chapter 9 chemical quantities - edl · world of chemistry 74 copyright houghton mifflin company....

World of Chemistry 74 Copyright Houghton Mifflin Company. All rights reserved.

Chapter 9 Chemical Quantities

1. Although we define mass as the “amount of matter in a substance,” the units in which we measure mass are a human invention. Atoms and molecules react on an individual particle-by-particle basis, and we have to count individual particles when doing chemical calculations.

2. Balanced chemical equations tell us in what proportions on a mole basis substances combine; since the molar masses of C(s) and O2(g) are different, 1 g of O2 could not represent the same number of moles as 1 g of C.

3. a. 2NO(g) + O2(g) 2NO2(g)

Two molecules of nitrogen monoxide combine with one molecule of oxygen gas, producing two molecules of nitrogen dioxide. Two moles of gaseous nitrogen monoxide combine with one mole of gaseous oxygen, producing two moles of gaseous nitrogen dioxide.

b. 2AgC2H3O2(aq) + CuSO4(aq) Ag2SO4(s) + Cu(C2H3O2)2(aq)

Note: The term “formula unit” is used in the following statement because the substances involved in the above reaction are ionic, and do not contain true molecules. Two formula units of silver acetate will react with one formula unit of copper(II) sulfate, precipitating one formula unit of silver sulfate and leaving one formula unit of copper(II) acetate in solution. Two moles of aqueous silver acetate react with one mole of aqueous copper(II) sulfate, to produce one mole of solid silver sulfate as a precipitate, and leaving one mole of copper(II) acetate in solution.

c. PCl3(l) + 3H2O(l) H3PO3(l) + 3HCl(g)

One molecule of phosphorus trichloride reacts with three molecules of water, producing one molecule of phosphorous acid and three molecules of gaseous hydrogen chloride. One mole of liquid phosphorus trichloride reacts with three moles of liquid water, producing one mole of liquid phosphorous acid and three moles of gaseous hydrogen chloride.

d. C2H6(g) + Cl2(g) C2H5Cl(g) + HCl(g)

One molecule of ethane (C2H6) reacts with one molecule of chlorine, producing one molecule of chloroethane (C2H5Cl) and one molecule of hydrogen chloride. One mole of gaseous ethane combines with one mole of chlorine gas, giving one mole of gaseous chloroethane and one mole of gaseous hydrogen chloride.

4. a. 3MnO2(s) + 4Al(s) 3Mn(s) + 2Al2O3(s)

Three formula units of manganese(IV) oxide react with four aluminum atoms, producing three manganese atoms and two formula units of aluminum oxide. Three moles of solid manganese(IV) oxide react with four moles of solid aluminum, to produce three moles of solid manganese and two moles of solid aluminum oxide.

Chemical Quantities 75

World of Chemistry Copyright Houghton Mifflin Company. All rights reserved.

b. B2O3(s) + 3CaF2(s) 2BF3(g) + 3CaO(s)

One molecule of diboron trioxide reacts with three formula units of calcium fluoride, producing two molecules of boron trifluoride and three formula units of calcium oxide. One mole of solid diboron trioxide reacts with three moles of solid calcium fluoride, to give two moles of gaseous boron trifluoride and three moles of solid calcium oxide.

c. 3NO2(g) + H2O(l) 2HNO3(aq) + NO(g)

Three molecules of nitrogen dioxide [nitrogen(IV) oxide] react with one molecule of water, to produce two molecules of nitric acid and one molecule of nitrogen monoxide [nitrogen(II) oxide]. Three moles of gaseous nitrogen dioxide react with one mole of liquid water, to produce two moles of aqueous nitric acid and one mole of nitrogen monoxide gas.

d. C6H6(g) + 3H2(g) C6H12(g)

One molecule of C6H6 (which is named benzene) reacts with three molecules of hydrogen, producing just one molecule of C6H12 (which is named cyclohexane). One mole of gaseous benzene reacts with three moles of hydrogen gas, giving one mole of gaseous cyclohexane.

5. False. The coefficients of the balanced chemical equation represent the ratios on a mole basis by which hydrogen peroxide decomposes.

6. For converting from a given number of moles of CH4 to the number of moles of oxygen needed for reaction, the correct mole ratio is

2 mol O2

1 mol CH4

For converting from a given number of moles of CH4 to the number of moles of product produced, the ratios are

1 mol CO2

1 mol CH4

and 2 mol H2O

1 mol CH4

7. 2Ag(s) + H2S(g) Ag2S(s) + H2(g)

1 mol Ag2S

2 mol Ag and

1 mol H2

2 mol Ag

8. a. 2FeO(s) + C(s) 2Fe(l) + CO2(g)

0.125 mol FeO 2 mol Fe

2 mol FeO = 0.125 mol Fe

0.125 mol FeO 1 mol CO2

2 mol FeO = 0.0625 mol CO2

76 Chapter 9


b. Cl2(g) + 2KI(aq) 2KCl(aq) + I2(s)

0.125 mol KI 2 mol KCl

2 mol KI = 0.125 mol KCl

0.125 mol KI 1 mol I2

2 mol KI = 0.0625 mol I2

c. Na2B4O7(s) + H2SO4(aq) + 5H2O(l) 4H3BO3(s) + Na2SO4(aq)

0.125 mol Na2B4O7

4 mol H3BO3

1 mol Na2B4O7

= 0.500 mol H3BO3

0.125 mol Na2B4O7

1 mol Na2SO4

1 mol Na2B4O7

= 0.125 mol Na2SO4

d. CaC2(s) + 2H2O(l) Ca(OH)2(s) + C2H2(g)

0.125 mol CaC2

1 mol Ca(OH)2

1 mol CaC2

= 0.125 mol Ca(OH)2

0.125 mol CaC2

1 mol C2H2

1 mol CaC2

= 0.125 mol C2H2

9. a. NH3(g) + HCl(g) NH4Cl(s)

molar mass of NH4Cl, 53.49 g

0.50 mol NH3

1 mol NH4Cl

1 mol NH3

= 0.50 mol NH4Cl

0.50 mol NH4Cl 53.49 g NH4Cl

1 mol NH4Cl = 27 g NH4Cl

b. CH4(g) + 4S(g) CS2(l) + 2H2S(g)

molar masses: CS2, 76.15 g; H2S, 34.09 g

0.50 mol S 1 mol CS2

4 mol S = 0.125 mol CS2 (= 0.13 mol CS2)

