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Chapter 5–1

Chapter 5 Chemical Reactions Solutions to In-Chapter Problems 5.1 The process is a chemical reaction because the reactants contain two gray spheres joined (indicating

H2) and two red spheres joined (indicating O2), while the product (H2O) contains a red sphere joined to two gray spheres (indicating O–H bonds).

5.2 The process is a physical change (freezing) since the particles in the reactants are the same as the

particles in the products. 5.3 Chemical equations are written with the reactants on the left and the products on the right

separated by a reaction arrow.

2 C8H18 + +

2 H2O2(aq) 2 H2O(l) O2(g) (4 H, 4 O)+a.

b.

c. 2 Na3PO4(aq) 3 MgCl2(aq)+ Mg3(PO4)2(s) 6 NaCl(aq) (3 Mg, 2 P, 8 O,+

25 O2 16 CO2 18 H2O (16 C, 50 O, 36 H)

reactants products

6 Na, 6 Cl)

5.4 To determine the number of each type of atom when a formula has both a coefficient and a subscript, multiply the coefficient by the subscript.

For 3 Al2(SO4)3: Al = 6 (3 × 2), S = 9 (3 × 3), O = 36 (3 × 3 × 4)

5.5 Write the chemical equation for the statement.

CH4(g) + 4 Cl2(g) CCl4(l) + 4 HCl(g)!

5.6 Balance the equation with coefficients one element at a time to have the same number of atoms

on each side of the equation. Follow the steps in Example 5.2.

2 H2 2 H2OO2+a.

[1] Place a 2 to balance O's.

[2] Place a 2 to balance H's.

b. 2 NO 2 NO2+ O2

c. CH4 2 Cl2 CH2Cl2 2 HCl+ +

5.7 Write the balanced chemical equation for carbon monoxide and oxygen reacting to form carbon

dioxide. The smallest set of whole numbers must be used.

2 CO + O2 2 CO2 5.8 Follow the steps in Example 5.2 to write the balanced chemical equation.

2 C2H6 + 7 O2 4 CO2 + 6 H2O

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Chemical Reactions 5–2

5.9 Write a balanced equation for the Haber process.

N2 + 3 H2 2 NH3 5.10 Balance the equations as in Example 5.2.

2 Al 3 H2SO4 Al2(SO4)3 3 H2+ +a.

b. 3 Na2SO3 2 H3PO4 3 H2SO3 2 Na3PO4+ + 5.11 One mole, abbreviated as mol, always contains an Avogadro’s number of particles (6.02 × 1023).

a, b, c, d: 6.02 × 1023 5.12 Multiply the number of moles by Avogadro’s number to determine the number of atoms.

Avogadro’s number is the conversion factor that relates moles to molecules, as in Example 5.3.

a. 2.00 mol × 6.02 × 1023 atoms/mol = 1.20 × 1024 atoms b. 6.00 mol × 6.02 × 1023 atoms/mol = 3.61 × 1024 atoms c. 0.500 mol × 6.02 × 1023 atoms/mol = 3.01 × 1023 atoms d. 25.0 mol × 6.02 × 1023 atoms/mol = 1.51 × 1025 atoms

5.13 Multiply the number of moles by Avogadro’s number to determine the number of molecules, as

in Example 5.3.

a. 2.5 mol × 6.02 × 1023 molecules/mol = 1.5 × 1024 molecules b. 0.25 mol × 6.02 × 1023 molecules/mol = 1.5 × 1023 molecules c. 0.40 mol × 6.02 × 1023 molecules/mol = 2.4 × 1023 molecules d. 55.3 mol × 6.02 × 1023 molecules/mol = 3.33 × 1025 molecules

5.14 Use Avogadro’s number as a conversion factor to relate molecules to moles.

6.02 x 1025 molecules6.02 x 1023 molecules

= 100. mol1 mol

xa.


= 0.0500 mol1 molxb.


= 15 mol1 mol

xc.


Chapter 5–3

5.15 To calculate the formula weight, multiply the number of atoms of each element by the atomic weight and add the results.

a. 1 Ca atom × 40.08 amu = 40.08 amu 1 C atom × 12.01 amu = 12.01 amu 3 O atoms × 16.00 amu = 48.00 amu Formula weight of CaCO3 100.09 amu b. 1 K atom × 39.10 amu = 39.10 amu 1 I atom × 126.9 amu = 126.9 amu Formula weight of KI 166.00 amu rounded to 166.0 amu

5.16 Calculate the molecular weight in two steps:

[1] Write the correct formula and determine the number of atoms of each element from the subscripts.

[2] Multiply the number of atoms of each element by the atomic weight and add the results.

a. 2 C atoms × 12.01 amu = 24.02 amu 6 H atoms × 1.008 amu = 6.048 amu 1 O atom × 16.00 amu = 16.00 amu Molecular weight of ethanol (C2H6O) 46.068 amu rounded to 46.07 amu

b. 6 H atoms × 1.008 amu = 6.048 amu 6 C atoms × 12.01 amu = 72.06 amu 1 O atom × 16.00 amu = 16.00 amu Molecular weight of phenol (C6H6O) 94.108 amu rounded to 94.11 amu

c. 1 H atom × 1.008 amu = 1.008 amu 2 C atom × 12.01 amu = 24.02 amu 1 Br atom × 79.90 amu = 79.90 amu 1 Cl atom × 35.45 amu = 35.45 amu 3 F atoms × 19.00 amu = 57.00 amu Molecular weight of halothane

(C2HBrClF3) 197.378 amu rounded to 197.38 amu

5.17

C20H24O10 20 C atoms × 12.01 amu = 240.2 amu 24 H atoms × 1.008 amu = 24.192 amu 10 O atoms × 16.00 amu = 160.0 amu Molecular weight of ginkgolide B: 424.392 amu = 424.4 g/mol

5.18 Convert the moles to grams using the molar mass as a conversion factor.

a. 0.500 mol of NaCl × 58.44 g/mol = 29.2 g c. 3.60 mol of C2H4 × 28.05 g/mol = 101 g b. 2.00 mol of KI × 166.0 g/mol = 332 g d. 0.820 mol of CH4O × 32.04 g/mol = 26.3 g



5.19 Convert grams to moles using the molar mass as a conversion factor.

100. g NaCl 1 mol = 1.71 mol58.44 g

xa.

