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CHAPTER 9

Chemical Quantities

The coefficients of the balanced chemical equation for a reaction give the relative numbers of
molecules of reactants and products that are involved in the reaction.

The coefficients of this balanced chemical equation indicate the relative numbers of molecules (or
moles) of each reactant that combine, as well as the number of molecules (or moles) of the
product formed.

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.

(e); Subscripts cannot be changed to balance the equation or else the identities of the reactants
and/or product are changed. The balanced equation is: N2(g) + 3H2(g)  2NH3(g). Nitrogen and
hydrogen will react regardless of how much is initially present (just one might be used up before
the other).

PCl3(l) + 3H2O(l)  H3PO3(aq) + 3HCl(g)

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

2XeF2(g) + 2H2O(l)  2Xe(g) + 4HF(g) + O2(g)

Two molecules of gaseous xenon difluoride react with two molecules of liquid water,
producing two gaseous xenon atoms, four molecules of gaseous hydrogen fluoride, and
one molecule of oxygen gas. Two moles of xenon difluoride reacts with two moles of
water, to produce two moles of xenon, four moles of hydrogen fluoride, and one mole of
oxygen.

S(s) + 6HNO3(aq)  H2SO4(aq) + 2H2O(l) + 6NO2(g)

One sulfur atom reacts with six molecules of aqueous nitric acid, producing one molecule
of aqueous sulfuric acid, two molecules of water, and six molecules of nitrogen dioxide
gas. One mole of sulfur reacts with six moles of nitric acid, to produce one mole of
sulfuric acid, two moles of water, and six moles of nitrogen dioxide.

2NaHSO3(s)  Na2SO3(s) + SO2(g) + H2O(l)

Two formula units of solid sodium hydrogen sulfite react to produce one formula unit of
solid sodium sulfite, one molecule of gaseous sulfur dioxide, and one molecule of liquid
water. Two moles of sodium hydrogen sulfite react to produce one mole of sodium
sulfite, one mole of sulfur dioxide, and one mole of water.

163

Chapter 9: Chemical Quantities

3MnO2(s) + 4Al(s)  3Mn(s) + 2Al2O3(s)

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

B2O3(s) + 3CaF2(s)  2BF3(g) + 3CaO(s)

One molecule (or one mole) of diboron trioxide reacts with three formula units (or three
moles) of calcium fluoride, producing two molecules (or two moles) of boron trifluoride
and three formula units (three moles) of calcium oxide.

3NO2(g) + H2O(l)  2HNO3(aq) + NO(g)

Three molecules (or three moles) of nitrogen dioxide react with one molecule (or one
mole) of water, producing two molecules (or two moles) of nitric acid and one molecule
(or one mole) of nitrogen monoxide.

C6H6(g) + 3H2(g)  C6H12(g)

One molecule (or one mole) of benzene (C6H6) reacts with three molecules (or three
moles) of hydrogen gas, producing one molecule (or one mole) of cyclohexane (C6H12).

False. The coefficients of the balanced chemical equation represent the ratios on a mole basis by
which carbon combines with oxygen.

Balanced chemical equations tell us in what molar ratios substances combine to form products,
not in what mass proportions they combine. How could a total of 3 g of reactants produce 2 g of
product?

4Al(s) + 3O2(g)  2Al2O3(s)

For converting from a given number of moles of aluminum metal to the number of moles of
oxygen needed for reaction, the correct mole ratio is

3 mol O

4 mol Al













For converting from a given number of moles of aluminum metal to the number of moles of
product produced, the mole ratio is

2 mol Al O

4 mol Al













10.

CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)

For converting from a given number of moles of methane to the number of moles of oxygen
required, the mole ratio is

For a given number of moles of methane reacting completely, the mole ratios used to
calculate the number of moles of each product are

164

Chapter 9: Chemical Quantities

For CO2:

For H2O:

11.

CO2(g) + 4H2(g)  CH4(g) + 2H2O(l)

0.500 mol CO2 

1 mol CH
1 mol CO

= 0.500 mol CH4

0.500 mol CO2 

2 mol H O

1 mol CO

= 1.00 mol H2O

BaCl2(aq) + 2AgNO3(aq)  2AgCl(s) + Ba(NO3)2(aq)

0.500 mol BaCl2 

2 mol AgCl

1 mol BaCl

= 1.00 mol AgCl

0.500 mol BaCl2 

3 2

1 mol Ba(NO )

1 mol BaCl

= 0.500 mol Ba(NO3)2

C3H8(g) + 5O2(g)  4H2O(l) + 3CO2(g)

0.500 mol C3H8 

4 mol H O

1 mol C H

= 2.00 mol H2O

0.500 mol C3H8 

3 mol CO

1 mol C H

= 1.50 mol CO2

3H2SO4(aq) + 2Fe(s)  Fe2(SO4)3(aq) + 3H2(g)

0.500 mol H2SO4 

4 3

1 mol Fe (SO )

3 mol H SO

= 0.167 mol Fe2(SO4)3

0.500 mol H2SO4 

3 mol H

3 mol H SO

= 0.500 mol H2

12.

Before doing the calculations, the equations must be balanced.

2FeO(s) + C(s)  2Fe(l) + CO2(g)

0.125 mol FeO ×

= 0.125 mol Fe

0.125 mol FeO ×

= 0.0625 mol CO2

Cl2(g) + 2KI(aq)  2KCl(aq) + I2(s)

0.125 mol KI ×

= 0.125 mol KCl

0.125 mol KI ×

= 0.0625 mol I2

2Na2B4O7(s) + 2H2SO4(aq) + 10H2O(l)  8H3BO3(s) + 2Na2SO4(aq)

165

Chapter 9: Chemical Quantities

0.125 mol Na2B4O7 ×

= 0.500 mol H3BO3

0.125 mol Na2B4O7 ×

= 0.125 mol Na2SO4

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

0.125 mol CaC2 ×

= 0.125 mol Ca(OH)2

0.125 mol CaC2 ×

= 0.125 mol C2H2

13.

