Use the NaOH concentration and the reaction of HCl with NaOH to determine HCl concentration:


1 mL L

−

= 0.0116875 mol NaOH

Moles of HCl = ( ) 1 mol HCl0.0116875 mol NaOH1 mol NaOH

= 0.0116875 mol HCl

Molarity of HCl = ( ) 31 1 mL0.0116875 mol HCl

100.mL 10 L−

= 0.116875 = 0.117 M HCl

4.133 Plan: Recall that the total ionic equation shows all soluble ionic substances dissociated into ions and the net ionic equation omits the spectator ions. Use the molar ratio in the balanced reaction to find the moles of acid and base. Divide the moles of acid and base by the volume to obtain the molarity. Solution:

a) Molecular: H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l) Total ionic: 2H+(aq) + SO4

2–(aq) + 2Na+(aq) + 2OH–(aq) → 2Na+(aq) + SO42–(aq) + 2H2O(l)

Net ionic: H+(aq) + OH–(aq) → H2O(l) (Na+ and SO4

2– are spectator ions.)

b) Moles of H2SO4 = ( )2

4 2 42

4

0.010 mol SO 1 mol H SO2 orange spheres

1 orange sphere 1 mol SO

−

−

= 0.020 mol H2SO4

` Moles of NaOH = ( )2 42 4

2 mol NaOH0.020 mol H SO1 mol H SO

= 0.040 mol NaOH

c) Molarity of H2SO4 = ( )2 4 31 1 mL0.020 mol H SO

25mL 10 L−

= 0.80 M H2SO4

Molarity of NaOH = ( ) 31 1 mL0.040 mol NaOH

25mL 10 L−

= 1.6 M NaOH

4.134 Plan: Write a balanced equation for the reaction. This is an acid-base reaction between HCl and the base

CO32– to form H2CO3 which then decomposes into CO2 and H2O. Find the moles of HCl from the molarity and

volume information and use the molar ratio in the balanced equation to find the moles and mass of dolomite. To find mass %, divide the mass of dolomite by the mass of soil and multiply by 100.

4-55

Solution: The balanced equation for this reaction is:

CaMg(CO3)2(s) + 4HCl(aq) → Ca2+(aq) + Mg2+(aq) + 2H2O(l) + 2CO2(g) + 4Cl–(aq)

Moles of HCl = ( )310 L 0.2516 mol HCl33.56 mL

1 mL 1 L

−

= 0.0084437 mol HCl

Mass CaMg(CO3)2 = ( ) 3 2 3 2

3 2

1 mol CaMg(CO ) 184.41 g CaMg(CO )0.0084437 mol HCl

4 mol HCl 1 mol CaMg(CO )

= 0.389276 g CaMg(CO3)2

Mass percent CaMg(CO3)2 = ( )3 2mass CaMg(CO )100

mass soil = ( )3 20.389276 g CaMg(CO )

10013.86 g soil

= 2.80863 = 2.809% CaMg(CO3)2 4.135 Plan: Write balanced chemical equations for the acid-base titration reactions. To find the concentration of HA, find the moles of NaOH used for its titration by multiplying the molarity of NaOH by the volume used in the titration and using the molar ratio to find the moles of HA; dividing moles of HA by its volume gives the molarity. Multiply the molarity of HA by the volume of HA in the acid mixture to find the moles of HA in the mixture. Use the molar ratio to find the volume of NaOH required to titrate this amount of HA. The total volume of NaOH used in the titration of the mixture minus the volume required to titrate HA is the volume of NaOH required to titrate HB. Use this volume and molarity of NaOH and the molar ratio to find the moles and then molarity of HB. The volume of HB in the acid mixture is the total volume minus the volume of HA. Solution:

The balanced chemical equations for HA or HB with sodium hydroxide are the same. For HA it is: HA(aq) + NaOH(aq) → NaA(aq) + H2O(l) To find the concentration of HA:


1 mL L

−

= 0.007909 mol NaOH

Moles of HA = ( ) 1 mol HA0.007909 mol NaOH1 mol NaOH

= 0.007909 mol HA

Molarity of HA = ( ) 31 1 mL0.007909 mol HA

43.5 mL 10 L−

= 0.1818248 = 0.182 M HA

The titration of the acid mixture involves the reaction of NaOH with both of the acids.

Moles of HA in the acid mixture = ( )310 L 0.1818248 mol HA37.2 mL

1 mL L

−

= 0.0067639 mol HA

Volume (mL) of NaOH required to titrate HA =

( ) 31 mol NaOH 1 L 1 mL0.0067639 mol HA

1 mol HA 0.0906 mol NaOH 10 L−

= 74.6565 mL NaOH

Volume of NaOH required to titrate HB = total NaOH volume – volume of NaOH required to titrate HA = 96.4 mL – 74.6565 mL = 21.7435 mL NaOH

Moles of HB = ( )310 L 0.0906 mol NaOH 1 mol HB21.7435 mL

1 mL L 1 mol NaOH

−

= 0.00196996 mol HB

Volume (mL) of HB = Volume of mixture – volume of HA = 50.0 mL – 37.2 mL = 12.8 mL

Molarity of HB = ( ) 31 1 mL0.00196996 mol HB

12.8 mL 10 L−

= 0.153903 = 0.154 M HB

4.136 Plan: For part a), assign oxidation numbers to each element; the oxidizing agent has an atom whose oxidation

4-56

number decreases while the reducing agent has an atom whose oxidation number increases. For part b), use the molar ratios, beginning with step 3, to find the moles of NO2, then moles of NO, then moles of NH3 required to produce the given mass of HNO3. Solution: a) Step 1 +3 +2 –3 +1 0 +2 –2 +1 –2 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(l)

N is oxidized from –3 in NH3 to +2 in NO; O is reduced from 0 in O2 to to –2 in NO. Oxidizing agent = O2 Reducing agent = NH3

Step 2 –4 +2 –2 0 +4 –2

2NO(g) + O2(g) → 2NO2(g) N is oxidized from +2 in NO to +4 in NO2; O is reduced from 0 in O2 to –2 in NO2.