25.5 g CH4

1 mol = 1.59 mol16.04 g

xb.

0.250 g C9H8O41 mol = 1.39 x 10–3 mol

180.2 gxc.

25.0 g H2O 1 mol = 1.39 mol18.02 g

xd.

5.20 Use conversion factors to determine the number of molecules in 1.00 g; two 500.-mg tablets =

1.00 g.

1.00 g penicillin =334.4 g penicillin

6.02 x 1023 moleculesx

Grams cancel.

1.80 x 1021 molecules of penicillinAnswer:

5.21 Use mole–mole conversion factors as in Example 5.5 and the equation below to solve the

problems.

N2(g) O2(g) 2 NO(g)!+

a. 3.3 mol N2 × (2 mol NO/1 mol N2) = 6.6 mol NO b. 0.50 mol O2 × (2 mol NO/1 mol O2) = 1.0 mol NO c. 1.2 mol N2 × (1 mol O2/1 mol N2) = 1.2 mol O2

5.22 Use mole–mole conversion factors as in Example 5.5 and the equation below to solve the

problems.

2 C2H6(g) 5 O2(g) 4 CO(g) 6 H2O(g)!+ +

a. 3.0 mol C2H6 × (5 mol O2/2 mol C2H6) = 7.5 mol O2 b. 0.50 mol C2H6 × (6 mol H2O/2 mol C2H6) = 1.5 mol H2O c. 3.0 mol CO × (2 mol C2H6/4 mol CO) = 1.5 mol C2H6


Chapter 5–5

5.23 [1] Convert the number of moles of reactant to the number of moles of product using a mole–mole conversion factor.

[2] Convert the number of moles of product to the number of grams of product using the product’s molar mass.

C6H12O6(aq) 2 C2H6O(aq) 2 CO2(g)+

mole–mole conversion factor

x

Moles C6H12O6 cancel.

=0.55 mol C6H12O6

Moles of reactant

Moles of product

1 mol C6H12O6

2 mol C2H6O1.1 mol C2H6O

molar mass conversion factor

Grams of product

Moles of product

x

Moles cancel.

=Answer

1.1 mol C2H6O 46.07 g C2H6O1 mol C2H6O

51 g C2H6O

a.


x

Moles C6H12O6 cancel.

=0.25 mol C6H12O6

Moles of reactant

Moles of product

1 mol C6H12O6

2 mol CO2 0.50 mol CO2


Grams of product

Moles of product

x

Moles cancel.

=Answer

0.50 mol CO244.01 g CO2

1 mol CO2

22 g CO2

b.

mole–mole

conversion factor

x

Moles C2H6O cancel.

=1.0 mol C2H6O

Moles of product

Moles of reactant

2 mol C2H6O1 mol C6H12O6 0.50 mol C6H12O6


Grams of reactant

Moles of reactant

x

Moles cancel.

=Answer

0.50 mol C6H12O6 180.2 g C6H12O6

1 mol C6H12O6

90. g C6H12O6

c.



5.24 Use the steps outlined in Answer 5.23 to answer the questions.

C2H6O(l) 3 O2(g) 2 CO2(g) 3 H2O(g)+ + a. 0.50 mol C2H6O × (2 mol CO2/1 mol C2H6O) = 1.0 mol CO2 1.0 mol CO2 × (44.01 g CO2/1 mol CO2) = 44 g CO2

b. 2.4 mol C2H6O × (3 mol H2O/1 mol C2H6O) = 7.2 mol H2O 7.2 mol H2O × (18.02 g H2O/1 mol H2O) = 130 g H2O c. 0.25 mol C2H6O × (3 mol O2/1 mol C2H6O) = 0.75 mol O2 0.75 mol O2 × (32.00 g O2/1 mol O2) = 24 g O2

5.25 Use conversion factors to solve the problems. Follow the steps in Sample Problem 5.14.

C7H6O3(s) + C2H4O2(l) C9H8O4(s) H2O(l)salicylic acid aspirinacetic acid

+

a. 55.5 g C7H6O3 × (1 mol C7H6O3/138.1 g C7H6O3) = 0.402 mol C7H6O3 0.402 mol C7H6O3 × (1 mol C9H8O4/1 mol C7H6O3) = 0.402 mol C9H8O4 0.402 mol C9H8O4 × (180.2 g C9H8O4/1 mol C9H8O4) = 72.4 g C9H8O4

b. 55.5 g C7H6O3 × (1 mol C7H6O3/138.1 g C7H6O3) = 0.402 mol C7H6O3

0.402 mol C7H6O3 × (1 mol C2H4O2/1 mol C7H6O3) = 0.402 mol C2H4O2

0.402 mol C2H4O2 × (60.05 g C2H4O2/1 mol C2H4O2) = 24.1 g C2H4O2

c. 55.5 g C7H6O3 × (1 mol C7H6O3/138.1 g C7H6O3) = 0.402 mol C7H6O3

0.402 mol C7H6O3 × (1 mol H2O/1 mol C7H6O3) = 0.402 mol H2O

0.402 mol H2O × (18.02 g H2O/1 mol H2O) = 7.24 g H2O

5.26 Use conversion factors to solve the problems. Follow the steps in Sample Problem 5.14.

N2 + O2 → 2 NO

a. 10.0 g N2 × (1 mol N2/28.02 g N2) = 0.357 mol N2 0.357 mol N2 (2 mol NO/1 mol N2) = 0.714 mol NO 0.714 mol NO × (30.01 g NO/1 mol NO) = 21.4 g NO b. 10.0 g O2 × (1 mol O2/32.00 g O2) = 0.313 mol O2 0.313 mol O2 × (2 mol NO/1 mol O2) = 0.626 mol NO 0.626 mol NO × (30.01 g NO/1 mol NO) = 18.8 g NO c. 10.0 g N2 × (1 mol N2/28.02 g N2) = 0.357 mol N2 0.357 mol N2 × (1 mol O2/1 mol N2) = 0.357 mol O2

0.357 mol O2 × (32.00 g O2/1 mol O2) = 11.4 g O2


Chapter 5–7

5.27 [1] Convert the number of moles of reactant to the number of moles of product using a mole–mole conversion factor.