AgNO3(aq) + LiOH(aq)  AgOH(s) + LiNO3(aq)

molar masses: AgOH, 124.91 g; LiNO3, 68.95 g

0.125 mol AgNO3 

1 mol AgOH

1 mol AgNO

= 0.125 mol AgOH

0.125 mol AgOH ×

124.91 g AgOH

1 mol AgOH

= 15.6 g AgOH

0.125 mol AgNO3 

1 mol LiNO

1 mol AgNO

= 0.125 mol LiNO3

0.125 mol LiNO3 ×

68.95 g LiNO

1 mol LiNO

= 8.62 g LiNO3

Al2(SO4)3(aq) + 3CaCl2(aq)  2AlCl3(aq) + 3CaSO4(s)

molar masses: AlCl3, 133.33 g; CaSO4, 136.15 g

0.125 mol Al2(SO4)3 

4 3

2 mol AlCl

1 mol Al (SO )

= 0.250 mol AlCl3

0.250 mol AlCl3 ×

133.33 g AlCl

1 mol AlCl

= 33.3 g AlCl3

0.125 mol Al2(SO4)3 

4 3

3 mol CaSO

1 mol Al (SO )

= 0.375 mol CaSO4

0.375 mol CaSO4 ×

136.15 g CaSO4

1 mol CaSO

= 51.1 g CaSO4

CaCO3(s) + 2HCl(aq)  CaCl2(aq) + CO2(g) + H2O(l)

molar masses: CaCl2, 110.98 g; CO2, 44.01 g; H2O, 18.02 g

0.125 mol CaCO3 

1 mol CaCl

1 mol CaCO

= 0.125 mol CaCl2

166

Chapter 9: Chemical Quantities

0.125 mol CaCl2 

110.98 g CaCl

1 mol CaCl

= 13.9 g CaCl2

0.125 mol CaCO3 

1 mol CO

1 mol CaCO

= 0.125 mol CO2

0.125 mol CO2 ×

44.01 g CO

1 mol CO

= 5.50 g CO2

0.125 mol CaCO3 

1 mol H O

1 mol CaCO

= 0.125 mol H2O

0.125 mol H2O ×

18.02 g H O

1 mol H O

= 2.25 g H2O

2C4H10(g) + 13O2(g)  8CO2(g) + 10H2O(g)

molar masses: CO2, 44.01 g; H2O, 18.02 g

0.125 mol C4H10 

8 mol CO

2 mol C H

= 0.500 mol CO2

0.500 mol CO2 ×

44.01 g CO

1 mol CO

= 22.0 g CO2

0.125 mol C4H10 

10 mol H O
2 mol C H

= 0.625 mol H2O

0.625 mol H2O ×

18.02 g H O

1 mol H O

= 11.3 g H2O

14.

NH3(g) + HCl(g)  NH4Cl(s)

molar mass: NH4Cl, 53.492 g

0.50 mol NH3 

= 0.50 mol NH4Cl

0.50 mol NH4Cl 

= 27 g NH4Cl

CH4(g) + 4S(s)  CS2(l) + 2H2S(g)

molar masses: CS2, 76.15 g; H2S, 34.086 g

0.50 mol S 

= 0.125 mol CS2

0.125 mol CS2 

= 9.5 g CS2

0.50 mol S 

= 0.25 mol H2S

167

Chapter 9: Chemical Quantities

0. 25 mol H2S 

= 8.5 g H2S

PCl3(l) + 3H2O(l)  H3PO3(aq) + 3HCl(aq)

molar masses: H3PO3, 81.994 g; HCl, 36.458 g

0.50 mol PCl3 

= 0.50 mol H3PO3

0.50 mol H3PO3 

= 41 g H3PO3

0.50 mol PCl3 

= 1.50 mol HCl

1.50 mol HCl 

= 55 g HCl

NaOH(s) + CO2(g)  NaHCO3(s)

molar masses: NaHCO3, 84.008 g

0.50 mol NaOH 

= 0.50 mol NaHCO3

0.50 mol NaHCO3 

= 42 g NaHCO3

15.

Cl2(g) + 2KI(aq)  2KCl(aq) + I2(s)

0.275 mol Cl2 

2 mol KI

1 mol Cl

= 0.550 mol KI

6Co(s) + P4(s)  2Co3P2(s)

0.275 mol Co 

1 mol P

6 mol Co

= 0.0458 mol P4

Zn(s) + 2HNO3(aq)  Zn(NO3)2(aq) + H2(g)

0.275 mol Zn 

2 mol HNO

1 mol Zn

= 0.550 mol HNO3

168

Chapter 9: Chemical Quantities

C5H12(l) + 8O2(g)  5CO2(g) + 6H2O(g)

0.275 mol C5H12 

8 mol O

1 mol C H

= 2.20 mol O2

16.

Before doing the calculations, the equations must be balanced.

4KO2(s) + 2H2O(l)  3O2(g) + 4KOH(s)

0.625 mol KOH 

3 mol O

4 mol KOH

= 0.469 mol O2

SeO2(g) + 2H2Se(g)  3Se(s) + 2H2O(g)

0.625 mol H2O 

3 mol Se

2 mol H O

= 0.938 mol Se

2CH3CH2OH(l) + O2(g)  2CH3CHO(aq) + 2H2O(l)

0.625 mol H2O 

2 mol CH CHO

2 mol H O

= 0.625 mol CH3CHO

Fe2O3(s) + 2Al(s)  2Fe(l) + Al2O3(s)

0.625 mol Al2O3 

2 mol Fe

1 mol Al O

= 1.25 mol Fe

17.

the molar mass of the substance

18.

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

19.

molar mass Si = 28.09 g

4.15 g Si ×

1 mol Si

28.09 g Si

= 0.148 mol Si

molar mass AuCl3 = 303.4 g

2.72 mg AuCl3 ×

1 mol AuCl

1 g

1000 mg

303.4 g AuCl



= 8.97 × 10–6 mol AuCl3

molar mass S = 32.07 g

1.05 kg S ×

1000 g

1 mol S

1 kg

32.07 g S



= 32.7 mol S

molar mass FeCl3 = 162.20 g

0.000901 g FeCl3 ×

1 mol FeCl

162.2 g FeCl

= 5.55 × 10–6 mol FeCl3

169

Chapter 9: Chemical Quantities

molar mass MgO = 40.31 g

5.62 × 103 g MgO ×

1 mol MgO

40.31 g MgO

= 139 mol MgO

20.

molar mass Ag = 107.9 g

2.01 × 10–2 g Ag ×

= 1.86 × 10–4 mol Ag

molar mass (NH4)2S = 68.154 g

45.2 mg (NH4)2S ×

= 6.63 × 10–4 mol (NH4)2S

molar mass U = 238.00 g

61.7

U ×

= 2.59 × 10–7 mol U

molar mass SO2 = 64.07 g

5.23 kg SO2 ×

= 81.6 mol SO2

molar mass Fe(NO3)3 = 241.88 g

272 g Fe(NO3)3 ×

= 1.12 mol Fe(NO3)3

21.

molar mass Ge = 72.59 g

2.17 mol Ge 

72.59 g Ge

1 mol Ge

= 158 g Ge

molar mass PbCl2 = 278.1 g;

4.24 millimol = 0.00424 mol

0.00424 mol PbCl2 

278.1 g PbCl

1 mol PbCl

= 1.18 g PbCl2

molar mass NH3 = 17.03 g

0.0971 mol NH3 

17.03 g NH

1 mol NH

= 1.65 g NH3

molar mass C6H14 = 86.17 g

4.26 × 103 mol C6H14 

86.17 g C H

1 mol C H

= 3.67 × 105 g C6H14

molar mass ICl = 162.35 g

1.71 mol ICl 

162.35 g ICl

1 mol ICl

= 278 g ICl

170

Chapter 9: Chemical Quantities

22.

molar mass of K3N = 131.31 g

0.341 mol K3N 

= 44.8 g K3N

molar mass of Ne = 20.18 g; 2.62 millimol = 0.00262 mol

0.00262 mol Ne 

= 0.0529 g Ne

molar mass of MnO = 70.94 g

0.00449 mol MnO 

= 0.319 g MnO

molar mass of SiO2 = 60.09 g

7.18 × 105 mol SiO2 

= 4.31 × 107 g SiO2

molar mass of FePO4 = 150.82 g

0.000121 mol FePO4 

= 0.0182 g FePO4

23.