0.125 mol CS2

76.15 g CS2

1 mol CS2

= 9.5 g CS2

0.50 mol S 2 mol H2S

4 mol S = 0.25 mol H2S

0.25 mol H2S34.09 g H2S

1 mol H2S = 8.5 g H2S

c. PCl3(l) + 3H2O(l) H3PO3(aq) + 3HCl(aq)



molar masses: H3PO3, 81.99 g; HCl, 36.46 g

0.50 mol PCl31 mol H3PO3

1 mol PCl3

= 0.50 mol H3PO3

0.50 mol H3PO3

81.99 g H3PO3

1 mol H3PO3

= 41 g H3PO3

0.50 mol PCl33 mol HCl

1 mol PCl3

= 1.5 mol HCl

1.50 mol HCl 36.46 g HCl

1 mol HCl = 54.7 = 55 g HCl

d. NaOH(s) + CO2(g) NaHCO3(s)

molar mass of NaHCO3, 84.01 g

0.50 mol NaOH 1 mol NaHCO3

1 mol NaOH = 0.50 mol NaHCO3

0.50 mol NaHCO3

84.01 g NaHCO3

1 mol NaHCO3

= 42 g NaHCO3

10. Before doing the calculations, the equations must be balanced.

a. 4KO2(s) + 2H2O(l) 3O2(g) + 4KOH(s)

0.625 mol KOH 3 mol O2

4 mol KOH = 0.469 mol O2

b. SeO2(g) + 2H2Se(g) 3Se(s) + 2H2O(g)

0.625 mol H2O3 mol Se

2 mol H2O = 0.938 mol Se

c. 2CH3CH2OH(l) + O2(g) 2CH3CHO(aq) + 2H2O(l)

0.625 mol H2O2 mol CH3CHO

2 mol H2O = 0.625 mol CH3CHO

d. Fe2O3(s) + 2Al(s) 2Fe(l) + Al2O3(s)

0.625 mol Al2O3

2 mol Fe

1 mol Al2O3

= 1.25 mol Fe

11. the molar mass of the substance

12. Stoichiometry is the process of using a chemical equation to calculate the relative masses of reactants and products involved in a reaction.

78 Chapter 9


13. a. molar mass of Ag = 107.9 g

2.01 10–2 g Ag 1 mol

107.9 g = 1.86 10–4 mol Ag

b. molar mass of (NH4)2S = 68.15 g

45.2 mg (NH4)2S1 g

1000 mg

1 mol

68.15 g = 6.63 10–4 mol (NH4)2S

c. molar mass of uranium = 238.0 g

61.7 g U 1 g

106

g

1 mol

238.0 g = 2.59 10–7 mol U

d. molar mass of SO2 = 64.07 g

5.23 kg SO2

1000 g

1 kg

1 mol

64.07 g = 81.6 mol SO2

e. molar mass of Fe(NO3)3 = 241.9 g

272 g Fe(NO3)3

1 mol

241.9 g = 1.12 mol Fe(NO3)3

f. molar mass of FeSO4 = 151.9 g

12.7 mg FeSO4

1 g

1000 mg

1 mol

151.9 g = 8.36 10–5 mol FeSO4

g. molar mass of LiOH = 23.95 g

6.91 103 g LiOH 1 mol

23.95 g = 288.5 = 289 mol LiOH

14. a. molar mass of CaCO3 = 100.1 g

2.21 10–4 mol CaCO3

100.1 g

1 mol = 0.0221 g CaCO3

b. molar mass of He = 4.003 g

2.75 mol He 4.003 g

1 mol = 11.0 g He

c. molar mass of O2 = 32.00 g

0.00975 mol O2

32.00 g

1 mol = 0.312 g O2

d. molar mass of CO2 = 44.01 g

7.21 millimol = 0.00721 mol

0.00721 mol CO2

44.01 g

1 mol = 0.317 g CO2



e. molar mass of FeS = 87.92 g

0.835 mol FeS 87.92 g

1 mol = 73.4 g FeS

f. molar mass of KOH = 56.11 g

4.01 mol KOH 56.11 g

1 mol = 225 g KOH

g. molar mass of H2 = 2.016 g

0.0219 mol H2

2.016 g

1 mol = 0.0442 g H2

15. Before any calculations are done, the equations must be balanced.

a. Mg(s) + CuCl2(aq) MgCl2(aq) + Cu(s)

molar mass of Mg = 24.31 g

25.0 g Mg 1 mol

24.31 g = 1.03 mol Mg

1.03 mol Mg 1 mol CuCl2

1 mol Mg = 1.03 mol CuCl2

b. 2AgNO3(aq) + NiCl2(aq) 2AgCl(s) + Ni(NO3)2(aq)

molar mass of AgNO3 = 169.9 g

25.0 g AgNO3

1 mol

169.9 g = 0.147 mol AgNO3

0.147 mol AgNO3

1 mol NiCl2

2 mol AgNO3

= 0.0735 mol NiCl2

c. NaHSO3(aq) + NaOH(aq) Na2SO3(aq) + H2O(l)

molar mass of NaHSO3 = 104.1 g

25.0 g NaHSO3

1 mol

104.1 g = 0.240 mol NaHSO3

0.240 mol NaHSO3

1 mol NaOH

1 mol NaHSO3

= 0.240 mol NaOH

d. KHCO3(aq) + HCl(aq) KCl(aq) + H2O(l) + CO2(g)

molar mass of KHCO3 = 100.1 g

25.0 g KHCO3

1 mol

100.1 g = 0.250 mol KHCO3

0.250 mol KHCO3

1 mol HCl

1 mol KHCO3

= 0.250 mol HCl

80 Chapter 9


16. Before any calculations are done, the equations must be balanced. Since the given and required quantities in this question are shown in milligrams, it is most convenient to perform the calculations in terms of millimoles of the substances involved. One millimole of a substance represents the molar mass of the substance expressed in milligrams.