[2] Convert the number of moles of product to the number of grams of product—the theoretical yield—using the product’s molar mass.

C(s) + O2(g) CO2(g)


x =3.50 mol C

Moles of reactant

Moles of product

1 mol C1 mol CO2 3.50 mol CO2

3.50 mol CO2

molar mass conversion factor Grams of

productMoles of product

x =

Answer part (a)

44.01 g CO2

1 mol CO2

154 g CO2

Theoretical yield

actual yield (g)=Percent yield

theoretical yield (g)x 100%

53.5 g154 g

x 100%= =Answer part (b)

34.7%

5.28 Use the steps in Answer 5.27 to solve the problem.


x =8.0 mol O2

Moles of reactant

Moles of product

3 mol O2

2 mol O3 5.3 mol O3

5.3 mol O3


Grams of product

Moles of product

x =

Answer part (a)

48.00 g O3

1 mol O3

250 g O3

Theoretical yield



155 g250 g

x 100%= =Answer part (b)

62%



5.29 Use the steps in Sample Problem 5.17 to answer the questions.

CCC

C CCHO NH2

CCC

C CCHO N C

H

H H

H H

HH

H H

OC HH

HHClC2H3ClO+ +

4-aminophenolacetyl

chloride acetaminophenmolar mass 151.2 g/molmolar mass 109.1 g/mol

=

Grams of reactant Moles of

reactant

80.0 g 4-aminophenol 1 mol 4-aminophenol

109.1 g 4-aminophenolx 0.733 mol 4-aminophenol



x =

Moles of reactant

Moles of product

0.733 mol 4-aminophenol1 mol 4-aminophenol

1 mol acetaminophen0.733 mol acetaminophen


Grams of product

Moles of product

x =151.2 g acetaminophen0.733 mol acetaminophen1 mol acetaminophen

111 g acetaminophenTheoretical yield

a.

actual yield (g)=Percent yieldtheoretical yield (g)

x 100%

111 gx 100%= =

Answer59.0%

65.5 g

b.


Chapter 5–9

5.30 Use conversion factors to solve the problem. Follow the steps in Answer 5.29.

=


reactant

324 g O2 1 mol O2

32.00 g O2x 10.1 mol O2



x =

Moles of reactant

Moles of product

10.1 mol O2 3 mol O2

2 mol O3 6.73 mol O3


Grams of product

Moles of product

x =48.00 g O36.73 mol O31 mol O3

323 g O3

Theoretical yield

a.


x 100%

323 gx 100%= =

Answer37.8%

122 g

b.

5.31 To determine the overall percent yield in a synthesis that has more than one step, multiply the percent yield for each step.

a. (0.90)10 × 100% = 35% b. (0.80)10 × 100% = 11% c. 0.50 × (0.90)9 × 100% = 19% d. 0.20 × 0.50 × 0.50 × 0.80 × 0.80 × 0.80 × 0.80 × 0.80 × 0.80 × 0.80 × 100% = 1.0%



5.32 Use the steps in Sample Problem 5.18 to answer the questions.

3 molecules of H2original quantity

? molecules of N2unknown quantity

1 molecule N2

3 molecules H21 molecule N2

3 molecules H2 or Choose this conversion factor to cancel molecules of H2.

1 molecule N2

3 molecules H23 molecules H2 x = 1 molecule of N2 is needed.

2 molecules of N2 are left over.2 molecules of NH3 are formed.

NH3

N2

N

H

H2 is the limiting reactant.

To determine the limiting reagent:

5.33

1 mol O2

2 mol H25.0 mol H2 x = 2.5 mol of O2 are needed.

Since 5.0 mol of O2 are present, H2 is the limiting reactant.

a.

1 mol O2



b.

c.

1 mol O2


Since only 2.0 mol of O2 are present, O2 is the limiting reactant.

1 mol O2

2 mol H22.0 mol H2 x = 1.0 mol of O2 is needed.


d.

5.34 Calculate the number of moles of product formed as in Sample Problem 5.19.

x =


1.5 mol H21 mol N2

3 mol H20.50 mol N2 needed

x =


1.5 mol H22 mol NH3

3 mol H21.0 mol NH3Answer

a. H2 is the limiting reactant.


Chapter 5–11

x =


1.0 mol H21 mol N2


x =


1.0 mol H22 mol NH3


b. H2 is the limiting reactant.

x =


2.0 mol H21 mol N2


x =


2.0 mol H22 mol NH3


c. H2 is the limiting reactant.

x =


7.5 mol H21 mol N2


x =


2.0 mol N22 mol NH3

1 mol N24.0 mol NH3Answer

d. N2 is the limiting reactant.

5.35 Convert the number of grams of each reactant to the number of moles using molar masses. Since

the mole ratio of O2 to N2 is 1:1, the limiting reactant has fewer moles.

=


reactant

12.5 g N2 1 mol N2

28.02 g N2x 0.446 mol N2

molar mass conversion factora.

=


reactant

15.0 g O2 1 mol O2

32.00 g O2x 0.469 mol O2


N2 is the limiting reactant.



=


reactant

14.0 g N2 1 mol N2

28.02 g N2x 0.500 mol N2

molar mass conversion factorb.