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

2Co(s) + 3F2(g)  2CoF3(s)

0.413 mol Co ×

3 mol F

2 mol Co

= 0.620 mol F2

2Al(s) + 3H2SO4(aq)  Al2(SO4)3(aq) + 3H2(g)

0.413 mol Al ×

3 mol H SO

2 mol Al

= 0.620 mol H2SO4

2K(s) + 2H2O(l)  2KOH(aq) + H2(g)

0.413 mol K ×

2 mol H O

2 mol K

= 0.413 mol H2O

4Cu(s) + O2(g)  2Cu2O(s)

0.413 mol Cu ×

1 mol O
4 mol Cu

= 0.103 mol O2

24.

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

2Al(s) + 3Br2(l)  2AlBr3(s)

molar mass Al = 26.98 g

0.557 g Al ×

3 mol Br

1 mol Al

26.98 g Al

2 mol Al



= 0.0310 mol Br2

171

Chapter 9: Chemical Quantities

Hg(s) + 2HClO4(aq)  Hg(ClO4)2(aq) + H2(g)

molar mass Hg = 200.6 g

0.557 g Hg ×

2 mol HClO

1 mol Hg

200.6 g

1 mol Hg



= 0.00555 mol HClO4

3K(s) + P(s)  K3P(s)

molar mass K = 39.10 g

0.557 g K ×

1 mol K

1 mol P

39.10 g K 3 mol K



= 0.00475 mol P

CH4(g) + 4Cl2(g)  CCl4(l) + 4HCl(g)

molar mass CH4 = 16.04 g

0.557 g CH4 ×

1 mol CH

4 mol Cl

16.04 g CH

1 mol CH



= 0.139 mol Cl2

25.

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

TiBr4(g) + 2H2(g)  Ti(s) + 4HBr(g)

molar mass H2 = 2.016 g; molar mass Ti = 47.90 g; molar mass of HBr = 80.91 g

12.5 g H2 

1 mol H

2.016 g H

= 6.20 mol H2

6.20 mol H2 

1 mol Ti

2 mol H

= 3.10 mol Ti

3.10 mol Ti 

47.90 g Ti

1 mol Ti

= 148 g Ti

6.20 mol H2 

4 mol HBr

2 mol H

= 12.4 mol HBr

12.4 mol HBr 

80.91 g HBr

1 mol HBr

= 1.00  103 g HBr

3SiH4(g) + 4NH3(g)  Si3N4(s) + 12H2(g)

molar mass SiH4 = 32.12 g; molar mass Si3N4 = 140.3 g; molar mass H2 = 2.016 g

12.5 g SiH4 

1 mol SiH

32.12 g SiH

= 0.389 mol SiH4

0.389 mol SiH4 

1 mol Si N

3 mol SiH

= 0.130 mol Si3N4

0.130 mol Si3N4 

140.3 g Si N

1 mol Si N

= 18.2 g Si3N4

172

Chapter 9: Chemical Quantities

0.389 mol SiH4 

12 mol H

3 mol SiH

= 1.56 mol H2

1.56 mol H2 

2.016 g H

1 mol H

= 3.14 g H2

2NO(g) + 2H2(g)  N2(g) + 2H2O(l)

molar mass H2 = 2.016 g; molar mass N2 = 28.02 g; molar mass H2O = 18.02 g

12.5 g H2 

1 mol H

2.016 g H

= 6.20 mol H2

6.20 mol H2 

1 mol N

2 mol H

= 3.10 mol N2

3.10 mol N2 

28.02 g N

1 mol N

= 86.9 g N2

6.20 mol H2 

2 mol H O

2 mol H

= 6.20 mol H2O

6.20 mol H2O 

18.02 g H O

1 mol H O

= 112 g H2O

Cu2S(s)  2Cu(s) + S(g)

molar mass Cu2S = 159.2 g; molar mass Cu = 63.55 g; molar mass S = 32.07 g

12.5 g Cu2S 

1 mol Cu S

159.2 g Cu S

= 0.0785 mol Cu2S

0.0785 mol Cu2S 

2 mol Cu

1 mol Cu S

= 0.157 mol Cu

0.157 mol Cu 

63.55 g Cu

1 mol Cu

= 9.98 g Cu

0.0785 mol Cu2S 

1 mol S

1 mol Cu S

= 0.0785 mol S

0.0785 mol S 

32.07 g S

1 mol S

= 2.52 g S

26.

molar masses: PCl3, 137.32 g; H2O, 18.016 g

20.0 g PCl3 ×

= 0.146 mol PCl3

0.146 mol PCl3 ×

= 0.437 mol H2O

173

Chapter 9: Chemical Quantities

0.437 mol H2O ×

= 7.87 g H2O

27.

The equation must first be balanced.

(NH4)2CO3(s)  2NH3(g) + CO2(g) + H2O(g

molar masses: (NH4)2CO3, 96.09 g; NH3, 17.03 g

1.25 g (NH4)2CO3 









1 mol NH

96.09 g NH

= 0.0130 mol (NH4)2CO3

0.0130 mol (NH4)2CO3 ×





2 mol NH

1 mol NH

= 0.0260 mol NH3

0.0260 mol NH3 

17.03 g NH

1 mol NH

= 0.443 g NH3

28.

The balanced equation for the reaction is:

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

molar masses: CaC2, 64.10 g; C2H2, 26.04 g

3.75 g CaC2 ×

1 mol CaC

64.10 g CaC

= 0.0585 mol CaC2

0.0585 mol CaC2×

1 mol C H
1 mol CaC

= 0.0585 mol C2H2

0.0585 mol C2H2 ×

26.04 g C H

1 mol C H

= 1.52 g C2H2

29.

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 CO

1 mol C

= 0.4163 mol CO2

0.4163 mol CO2 

44.01 g CO

1 mol CO

= 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

174

Chapter 9: Chemical Quantities

30.

2NaHCO3(s)  Na2CO3(s) + H2O(g) + CO2(g)

molar masses: NaHCO3, 84.01 g; Na2CO3, 106.0 g

1.52 g NaHCO3 ×

1 mol NaHCO

84.01 g NaHCO

= 0.01809 mol NaHCO3

0.01809 mol NaHCO3 ×

1 mol Na CO

2 mol NaHCO

= 0.009047 mol Na2CO3

0.009047 mol Na2CO3 ×

106.0 g Na CO

1 mol Na CO

= 0.959 g Na2CO3

31.