a. FeSO4(aq) + K2CO3(aq) FeCO3(s) + K2SO4(aq)

millimolar masses: FeSO4, 151.9 mg; FeCO3, 115.9 mg; K2SO4, 174.3 mg

10.0 mg FeSO4

1 mmol FeSO4

151.9 mg FeSO4

= 0.0658 mmol FeSO4

0.0658 mmol FeSO4

1 mmol FeCO3

1 mmol FeSO4

115.9 mg FeCO3

1 mmol FeCO3

= 7.63 mg FeCO3

0.0658 mmol FeSO4

1 mmol K2SO

4

1 mmol FeSO4

174.3 mg K2SO

4

1 mmol K2SO

4

= 11.5 mg K2SO4

b. 4Cr(s) + 3SnCl4(l) 4CrCl3(s) + 3Sn(s)

millimolar masses: Cr, 52.00 mg; CrCl3, 158.4 mg; Sn, 118.7 mg

10.0 mg Cr 1 mmol Cr

52.00 mg Cr = 0.192 mmol Cr

0.192 mmol Cr 4 mmol CrCl

3

4 mmol Cr

158.4 mg CrCl3

1 mmol CrCl3

= 30.4 mg CrCl3

0.192 mmol Cr 3 mmol Sn

4 mmol Cr

118.7 mg Sn

1 mmol Sn = 17.1 mg Sn

c. 16Fe(s) + 3S8(s) 8Fe2S3(s)

millimolar masses: S8, 256.6 mg; Fe2S3, 207.9 mg

10.0 mg S8

1 mmol S8

256.6 mg S8

= 0.0390 mmol S8

0.0390 mmol S8

8 mmol Fe2S

3

3 mmol S8

207.9 mg Fe2S

3

1 mmol Fe2S

3

= 21.6 mg Fe2S3

d. 3Ag(s) + 4HNO3(aq) 3AgNO3(aq) + 2H2O(l) + NO(g)

millimolar masses: HNO3, 63.0 mg; AgNO3, 169.9 mg

H2O, 18.0 mg; NO, 30.0 mg

10.0 mg HNO3

1 mmol HNO3

63.0 mg HNO3

= 0.159 mmol HNO3

0.159 mmol HNO3

3 mmol AgNO3

4 mmol HNO3

169.9 mg AgNO3

1 mmol AgNO3

= 20.3 mg AgNO3



0.159 mmol HNO3

2 mmol H2O

4 mmol HNO3

18.0 mg H2O

1 mmol H2O

= 1.43 mg H2O

0.159 mmol HNO3

1 mmol NO

4 mmol HNO3

30.0 mg NO

1 mmol NO = 1.19 mg NO

17. 2H2(g) + O2(g) 2H2O(g)

molar masses: H2, 2.016 g; H2O, 18.02 g

56.0 g H2

1 mol H2

2.016 g H2

= 27.77 mol H2

27.77 mol H2

2 mol H2O

2 mol H2

= 27.77 mol H2O

27.77 mol H2O18.02 g H2O

1 mol H2O = 500. g H2O

18. 2H2(g) + O2(g) 2H2O(g)

molar masses of O2 = 32.00 g

0.0275 mol H2

1 mol O2

2 mol H2

= 0.01375 = 0.0138 mol O2

0.01375 mol O2

32.00 g O2

1 mol O2

= 0.440 g O2

19. molar masses: C, 12.01 g; CO, 28.01 g; CO2, 44.01 g

5.00 g C 1 mol C

12.01 g C = 0.4163 mol C

carbon dioxide: C(s) + O2(g) CO2(g)

0.4163 mol C 1 mol CO2

1 mol C = 0.4163 mol CO2

0.4163 mol CO2

44.01 g CO2

1 mol CO2

= 18.3 g CO2

carbon monoxide: 2C(s) + O2(g) 2CO(g)

0.4163 mol C 2 mol CO

2 mol C = 0.4163 mol CO

0.4163 mol CO 28.01 g CO

1 mol CO = 11.7 g CO

82 Chapter 9


20. 2Fe(s) + 3Cl2(g) 2FeCl3(s)

millimolar masses: Fe, 55.85 mg; FeCl3, 162.2 mg

15.5 mg Fe 1 mmol Fe

55.85 mg Fe = 0.2775 mmol Fe

0.2775 mmol Fe 2 mmol FeCl3

2 mmol Fe = 0.2775 mmol FeCl3

0.2775 mmol FeCl3

162.2 mg FeCl3

1 mmol FeCl3

= 45.0 mg FeCl3

21. 2H2O2(aq) 2H2O(l) + O2(g)

molar masses: H2O2, 34.02 g; O2, 32.00 g

10.00 g H2O2

1 mol H2O2

34.02 g H2O2

= 0.2939 mol H2O2

0.2939 mol H2O2

1 mol O2

2 mol H2O2

= 0.1470 mol O2

0.1470 mol O2

32.00 g O2

1 mol O2

= 4.704 g O2

22. Cu(s) + S(s) CuS(s)

molar masses: Cu, 63.55 g; S, 32.07 g

1.25 g Cu 1 mol

63.55 g = 1.97 10–2 mol Cu

1.97 10–2 mol Cu 1 mol S

1 mol Cu = 1.97 10–2 mol S

1.97 10–2 mol S 32.07 g

1 mol = 0.631 g S

23. 2NH4NO3(s) 2N2(g) + O2(g) + 4H2O(g)

molar masses: NH4NO3, 80.05 g; N2, 28.02 g; O2, 32.00 g; H2O, 18.02 g

1.25 g NH4NO3

1 mol NH4NO3

80.05 g NH4NO3

= 0.0156 mol NH4NO3

0.0156 mol NH4NO3

2 mol N2

2 mol NH4NO3

= 0.0156 mol N2

0.0156 mol N2

28.02 g N2

1 mol N2

= 0.437 g N2



0.0156 mol NH4NO3

1 mol O2

2 mol NH4NO3

= 0.00780 mol O2

0.00780 mol O2

32.00 g O2

1 mol O2

= 0.250 g O2

0.0156 mol NH4NO3

4 mol H2O

2 mol NH4NO3

= 0.0312 mol H2O

0.0312 mol H2O18.02 g H2O

1 mol H2O = 0.562 g H2O

As a check, note that 0.437 g + 0.250 g + 0.562 g = 1.249 g = 1.25 g.

24. 2Mg(s) + O2(g) 2MgO(s)

molar masses: Mg, 24.31 g; MgO, 40.31 g

1.25 g Mg 1 mol

24.31 g = 5.14 10–2 mol Mg

5.14 10–2 mol Mg 2 mol MgO

2 mol Mg = 5.14 10–2 mol MgO

5.14 10–2 mol MgO 40.31 g

1 mol = 2.07 g MgO

25. From the balanced equation:

Cl2 + 2KI I2 + 2KCl

We can calculate the following:

4.50 103 g Cl2

1 mol Cl2

70.9 g Cl2

1 mol I2

1 mol Cl2

253.8 g I2

1 mol I2

= 1.61 104 g I2

26. From the balanced equation:

Cl2 + F2 2ClF

We can calculate the following:

5.00 10–3 g ClF 1 mol ClF

54.45 g ClF

1 mol Cl2

2 mol ClF

70.9 g Cl2

1 mol Cl2

= 3.26 10–3 g Cl2

27. Start by balancing the equations:

C3H8 + 5O2 3CO2 + 4H2O

2C4H10 + 13O2 8CO2 + 10H2O

If we had, for example, 10.00 g of C3H8 and 10.00 g of C4H10, we can calculate the mass of oxygen required for each reaction and compare the results.