=


reactant

13.0 g O2 1 mol O2

32.00 g O2x 0.406 mol O2


O2 is the limiting reactant.

5.36 Calculate the number of moles of product formed based on the limiting reactant. Then convert

moles to grams using molar mass.

0.446 mol N22 mol NO1 mol N2

x = 0.892 mol NO

0.892 mol NO 30.01 g1 mol NO

x = 26.8 g NO

a.

0.406 mol O22 mol NO1 mol O2

x = 0.812 mol NO

0.812 mol NO 30.01 g1 mol NO

x = 24.4 g NO

b.

5.37

0.313 mol O22 mol H2O1 mol O2

x = 0.626 mol H2O

0.626 mol H2O 18.02 g1 mol H2O

x = 11.3 g H2O

=


reactant

5.00 g H21 mol H2

2.016 g H2

x 2.48 mol H2


=10.0 g O21 mol O2

32.00 g O2x 0.313 mol O2 O2 is the limiting reactant.


Chapter 5–13

5.38 A compound that gains electrons is reduced. A compound that loses electrons is oxidized.

2 H+(aq)Zn(s) + Zn2+(aq) + H2(g)a.

b. Fe3+(aq) Al(s) Al3+(aq) Fe(s)+ +

(oxidized) (reduced)

Zn2+Zn + 2 e–

H22 H+ + 2 e–

(oxidized)(reduced)

Al3+Al + 3 e–

FeFe3+ + 3 e–

c. 2 I– Br2 I2 2 Br–+ +

d. 2 AgBr 2 Ag Br2+

(oxidized) (reduced)

I22 I– + 2 e–

2 Br–Br2 + 2 e–

(oxidized) (reduced) 2 Br– 2 Ag+

Br22 Br– + 2 e–

2 Ag2 Ag+ + 2 e–

5.39 A compound that gains electrons while causing another compound to be oxidized is called an

oxidizing agent. A compound that loses electrons while causing another compound to be reduced is called a reducing agent.

a. Zn reducing agent, H+ oxidizing agent c. I– reducing agent, Br2 oxidizing agent b. Fe3+ oxidizing agent, Al reducing agent d. Br– reducing agent, Ag+ oxidizing agent

5.40

Zn is oxidized, and Hg2+ is reduced.

Zn2+Zn + 2 e– HgHg2+ + 2 e–

Zn loses electrons and is oxidized. Hg2+ gains electrons and is reduced.

5.41 H2 is oxidized since it gains an O atom and C2H4O2 is reduced since it gains hydrogen.

+C2H4O2 C2H6O H2O+ 2 H2

gains O atomoxidized

gains H atoms

reduced

5.42 Zn is the reducing agent and Hg2+ is the oxidizing agent.

oxidized

reduced

Zn + Hg2+ Zn2+ Hg+



Solutions to End-of-Chapter Problems 5.43 The process is a chemical reaction because the spheres in the reactants are joined differently than

the spheres in the products.

2 O3 2 CO2 + 2 O22 CO + (not balanced) 5.44 a. The transformation of [1] to [2] is a chemical reaction because the spheres in the reactants (AB)

are joined differently than the spheres in the products (A2 and B2). b. The transformation of [1] to [3] is a physical change because the spheres are joined the same

(AB) but they are now closer together indicating a physical state change from gas to liquid. 5.45 The difference between a coefficient and a subscript is that the coefficient indicates the number of

molecules or moles undergoing reaction, whereas the subscript indicates the number of atoms of each element in a chemical formula.

5.46 It is not possible to change the subscripts of a chemical formula to balance an equation because

changing the subscripts changes the identity of the compound. 5.47 Add up the number of atoms on each side of the equation and then label the equations as balanced

or not balanced.

2 HCl(aq) + Ca(s) CaCl2(aq) H2(g)

TiCl4 2 H2O TiO2 HCl

Al(OH)3 H3PO4 AlPO4 3 H2O

+

+

+

+

+

a.

b.

c.

2 H, 2 Cl, 1 Ca: both sides, therefore balanced

1 Ti, 4 Cl, 4 H, 2 O 1 Ti, 1 Cl, 1 H, 2 O

NOT balanced

1 Al, 1 P, 7 O, 6 H: both sides, therefore balanced 5.48 Add up the number of atoms on each side of the equation and then label the equations as balanced or not balanced.

3 NO2 + H2O HNO3 2 NO

2 H2S 3 O2 H2O 2 SO2

Ca(OH)2 2 HNO3 2 H2O Ca(NO3)2

+

+

+

+

+

a.

b.

c.

3 N, 7 O, 2H

4 H, 2 S, 6 O 2 H, 2 S, 5 ONOT balanced

1 Ca 8 O, 4 H, 2 N: both sides, therefore balanced

3 N, 5 O, 1 HNOT balanced

5.49 Write the balanced equation using the colors of the spheres to identify the atoms (gray = hydrogen

and green = chlorine).

Cl2H2 + 2 HCl


Chapter 5–15

5.50 Write the balanced equation using the colors of the spheres to identify the atoms (red = oxygen and blue = nitrogen).

NOO2 + NO3 5.51 Balance the equation with coefficients one element at a time to have the same number of atoms on

each side of the equation. Follow the steps in Example 5.2.

+4 Al(OH)3 6 H2SO4 2 Al2(SO4)3 12 H2O

CH4(g) 4 Cl2(g) CCl4(g) 4 HCl(g)

Ni(s) 2 HCl(aq) NiCl2(aq) H2(g)

2 KClO3 2 KCl 3 O2

Al2O3 6 HCl 2 AlCl3 3 H2O

a.

b.

c.

d.

e.

+ +

++

+

+ +

+ 5.52 Balance the equation with coefficients one element at a time so that there are the same numbers of

atoms on each side of the equation. Follow the steps in Example 5.2.