2Fe(s) + 3Cl2(g)  2FeCl3(s)

millimolar masses: iron, 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 FeCl

2 mmol Fe

= 0.2775 mmol FeCl3

0.2775 mmol FeCl3 

162.2 mg FeCl

1 mmol FeCl

= 45.0 mg FeCl3

32.

C6H12O6(aq)  2C2H5OH(aq) + 2CO2(g)

molar masses: C6H12O6, 180.2 g; C2H5OH, 46.07 g

5.25 g C6H12O6 ×

1 mol C H O

180.2 g C H O

= 0.02913 mol 6 12 6

C H O

0.02913 mol 6 12 6

C H O ×

2 mol C H OH
1 mol C H O

= 0.5826 mol C2H5OH

0.5286 mol C2H5OH ×

46.07 g C H OH

1 mol C H OH

= 2.68 g ethyl alcohol

33.

H2SO3(aq)  H2O(l) + SO2(g)

molar masses: H2SO3, 82.09 g; SO2, 64.07 g

4.25 g H2SO3 ×

1 mol H SO

82.09 g H SO

= 0.05177 mol H2SO3

0.05177 mol H2SO3 ×

1 mol SO

1 mol H SO

= 0.05177 mol SO2

0.05177 mol SO2 ×

64.07 g SO

1 mol SO

= 3.32 g SO2

175

Chapter 9: Chemical Quantities

34.

The balanced equation for the reaction is:

2H2O2(aq)  2H2O(l) + O2(g)

molar masses: H2O2, 34.016 g; O2, 32.00 g

10.00 g H2O2 ×

= 0.2940 mol H2O2

0.2940 mol H2O2 ×

= 0.1470 mol O2

0.1470 mol O2 ×

= 4.704 g O2

35.

P4(s) + 5O2(g)  2P2O5(s)

molar masses: P4, 123.88 g; O2, 32.00 g

4.95 g P4 ×

1 mol P

123.88 g P

= 0.03996 mol P4

0.03996 mol P4 ×

5 mol O

1 mol P

= 0.1998 mol O2

0.1998 mol O2 ×

32.00 g O

1 mol O

= 6.39 g O2

36.

4HgS(s) + 4CaO(s)  4Hg(l) + 3CaS(s) + CaSO4(s)

molar masses: HgS, 232.7 g Hg, 200.6 g; 10.0 kg = 1.00 × 104 g

1.00 × 104 g HgS ×

1 mol HgS

232.7 g HgS

= 42.97 mol HgS

42.97 mol HgS ×

4 mol Hg

4 mol HgS

= 42.97 mol Hg

42.97 mol Hg ×

200.6 g Hg

1 mol Hg

= 8.62 × 103 g Hg = 8.62 kg Hg

37.

2NH4NO3(s)  2N2(g) + O2(g) + 4H2O(g)

molar masses: NH4NO3, 80.05 g; N2, 28.02 g; 02, 32.00 g; H2O, 18.02 g

1.25 g NH4NO3 

1 mol NH NO

80.05 g NH NO

= 0.0156 mol NH4NO3

0.0156 mol NH4NO3 

2 mol N

2 mol NH NO

= 0.0156 mol N2

176

Chapter 9: Chemical Quantities

0.0156 mol N2 

28.02 g N

1 mol N

= 0.437 g N2

0.0156 mol NH4NO3 

1 mol O

2 mol NH NO

= 0.00780 mol O2

0.00780 mol O2 

32.00 g O

1 mol O

= 0.250 g O2

0.0156 mol NH4NO3 

4 mol H O

2 mol NH NO

= 0.0312 mol H2O

0.0312 mol H2O 

18.02 g H O

1 mol H O

= 0.562 g H2O

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

38.

C12H22O11(s)  12C(s) + 11H2O(g)

molar masses: C12H22O11, 342.3 g; C, 12.01

1.19 g C12H22O11 ×

1 mol C H O

342.3 g C H O

= 3.476 × 10–3 mol 12 22 11

C H O

3.476 × 10–3 mol

C H O ×

12 mol C

1 mol C H O

= 0.04172 mol C

0.04172 mol C ×

12.01 g C

1 mol C

= 0.501 g C

39.

SOCl2(l) + H2O(l)  SO2(g) + 2HCl(g)

molar masses: SOCl2, 119.0 g; H2O, 18.02 g

35.0 g SOCl2 

1 mol SOCl

119.0 g SOCl

= 0.294 mol SOCl2

0.294 mol SOCl2 

1 mol H O

1 mol SOCl

= 0.294 mol H2O

0.294 mol H2O 

18.02 g H O

1 mol H O

= 5.30 g H2O

40.

The balanced equation is:

2C8H18 + 25O2  16CO2 + 18H2O

molar masses: C8H18, 114.22 g; CO2, 44.01 g

1 lb of CO2 = 453.59 g CO2

453.59 g CO2 ×

1 mol CO

44.01 g CO

= 10.31 mol CO2

From the balanced chemical equation, we can calculate the number of moles and number of
grams of pure octane that would be required to produce 10.31 mol CO2.

177

Chapter 9: Chemical Quantities

10.31 mol CO2 ×

2 mol C H

16 mol CO

= 1.288 mol C8H18

1.288 mol C8H18 ×

114.22 g C H

1 mol C H

= 147.2 g C8H18

From the density of C8H18 we can calculate the volume of 147.2 g C8H18.

147.2 g C8H18 ×

1 mL C H

0.75 g C H

= 196.3 mL C8H18 (2.0 × 102 mL to two significant figures)

From the preceding, we know that to travel 1 mile, we need approximately 200 mL of octane

1 mi

1000 mL 3.7854 L

196.3 mL

1 L

1 gal



= approximately 19 mi/gal

41.

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. Specific answer depends on student response.

42.

43.

(c); The limiting reactant must be determined by the starting masses of each reactant, molar
masses of each reactant, and the ratio in which they react in the balanced chemical equation.

44.

(a); When checking to see if mass is conserved in a reaction, add the masses of the two reactants
before the reaction takes place. This should be equal to the total mass of products after the
reaction plus whatever excess reactant is leftover.

45.