84 Chapter 9


10.00 g C3H8

1 mol C3H

8

44.09 g C3H

8

5 mol O2

1 mol C3H

8

32.00 g O2

1 mol O2

= 36.29 g O2

10.00 g C4H10

1 mol C4H

10

58.12 g C4H

10

13 mol O2

2 mol C4H

10

32.00 g O2

1 mol O2

= 35.79 g O2

Therefore, more O2 would be required for the combustion of C3H8. The same result is obtained with masses other than 10.00 g.

28. Start with the balanced equations:

CH4 + 2O2 CO2 + 2H2O

4NH3 + 5O2 4NO + 6H2O

The amount of water produced from 1.00 g CH4 is:

1.00 g CH4

1 mol CH4

16.042 g CH4

2 mol H2O

1 mol CH4

= 0.125 mol H2O

The mass of NH3 needed to produce 0.125 mol H2O is:

0.125 mol H2O4 mol NH

3

6 mol H2O

17.034 g NH3

1 mol NH3

= 1.42 g NH3

29. The limiting reactant is the reactant that limits the amounts of products that can form in a chemical reaction. All given reactants are necessary for the production of products: if the limiting reactant has been consumed, then there is none of this reactant present for reaction.

30. We start with 6 molecules N2 and 6 molecules H2, and these react according to the balanced equation:

N2 + 3H2 2NH3

We can determine amounts of product and leftover reactant from this information given.

N2 + 3H2 2NH3

start 6 6 0 react –2 –6 +4

end 4 0 4

Note that these react in the same ratio as given in the balanced equation (that is, 2:6:4 = 1:3:2). The pictures should show 4 molecules N2 and 4 molecules NH3.

31. To determine the limiting reactant, first calculate the number of moles of each reactant present. Then determine how these numbers of moles correspond to the stoichiometric ratio indicated by the balanced chemical equation for the reaction.

32. The theoretical yield of a reaction represents the stoichiometric amount of product that should form if the limiting reactant for the process is completely consumed.



33. A reactant is present in excess if there is more of that reactant present than is needed to combine with the limiting reactant for the process. By definition, the limiting reactant cannot be present in excess. An excess of any reactant does not affect the theoretical yield for a process: the theoretical yield is determined by the limiting reactant.

34. a. 2Al(s) + 6HCl(aq) 2AlCl3(aq) + 3H2(g)

Molar masses: Al, 26.98 g; HCl, 36.46 g; AlCl3, 133.3 g; H2, 2.016 g

15.0 g Al 1 mol

26.98 g = 0.556 mol Al

15.0 g HCl 1 mol

36.46 g = 0.411 mol HCl

Since HCl is needed to react with Al in a 6:2 (i.e., 3:1) molar ratio, it seems pretty certain that HCl is the limiting reactant. To prove this, we can calculate the quantity of Al that would react with the given number of moles of HCl.

0.411 mol HCl 2 mol Al

6 mol HCl = 0.137 mol Al

By this calculation we have shown that all the HCl present will be needed to react with only 0.137 mol Al (out of the 0.556 mol Al present). Therefore, HCl is the limiting reactant, and Al is present in excess. The calculation of the masses of products produced is based on the number of moles of the limiting reactant.

0.411 mol HCl 2 mol AlCl

3

6 mol HCl

133.3 g

1 mol = 18.3 g AlCl3

0.411 mol HCl 3 mol H

2

6 mol HCl

2.016 g

1 mol = 0.414 g H2

b. 2NaOH(aq) + CO2(g) Na2CO3(aq) + H2O(l)

molar masses: NaOH, 40.00 g; CO2, 44.01 g; Na2CO3, 105.99 g; H2O, 18.02 g

15.0 g NaOH 1 mol

40.00 g = 0.375 mol NaOH

15.0 g CO2

1 mol

44.01 g = 0.341 mol CO2

For the 0.375 mol NaOH, let's calculate if there is enough CO2 present to react:

0.375 mol NaOH 1 mol CO2

2 mol NaOH = 0.1875 mol CO2 (0.188 mol)

86 Chapter 9


We have present more CO2 (0.341 mol) than is needed to react with the given quantity of NaOH. NaOH is therefore the limiting reactant, and CO2 is present in excess. The quantities of products resulting are based on the complete conversion of the limiting reactant (0.375 mol NaOH):

0.375 mol NaOH 1 mol Na

2CO

3

2 mol NaOH

105.99 g

1 mol = 19.9 g Na2CO3

0.375 mol NaOH 1 mol H

2O

2 mol NaOH

18.02 g

1 mol = 3.38 g H2O

c. Pb(NO3)2(aq) + 2HCl(aq) PbCl2(s) + 2HNO3(aq)

Molar masses: Pb(NO3)2, 331.2 g; HCl, 36.46 g; PbCl2, 278.1 g; HNO3, 63.02 g

15.0 g Pb(NO3)2

1 mol

331.2 g = 0.0453 mol Pb(NO3)2

15.0 g HCl 1 mol

36.46 g = 0.411 mol HCl

With such a large disparity between the numbers of moles of the reactants, it’s probably a sure bet that Pb(NO3)2 is the limiting reactant. To confirm this, we can calculate how many mol of HCl are needed to react with the given amount of Pb(NO3)2.

0.0453 mol Pb(NO3)2

2 mol HCl

1 mol Pb(NO3)2

= 0.0906 mol HCl

We have considerably more HCl present than is needed to react completely with the Pb(NO3)2. Therefore, Pb(NO3)2 is the limiting reactant, and HCl is present in excess. The quantities of products produced are based on the limiting reactant being completely consumed.