+2 H3PO4 3 Ca(OH)2 Ca3(PO4)2 6 H2O

2 CO(g) O2(g) 2 CO2(g)

Mg(s) 2 HBr(aq) MgBr2(aq) H2(g)

2 PbS(s) 2 PbO(s)3 O2(g)

H2SO4 2 NaOH Na2SO4 2 H2O

a.

b.

c.

d.

e.

+ +

+

+

+ +

+

+ 2 SO2(g)

5.53 Follow the steps in Example 5.2 and balance the equations.

a. 2 C6H6 + 15 O2 12 CO2 + 6 H2O

b. C7H8 + 9 O2 7 CO2 + 4 H2Oc. 2 C8H18 + 25 O2 16 CO2 + 18 H2O

5.54 Follow the steps in Example 5.2 and balance the equation.

2 C5H12O + 15 O2 10 CO2 + 12 H2O 5.55 Follow the steps in Example 5.2 and balance the equation.

2 S(s) + 3 O2(g) 2 H2O(l) 2 H2SO4(l)+ 5.56 Follow the steps in Example 5.2 and balance the equation.

MgCl2 + 2 NaOH Mg(OH)2 + 2 NaCl



5.57 Fill in the molecules of the products using the balanced equation and following the law of conservation of mass. Each side must have the same number of O and C atoms.

reactants products

CO2O2

O3 CO

5.58 Fill in the molecules of the products using the balanced equation and following the law of

conservation of mass. Each side must have the same number of C, O, and N atoms.

reactants products

CO2N2CO NO

5.59 To calculate the formula weight, multiply the number of atoms of each element by the atomic

weight and add the results. The formula weight in amu is equal to the molar mass in g/mol.

a. 1 Na atom × 22.99 amu = 22.99 amu 1 N atom × 14.01 amu = 14.01 amu 2 O atoms × 16.00 amu = 32.00 amu Formula weight of NaNO2 69.00 amu = 69.00 g/mol

b. 2 Al atom × 26.98 amu = 53.96 amu 3 S atoms × 32.07 amu = 96.21 amu 12 O atoms × 16.00 amu = 192.0 amu Formula weight of Al2(SO4)3 342.17 amu rounded to 342.2 amu = 342.2 g/mol

c. 6 C atom × 12.01 amu = 72.06 amu 8 H atoms × 1.008 amu = 8.064 amu 6 O atoms × 16.00 amu = 96.00 amu Formula weight of C6H8O6 176.124 amu rounded to 176.12 amu = 176.12 g/mol


Chapter 5–17

5.60 To calculate the formula weight, multiply the number of atoms of each element by the atomic weight and add the results. The formula weight in amu is equal to the molar mass in g/mol.

a. 1 Mg atom × 24.30 amu = 24.30 amu 1 S atom × 32.07 amu = 32.07 amu 4 O atoms × 16.00 amu = 64.00 amu Formula weight of MgSO4 120.37 amu = 120.37 g/mol

b. 3 Ca atoms × 40.08 amu = 120.24 amu 2 P atoms × 30.97 amu = 61.94 amu 8 O atoms × 16.00 amu = 128.00 amu Formula weight of Ca3(PO4)2 310.18 amu = 310.18 g/mol

c. 16 C atoms × 12.01 amu = 192.16 amu 16 H atoms × 1.01 amu = 16.16 amu 1 Cl atom × 35.45 amu = 35.45 amu 1 N atom × 14.01 amu = 14.01 amu 2 O atoms × 16.00 amu = 32.00 amu 2 S atoms × 32.07 amu = 64.14 amu Formula weight of C16H16ClNO2S 353.92 amu = 353.92 g/mol

5.61 Determine the molecular formula of L-dopa. Then calculate the formula weight and molar mass

as in Answer 5.59.

C CH

HCH

NH2

CO

OH

CCC

C C

HHO

HO

H HL-dopa

a. molecular formula = C9H11NO4!!b. formula weight = 197.2 amu!c. molar mass = 197.2 g/mol

5.62 Determine the molecular formula of niacin. Then calculate the formula weight and molar mass as

in Answer 5.59

C CO

OH

CCC

N C

HH

H

Hniacin

a. molecular formula = C6H5NO2!!b. formula weight = 123.1 amu!c. molar mass = 123.1 g/mol

5.63 Convert all of the units to moles, and then compare the atomic mass or formula weight to determine

the quantity with the larger mass.

a. 1 mol of Fe atoms (55.85 g/mol) < 1 mol of Sn atoms (118.7 g/mol) b. 1 mol of C atoms (12.01 g/mol) < 6.02 × 1023 N atoms = 1 mol N atoms (14.01 g/mol) c. 1 mol of N atoms (14.01 g/mol) < 1 mol of N2 molecules = 2 mol N atoms (28.02 g/mol N2) d. 1 mol of CO2 molecules (44.01 g/mol) > 3.01 × 1023 N2O molecules = 0.500 mol N2O (44.02 g/mol N2O) = 22.01 g N2O



5.64 Convert all of the units to moles, and then compare the atomic mass or formula weight to determine the quantity with the larger mass.

a. 1 mol of Si atoms (28.08 g/mol) < 1 mol of Ar atoms (39.95 g/mol) b. 1 mol of He atoms (4.00 g/mol) > 6.02 × 1023 H atoms = 1 mol H atoms (1.01 g/mol) c. 1 mol of Cl atoms (35.45 g/mol) < 1 mol of Cl2 molecules = 2 mol Cl atoms (70.90 g/mol Cl2) d. 1 mol of C2H4 molecules (28.06 g/mol) > 3.01 × 1023 C2H4 molecules = 0.500 mol C2H4 (28.06

g/mol C2H4) = 14.03 g C2H4 5.65 Calculate the molar mass of each compound as in Answer 5.59, and then multiply by 5.00 mol.

a. HCl = 182 g c. C2H2 = 130. g b. Na2SO4 = 710. g d. Al(OH)3 = 390. g

5.66 Calculate the molar mass of each compound as in Answer 5.59, and then multiply by 0.50 mol.

a. NaOH = 20. g c. C3H6 = 21 g b. CaSO4 = 68 g d. Mg(OH)2 = 29 g

5.67 Convert the grams to moles using the molar mass as a conversion factor.

0.500 g 1 mol = 1.46 x 10–3 mol342.3 g

xa.