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

molar masses: Na2B4O7, 201.2 g; H2SO4, 98.09 g; H2O, 18.02 g

5.00 g Na2B4O7 

1 mol

201.2 g

= 0.0249 mol Na2B4O7

5.00 g H2SO4 

1 mol

98.09 g

= 0.0510 mol H2SO4

5.00 g H2O 

1 mol

18.02 g

= 0.277 mol H2O

Na2B4O7 is the limiting reactant.

mol H2SO4 remaining unreacted = 0.0510 – 0.0249 = 0.0261 mol

mol H2O remaining unreacted = 0.277 – 5(0.0249) = 0.153 mol

mass of H2SO4 remaining = 0.0261 mol 

98.09 g

1 mol

= 2.56 g H2SO4

mass of H2O remaining = 0.153 mol 

18.02 g

1 mol

= 2.76 g H2O

178

Chapter 9: Chemical Quantities

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

molar masses: CaC2, 64.10 g; H2O, 18.02 g

5.00 g CaC2 

1 mol

64.10 g

= 0.0780 mol CaC2

5.00 g H2O 

1 mol

18.02 g

= 0.277 mol H2O

CaC2 is the limiting reactant; water is present in excess.

mol of H2O remaining = 0.277 – 2(0.0780) = 0.121 mol H2O

mass of H2O remaining = 0.121 mol 

18.02 g

1 mol

= 2.18 g H2O

2NaCl(s) + H2SO4(l)  2HCl(g) + Na2SO4(s)

molar masses: NaCl, 58.44 g; H2SO4, 98.09 g

5.00 g NaCl 

1 mol

58.44 g

= 0.0856 mol NaCl

5.00 g H2SO4 

1 mol

98.09 g

= 0.0510 mol H2SO4

NaCl is the limiting reactant; H2SO4 is present in excess.

mol H2SO4 that reacts = 0.5(0.0856) = 0.0428 mol H2SO4

mol H2SO4 remaining = 0.0510 – 0.0428 = 0.0082 mol

mass of H2SO4 remaining = 0.0082 mol 

98.09 g

1 mol

= 0.80 g H2SO4

SiO2(s) + 2C(s)  Si(l) + 2CO(g)

molar masses: SiO2, 60.09 g; C, 12.01 g

5.00 g SiO2 

1 mol

60.09 g

= 0.0832 mol SiO2

5.00 g C 

1 mol

12.01 g

= 0.416 mol C

SiO2 is the limiting reactant; C is present in excess.

mol C remaining = 0.416 – 2(0.0832) = 0.250 mol

mass of C remaining = 0.250 mol 

12.01 g

1 mol

= 3.00 g C

179

Chapter 9: Chemical Quantities

46.

S(s) + 2H2SO4(aq)  3SO2(g) + 2H2O(l)

Molar masses: S, 32.07 g; H2SO4, 98.09 g; SO2, 64.07 g; H2O, 18.02 g

5.00 g S 

1 mol

32.07 g

= 0.1559 mol S

5.00 g H2SO4 

1 mol

98.09 g

= 0.05097 mol H2SO4

According to the balanced chemical equation, we would need twice as much sulfuric acid
as sulfur for complete reaction of both reactants. We clearly have much less sulfuric acid
present than sulfur: sulfuric acid is the limiting reactant. The calculation of the masses of
products produced is based on the number of moles of the sulfuric acid.

0.05097 mol H2SO4 

3 mol SO

64.07 g SO

2 mol H SO

1 mol SO



= 4.90 g SO2

0.05097 mol H2SO4 

2 mol H O

18.02 g H O

2 mol H SO

1 mol H O



= 0.918 g H2O

MnO2(s) + 2H2SO4(aq)  Mn(SO4)2 + 2H2O(l)

molar masses: MnO2, 86.94 g; H2SO4 98.09 g; Mn(SO4)2, 247.1 g; H2O, 18.02 g

5.00 g MnO2 

1 mol

86.94 g

= 0.05751 mol MnO2

5.00 g H2SO4 

1 mol

98.09 g

= 0.05097 mol H2SO4

According to the balanced chemical equation, we would need twice as much sulfuric acid
as manganese(IV) oxide for complete reaction of both reactants. We do not have this
much sulfuric acid, so sulfuric acid must be the limiting reactant. The amount of each
product produced will be based on the sulfuric acid reacting completely.

0.05097 mol H2SO4 

4 2

1 mol Mn(SO )

247.1 g Mn(SO )

2 mol H2SO4

1 mol Mn(SO )



= 6.30 g Mn(SO4)2

0.05097 mol H2SO4 

2 mol H O

18.02 g H O

2 mol H SO

1 mol H O



= 0.918 g H2O

2H2S(g) + 3O2(g)  2SO2(g) + 2H2O(l)

Molar masses: H2S, 34.09 g; O2, 32.00 g; SO2, 64.07 g; H2O, 18.02 g

5.00 g H2S 

1 mol

34.09 g

= 0.1467 mol H2S

5.00 g O2 

1 mol

32.00 g

= 0.1563 mol O2

180

Chapter 9: Chemical Quantities

According to the balanced equation, we would need 1.5 times as much O2 as H2S for
complete reaction of both reactants. We don’t have that much O2, so O2 must be the
limiting reactant that will control the masses of each product produced.

0.1563 mol O2 

2 mol SO

64.07 g SO

3 mol O

1 mol SO



= 6.67 g SO2

0.1563 mol O2 

2 mol H O 18.02 g H O

3 mol O

1 mol H O



= 1.88 g H2O

3AgNO3(aq) + Al(s)  3Ag(s) + Al(NO3)3(aq)

Molar masses: AgNO3, 169.9 g; Al, 26.98 g; Ag, 107.9 g; Al(NO3)3, 213.0 g

5.00 g AgNO3 

1 mol

169.9 g

= 0.02943 mol AgNO3

5.00 g Al 

1 mol

26.98 g

= 0.1853 mol Al

According to the balanced chemical equation, we would need three moles of AgNO3 for
every mole of Al for complete reaction of both reactants. We in fact have fewer moles of
AgNO3 than aluminum, so AgNO3 must be the limiting reactant. The amount of product
produced is calculated from the number of moles of the limiting reactant present:

0.02943 mol AgNO3 

3 mol Ag

107.9 g Ag

3 mol AgNO

1 mol Ag



= 3.18 g Ag

0.02943 mol AgNO3 

3 3

1 mol Al(NO )

213.0 g Al(NO )

3 mol AgNO

1 mol Al(NO )



= 2.09 g

47.

Before any calculations are attempted, the equations must be balanced.

C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)

molar masses: C3H8, 44.09 g; O2, 32.00 g; CO2, 44.01 g; H2O, 18.02 g

10.0 g C3H8 

1 mol

44.09 g

= 0.2268 mol C3H8

10.0 g O2 

1 mol

32.00 g

= 0.3125 mol O2

For 0.2268 mol C3H8, the amount of O2 that would be needed is

0.2268 mol C3H8 

5 mol O

1 mol C H

= 1.134 mol O2

Because we do not have this amount of O2, then O2 is the limiting reactant.

0.3125 mol O2 

3 mol CO

44.01 g CO

5 mol O

1 mol CO



= 8.25 g CO2

181

Chapter 9: Chemical Quantities

0.3125 mol O2 

4 mol H O 18.02 g H O

5 mol O

1 mol H O



= 4.51 g H2O

2Al(s) + 3Cl2(g)  2AlCl3(s)

molar masses: Al, 26.98 g; Cl2, 70.90 g; AlCl3, 133.3 g

10.0 g Al 

1 mol

26.98 g

= 0.3706 mol Al

10.0 g Cl2 

1 mol

70.90 g

= 0.1410 mol Cl2

For 0.1410 mol Cl2(g), the amount of Al(s) required is

0.1410 mol Cl2 

2 mol Al

3 mol Cl

= 0.09400 mol Al

We have far more than this amount of Al(s) present, so Cl2(g) must be the limiting
reactant that will control the amount of AlCl3 which forms.