Oxidizing agent = O2 Reducing agent = NO Step 3 –4 +2 –6 +4 –2 +1 –2 +1+5 –2 +2 –2 3NO2(g) + H2O(l) → 2HNO3(l) + NO(g)

N is oxidized from +4 in NO2 to +5 in HNO3; N is reduced from +4 in NO2 to +2 in NO. Oxidizing agent = NO2 Reducing agent = NO2

b) Moles of NO2 = ( )3

4 3 23

3 3

1 mol HNO10 g 3 mol NO3.0x10 kg HNO 1 kg 63.02 g HNO 2 mol HNO

= 7.14059x105 mol NO2

Moles of NO = ( )52

2

2 mol NO7.14059x10 mol NO2 mol NO

= 7.14059x105 mol NO

Moles of NH3 = ( )5 34 mol NH7.14059x10 mol NO4 mol NO

= 7.14059x105 mol NH3

Mass (kg) of NH3 = ( )5 33 3

3

17.03 g NH 1 kg7.14059x10 mol NH1 mol NH 10 g

= 1.21604x104 = 1.2x104 kg NH3

4.137 Plan: In part a), use the density of the alloy to find the volume of a 0.263-g sample of alloy. That volume is the

sum of the volume of Mg and Al in the alloy. Letting x = mass of Mg and 0.263 – x = mass of Al, find the volume of each metal and set that equal to the total volume of the alloy. In part b), write balanced displacement reactions in which Mg and Al displace hydrogen from the HCl to produce H2. Use the molar ratios to find the masses of Mg and Al that must be present to produce the given amount of H2. In part c), write balanced reactions for the formation of MgO and Al2O3 and use molar ratios to find the masses of Mg and Al that must be present in the sample to produce the given amount of oxide. Solution: a) Let x = mass of Mg and 0.263 – x = mass of Al

Volume (cm3) of alloy = ( )31 cm0.263 g alloy

2.40 g alloy

= 0.10958 cm3

Volume of alloy = volume of Mg + volume of Al

0.10958 cm3 = ( ) ( )3 31 cm Mg 1 cm Alx g Mg + (0.263 x) g Al

1.74 g Mg 2.70 g Al

−

0.10958 cm3 = 0.574713x + 0.097407 – 0.37037x 0.012173 = 0.204343x x = 0.05957 g Mg

Mass percent Mg = ( ) ( )mass of Mg 0.05957 g Mg100 = 100mass of alloy sample 0.263 g sample alloy

= 22.6502 = 22.7% Mg

4-57

b) Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g) 2Al(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2(g)

Let x = mass of Mg and 0.263 – x = mass of Al Moles of H2 produced = moles of H2 from Mg + moles of H2 from Al

1.38x10–2 mol H2 = ( ) ( )2 21 mol H 3 mol H1 mol Mg 1 mol Alx g Mg + (0.263 x) g Al24.31 g Mg 1 mol Mg 26.98 g Al 2 mol Al

−

1.38x10–2 mol H2 = 0.041135x + 0.014622 – 0.055597x 8.22x10–4 = 0.014462x x = 0.05684 g Mg


= 21.6122 = 21.6% Mg

c) 2Mg(s) + O2(g) → 2MgO(s) 4Al(s) + 3O2(g) → 2Al2O3(s)

Let x = mass of Mg and 0.263 – x = mass of Al Mass of oxide produced = mass of MgO from Mg + mass of Al2O3 from Al 0.483 g oxide =

( ) ( ) 2 3 2 3

2 3

2 mol Al O 101.96 g Al O1 mol Mg 2 mol MgO 40.31 g MgO 1 mol Alx g Mg + (0.263 x) g Al24.31 g Mg 2 mol Mg 1 mol MgO 26.98 g Al 4 mol Al 1 mol Al O

−

0.483 g = 1.6582x + 0.49695 – 1.88955x 0.01395 = 0.23135x x = 0.060298 g Mg


= 22.927 = 22.9% Mg

4.138 Plan: An oxidation-reduction reaction is one in which oxidation numbers of atoms change from reactant to product. If there is no change in oxidation numbers, the reaction is not an oxidation-reduction reaction. Solution: a) Reaction (1):

0 0 +1 –1 0 Cl(g) + O3(g) → ClO(g) + O2(g) Cl is oxidized (0 to +1), and O is reduced (0 to –1). This is an oxidation-reduction reaction.

Reaction (2): +1 –1 +1 –1 +1 –1

ClO(g) + ClO(g) → Cl2O2(g) No atom has a change in oxidation number; this is not an oxidation-reduction reaction. Reaction (3):

+1 –1 0 0 Cl2O2(g) light→ 2Cl(g) + O2(g)

Cl is reduced (+1 to 0) and O is oxidized (–1 to 0). This is an oxidation-reduction reaction. Reaction (4): +1 –1 0 0 0 ClO(g) + O(g) → Cl(g) + O2(g)

Cl is reduced (+1 to 0) and O is oxidized (–1 to 0). This is an oxidation-reduction reaction. The oxidation-reduction reactions are 1, 3, and 4. b) Combining the equations requires multiplying equation 1 by 2, and then adding the three equations. 1) 2Cl(g) + 2O3(g) → 2ClO(g) + 2O2(g) 2) ClO(g) + ClO(g) → Cl2O2(g) 3) Cl2O2(g) light→ 2Cl(g) + O2(g)

2O3(g) → 3O2(g) 4.139 Plan: Write a balanced equation and use the molar ratio between Na2O2 and CO2 to convert the amount of

Na2O2 given to the amount of CO2 that reacts with that amount. Convert that amount of CO2 to liters of air.

4-58

Solution: The reaction is: 2Na2O2(s) + 2CO2(g) → 2Na2CO3(s) + O2(g).

Mass (g) of CO2 = ( ) 2 2 2 22 2

2 2 2 2 2

1 mol Na O 2 mol CO 44.01 g CO80.0 g Na O

77.98 g Na O 2 mol Na O 1 mol CO

= 45.1500 g CO2

Volume (L) of air = ( )22

L air45.150 g CO0.0720 g CO

= 627.08 = 627 L air

4.140 The salts are NaBr•2H2O and MgBr2•6H2O

NaBr(s) + AgNO3(aq) → AgBr(s) + NaNO3(aq) MgBr2(s) + 2AgNO3(aq) → 2AgBr(s) + Mg(NO3)2(aq)

Total moles of H2O in the sample = ( )3

22

2

1 mol H O10 g252.1 mg H O1 mg 18.02 g H O

−

= 0.013990 mol H2O

0.013990 mol H2O = mol H2O from NaBr•2H2O + mol H2O from MgBr2•6H2O. Let x = mol H2O from NaBr•2H2O and 0.013990 – x = mol H2O from MgBr2•6H2O.