5.00 g 1 mol = 0.0146 mol342.3 g

xb.

25.0 g 1 mol = 0.0730 mol342.3 g

xc.

0.0250 g 1 mol = 7.30 x 10–5 mol342.3 g

xd.

5.68 Convert the grams to moles using the molar mass as a conversion factor.

0.500 g 1 mol = 2.77x 10–3 mol180.2 g

xa.

5.00 g 1 mol = 0.0277 mol180.2 g

xb.

25.0 g 1 mol = 0.139 mol180.2 g

xc.

0.0250 g 1 mol = 1.39 x 10–4 mol180.2 g

xd.

5.69 Multiply the number of moles by Avogadro’s number to determine the number of molecules, as in

Example 5.3.

a. 2.00 mol × 6.02 × 1023 molecules/mol = 1.20 × 1024 molecules b. 0.250 mol × 6.02 × 1023 molecules/mol = 1.51 × 1023 molecules c. 26.5 mol × 6.02 × 1023 molecules/mol = 1.60 × 1025 molecules d. 222 mol × 6.02 × 1023 molecules/mol = 1.34 × 1026 molecules e. 5.00 × 105 mol × 6.02 × 1023 molecules/mol = 3.01 × 1029 molecules


Chapter 5–19

5.70 Use Avogadro’s number to convert the number of molecules to moles.

5.00 x 1019 molecules = 8.31 x 10–5 g6.02 x 1023 molecules

1 molxa.

6.51 x 1028 molecules = 1.08 x 105 g6.02 x 1023 molecules

1 molxb.

8.32 x 1021 molecules = 1.38 x 10-2 g6.02 x 1023 molecules

1 molxc.

3.10 x 1020 molecules = 5.15 x 10-4 g6.02 x 1023 molecules

1 molxd.

5.71 Use the molar mass as a conversion factor to convert the moles to grams. Use Avogadro’s number

to convert the number of molecules to moles.

3.60 mol1 mol

= 324 g90.08 g

xa.

0.580 mol

1 mol= 52.2 g

90.08 gxb.

7.3 x 1024 molecules1 mol

= 1.1 x 103 g90.08 g

xc.6.02 x 1023 molecules

1 molx


= 9.82 g90.08 g

xd.6.02 x 1023 molecules

1 molx

5.72 Use the molar mass as a conversion factor to convert the moles to grams. Use Avogadro’s number

to convert the number of molecules to moles.

3.6 mol1 mol

= 1.4 x 103 g384.7 g

xa.

0.58 mol

1 mol= 2.2 x 102 g

384.7gxb.


= 4.7 x 103 g384.7 g

xc.6.02 x 1023 molecules

1 molx


= 41.9 g384.7 g

xd.6.02 x 1023 molecules

1 molx



5.73 C C HH 5 O2 2 H2O4 CO22 + +

acetylene

a. 12.5 moles of O2 are needed to react completely with 5.00 mol of C2H2. 5.00 mol C2H2 × (5 mol O2/2 mol C2H2) = 12.5 mol O2

b. 12 moles of CO2 are formed from 6.0 mol of C2H2. 6.0 mol C2H2 × (4 mol CO2/2 mol C2H2) = 12 mol CO2 c. 0.50 moles of H2O are formed from 0.50 mol of C2H2. 0.50 mol C2H2 × (2 mol H2O/2 mol C2H2) = 0.50 mol H2O d. 0.40 moles of C2H2 are needed to form 0.80 mol of CO2. 0.80 mol CO2 × (2 mol C2H2/4 mol CO2) = 0.40 mol C2H2

5.74 2 H2O(l) 2 NaOH(aq)+2 Na(s) H2(g)+

a. 3.0 moles of H2O are needed to react completely with 3.0 mol of Na. 3.0 mol Na × (2 mol H2O/2 mol Na) = 3.0 mol H2O

b. 0.19 moles of H2 are formed from 0.38 mol of Na. 0.38 mol Na × (1 mol H2/2 mol Na) = 0.19 mol H2 c. 1.82 moles of H2 are formed from 3.64 mol of H2O. 3.64 mol H2O × (1 mol H2/2 mol H2O) = 1.82 mol H2

5.75 Use conversion factors as in Example 5.6 to solve the problems.

a. 220 g of CO2 are formed from 2.5 mol of C2H2. b. 44 g of CO2 are formed from 0.50 mol of C2H2. c. 4.5 g of H2O are formed from 0.25 mol of C2H2. d. 240 g of O2 are needed to react with 3.0 mol of C2H2.

5.76 Use conversion factors as in Example 5.6 to solve the problems.

a. 120 g of NaOH are formed from 3.0 mol of Na. b. 0.30 g of H2 are formed from 0.30 mol of Na. c. 3.6 g of H2O are needed to react with 0.20 mol of Na.

5.77 Use the equation to determine the percent yield.


x 100%

9.0 g12.0 g

x 100% = 75%=


Chapter 5–21

5.78 Use the equation to determine the percent yield.


x 100%

17.0 g 20.0 g

x 100% = 85.0 %=

5.79 Use the following equations to determine the percent yield.


x =

Moles of reactant

Moles of product

1 mol CH4

1 mol CHCl3 0.200 mol CHCl3


Grams of product

Moles of product

x =119.4 g CHCl31 mol CHCl3

23.9 g CHCl3Theoretical yield

3.20 g CH4


x =

Grams of reactant

Moles of reactant

16.04 g CH4

1 mol CH4 0.200 mol CH4

0.200 mol CH4

0.200 mol CHCl3

a.


x 100%

x 100%15.0 g CHCl323.9 g CHCl3

= 62.8%

b.