0.0453 mol Pb(NO3)2

1 mol PbCl2

1 mol Pb(NO3)

2

278.1 g

1 mol = 12.6 g PbCl2

0.0453 mol Pb(NO3)2

2 mol HNO3

1 mol Pb(NO3)

2

63.02 g

1 mol = 5.71 g HNO3

d. 2K(s) + I2(s) 2KI(s)

Molar masses: K, 39.10 g; I2, 253.8 g; KI, 166.0 g

15.0 g K 1 mol

39.10 g = 0.384 mol K

15.0 g I2

1 mol

253.8 g = 0.0591 mol I2



Since, from the balanced chemical equation, we need twice as many moles of K as moles of I2, and since there is so little I2 present, it’s a safe bet that I2 is the limiting reactant. To confirm this, we can calculate how many moles of K are needed to react with the given amount of I2 present:

0.0591 mol I2

2 mol K

1 mol I2

= 0.1182 mol K

Clearly we have more potassium present than is needed to react with the small amount of I2 present. I2 is therefore the limiting reactant, and potassium is present in excess. The amount of product produced is calculated from the number of moles of the limiting reactant present:

0.0591 mol I2

2 mol KI

1 mol I2

166.0 g

1 mol = 19.6 g KI

35. a. 2NH3(g) + 2Na(s) 2NaNH2(s) + H2(g)

Molar masses: NH3, 17.03 g; Na, 22.99 g; NaNH2, 39.02 g

50.0 g NH3

1 mol

17.03 g = 2.94 mol NH3

50.0 g Na 1 mol

22.99 g = 2.17 mol Na

Since the coefficients of NH3 and Na are the same in the balanced chemical equation for the reaction, the two reactants combine in a 1:1 molar ratio. Therefore, Na is the limiting reactant, which will control the amount of product produced.

2.17 mol Na 2 mol NaNH

2

2 mol Na

39.02 g

1 mol = 84.7 g NaNH2

b. BaCl2(aq) + Na2SO4(aq) BaSO4(s) + 2NaCl(aq)

Molar masses: BaCl2, 208.2 g; Na2SO4, 142.1 g; BaSO4, 233.4 g

50.0 g BaCl2

1 mol

208.2 g = 0.240 mol BaCl2

50.0 g Na2SO4

1 mol

142.1 g = 0.352 mol Na2SO4

Since the coefficients of BaCl2 and Na2SO4 are the same in the balanced chemical equation for the reaction, the reactant having the smaller number of moles present (BaCl2) must be the limiting reactant, which will control the amount of product produced.

0.240 mol BaCl2

1 mol BaSO4

1 mol BaCl2

233.4 g

1 mol = 56.0 g BaSO4

88 Chapter 9


c. SO2(g) + 2NaOH(aq) Na2SO3(aq) + H2O(l)

Molar masses: SO2, 64.07 g; NaOH, 40.00 g; Na2SO3, 126.1 g

50.0 g SO2

1 mol

64.07 g = 0.780 mol SO2

50.0 g NaOH 1 mol

40.00 g = 1.25 mol NaOH

From the balanced chemical equation for the reaction, every time one mol of SO2

reacts, two mol of NaOH is needed. For 0.780 mol of SO2, 2(0.780 mol) = 1.56 mol of NaOH would be needed. We do not have sufficient NaOH to react with the SO2

present: therefore, NaOH is the limiting reactant, and controls the amount of product obtained.

1.25 mol NaOH 1 mol Na

2SO

3

2 mol NaOH

126.1 g

1 mol = 78.8 g Na2SO3

d. 2Al(s) + 3H2SO4(l) Al2(SO4)3(s) + 3H2(g)

Molar masses: Al, 26.98 g; H2SO4, 98.09 g; Al2(SO4)3, 342.2 g

50.0 g Al 1 mol

26.98 g = 1.85 mol Al

50.0 g H2SO4

1 mol

98.09 g = 0.510 mol H2SO4

Since the amount of H2SO4 present is smaller than the amount of Al, let’s see if H2SO4 is the limiting reactant by calculating how much Al would react with the given amount of H2SO4.

0.510 mol H2SO4

2 mol Al

3 mol H2SO4

= 0.340 mol Al

Since all the H2SO4 present would react with only a small portion of the Al present, H2SO4 is therefore the limiting reactant and will control the amount of product obtained.

0.510 mol H2SO4

1 mol Al2(SO

4)

3

3 mol H2SO

4

342.2 g

1 mol = 58.2 g Al2(SO4)3

36. a. CO(g) + 2H2(g) CH3OH(l)

CO is the limiting reactant; 11.4 mg CH3OH

b. 2Al(s) + 3I2(s) 2AlI3(s)

I2 is the limiting reactant; 10.7 mg AlI3

c. Ca(OH)2(aq) + 2HBr(aq) CaBr2(aq) + 2H2O(l)

HBr is the limiting reactant; 12.4 mg CaBr2; 2.23 mg H2O



d. 2Cr(s) + 2H3PO4(aq) 2CrPO4(s) + 3H2(g)

H3PO4 is the limiting reactant; 15.0 mg CrPO4; 0.309 mg H2

37. Cl2(g) + 2NaI(aq) 2NaCl(aq) + I2(s)

Br2(l) + 2NaI(aq) 2NaBr(aq) + I2(s)

molar masses: Cl2, 70.90 g; Br2, 159.8 g; I2, 253.8 g; NaI, 149.9 g

5.00 g Cl2

1 mol Cl2

70.90 g Cl2

= 0.0705 mol Cl2

25.0 g NaI 1 mol NaI

149.9 g NaI = 0.167 mol NaI

Since we would need 2(0.0705) = 0.141 mol of NaI for the Cl2 to react completely, and we have more than this amount of NaI, then Cl2 must be the limiting reactant.

0.0705 mol Cl2

1 mol I2

1 mol Cl2

= 0.0705 mol I2

0.0705 mol I2

253.8 g I2

1 mol I2

= 17.9 g I2

5.00 g Br2

1 mol Br2

159.8 g Br2

= 0.0313 mol Br2

25.0 g NaI 1 mol NaI

149.9 g NaI = 0.167 mol NaI

Since we would need 2(0.0313) = 0.0626 mol of NaI for the Br2 to react completely, and we have more than this amount of NaI, then Br2 must be the limiting reactant.

0.0313 mol Br2

1 mol I2

1 mol Br2

= 0.0313 mol I2

0.0313 mol I2

253.8 g I2

1 mol I2

= 7.94 g I2

38. 4Fe(s) + 3O2(g) 2Fe2O3(s)

Molar masses: Fe, 55.85 g; Fe2O3, 159.7 g

1.25 g Fe 1 mol

55.85 g = 0.0224 mol Fe present

Calculate how many mol of O2 are required to react with this amount of Fe.

0.0224 mol Fe 3 mol O2

4 mol Fe = 0.0168 mol O2

90 Chapter 9


Since we have more O2 than this, Fe must be the limiting reactant.