Moles of NaBr = ( )22

1 mol NaBrx mol H O2 mol H O

= 0.5000x mol NaBr

Moles of MgBr2 = ( ) 22

2

1 mol MgBr0.013990 x mol H O

6 mol H O

−

= 0.002332 – 0.166667x mol MgBr2

Moles of AgBr produced = moles of AgBr from NaBr + moles of AgBr from MgBr2

6.00x10–3 mol AgBr = ( ) 1 mol AgBr0.5000x mol NaBr1 mol NaBr

+ ( )22

2 mol AgBr(0.002332 0.166667x) mol MgBr1 mol MgBr

−

6.00x10–3 mol AgBr = 0.50000x + 0.004664 – 0.333334x 0.001336 = 0.166666x x = 0.008016 mol Moles of NaBr•2H2O = 0.5000x = 0.5000(0.008016 mol) = 0.004008 mol NaBr•2H2O

Mass of NaBr•2H2O = ( ) 22

2

138.93 g NaBr•2H O0.004008 mol NaBr•2H O1 mol NaBr•2H O

= 0.5568 g NaBr•2H2O

Moles of MgBr2•6H2O = 0.002332 – 0.166667x = 0.002332 – 0.166667(0.008016 mol) = 9.960 x 10–4 mol MgBr2•6H2O

Mass of MgBr2•6H2O = ( )4 2 22 2

2 2

292.23 g MgBr •6H O9.960 x 10 mol MgBr •6H O1 mol MgBr •6H O

−

= 0.2911 g MgBr2•6H2O

Total mass of hydrates = 0.5568 g NaBr•2H2O + 0.2911 g MgBr2•6H2O = 0.8479 g

Mass percent of NaBr•2H2O = ( ) ( )2 2mass of NaBr•2H O 0.5568 g NaBr•2H O100 = 100total mass 0.8479 g

= 65.668 = 65.7% NaBr•2H2O

Mass percent of MgBr2•6H2O = ( ) ( )2 2 2 2mass of MgBr •6H O 0.2911 g MgBr •6H O100 = 100total mass 0.8472 g

= 34.662 = 34.7% MgBr2•6H2O 4.141 Plan: Use the mass percents to find the mass of each component needed for 1.00 kg of glass. Write balanced

reactions for the decomposition of sodium carbonate and calcium carbonate to produce an oxide and CO2. Use the molar ratios in these reactions to find the moles and mass of each carbonate required to produce the amount of oxide in the glass sample. Solution: A 1.00-kg piece of glass of composition 75% SiO2, 15% Na2O, and 10% CaO would contain:

Mass (kg) of SiO2 = ( ) 275% SiO1.00 kg glass

100% glass

= 0.75 kg SiO2

4-59

Mass (kg) of Na2O = ( ) 215% Na O1.00 kg glass

100% glass

= 0.15 kg Na2O

Mass (kg) of CaO = ( ) 10.% CaO1.00 kg glass100% glass

= 0.10 kg CaO

In this example, the SiO2 is added directly while the sodium oxide comes from decomposition of sodium carbonate and the calcium oxide from decomposition of calcium carbonate:

Na2CO3(s) → Na2O(s) + CO2(g) CaCO3(s) → CaO(s) + CO2(g)

Mass (kg) of Na2CO3 =

( )3

2 3 2 322 3

2 2 2 3

1 mol Na CO 105.99 g Na CO1 mol Na O10 g 1 kg0.15 kg Na O1 kg 61.98 g Na O 1 mol Na O 1 mol Na CO 10 g

= 0.25651 = 0.26 kg Na2CO3

Mass (kg) of CaCO3 = ( )3

3 33

3

1 mol CaCO 100.09 g CaCO10 g 1 mol CaO 1 kg0.10 kg CaO1 kg 56.08 g CaO 1 mol CaO 1 mol CaCO 10 g

= 0.178477 = 0.18 kg CaCO3 4.142 Plan: For part a), assign oxidation numbers to each element; the substance that has reduced has a decrease in

oxidation number as it gains electrons while the substance that has oxidized has an increase in oxidation number as it loses electrons. For part b), calculate the moles of Na2S2O3 by multiplying its molarity by the volume required in the titration. Then use the molar ratios in the balanced equations to find the moles and mass of oxygen in the sample. Solution: a) Step 1 +2 –8 +2 +2 –8 0 +1 –1 +1 +6 –2 0 +1 –2 +1 +6 –2 O2(aq) + 4KI(aq) + 2H2SO4(aq) → 2I2(aq) + 2H2O(l) + 2K2SO4(aq) The O.N. of iodine increases from –1 in KI to 0 in I2; I in KI is oxidized. Step 2 +2 +4 –6 +2 +10 –12 0 +1 +2 –2 +1 +2.5 –2 +1 –1 I2(aq) + 2Na2S2O3(aq) → Na2S4O6(aq) + 2NaI(aq) The O.N. of iodine decreases from 0 in I2 to –1 in NaI; I2 is reduced.

b) Moles of Na2S2O3 = ( )3

2 2 30.0105 mol Na S O10 L15.75 mL1 mL 1 L

−

= 1.65375x10–4 mol Na2S2O3

Moles of I2 (Step 2) = ( )4 22 2 3

2 2 3

1 mol I1.65375x10 mol Na S O2 mol Na S O

−

= 8.26875x10–5 mol I2

Moles of O2 (Step 1) = ( )5 22

2

1 mol O8.26875x10 mol I2 mol I

−

= 4.134375x10–5 mol O2

Mass (g) of O2 = ( )5 22

2

32.00 g O4.134375x10 mol O1 mol O

−

= 0.001323 g = 0.00132 g O2

4.143 Plan: To determine the reactant in excess, convert reactant masses to moles, and use molar ratios to see which

reactant makes the smaller (“limiting”) amount of product. Use the limiting reactant to calculate the amount of product formed. Use the molar ratio to find the amount of excess reactant required to react with the limiting reactant; the amount of excess reactant that remains is the initial amount of excess reactant minus the amount required for the reaction. Multiply moles of products and excess reactant by Avogadro’s number to obtain number of molecules.