0.1410 mol Cl2 

2 mol AlCl

133.3 g AlCl

3 mol Cl

1 mol AlCl



= 12.5 g AlCl3

2NaOH(s) + CO2(g)  Na2CO3(s) + H2O(l)

molar masses: NaOH, 40.00 g; CO2, 44.01 g; Na2CO3, 106.0 g; H2O, 18.02 g

10.0 g NaOH 

1 mol

40.00 g

= 0.2500 mol NaOH

10.0 g CO2 

1 mol

44.01 g

= 0.2272 mol CO2

Without having to calculate, according to the balanced chemical equation, we would need
twice as many moles of NaOH as CO2 for complete reaction. For the amounts calculated
above, there is not nearly enough NaOH present for the amount of Cl2 used: NaOH is the
limiting reactant.

0.2500 mol NaOH 

1 mol Na CO

106.0 g Na CO

2 mol NaOH

1 mol Na CO



= 13.3 g Na2CO3

0.2500 mol NaOH 

1 mol H O

18.02 g H O

2 mol NaOH

1 mol H O



= 2.25 g H2O

NaHCO3(s) + HCl(aq)  NaCl(aq) + H2O(l) + CO2(g)

molar masses: NaHCO3, 84.01 g; HCl, 36.46 g; NaCl, 58.44 g; H2O, 18.02 g; CO2, 44.01
g

10.0 g NaHCO3 

1 mol

84.01 g

= 0.1190 mol NaHCO3

10.0 g HCl 

1 mol

36.46 g

= 0.2742 mol HCl

182

Chapter 9: Chemical Quantities

Because the coefficients of NaHCO3(s) and HCl(aq) are both one in the balanced
chemical equation for the reaction, there is not enough NaHCO3 present to react with the
amount of HCl present: the 0.1190 mol NaHCO3 present is the limiting reactant. Because
all the coefficients of the products are also each one, then if 0.1190 mol NaHCO3 reacts
completely (with 0.1190 mol HCl), 0.1190 mol of each product will form.

0.1190 mol NaCl 

58.44 g

1 mol

= 6.95 g NaCl

0.1190 mol H2O 

18.02 g

1 mol

= 2.14 g H2O

0.1190 mol CO2 

44.01 g

1 mol

= 5.24 g CO2

48.

CS2(l) + 3O2(g)  CO2(g) + 2SO2(g)

Molar masses: CS2, 76.15 g; O2, 32.00 g; CO2, 44.01 g

1.00 g CS2 

1 mol

76.15 g

= 0.01313 mol CS2

1.00 g O2 

1 mol

32.00 g

= 0.03125 mol O2

From the balanced chemical equation, we would need three times as much oxygen as
carbon disulfide for complete reaction of both reactants. We do not have this much
oxygen, and so oxygen must be the limiting reactant.

0.03125 mol O2 

1 mol CO

44.01 g CO

3 mol O

1 mol CO



= 0.458 g CO2

2NH3(g) + CO2(g)  CN2H4O(s) + H2O(l)

Molar masses: NH3, 17.03 g; CO2, 44.01 g; H2O, 18.02 g

1.00 g NH3 

1 mol

17.03 g

= 0.05872 mol NH3

1.00 g CO2 

1 mol

44.01 g

= 0.02272 mol CO2

The balanced chemical equation tells us that we would need twice as many moles of
ammonia as carbon dioxide for complete reaction of both reactants. We have more than
this amount of ammonia present, so the reaction will be limited by the amount of carbon
dioxide present.

0.02272 mol CO2 

1 mol H O 18.02 g H O
1 mol CO

1 mol H O



= 0.409 g H2O

H2(g) + MnO2(s)  MnO(s) + H2O(l)

Molar masses: H2, 2.016 g; MnO2, 86.94 g; H2O, 18.02 g

183

Chapter 9: Chemical Quantities

1.00 g H2 

1 mol

2.016 g

= 0.496 mol H2

1.00 g MnO2 

1 mol

86.94 g

= 0.0115 mol MnO2

Because the coefficients of both reactants in the balanced chemical equation are the same,
we would need equal amounts of both reactants for complete reaction. Therefore,
manganese(IV) oxide must be the limiting reactant and controls the amount of product
obtained.

0.0115 mol MnO2 

1 mol H O

18.02 g H O

1 mol MnO

1 mol H O



= 0.207 g H2O

I2(s) + Cl2(g)  2ICl(g)

Molar masses: I2, 253.8 g; Cl2, 70.90 g; ICl, 162.35 g

1.00 g I2 

1 mol

253.8 g

= 0.00394 mol I2

1.00 g Cl2 

1 mol

70.90 g

= 0.0141 mol Cl2

From the balanced chemical equation, we would need equal amounts of I2 and Cl2 for
complete reaction of both reactants. As we have much less iodine than chlorine, iodine
must be the limiting reactant.

0.00394 mol I2 

2 mol ICl 162.35 g ICl

1 mol I

1 mol ICl



= 1.28 g ICl

49.

UO2(s) + 4HF(aq)  UF4(aq) + 2H2O(l)

UO2 is the limiting reactant; 1.16 g UF4, 0.133 g H2O

2NaNO3(aq) + H2SO4(aq)  Na2SO4(aq) + 2HNO3(aq)

NaNO3 is the limiting reactant; 0.836 g Na2SO4; 0.741 g HNO3

Zn(s) + 2HCl(aq)  ZnCl2(aq) + H2(g)

HCl is the limiting reactant; 1.87 g ZnCl2; 0.0276 g H2

B(OH)3(s) + 3CH3OH(l)  B(OCH3)3(s) + 3H2O(l)

CH3OH is the limiting reactant; 1.08 g B(OCH3)3; 0.562 g H2O

50.

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

HCl is the limiting reactant; 18.3 g AlCl3; 0.415 g H2

2NaOH(aq) + CO2(g)  Na2CO3(aq) + H2O(l)

NaOH is the limiting reactant; 19.9 g Na2CO3; 3.38 g H2O

Pb(NO3)2(aq) + 2HCl(aq)  PbCl2(s) + 2HNO3(aq)

Pb(NO3)2 is the limiting reactant; 12.6 g PbCl2; 5.71 g HNO3

184

Chapter 9: Chemical Quantities

2K(s) + I2(s)  2KI(s)

I2 is the limiting reactant; 19.6 g KI

51.

Pb(C2H3O2)2(aq) + H2O(l) + CO2(g)  PbCO3(s) + 2HC2H3O2(aq)

molar masses: Pb(C2H3O2)2, 325.3 g; CO2, 44.01 g; PbCO3, 267.21 g

1.25 g Pb(C2H3O2)2 ×

1 mol

325.3 g

= 0.00384 mol Pb(C2H3O2)2

5.95 g CO2 ×

1 mol

44.01 g

= 0.135 mol CO2

Pb(C2H3O2)2 is the limiting reactant, which determines the yield of product.