0.0224 mol Fe 2 mol Fe2O3

4 mol Fe = 0.0112 mol Fe2O3

0.0112 mol Fe2O3

159.7 g Fe2O3

1 mol Fe2O3

= 1.79 g Fe2O3

39. Ca2+(aq) + Na2C2O4(aq) CaC2O4(s) + 2Na+(aq)

molar masses: Ca2+, 40.08 g; Na2C2O4, 134.0 g

15 g Ca2+1 mol Ca

2+

40.08 g Ca2+ = 0.37 mol Ca2+

15 g Na2C2O4

1 mol Na2C2O4

134.0 g Na2C2O4

= 0.11 mol Na2C2O4

Since the balanced chemical equation tells us that one oxalate ion is needed to precipitate each calcium ion, from the number of moles calculated to be present, it should be clear that not nearly enough sodium oxalate ion has been added to precipitate all the calcium ion in the sample.

40. The actual yield for a reaction is the quantity of product actually isolated from the reaction vessel. The theoretical yield represents the mass of products that should be produced by the reaction if the limiting reactant is fully consumed. The percent yield represents what fraction

of the amount of product that should have been collected was actually collected.

41. If the reaction is performed in a solvent, the product may have a substantial solubility in the solvent; the reaction may come to equilibrium before the full yield of product is achieved; loss of product may occur through operator error.

42. Percent yield = actual yield

theoretical yield 100 =

1.23 g

1.44 g 100 = 85.4%

43. S8(s) + 8Na2SO3(aq) + 40H2O(l) 8Na2S2O3•5H2O

molar masses: S8, 256.6 g; Na2SO3, 126.1 g; Na2S2O3•5H2O, 248.2 g

3.25 g S8

1 mol S8

256.6 g S8

= 0.01267 mol S8

13.1 g Na2SO3

1 mol Na2SO3

126.1 g Na2SO3

= 0.1039 mol Na2SO3

S8 is the limiting reactant.

0.01267 mol S8

8 mol Na2S2O3 • 5H2O

1 mol S8

= 0.1014 mol Na2S2O3•5H2O



0.1014 mol Na2S2O3•5H2O248.2 g Na2S2O3 • 5H2O

1 mol Na2S2O3 • 5H2O = 25.2 g Na2S2O3•5H2O

Percent yield = actual yield


5.26 g

25.2 g 100 = 20.9%

44. 2LiOH(s) + CO2(g) Li2CO3(s) + H2O(g)

molar masses: LiOH, 23.95 g; CO2, 44.01 g

155 g LiOH 1 mol LiOH

23.95 g LiOH

1 mol CO2

2 mol LiOH

44.01 g CO2

1 mol CO2

= 142 g CO2

Since the cartridge has only absorbed 102 g CO2 out of a total capacity of 142 g CO2, the cartridge has absorbed

102 g

142 g 100 = 71.8% of its capacity

45. Xe(g) + 2F2(g) XeF4(s)

molar masses: Xe, 131.3 g; F2, 38.00 g; XeF4, 207.3 g

130. g Xe 1 mol Xe

131.3 g Xe = 0.9901 mol Xe

100. g F2

1 mol F2

38.00 g F2

= 2.632 mol F2

Xe is the limiting reactant.

0.9901 mol Xe 1 mol XeF4

1 mol Xe = 0.9901 mol XeF4

0.9901 mol XeF4

207.3 g XeF4

1 mol XeF4

= 205 g XeF4



145 g

205 g 100 = 70.7%

46. CaCO3(s) + 2HCl(g) CaCl2(s) + CO2(g) + H2O(g)

molar masses: CaCO3, 100.1 g; HCl, 36.46 g; CaCl2, 111.0 g

155 g CaCO3

1 mol CaCO3

100.1 g CaCO3

= 1.548 mol CaCO3

250. g HCl 1 mol HCl

36.46 g HCl = 6.857 mol HCl

CaCO3 is the limiting reactant.

92 Chapter 9


1.548 mol CaCO3

1 mol CaCl2

1 mol CaCO3

= 1.548 mol CaCl2

1.548 mol CaCl2

111.0 g CaCl2

1 mol CaCl2

= 172 g CaCl2



142 g

172 g 100 = 82.6%

47. NaCl(aq) + NH3(aq) + H2O(l) + CO2(s) NH4Cl(aq) + NaHCO3(s)

molar masses: NH3, 17.03 g; CO2, 44.01 g; NaHCO3, 84.01 g

10.0 g NH3

1 mol NH3

17.03 g NH3

= 0.5872 mol NH3

15.0 g CO2

1 mol CO2

44.01 g CO2

= 0.3408 mol CO2

CO2 is the limiting reactant.

0.3408 mol CO2

1 mol NaHCO3

1 mol CO2

= 0.3408 mol NaHCO3

0.3408 mol NaHCO3

84.01 g NaHCO3

1 mol NaHCO3

= 28.6 g NaHCO3

48. Fe(s) + S(s) FeS(s)

molar masses: Fe, 55.85 g; S, 32.07 g; FeS, 87.92 g

5.25 g Fe 1 mol Fe

55.85 g Fe = 0.0940 mol Fe

12.7 g S 1 mol S

32.07 g S = 0.396 mol S

Fe is the limiting reactant.

0.0940 mol Fe 1 mol FeS

1 mol Fe

87.92 g FeS

1 mol FeS = 8.26 g FeS produced

49. C6H12O6(s) + 6O2(g) 6CO2(g) + 6H2O(g)

molar masses: glucose, 180.2 g; CO2, 44.01 g

1.00 g glucose 1 mol glucose

180.2 g glucose = 5.549 10–3 mol glucose

5.549 10–3 mol glucose 6 mol CO2

1 mol glucose = 3.33 10–2 mol CO2

3.33 10–2 mol CO2

44.01 g CO2

1 mol CO2

= 1.47 g CO2



50. mass of Cl– present = 1.054 g sample 10.3 g Cl

100.0 g sample = 0.1086 g Cl–

molar masses: Cl–, 35.45 g; AgNO3, 169.9 g; AgCl, 143.4 g

0.1086 g Cl–1 mol Cl

35.45 g Cl = 3.063 10–3 mol Cl–

3.063 10–3 mol Cl– 1 mol AgNO3

1 mol Cl = 3.063 10–3 mol AgNO3

3.063 10–3 mol AgNO3

169.9 g AgNO3

1 mol AgNO3

= 0.520 g AgNO3 required

3.063 10–3 mol Cl– 1 mol AgCl

1 mol Cl- = 3.063 10–3 mol AgCl

3.063 10–3 mol AgCl 143.4 g AgCl

1 mol AgCl = 0.439 g AgCl produced

51. For O2:5 mol O2

1 mol C3H8

For CO2:3 mol CO2

1 mol C3H8

For H2O: 4 mol H2O

1 mol C3H8

52. a. 2H2O2(l) 2H2O(l) + O2(g)