4-60

Solution:

a) Moles of C2H5Cl if C2H4 is limiting = ( )3

2 52 42 4

2 4 2 4

1 mol C H Cl1 mol C H10 g0.100 kg C H1 kg 28.05 g C H 1 mol C H

= 3.56506 mol C2H5Cl

Moles of C2H5Cl if HCl is limiting = ( )3

2 51 mol C H Cl10 g 1 mol HCl0.100 kg HCl1 kg 36.46 g HCl 1 mol HCl

= 2.74273 mol C2H5Cl The HCl is limiting.

Moles HCl remaining = 0 mol

Moles of C2H4 initially present = ( )3

2 42 4

2 4

1 mol C H10 g0.100 kg C H1 kg 28.05 g C H

= 3.56506 mol C2H4

Moles of C2H4 that react = ( )3

2 41 mol C H10 g 1 mol HCl0.100 kg HCl1 kg 36.46 g HCl 1 mol HCl

= 2.74273 mol C2H4

Moles of C2H4 remaining = initial moles – reacted moles = 3.56506 mol – 2.74273 mol = 0.82233 mol C2H4 Moles of C2H5Cl formed = 2.74273 mol C2H5Cl Total moles of gas = moles HCl + moles C2H4 + moles C2H5Cl = 0 mol + 0.82233 mol + 2.74273 mol = 3.56506 mol

Molecules of gas = ( )236.022x10 molecules3.56506 mol gas

1 mol gas

= 2.146879x1024 = 2.15x1024 molecules

b) This will still be based on the HCl as the limiting reactant.

Initial moles of HCl = ( )310 g 1 mol HCl0.100 kg HCl

1 kg 36.46 g HCl

= 2.74273 mol HCl

Moles of HCl remaining = intial moles/2 = (2.74273 mol HCl)/2 = 1.371365 mol HCl

Moles of C2H4 reacting with half of HCl = ( ) 2 41 mol C H1.371365 mol HCl

1 mol HCl

= 1.371365 mol C2H4

Moles of C2H4 remaining = initial moles – reacted moles = 3.56506 mol – 1.371365 mol = 2.193695 mol C2H4

Moles of C2H5Cl formed = ( ) 2 51 mol C H Cl1.371365 mol HCl

1 mol HCl

= 1.371365 mol C2H5Cl

Total moles of gas = moles HCl + moles C2H4 + moles C2H5Cl = 1.371365 mol + 2.193695 mol + 1.371365 mol = 4.936425 = 4.94 mol total

4.144 Plan: Write balanced reactions and use molar ratios to find the moles of product. To find mass %, divide the mass

of thyroxine by the mass of extract and multiply by 100. Solution: a) There is not enough information to write complete chemical equations, but the following equations can be

written: C15H11I4NO4(s) + Na2CO3(s) → 4I–(aq) + other products I–(aq) + Br2(l) + HCl(aq) → IO3

–(aq) + other products

Moles of IO3– = ( ) 3

15 11 4 415 11 4 4

1 mol IO4 mol I1 mol C H I NO1 mol C H I NO 1 mol I

−−

−

= 4 moles IO3– are produced

b) +5 –2 +1 –1 0 +1 –2 IO3

–(aq) + H+(aq) + I–(aq) → I2(aq) + H2O(l) This is a difficult equation to balance because the iodine species are both reducing and oxidizing. The O.N. of iodine decreases from +5 in IO3

– to 0 in I2 and so gains 5 electrons; the O.N. of iodine increases from –1 in I– to 0 in I2 and so loses 1 electron. Start balancing the equation by placing a coefficient of 5 in front of I–(aq), so the electrons lost equal the electrons gained. Do not place a 5 in front of I2(aq), because not all of the I2(aq) comes

4-61

from oxidation of I–(aq). Some of the I2(aq) comes from the reduction of IO3–(aq). Place a coefficient of 3 in

front of I2(aq) to correctly balance iodine: IO3

–(aq) + H+(aq) + 5I–(aq) → 3I2(aq) + H2O(l) The reaction is now balanced from a redox standpoint, so finish balancing the reaction by balancing the oxygen and hydrogen.

IO3–(aq) + 6H+(aq) + 5I–(aq) → 3I2(aq) + 3H2O(l)

IO3– is the oxidizing agent, and I– is the reducing agent.

Moles of I2 produced per mole of thyroxine = ( ) 3 215 11 4 4

15 11 4 4 3

4 mol IO 3 mol I1 mol C H I NO

1 mol C H I NO 1 mol IO

−

−

= 12 moles of I2 are produced per mole of thyroxine c) Using Thy to represent thyroxine: The balanced equation for this reaction is I2(aq) + 2S2O3

2–(aq) → 2I–(aq) + S4O62–(aq).

Moles of S2O32– = ( )

232 30.1000 mol S O10 L17.23 mL

1 mL L

−−

= 0.001723 mol S2O32–

Moles of Thy = ( )2 22 3 2

22 3

1 mol I 1 mol Thy0.001723 mol S O12 mol I2 mol S O

−−

= 7.179167x10–5 mol Thy

Mass (g) of Thy = ( )5 776.8 g Thy7.179167x10 mol Thy1 mol Thy

−

= 0.055767769 g Thy

Mass % Thy = ( ) ( )mass of Thy 0.055767769 g Thy100 = 100mass of extract 0.4332 g

= 12.8734 = 12.87% thyroxine

4.145 Plan: Write the balanced equation for the reaction between the base KOH and the oleic acid. For each

sample, obtain moles of added KOH by multiplying the molarity and volume of the KOH. Use the molar ratio in the equation to obtain the moles and then mass of acid. To find the mass percent, divide the mass of acid by the mass of the sample and multiply by 100. Solution: Let HA represent oleic acid. The titration reaction is HA + KOH → H2O + KA. a) Sample 1

Moles of KOH = ( )3 10 L 0.050 mol KOH19.60 mL

1 mL 1 L

−

= 9.8x10–4 mol KOH

Mass (g) of oleic acid = ( )4 1 mol HA 282.45 g HA9.8x10 mol KOH1 mol KOH 1 mol HA

−

= 0.276801 g HA

Mass % oleic acid = ( ) ( )mass of oleic acid 0.276801 g 100 = 100mass of sample 10.00 g

= 2.76801 = 2.8% oleic acid

Sample 2


1 mL 1 L

−

= 9.8x10–4 mol KOH


−

= 0.2796255 g HA


= 2.796255 = 2.8% oleic acid

Samples 3 and 4


1 mL 1 L

−

= 1.0x10–3 mol KOH

4-62


−

= 0.28245 g HA


= 2.8245 = 2.8% oleic acid

b) The variation in acidity is a systematic error. The acidity values steadily increased over time as the KOH solution warmed and its volume increased. c) The actual acidity is 2.8245% as in Samples 3 and 4. This can be demonstrated by performing the titration several more times with the ethanol solution at 25°C.