0.00384 mol Pb(C2H3O2)2 ×





1 mol PbCO

267.21 g PbCO

1 mol Pb C H O

1 mol PbCO



= 1.03 g PbCO3

52.

CuO(s) + H2SO4(aq)  CuSO4(aq) + H2O(l)

molar masses: CuO, 79.55 g; H2SO4, 98.09 g

2.49 g CuO 

1 mol CuO

79.55 g CuO

= 0.0313 mol CuO

5.05 g H2SO4 

1 mol H SO

98.09 g H SO

= 0.0515 mol H2SO4

Since the reaction is of 1:1 stoichiometry, CuO must be the limiting reactant since it is present in
the lesser amount on a molar basis.

53.

PbO(s) + C(s)  Pb(l) + CO(g)

molar masses: PbO, 223.2 g; C, 12.01 g; Pb, 207.2 g

50.0  103 g PbO 

1 mol

223.2 g

= 224.0 mol PbO

50.0  103 g C 

1 mol

12.01 g

= 4163 mol C

PbO is the limiting reactant.

224.0 mol PbO 

1 mol Pb

207.2 g Pb

1 mol PbO

1 mol Pb



= 4.64  104 g = 46.4 kg Pb

54.

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

185

Chapter 9: Chemical Quantities

0.0224 mol Fe 

3 mol O

4 mol Fe

= 0.0168 mol O2

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

0.0224 mol Fe 

2 mol Fe O

159.7 g Fe O

4 mol Fe

1 mol Fe O



= 1.79 g Fe2O3

55.

Ag+(aq) + Cl–(aq)  AgCl(s)

molar masses: AgNO3, 169.91 g; NaCl, 58.44 g

The number of moles of silver ion present will be the same as the number of moles of silver
nitrate taken because each formula unit of silver nitrate contains one silver ion.

1.15 g AgNO3 

1 mol AgNO

169.91 g AgNO

= 0.00677 mol AgNO3 = 0.00677 mol Ag+ ion

The number of moles of chloride ion present will be the same as the number of moles of sodium
chloride taken, because each formula unit of sodium chloride contains one sodium ion.

5.45 g NaCl 

1 mol NaCl

58.44 g NaCl

= 0.0933 mol NaCl = 0.0933 mol Cl– ion

Because the balanced chemical equation indicates a 1:1 stoichiometry for the reaction, there is not
nearly enough silver ion present (0.00677 mol) to precipitate the amount of chloride ion in the
sample (0.0933 mol).

56.

CaCl2(aq) + Na2SO4(aq)  CaSO4(s) + 2NaCl(aq)

molar masses: CaCl2, 110.98 g; Na2SO4, 142.05 g

5.21 g CaCl2 

1 mol CaCl

110.98 g CaCl

= 0.0469 mol CaCl2 = 0.0469 mol Ca2+ ion

4.95 g Na2SO4 

1 mol Na SO

142.05 g Na SO

= 0.0348 mol Na2SO4 = 0.0348 mol SO42– ion

Because the balanced chemical equation indicates a 1:1 stoichiometry for the reaction, there is not
nearly enough sulfate ion present (0.0348 mol) to precipitate the amount of calcium ion in the
sample (0.0469 mol). Sodium sulfate (sulfate ion) is the limiting reactant. Calcium chloride
(calcium ion) is present in excess.

57.

BaO2(s) + 2HCl(aq)  H2O2(aq) + BaCl2(aq)

molar masses: BaO2, 169.3 g; HCl, 36.46 g; H2O2, 34.02 g

1.50 g BaO2 

1 mol

169.3 g

= 8.860  10–3 mol BaO2

25.0 mL solution 

0.0272 g HCl

1 mL solution

= 0.680 g HCl

0.680 g HCl 

1 mol

36.46 g

= 1.865  10–2 mol HCl

186

Chapter 9: Chemical Quantities

BaO2 is the limiting reactant.

8.860  10–3 mol BaO2 

1 mol H O

34.02 g H O

1 mol BaO2

1 mol H O



= 0.301 g H2O2

58.

SiO2(s) + 3C(s)  2CO(g) + SiC(s)

molar masses: SiO2, 60.09 g; SiC, 40.10 g; 1.0 kg = 1.0  103 g

1.0  103 g SiO2 

1 mol

60.09 g

= 16.64 mol SiO2

From the balanced chemical equation, if 16.64 mol of SiO2 were to react completely (an excess of
carbon is present), then 16.64 mol of SiC should be produced (the coefficients of SiO2 and SiC
are the same).

16.64 mol SiC 

40.01 g

1 mol

= 6.7  102 g SiC = 0.67 kg SiC

59.

The theoretical yield represents the yield we calculate from the stoichiometry of the reaction and
the masses of reactants taken for the experiment. The actual yield is what is actually obtained in
an experiment. The percent yield is the ratio of what is actually obtained to the theoretical amount
that could be obtained, converted to a percent basis.

60.

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 (see
Chapter 16); loss of product may occur through operator error.

61.

Percent yield =

actual yield

theoretical yield

 100 =

1.23 g
1.44 g

 100 = 85.4%

62.

2NaN3(s)  2Na(s) + 3N2(g)

molar mass: NaN3, 65.02 g; Na, 22.99 g

10.5 g NaN3 ×

= 0.161 mol NaN3

0.161 mol NaN3 ×

= 0.161 mol Na

0.161 mol Na ×

= 3.71 g Na theoretical yield

% yield =

× 100 = 76.5%

63.

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

256.6 g

= 0.01267 mol S8

187

Chapter 9: Chemical Quantities

13.1 g Na2SO3 

1 mol

126.1 g

= 0.1039 mol Na2SO3

S8 is the limiting reactant.

0.01267 mol S8 

2 2

8 mol Na S O 5H O

1 mol S

= 0.1014 mol Na2S2O3 5H2O

0.1014 mol Na2S2O3 5H2O 

= 25.2 g Na2S2O3 5H2O

Percent yield =

actual yield

theoretical yield

 100 =

5.26 g
25.2 g

 100 = 20.9%

64.

2LiOH(s) + CO2(g)  Li2CO3(s) + H2O(g)

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

155 g LiOH ×

1 mol CO

44.01 g CO

1 mol LiOH

23.95 g LiOH 2 mol LiOH

1 mol CO



= 142 g CO2

As 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.

65.

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

131.3 g

= 0.9901 mol Xe

100. g F2 

1 mol

38.00 g

= 2.632 mol F2

Xe is the limiting reactant.

0.9901 mol Xe 

1 mol XeF

207.3 g XeF

1 mol Xe

1 mol XeF



= 205 g XeF4

Percent yield =

actual yield

theoretical yield

 100 =

145 g

205 g

 100 = 70.7 % of theory

66.