0.50 mol H2O2

2 mol H2O

2 mol H2O2

= 0.50 mol H2O

0.50 mol H2O2

1 mol O2

2 mol H2O2

= 0.25 mol O2

b. 2KClO3(s) 2KCl(s) + 3O2(g)

0.50 mol KClO3

2 mol KCl

2 mol KClO3

= 0.50 mol KCl

0.50 mol KClO3

3 mol O2

2 mol KClO3

= 0.75 mol O2

c. 2Al(s) + 6HCl(aq) 2AlCl3(aq) + 3H2(g)

0.50 mol Al 2 mol AlCl3

2 mol Al = 0.50 mol AlCl3

0.50 mol Al 3 mol H

2

2 mol Al = 0.75 mol H2

94 Chapter 9


d. C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l)

0.50 mol C3H8

3 mol CO2

1 mol C3H8

= 1.5 mol CO2

0.50 mol C3H8

4 mol H2O

1 mol C3H8

= 2.0 mol H2O

53. a. NH3(g) + HCl(g) NH4Cl(s)

molar mass of NH3 = 17.0 g

1.00 g NH3

1 mol NH3

17.0 g NH3

= 0.0588 mol NH3

0.0588 mol NH3

1 mol NH4Cl

1 mol NH3

= 0.0588 mol NH4Cl

b. CaO(s) + CO2(g) CaCO3(s)

molar mass CaO = 56.1 g

1.00 g CaO 1 mol CaO

56.1 g CaO = 0.0178 mol CaO

0.0178 mol CaO 1 mol CaCO3

1 mol CaO = 0.0178 mol CaCO3

c. 4Na(s) + O2(g) 2Na2O(s)

molar mass Na = 22.99 g

1.00 g Na 1 mol Na

22.99 g Na = 0.0435 mol Na

0.0435 mol Na 2 mol Na2O

4 mol Na = 0.0217 mol Na2O

d. 2P(s) + 3Cl2(g) 2PCl3(l)

molar mass P = 30.97 g

1.00 g P 1 mol P

30.97 g P = 0.0323 mol P

0.0323 mol P 2 mol PCl3

2 mol P = 0.0323 mol PCl3

54. 2Na2O2(s) + 2H2O(l) 4NaOH(aq) + O2(g)

molar masses: Na2O2, 77.98 g; O2, 32.00 g

3.25 g Na2O2

1 mol Na2O2

77.98 g Na2O2

= 0.0417 mol Na2O2



0.0417 mol Na2O2

1 mol O2

2 mol Na2O2

= 0.0209 mol O2

0.0209 mol O2

32.00 g O2

1 mol O2

= 0.669 g O2

55. 2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(g)

molar masses: C2H2, 26.04 g; O2, 32.00 g 150 g = 1.5 102 g

1.5 102 g C2H2

1 mol C2H2

26.04 g C2H2

= 5.760 mol C2H2

5.760 mol C2H2

5 mol O2

2 mol C2H2

= 14.40 mol O2

14.40 mol O2

32.00 g O2

1 mol O2

= 4.6 102 g O2

56. a. C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(l)

molar masses: C2H5OH, 46.07 g; O2, 32.00 g; CO2, 44.01 g

25.0 g C2H5OH1 mol C2H5OH

46.07 g C2H5OH = 0.5427 mol C2H5OH

25.0 g O2

1 mol O2

32.00 g O2

= 0.7813 mol O2

Since there is less C2H5OH present on a mole basis, see if this substance is the limiting reactant.

0.5427 mol C2H5OH3 mol O2

1 mol C2H5OH = 1.6281 mol O2

From the above calculation, C2H5OH must not be the limiting reactant (even though there is a smaller number of moles of C2H5OH present) since more oxygen than is present would be required to react completely with the C2H5OH present. Oxygen is the limiting reactant.

0.7813 mol O2

2 mol CO2

3 mol O2

= 0.5209 mol CO2

0.5209 mol CO2

44.01 g CO2

1 mol CO2

= 22.9 g CO2

b. N2(g) + O2(g) 2NO(g)

molar masses: N2, 28.02 g; O2, 32.00 g; NO, 30.01 g

25.0 g N2

1 mol N2

28.02 g N2

= 0.8922 mol N2

96 Chapter 9


25.0 g O2

1 mol O2

32.00 g O2

= 0.7813 mol O2

Since the coefficients of N2 and O2 are the same in the balanced chemical equation for the reaction, an equal number of moles of each substance would be necessary for complete reaction. Since there is less O2 present on a mole basis, O2 must be the limiting reactant.

0.7813 mol O2

2 mol NO

1 mol O2

= 1.5626 mol NO

1.5626 mol NO 30.01 g NO

1 mol NO = 46.9 g NO

c. 2NaClO2(aq) + Cl2(g) 2ClO2(g) + 2NaCl(aq)

molar masses: NaClO2, 90.44 g; Cl2, 70.90 g; NaCl, 58.44 g

25.0 g NaClO2

1 mol NaClO2

90.44 g NaClO2

= 0.2764 mol NaClO2

25.0 g Cl2

1 mol Cl2

70.90 g Cl2

= 0.3526 mol Cl2

See if NaClO2 is the limiting reactant.

0.2764 mol NaClO2

1 mol Cl2

1 mol NaClO2

= 0.1382 mol Cl2

Since 0.2764 mol of NaClO2 would require only 0.1382 mol Cl2 to react completely (and since we have more than this amount of Cl2), then NaClO2 must indeed be the limiting reactant.

0.2764 mol NaClO2

2 mol NaCl

2 mol NaClO2

= 0.2764 mol NaCl

0.2764 mol NaCl 58.44 g NaCl

1 mol NaCl = 16.2 g NaCl

d. 3H2(g) + N2(g) 2NH3(g)

molar masses: H2, 2.016 g; N2, 28.02 g; NH3, 17.03 g

25.0 g H2

1 mol H2

2.016 g H2

= 12.40 mol H2

25.0 g N2

1 mol N2

28.02 g N2

= 0.8922 mol N2

See if N2 is the limiting reactant.

0.8922 mol N2

3 mol H2

1 mol N2

= 2.677 mol H2



N2 is clearly the limiting reactant, since there is 12.40 mol H2 present (a large excess).