4.146 Plan: Check the solubility rules to see if the compounds are soluble or insoluble. Any insoluble ionic

compound(s) will appear as solid in the correct scene, while any soluble ionic compound(s) will appear as dissociated ions. Find molarity by dividing moles by the volume of the solution in liters. Solution: a) Of the three ionic compounds, only AgCl is insoluble. CuCl2 and MgCl2 should appear as ions and AgCl should appear as a solid. Scene A best represents the mixture. There is AgCl solid and Mg2+, Cu2+, and Cl– ions in the proper mole ratio in A. Scene B shows solid MgCl2 and Scenes C and D show solid CuCl2. Since MgCl2 and CuCl2 are soluble in water, they would be present as ions only. b) There are 12 spheres representing ions in Scene A.

Moles of dissolved ions = ( )35.0x10 mol12 spheres

1 sphere

−

= 0.060 moles of ions

Molarity of dissolved ions = moles of ionsvolume of solution

= 3

0.060 moles of ions 1 mL50.0 mL 10 L−

= 1.2 M

c) Mass (g) of AgCl(s) = ( )35.0x10 mol Ag 1 mol AgCl 143.4 g AgCl4 Ag spheres

1 sphere 1 mol Ag 1 mol AgCl

−

= 2.868 = 2.9 g AgCl

4.147 Plan: Balance the equation to obtain the correct molar ratios. Use the mass percents to find the mass of each reactant in a 1.00 g sample, convert the mass of each reactant to moles, and use the molar ratios to find the limiting reactant and the amount of CO2 produced. Convert moles of CO2 produced to volume using the given conversion factor. Solution:

a) Here is a suggested method for balancing the equation. — Since PO4

2– remains as a unit on both sides of the equation, treat it as a unit when balancing. — On first inspection, one can see that Na needs to be balanced by adding a “2” in front of NaHCO3. This then affects the balance of C, so add a “2” in front of CO2. — Hydrogen is not balanced, so change the coefficient of water to “2,” as this will have the least impact on the other species. — Verify that the other species are balanced.

Ca(H2PO4)2(s) + 2NaHCO3(s) →∆ 2CO2(g) + 2H2O(g) + CaHPO4(s) + Na2HPO4(s) Determine whether Ca(H2PO4)2 or NaHCO3 limits the production of CO2. In each case calculate the moles of CO2 that might form.

Mass (g) of NaHCO3 = ( ) 31.0%1.00 g100%

= 0.31 g NaHCO3

Mass (g) of Ca(H2PO4)2 = ( ) 35.0%1.00 g100%

= 0.35 g Ca(H2PO4)2

Moles of CO2 if NaHCO3 is limiting = ( ) 3 23

3 3

1 mol NaHCO 2 mol CO0.31 g NaHCO

84.01 g NaHCO 2 mol NaHCO

= 3.690x10–3 mol CO2

4-63

Moles of CO2 if Ca(H2PO4)2 is limiting = ( ) 2 4 2 22 4 2

2 4 2 2 4 2

1 mol Ca(H PO ) 2 mol CO0.35g Ca(H PO )

234.05 g Ca(H PO ) 1 mol Ca(H PO )

= 2.9908x10–3 mol CO2 Since Ca(H2PO4)2 produces the smaller amount of product, it is the limiting reactant and 3.0x10–3 mol CO2 will be produced.

b) Volume (L) of CO2 = ( )32

2

37.0 L2.9908x10 mol CO1 mol CO

−

= 0.1106596 = 0.11 L CO2

4.148 Plan: Write a balanced acid-base reaction. Find the total moles of NaOH used by multiplying its molarity and volume in liters and use the molar ratio in the reaction to find the moles of HNO3. Divide moles of HNO3 by its volume to obtain the molarity. Use the molarity and volume information to find the moles of NaOH initially added and the moles of HNO3 initially present. The difference of these two values is the moles of excess NaOH. Solution: The chemical equation is:

HNO3(g) + NaOH(aq) → NaNO3(aq) + H2O(l) a) It takes a total of (20.00 + 3.22) mL = 23.22 mL NaOH to titrate a total of (50.00 + 30.00) mL = 80.00 mL of acid.


1 mL L

−

= 0.0011656 mol NaOH

Moles of HNO3 = ( ) 31 mol HNO0.0011656 mol NaOH

1 mol NaOH

= 0.0011656 mol HNO3

Molarity of HNO3 = ( )3 31 1 mL0.0011656 mol HNO

80.00 mL 10 L−

= 0.01457055 = 0.0146 M HNO3

b) First calculate the moles of the acid and base initially present. The difference will give the excess NaOH.


1 mL L

−

= 1.004x10–3 mol NaOH

Moles of HNO3 = ( )3

30.01457055 mol HNO10 L50.00 mL1 mL L

−

= 7.285275x10–4 mol HNO3

Moles of NaOH required to titrate 7.285275x10–4 mol HNO3 = 7.285275x10–4 mol NaOH Moles excess NaOH = moles of added NaOH – moles of NaOH required for reaction = 1.004x10–3 mol NaOH – 7.285275x10–4 mol NaOH

= 2.754725x10–4 = 2.8x10–4 mol NaOH 4.149 Plan: To determine the empirical formula, find the moles of each element present and divide by the

smallest number of moles to get the smallest ratio of atoms. To find the molecular formula, divide the molar mass by the mass of the empirical formula to find the factor by which to multiple the empirical formula. Write the balanced acid-base reaction for part c) and use the molar ratio in that reaction to find the mass of bismuth(III) hydroxide. Solution: a) Determine the moles of each element present. The sample was burned in an unknown amount of O2, therefore,

the moles of oxygen must be found by a different method.