=



5.80 Use the following equations to determine the percent yield.


x =

Moles of reactant

Moles of product

2 mol CH4O2 mol CO2 1.50 mol CO2


Grams of product

Moles of product

x =44.0 g CO2

1 mol CO2

66.0 g CO2

Theoretical yield

48.0 g CH4O


x =

Grams of reactant

Moles of reactant

32.0 g CH4O1 mol CH4O

1.50 mol CH4O

1.50 mol CH4O

1.50 mol CO2

a.



x 100%48.0 g CO2

66.0 g CO2= 72.7%

b.

=

5.81

1 molecule B1 molecule A

4 molecules A x = 4 molecules of B are needed.Molecule A is in excess.Molecule B is the limiting reactant.

a.

1 molecule B

2 molecules A4 molecules A x = 2 molecules of B are needed.

Molecule B is in excess.Molecule A is the limiting reactant.

b.

2 molecules B1 molecule A

4 molecules A x = 8 molecules of B are needed.Molecule A is in excess.Molecule B is the limiting reactant.

c.


Chapter 5–23

5.82

1 molecule B22 molecules A2

4 molecules A2 x = 2 molecules of B2 are needed.Molecule A2 is in excess.Molecule B2 is the limiting reactant.

reactants products

A2BA2A2 B2

5.83

1 mol O2

2 mol NO1.0 mol NO x = 0.50 mol of O2 is needed.

O2 is in excess.NO is the limiting reactant.

a.

1 mol O2


NO is in excess.O2 is the limiting reactant.

b.

1 mol NO30.01 g NO

10.0 g NO x = 0.333 mol NOc.

1 mol O2



1 mol O2

32.00 g O210.0 g O2 x = 0.313 mol O2

1 mol NO30.01 g NO

28.0 g NO x = 0.933 mol NOd.

1 mol O2



1 mol O2

32.00 g O216.0 g O2 x = 0.500 mol O2



5.84

1 mol O2



a.

1 mol O2



b.

1 mol NO30.01 g NO

15.0 g NO x = 0.500 mol NOc.

1 mol O2



1 mol O2

32.00 g O210.0 g O2 x = 0.313 mol O2

1 mol NO30.01 g NO

10.0 g NO x = 0.333 mol NOd.

1 mol O2



1 mol O2

32.00 g O2 4.0 g O2 x = 0.125 mol O2

5.85 Use the limiting reactant from Problem 5.83 to determine the amount of product formed. The

conversion of moles of limiting reagent to grams of product is combined in a single step.

2 mol NO2

2 mol NO1.0 mol NO x = 46 g NO2a. 46.01 g NO2

1 mol NO2x

2 mol NO2

1 mol O20.50 mol O2 x = 46 g NO2b. 46.01 g NO2

1 mol NO2x

2 mol NO2

2 mol NO0.333 mol NO x = 15.3 g NO2c. 46.01 g NO2

1 mol NO2x

2 mol NO2

2 mol NO0.933 mol NO x = 42.9 g NO2d. 46.01 g NO2

1 mol NO2x


Chapter 5–25

5.86 Use the limiting reactant from Problem 5.84 to determine the amount of product formed. The conversion of moles of limiting reagent to grams of product is combined in a single step.

2 mol NO2

2 mol NO2.0 mol NO x = 92 g NO2a.

46.01 g NO2

1 mol NO2x

2 mol NO2

1 mol O22.0 mol O2 x = 180 g NO2b. 46.01 g NO2

1 mol NO2x

2 mol NO2

2 mol NO0.500 mol NO x = 23.0 g NO2c.

46.01 g NO2

1 mol NO2x

2 mol NO2

1 mol O20.125 mol O2 x = 11.5 g NO2d.

46.01 g NO2

1 mol NO2x

5.87

1 mol C2H4

28.05 g C2H4

8.00 g C2H4 x = 0.285 mol C2H4

1 mol HCl36.46 g HCl

12.0 g HCl x = 0.329 mol HCl

a.

Since the mole ratio in the balanced equation is 1:1, the reactant with the smaller number of moles is the limiting reactant: C2H4.

b.

1 mol C2H5Cl1 mol C2H4

0.285 mol C2H4 x = 0.285 mol C2H5Clc.

64.51 g C2H5Cl1 mol C2H5Cl

0.285 mol C2H5Cl x = 18.4 g C2H5Cld.

e. = 57.6 % percent yield10.6 g C2H5Cl18.4 g C2H5Cl

x 100%

5.88

1 mol CH4

16.05 g CH4

5.00 g CH4 x = 0.312 mol CH4

1 mol Cl270.90 g Cl2

15.0 g Cl2 x = 0.212 mol Cl2

a.

Since the mole ratio in the balanced equation is 1:2, (2)(0.312 mol) = 0.624 mol Cl2 would be needed to react with all of the CH4. There are only 0.212 mol Cl2, however, so Cl2 is the limiting reactant.

b.

1 mol CH2Cl21 mol Cl2

0.212 mol Cl2 x = 0.212 mol CH2Cl2c.



84.93 g CH2Cl21 mol CH2Cl2

0.212mol CH2Cl2 x = 18.0 g CH2Cl2d.

e. = 86.7 % percent yield15.6 g CH2Cl218.0 g CH2Cl2

x 100%

5.89 A substance that is oxidized loses electrons, whereas an oxidizing agent gains electrons (it is

reduced). 5.90 A substance that is reduced gains electrons, whereas a reducing agent loses electrons (it is

oxidized). 5.91 The species that is oxidized loses one or more electrons. The species that is reduced gains one or

more electrons.