0.8922 mol N2

2 mol NH3

1 mol N2

= 1.784 mol NH3

1.784 mol NH3

17.03 g NH3

1 mol NH3

= 30.4 g NH3

57. N2H4(l) + O2(g) N2(g) + 2H2O(g)

molar masses: N2H4, 32.05 g; O2, 32.00 g; N2, 28.02 g; H2O, 18.02 g

20.0 g N2H4

1 mol N2H2

32.05 g N2H2

= 0.624 mol N2H4

20.0 g O2

1 mol O2

32.00 g O2

= 0.625 mol O2

The two reactants are present in nearly the required ratio for complete reaction (due to the 1:1 stoichiometry of the reaction and the very similar molar masses of the substances). We will consider N2H4 as the limiting reactant in the following calculations.

0.624 mol N2H4

1 mol N2

1 mol N2H4

= 0.624 mol N2

0.624 mol N2

28.02 g N2

1 mol N2

= 17.5 g N2

0.624 mol N2H4

2 mol H2O

1 mol N2H4

= 1.248 mol H2O = 1.25 mol H2O

1.248 mol H2O18.02 g H2O

1 mol H2O = 22.5 g H2O

58. 12.5 g theoretical 40 g actual

100 g theoretical = 5.0 g

59. We are concerned with the balanced equation:

X4 + 12HCl 4XCl3 + 6H2

We are given the mass of X4 (248 g). If we can determine the number of moles of X4, we can determine its molar mass, and therefore its identity. We can do this from the mass of H2.

24.0 g H2

1 mol H2

2.016 g H2

1 mol X4

6 mol H2

= 1.98 mol X4

Thus,

248 g X4

1.98 mol X4

= 125 g/mol (molar mass of X4)

98 Chapter 9


Element X has an atomic mass of about 125

4 = 31.3 g/mol.

This is closest to phosphorus (30.97 g/mol).

60. We know the following:

x + y xy and x + 3z xz3

5 g 15 g 3 g 18 g

Thus, the mole ratio between x and y is 1:1 and the mole ratio between x and z is 1:3, respectively.

So, the relative masses of x:y are 1:3, and the relative masses of x:z are 3:6 or 1:2 (if x = 3 g, 3z = 18 g or z = 6 g).

Thus, x:y:z = 1:3:2.

If y = 60 g/mol, x = 20 g/mol, and z = 40 g/mol.

61. This is a “double” limiting reactant problem. First, determine the maximum yield of P4O10

from the equation:

P4 + 5O2 P4O10

20.0 g P4

1 mol P4

123.88 g P4

1 mol P4O

10

1 mol P4

283.88 g P4O

10

1 mol P4O

10

= 45.8 g P4O10

30.0 g O2

1 mol O2

32.00 g O2

1 mol P4O

10

5 mol O2

283.88 g P4O

10

1 mol P4O

10

= 53.2 g P4O10

Thus, the maximum yield of P4O10 is 45.8 g.

Determine the yield of H3PO4 from the equation:

P4O10 + 6H2O 4H3PO4

45.8 g P4O10

1 mol P4O

10

283.88 g P4O

10

4 mol H3PO

4

1 mol P4O

10

97.994 g H3PO

4

1 mol H3PO

4

= 63.2 g H3PO4

15.0 g H2O1 mol H

2O

18.016 g H2O

4 mol H3PO

4

6 mol H2O

97.994 g H3PO

4

1 mol H3PO

4

= 54.4 g H3PO4

Thus, the maximum yield of H3PO4 is 54.4 g.

62. a. Neither is limiting. 1 mol N2

2 mol NH3

1 mol N2

= 2 moles NH3

3 mol H2

2 mol NH3

3 mol H2

= 2 moles NH3

b. H2 is limiting. 3 mol H2

2 mol NH3

3 mol H2

= 2 moles NH3

c. H2 is limiting. 3 mol H2

2 mol NH3

3 mol H2

= 2 moles NH3

d. So, choice d (each would produce the same amount of product) is the correct answer.



63. Choice “c” is correct. Since the molar mass of O2 is larger than the molar mass of C2H6, an equal mass of each would result in a fewer number of moles of O2 than C2H6. If this were the case, O2 must be limiting since it is needed in a greater amount (a 7:2 mol ratio).

64. 50.0 g CaO 1 mol CaO56.08 g CaO

5 mol C2 mol CaO

= 2.23 mol C required

50.0 g C 1 mol C12.01 g C

= 4.16 mol C

Leftover C = 4.16 mol – 2.23 mol = 1.93 mol C 12.01 g C

1 mol C = 23.2 g C left over

(CaO is limiting.)

65. 13.97 g KClO3

1 mol KClO3

122.55 g KClO3

2 mol KCl2 mol KClO3

74.55 g KCl

1 mol KCl= 8.498 g KCl


theoretical yield 100% =

6.23 g

8.498 g 100% = 73.3% yield

66. False. Amounts used can vary. However, according to the equation, for every 28.02 g of N2

(2 mol) we need 6.048 g of H2 (3 mol). So, for example, if we have 28.02 g of N2 and 5.5 g of H2, the H2 is limiting. If we have 28.02 g of N2 and 6.5 g of H2, the N2 is limiting.

67. a. As we add more sodium, we get more product because sodium is the limiting reactant. However, eventually sodium is in excess (and chlorine is limiting), so the

amount of product does not increase with an increase in sodium.

b. 20.0 g Na 1 mol Na22.99 g Na

2 mol NaCl2 mol Na

58.44 g NaCl

1 mol NaCl = 50.8 g NaCl

c. If we look at the graph, we can see that 40.0 g of sodium reacts in a stoichiometric ratio with chlorine

(that is, there is no limiting reactant). Because of this, we can use 40.0 g of sodium to determine the amount of Cl2. Since the problem states that the mass of Cl2 is the same in all containers, each container will have this amount of Cl2.

40.0 g Na 1 mol Na22.99 g Na

1 mol Cl2

2 mol Na

70.9 g Cl2

1 mol Cl2

= 61.7 g Cl2

d. We should add “since the amount of product is equal when 40.0 g or 50.0 g of sodium is used, 50.0 g of sodium must be an excess amount.” Since we know from part c how much Cl2 is in the container, we use the amount of Cl2 (the limiting reactant) to determine the amount of product.

61.7 g Cl2

1 mol Cl2

70.9 g Cl2

2 mol NaCl1 mol Cl2

58.44 g NaCl

1 mol NaCl= 101.7 g NaCl = 102 g NaCl

e. In part b, sodium is limiting, and 30.85 g Cl2 is left over (half of the Cl2 is left over). In part d, Cl2 is limiting and 10.0 g Na is left over (50.0 g – 40.0 g).

chapter 9 chemical quantities - edl · world of chemistry 74 copyright houghton mifflin company....

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