Moles of C = ( ) 22

2 2

1 mol CO 1 mol C0.1880 g CO44.01 g CO 1 mol CO

= 4.271756x10–3 mol C

Moles of H = ( ) 22

2 2

1 mol H O 2 mol H0.02750 g H O18.02 g H O 1 mol H O

= 3.052164x10–3 mol H

Moles of Bi = ( ) 2 32 3

2 3 2 3

1 mol Bi O 2 mol Bi0.1422 g Bi O466.0 g Bi O 1 mol Bi O

= 6.103004x10–4 mol Bi

4-64

Subtracting the mass of each element present from the mass of the sample will give the mass of oxygen originally present in the sample. This mass is used to find the moles of oxygen.

Mass (g) of C = ( )3 12.01 g C4.271756x10 mol C 1 mol C

−

= 0.0513038 g C

Mass (g) of H = ( )3 1.008 g H3.052164x10 mol H 1 mol H

−

= 0.0030766 g H

Mass (g) of Bi = ( )4209.0 g Bi

6.103004x10 mol Bi 1 mol Bi−

= 0.127553 g Bi

Mass (g) of O = mass of sample – (mass C + mass H + mass Bi) = 0.22105 g sample – (0.0513038 g C + 0.0030766 g H + 0.127553 g Bi) = 0.0391166 g O

Moles of O = ( )

O g 16.00

O mol 1 O g 0.0391166 = 2.44482x10–4 mol O

Divide each of the moles by the smallest value (moles Bi).

C = 3

44.271756x106.103004x10

−

− = 7 H = 3

43.052164x106.103004x10

−

− = 5

O = 3

42.4448x10

6.103004x10

−

− = 4 Bi = 4

46.103004x106.103004x10

−

− = 1

Empirical formula = C7H5O4Bi b) The empirical formula mass is 362 g/mol. Therefore, there are 1086/362 = 3 empirical formula units per molecular formula making the molecular formula = 3 x C7H5O4Bi = C21H15O12Bi3. c) Bi(OH)3(s) + 3HC7H5O3(aq) → Bi(C7H5O3)3(s) + 3H2O(l)

d) Moles of C21H15O12Bi3 = ( )3

21 15 12 321 15 12 3

21 15 12 3

1 mol C H O Bi10 g0.600 mg C H O Bi1 mg 1086 g C H O Bi

−

= 5.52486x10–4 mol C21H15O12Bi3 Mass (mg) of Bi(OH)3 =

( )7 3 321 15 12 3 3

21 15 12 3 3

1 mol Bi(OH) 260.0 g Bi(OH)3 mol Bi 1 mg 100%5.52486x10 mol C H O Bi1 mol C H O Bi 1 mol Bi 1 mol Bi(OH) 88.0%10 g

−−

= 0.48970 = 0.490 mg Bi(OH)3 4.150 Plan: Use the solubility rules to predict the products of this reaction. For the total ionic equation, write all soluble

ionic substances as dissociated ions. Ions not involved in the precipitate are spectator ions and are not included in the net ionic equation. Find the moles of dissolved ions and divide each by the volume in liters to find the concentration. The volume of the final solution is the sum of the volumes of the two reactant solutions. Solution: a) According to the solubility rules, all chloride compounds are soluble and most common carbonate compounds are insoluble. CaCO3 is the precipitate. Molecular equation: Na2CO3(aq) + CaCl2(aq) → CaCO3(s) + 2NaCl(aq)

Total ionic equation: 2Na+(aq) + CO32–(aq) + Ca2+(aq) + 2Cl–(aq) → CaCO3(s) + 2Na+(aq) + 2Cl–(aq)

Net ionic equation: CO32–(aq) + Ca2+(aq) → CaCO3(s)

b) Ca2+ and CO32– combine in a 1:1 ratio in CaCO3. There are two spheres of Ca2+ and three spheres of CO3

2– ion. Since there are fewer spheres of Ca2+, Ca2+ is the limiting reactant.

Mass of CaCO3 = ( )2

2 3 32

3

1 mol CaCO 100.09 g CaCO0.050 mol Ca 2 Ca spheres1 sphere 1 mol CaCO1 mol Ca

++

+

= 10.009 = 10. g CaCO3 c) Original moles:

Moles of Na+ = ( ) 0.050 mol Na 6 Na spheres1 sphere

++

= 0.30 mol Na+

4-65

Moles of CO32– = ( )

22 3

30.050 mol CO

3 CO spheres1 sphere

−−

= 0.15 mol CO32–


2 0.050 mol Ca 2 Ca spheres1 sphere

++

= 0.10 mol Ca2+

Moles of Cl– = ( ) 0.050 mol Cl 4 Cl spheres1 sphere

−−

= 0.20 mol Cl–

The moles of Na+ and Cl– do not change. The moles of Ca2+ goes to zero, and removes 0.10 mol of CO32–.

Moles of remaining CO32– = 0.15 mol CO3

2– – 0.10 mol = 0.050 mol CO32–

Volume of final solution = 250. mL + 250. mL = 500. mL

Molarity of Na+ = 30.30 mol 1 mL500. mL 10 L−

= 0.60 M Na+

Molarity of Cl– = 3

0.20 mol 1 mL500. mL 10 L−

= 0.40 M Cl–

Molarity of CO32– = 3

0.050 mol 1 mL500. mL 10 L−

= 0.10 M CO32–

4.151 Plan: Write balanced equations. Use the density to convert volume of fuel to mass of fuel and then use the

molar ratios to convert mass of each fuel to the mass of oxygen required for the reaction. Use the conversion factor given to convert mass of oxygen to volume of oxygen. Solution: a) Complete combustion of hydrocarbons involves heating the hydrocarbon in the presence of oxygen to produce

carbon dioxide and water. Ethanol: C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l) Gasoline: 2C8H18(l) + 25O2(g) → 16CO2(g) + 18H2O(g) b) The mass of each fuel must be found:

Mass (g) of gasoline = ( ) 390% 1 mL 0.742 g1.00 L

100% 1 mL10 L−

= 667.8 g gasoline

Mass (g) of ethanol = ( ) 310% 1 mL 0.789 g1.00 L

100% 1 mL10 L−

= 78.9 g ethanol

Mass (g) of O2 to react with gasoline = ( ) 8 18 2 28 18

8 18 8 18 2

1 mol C H 25 mol O 32.00 g O667.8 g C H

114.22 g C H 2 mol C H 1 mol O

= 2338.64 g O2

Mass (g) of O2 to react with ethanol = ( ) 2 5 2 22 5

2 5 2 5 2

1 mol C H OH 3 mol O 32.00 g O78.9 g C H OH

46.07 g C H OH 1 mol C H OH 1 mol O

= 164.41 g O2 Total mass (g) of O2 = 2338.64 g O2 + 164.41 g O2 = 2503.05 = 2.50x103 g O2

c) Volume (L) of O2 = ( ) 22

2 2

1 mol O 22.4 L2503.05 g O32.00 g O 1 mol O

= 1752.135 = 1.75x103 L O2

d) Volume (L) of air = ( )2100%1752.135 L O20.9%

= 8383.42 = 8.38x103 L air

4.152 Plan: Write balanced reactions for the complete combustion of gasoline and for the incomplete combustion. Use

molar ratios to find the moles of CO2 and moles of CO produced. Obtain the number of molecules of each gas by multiplying moles by Avogadro’s number. Solution:

4-66

a) Complete combustion: 1. 2C8H18(l) + 25O2(g) → 16CO2(g) + 18H2O(g) Incomplete combustion: 2. 2C8H18(l) + 17O2(g) → 16CO(g) + 18H2O(g)

Assuming a 100-g sample of gasoline, 95%, or 95.0 g, will react by equation 1, and 5.0%, or 5.0 g, will react by equation 2.

Molecules of CO2 = ( )23

8 18 2 28 18

8 18 8 18 2

1 mol C H 16 mol CO 6.022 x10 CO95.0 g C H114.22 g C H 2 mol C H 1 mol CO

= 4.00693x1024 molecules CO2

Molecules of CO = ( )23

8 188 18

8 18 8 18

1 mol C H 16 mol CO 6.022x10 CO5.0 g C H114.22 g C H 2 mol C H 1 mol CO

= 2.10891x1023 molecules CO

Ratio of CO2 to CO molecules = 24

223

4.00693x10 CO molecules2.10891x10 CO molecules

= 18.99998 = 19

b) Again, we may assume 100 g of gasoline.

Mass (g) of CO2 = ( ) 8 18 2 28 18

8 18 8 18 2

1 mol C H 16 mol CO 44.01 g CO95.0 g C H

114.22 g C H 2 mol C H 1 mol CO

= 292.83 g CO2

Mass (g) of CO = ( ) 8 188 18

8 18 8 18

1 mol C H 16 mol CO 28.01 g CO5.0 g C H114.22 g C H 2 mol C H 1 mol CO

= 9.8091 g CO

Mass ratio of CO2 to CO = 2292.83 g CO 9.8091 g CO

= 29.85289 = 30

c) Let x = fraction of CO2 and y = fraction of CO. For a 1/1 mass ratio of CO2 to CO, (x)(44.01)y(28.01)

= 1, where

44.01 g/mol is the molar mass of CO2 and 28.01 g/mol is the molar mass of CO. x + y = 1 or y = 1 – x

Substituting: (x)(44.01)(1 x)(28.01)−

= 1

44.01x = 28.01 – 28.01x 72.02x = 28.01 x = 0.39 and y = 1 – 0.39 = 0.61 Thus, 61% of the gasoline must form CO. 4.153 Plan: From the molarity and volume of the base NaOH, find the moles of NaOH and use the molar ratios

from the two balanced equations to convert the moles of NaOH to moles of HBr to moles of vitamin C. Use the molar mass of vitamin C to convert moles to grams. Solution:

Moles of NaOH = ( )310 L 0.1350 mol NaOH43.20 mL NaOH

1 mL 1 L

−

= 0.005832 mol NaOH

Mass (g) of vitamin C = ( ) 6 8 6 6 8 63

6 8 6

1 mol C H O 176.12 g C H O1 mol HBr 1 mg0.005832 mol NaOH1 mol NaOH 2 mol HBr 1 mol C H O 10 g−

= 513.5659 = 513.6 mg C6H8O6 Yes, the tablets have the quantity advertised. 4.154 Plan: Convert mass of NaCl to moles of salt and then to moles of ions using the molar ratio in the formula. Use the molar ratio of ions in MgCl2 to find the mass of MgCl2 that contains the same number of moles of ions as the NaCl. Do the same for the CaS. Solution:

4-67

a) Moles of ions in NaCl = ( ) 1 mol NaCl 2 mol ions12.4 g NaCl58.44 g NaCl 1 mol NaCl

= 0.4243669 mol ions

Mass (g) of MgCl2 = ( ) 2 2

2

1 mol MgCl 95.21 g MgCl0.4243669 mol ions

3 mol ions 1 mol MgCl

= 13.46799 = 13.5 g MgCl2

b) Mass (g) of CaS = ( ) 1 mol CaS 72.14 g CaS0.4243669 mol ion2 mol ions 1 mol CaS

= 15.3069 = 15.3 g CaS

c) The CaS solution dissolves the most protein. All three solutions have equal ion concentrations, but CaS will form two divalent ions. 4.155 Plan: For part a), assign oxidation numbers to each element. The reactant that is the reducing agent contains an

atom that is oxidized (O.N. increases from the left side to the right side of the equation). The reactant that is the oxidizing agent contains an atom that is reduced (O.N. decreases from the left side to the right side of the equation). Use the molar ratios in the balanced equation to convert mass of ammonium perchlorate to moles of product and to moles of Al required in the reaction. Use the density values to convert masses to volumes. Solution: a) +4 –8 +6 –6 –3 +2

–3 +1+7 –2 0 +3 –2 +3 –1 +1 –2 +2 –2 3NH4ClO4(s) + 3Al(s) catalyst→Al2O3(s) +AlCl3(s) + 6H2O(g) + 3NO(g) The O.N. of chlorine decreases from +7 in NH4ClO4 to –1 in AlCl3 and is reduced; the O.N. of Al increases from 0 in Al to +3 in the products and is oxidized. The oxidizing agent is ammonium perchlorate and the reducing agent is aluminum.

b) Moles of gas = ( )3

4 44 4

4 4 4 4

1 mol NH ClO10 g 9 mol gas50.0 kg NH ClO1 kg 117.49 g NH ClO 3 mol NH ClO

= 1276.70 = 1.28x103 mol gas c) Initial volume:

Volume (L) of NH4ClO4 = ( )3 310 g 1 cc 1 mL 10 L50.0 kg

1 kg 1.95 g 1 cc 1 mL

−

= 25.6410 L

Mass of Al = ( )3

4 44 4

4 4 4 4

1 mol NH ClO10 g 3 mol Al 26.98g Al50.0 kg NH ClO1 kg 117.49 g NH ClO 3 mol NH ClO 1 mol Al

= 11481.828 g Al

Volume (L) of Al = ( )31 cc 1 mL 10 L11481.828 g Al

2.70 g Al 1 cc 1 mL

−

= 4.2525 L

Initial volume = 25.6410 L + 4.2525 L = 29.8935 L Final volume:

Mass (g) of Al2O3 = ( )3

2 3 2 34 44 4

4 4 4 4 2 3

1 mol Al O 101.96 g Al O1 mol NH ClO10 g50.0 kg NH ClO1 kg 117.49 g NH ClO 3 mol NH ClO 1 mol Al O

= 14463.64 g Al2O3

Volume (L) of Al2O3 = ( )3

2 32 3

1 cc 1 mL 10 L14463.674 g Al O3.97 g Al O 1 cc 1 mL

−

= 3.6432 L

Mass (g) of AlCl3 = ( )3

3 34 44 4

4 4 4 4 3

1 mol AlCl 133.33 g AlCl1 mol NH ClO10 g50.0 kg NH ClO1 kg 117.49 g NH ClO 3 mol NH ClO 1 mol AlCl

= 18913.67 g AlCl3

Volume (L) of AlCl3 = ( )3

33

1 cc 1 mL 10 L18913.67 g AlCl2.44 g AlCl 1 cc 1 mL

−

= 7.7515 L

4-68

Volume (L) of gas = ( ) 22.4 L1276.70 mol gas1 mol gas

= 28598.08 L

Final volume = 3.6432 L + 7.7515 L + 28598.08 L = 28609.4747 L Volume change = Final volume – initial volume = (28609.4747 L) – (29.8935 L) = 28579.5812 = 2.86x104 L The volumes of all solids (before and after) are insignificant. 4.156 Plan: Write soluble ionic substances as dissociated ions and omit spectator ions in the net ionic equations.

Solution: 1) This is a combination redox reaction. Cu(s) + Br2(aq) → CuBr2(aq)

Net ionic: Cu(s) + Br2(aq) → Cu2+(aq) + 2Br–(aq) 2) This is a precipitation reaction; most common hydroxide compounds are insoluble.

CuBr2(aq) + 2NaOH(aq) → Cu(OH)2(s) + 2NaBr(aq) Cu2+(aq) + 2Br–(aq) + 2Na+(aq) + 2OH–(aq) → Cu(OH)2(s) + 2Na+(aq) + 2Br–(aq) Net ionic: Cu2+(aq) + 2OH–(aq) → Cu(OH)2(s) 3) Cu(OH)2(s) → CuO(s) + H2O(l)

Net ionic: Cu(OH)2(s) → CuO(s) + H2O(l) 4) CuO(s) + 2HNO3(aq) → Cu(NO3)2(aq) + H2O(l)

CuO(s) + 2H+(aq) + 2NO3–(aq) → Cu2+(aq) + 2NO3

–(aq) + H2O(l) Net ionic: CuO(s) + 2H+(aq) → Cu2+(aq) + H2O(l) 5) This is a precipitation reaction; most common phosphate compounds are insoluble.

3Cu(NO3)2(aq) + 2Na3PO4(aq) → Cu3(PO4)2(s) + 6NaNO3(aq) 3Cu2+(aq) + 6NO3

–(aq) + 6Na+(aq) + 2PO43–(aq) → Cu3(PO4)2(s) + 6Na+(aq) + 6NO3

–(aq) Net ionic: 3Cu2+(aq) + 2PO4

3–(aq) → Cu3(PO4)2(s) 6) Cu3(PO4)2(s) + 3H2SO4(aq) → 3CuSO4(aq) + 2H3PO4(aq) Cu3(PO4)2(s) + 6H+(aq) + 3SO4

2–(aq) → 3Cu2+(aq) + 3SO42–(aq) + 2H3PO4(aq)

Net ionic: Cu3(PO4)2(s) + 6H+(aq) → 3Cu2+(aq) + 2H3PO4(aq) 7) This is a displacement reaction.

CuSO4(aq) + Zn(s) → Cu(s) + ZnSO4(aq) Cu2+(aq) + SO4

2–(aq) + Zn(s) → Cu(s) + Zn2+(aq) + SO42–(aq)

Cu2+(aq) + Zn(s) → Cu(s) + Zn2+(aq) 4.157 Plan: Write a balanced equation for this acid-base reaction. Find the moles of acid and base and use the molar

ratio in the balanced reaction to determine if stoichiometric amounts of each have been combined. To determine the molarity of B, divide the moles present by the volume in liters. For part c), find the volume of Solution B required for reaction and subtract from that the current volume to determine the additional volume needed. Solution: a) The balanced equation is Ca(OH)2(aq) + 2HCl(aq) → CaCl2(aq) + 2H2O(l).

Moles of Ca(OH)2 = ( )2

2 22

1 mol Ca(OH)0.0010 mol Ca4 spheres Ca1 sphere 1 mol Ca

++

+

= 0.0040 mol Ca(OH)2

Moles of HCl = ( ) 0.0010 mol H 1 mol HCl6 spheres H1 sphere 1 mol H

++

+

= 0.0060 mol HCl

Moles of HCl required for the titration = ( )22

2 mol HCl0.0040 mol Ca(OH)1 mol Ca(OH)

= 0.0080 mol HCl

Since 0.0060 mol of HCl has been added and 0.0080 mol HCl are required for complete titration, the equivalence point of the titration has not been reached.

b) Molarity = 3

mol HCl 0.0060 mol HCl 1 mL = volume of solution 25.0 mL 10 L−

= 0.24 M

4-69

c) Moles of HCl required = ( )22

2 mol HCl0.0040 mol Ca(OH)1 mol Ca(OH)

= 0.0080 mol HCl

Volume (L) of HCl = ( ) 31 L solution 1 mL0.0080 mol HCl

0.24 mol HCl 10 L−

= 33.3 mL

Additional volume = volume needed – volume added = 33.3 mL − 25.0 ml = 8.333 = 8.3 mL

chapter 4 three major classes of chemical · pdf file · 2014-10-26chapter 4 three...

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