2 Na Cl2

Fe Cu2+

Cl2 2 I–

a.

b.

c.

oxidized reducedFe2+Fe + 2 e– CuCu2+ + 2 e–

oxidizedreducedI22 I– + 2 e– 2 Cl–Cl2 + 2 e–

2 Na 2 Na+ + 2 e– 2 Cl–Cl2 + 2 e–

oxidized reduced 5.92 The species that is oxidized loses one or more electrons. The species that is reduced gains one or

more electrons.

4 Na O2

Mg Fe2+

Cu2+

a.

b.

c.

oxidized reducedMg2+Mg + 2 e– FeFe2+ + 2 e–

oxidizedreducedSn2+Sn + 2 e– CuCu2+ + 2 e–

4 Na 4 Na+ + 4 e– 2 O2–O2 + 4 e–

oxidized reduced

Sn

5.93 The oxidizing agent gains electrons (it is reduced). The reducing agent loses electrons (it is

oxidized).

Zn Ag+

oxidizedreducing agent

reducedoxidizing agent

+Zn Ag2O+ ZnO 2 Ag

5.94 The oxidizing agent gains electrons (it is reduced). The reducing agent loses electrons (it is

oxidized).

Cd Ni4+

oxidizedreducing agent

reducedoxidizing agent

+Cd Ni4++ Cd2+ Ni2+


Chapter 5–27

5.95 Acetylene is reduced because it gains hydrogen atoms. 5.96 Cl2 is reduced because it gains electrons. 5.97 Write the balanced equation and the half reactions.

2 Mg + O2 2 MgO

2 Mg2+2 Mg + 4 e– O2 2 O2–+ 4 e–

5.98 Write the balanced equation and the half reactions.

4 Al + 3 O2 2 Al2O3

4 Al3+4 Al + 12 e– 3 O2 6 O2–+ 12 e–

5.99 Refer to prior solutions to answer each part.

C12H22O11(s) H2O(l) C2H6O(l) CO2(g)+ +sucrose ethanol

a. Calculate the molar mass as in Answer 5.59; the molar mass of sucrose = 342.3 g/mol. b. Follow the steps in Example 5.2.

C12H22O11(s) H2O(l) 4 C2H6O(l) 4 CO2(g)+ +

c. 8 mol of ethanol are formed from 2 mol of sucrose. d. 10 mol of water are needed to react with 10 mol of sucrose. e. 101 g of ethanol are formed from 0.550 mol of sucrose. f. 18.4 g of ethanol are formed from 34.2 g of sucrose. g. 9.21 g ethanol h. 13.6%

5.100 Refer to prior solutions to answer each part.

a. Calculate the molar mass as in Answer 5.59; the molar mass of diethyl ether = 74.1 g/mol. b. Follow the steps in Example 5.2.

2 C2H6O(s) C4H10O(l) H2O (l)+ethanol diethyl ether

c. 1 mol of diethyl ether is formed from 2 mol of ethanol. d. 5 mol of water are formed from 10 mol of ethanol. e. 20. g of diethyl ether are formed from 0.55 mol of ethanol. f. 3.70 g of diethyl ether are formed from 4.60 g of ethanol. g. 1.85 g diethyl ether h. 97.3%



5.101

500 tablets 200. mg ibuprofen1 tablet 1000 mg

1 g 1 mol ibuprofen206.3 g ibuprofen

x x x = 0.485 mol ibuprofena.

0.485 mol ibuprofenb. 6.02 x 1023 molecules

1 molx = 2.92 x 1023 molecules

5.102

500. mg Mg(OH)21000 mg

1 g 1 mol Mg(OH)2

58.33 g Mg(OH)2

x x = 8.57 x 10-3 mol Mg(OH)2

500. mg Al(OH)31000 mg

1 g 1 mol Al(OH)3

78.01 g Al(OH)3

x x = 6.41x 10-3 mol Al(OH)3

5.103

20 cig1 cig 1000 mg

1 g 1 mol nicotine162.3 g nicotine

x x x = 2.38 x 10–4 mol nicotine


x = 1.43 x 1020 molecules

1.93 mg

2.38 x 10–4 mol nicotine

a.

b.

5.104

5 lb

1 lb 342.3 g1 mol

x x = 7 mol sucrose 454 g

5.105

2400 mg1000 mg

1 g 1 mol22.99 g

x x 6.02 x 1023 ions1 mol

x = 6.3 x 1022 ions

5.106

250 g18.0 g

x = 14 mol water1 mol

14 mol 6.022 x 1023 moleculesx = 8.4 x 1024 molecules water1 mol


Chapter 5–29

5.107

H2O+CC

CC C

CCC

CC C

C CCCl3

HCCl

C CC

CCCl

H H

HH H H

HH

Cl

H H

H

HH

C14H9Cl5

C2HCl3O +2

DDT112.6 g/mol

chlorobenzene

a. Calculate the molar mass as in Answer 5.59; the molar mass of DDT = 354.5 g/mol. b. 18 g of DDT would be formed from 0.10 mol of chlorobenzene. c. 17.8 g is the theoretical yield of DDT in grams from 11.3 g of chlorobenzene. d. 84.3%

5.108 Refer to prior solutions to answer each part. a. Calculate the molar mass as in Answer 5.59; the molar mass of linolenic acid = 278.5 g/mol.

C18H30O2 C18H36O23 H2+b. 2 C18H30O2 36 CO2 + 30 H2O49 O2+c.

d. 10.2 grams of C18H36O2 will be formed. 5.109 Use conversion factors to answer the questions about dioxin.

70. kg1 kg

3.0 x 10–2 mg 1 g1000 mg

x x

1 mol322.0 g

x = 3.9 x 1018 molecules

= 2.1 x 10–3 g dioxin

2.1 x 10–3 g dioxin6.02 x 1023 molecules

1 molx

a.

b.

5.110 Pb is the reducing agent. It is oxidized from Pb to Pb2+. PbO2 is the oxidizing agent. Pb is

reduced from +4 in PbO2 to +2 in PbSO4.


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