2NH3(g) + 3CuO(s)  N2(g) + 3Cu(s) + 3H2O(g)

molar masses: NH3, 17.034 g; CuO, 79.55 g; Cu, 63.55 g

18.1 g NH3 

= 1.06 mol NH3

188

Chapter 9: Chemical Quantities

90.4 g BaCl2 

= 1.14 mol CuO

CuO is the limiting reactant.

1.14 mol CuO 



= 72.4 g Cu

Percent yield =

actual yield

theoretical yield

 100 =

 100 = 62.6%

67.

Ca(HCO3)2(aq)  CaCO3(s) + CO2(g) + H2O(l)

millimolar masses: Ca(HCO3)2, 162.1 mg; CaCO3, 100.1 mg

2.0  10–3 mg Ca(HCO3)2 

1 mmol

162.1 mg

= 1.23  10–5 mmol Ca(HCO3)2

1.23  10–5 mmol Ca(HCO3)2 

3 2

1 mmol CaCO

1 mmol Ca(HCO )

= 1.23  10–5 mmol CaCO3

1.23  10–5 mmol 

100.1 mg

1 mmol

= 1.2  10–3 mg = 1.2  10–6 g CaCO3

68.

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

17.03 g

= 0.5872 mol NH3

15.0 g CO2 

1 mol

44.01 g

= 0.3408 mol CO2

CO2 is the limiting reactant.

0.3408 mol CO2 

1 mol NaHCO

1 mol CO

= 0.3408 mol NaHCO3

0.3408 mol NaHCO3 

84.01 g

1 mol

= 28.6 g NaHCO3

69.

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

55.85 g

= 0.0940 mol Fe

12.7 g S 

1 mol

32.07 g

= 0.396 mol S

Fe is the limiting reactant.

189

Chapter 9: Chemical Quantities

0.0940 mol Fe 

1 mol FeS 87.92 g FeS

1 mol Fe

1 mol FeS



= 8.26 g FeS produced

70.

C6H12O6(s) + 6O2(g)  6CO2(g) + 6H2O(g)

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

1.00 g glucose  = 5.549  10–3 mol glucose

5.549  10–3 mol glucose 

6 mol CO

1 mol glucose

= 3.33  10–2 mol CO2

3.33  10–2 mol CO2 

44.01 g

1 mol

= 1.47 g CO2

71.

Cu(s) + S(s)  CuS(s)

molar masses: Cu, 63.55 g; S, 32.07 g; CuS, 95.62 g

31.8 g Cu 

1 mol

63.55 g

= 0.5004 mol Cu

50.0 g S 

1 mol

32.07 g

= 1.559 mol S

Cu is the limiting reactant.

0.5004 mol Cu 

1 mol CuS

1 mol Cu

= 0.5004 mol CuS

0.5004 mol CuS 

95.62 g

1 mol

= 47.8 g CuS

% yield =

40.0 g
47.8 g

 100 = 83.7%

72.

Ba2+(aq) + SO42–(aq)  BaSO4(s)

millimolar ionic masses: Ba2+, 137.3 mg; SO42–, 96.07 mg; BaCl2, 208.2 mg

150 mg SO42– 

1 mmol

96.07 mg

= 1.56 millimol SO42–

As barium ion and sulfate ion react on a 1:1 stoichiometric basis, then 1.56 millimol of barium
ion is needed, which corresponds to 1.56 millimol of BaCl2

1.56 millimol BaCl2 

208.2 mg

1 mmol

= 325 milligrams BaCl2 needed

73.

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

190

Chapter 9: Chemical Quantities

0.1086 g Cl– 

1 mol

35.45 g

= 3.063  10–3 mol Cl–

3.063  10–3 mol Cl– 

1 mol AgNO

1 mol Cl

= 3.063  10–3 mol AgNO3

3.063  10–3 mol AgNO3 

169.9 g

1 mol

= 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

1 mol

= 0.439 g AgCl produced

74.

UO2(s) + 4HF(aq)  UF4(aq) + 2H2O(l)

One molecule (formula unit) of uranium(IV) oxide will combine with four molecules of
hydrofluoric acid, producing one uranium(IV) fluoride molecule and two water
molecules. One mole of uranium(IV) oxide will combine with four moles of hydrofluoric
acid to produce one mole of uranium(IV) fluoride and two moles of water.

2NaC2H3O2(aq) + H2SO4(aq)  Na2SO4(aq) + 2HC2H3O2(aq)

Two molecules (formula units) of sodium acetate react exactly with one molecule of
sulfuric acid, producing one molecule (formula unit) of sodium sulfate and two molecules
of acetic acid. Two moles of sodium acetate will combine with one mole of sulfuric acid,
producing one mole of sodium sulfate and two moles of acetic acid.

Mg(s) + 2HCl(aq)  MgCl2(aq) + H2(g)

One magnesium atom will react with two hydrochloric acid molecules (formula units) to
produce one molecule (formula unit) of magnesium chloride and one molecule of
hydrogen gas. One mole of magnesium will combine with two moles of hydrochloric
acid, producing one mole of magnesium chloride and one mole of gaseous hydrogen.

B2O3(s) + 3H2O(l)  2B(OH)3(aq)

One molecule of diboron trioxide will react exactly with three molecules of water,
producing two molecules of boron trihydroxide (boric acid). One mole of diboron
trioxide will combine with three moles of water to produce two moles of boron
trihydroxide (boric acid).

75.

False. For 0.40 mol of Mg(OH)2 to react, 0.80 mol of HCl will be needed. According to the
balanced equation, for a given amount of Mg(OH)2, twice as many moles of HCl is needed.

76.

For O2:

5 mol O

1 mol C H













For CO2:

3 mol CO

1 mol C H













For H2O:

4 mol H O

1 mol C H













77.

2H2O2(l)  2H2O(l) + O2(g)

0.50 mol H2O2 

2 mol H O

= 0.50 mol H2O

191

Chapter 9: Chemical Quantities

0.50 mol H2O2 

1 mol O

2 mol H O

= 0.25 mol O2

2KClO3(s)  2KCl(s) + 3O2(g)

0.50 mol KClO3 

2 mol KCl

2 mol KClO

= 0.50 mol KCl

0.50 mol KClO3 

3 mol O

2 mol KClO

= 0.75 mol O2

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

0.50 mol Al 

2 mol AlCl

2 mol Al

= 0.50 mol AlCl3

0.50 mol Al 

3 mol H

2 mol Al

= 0.75 mol H2

C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l)

0.50 mol C3H8 

3 mol CO

1 mol C H

= 1.5 mol CO2

0.50 mol C3H8 

4 mol H O

1 mol C H

= 2.0 mol H2O

78.

NH3(g) + HCl(g)  NH4Cl(s)

molar mass of NH3 = 17.01 g

1.00 g NH3 

1 mol

17.01 g

= 0.0588 mol NH3

0.0588 mol NH3 

1 mol NH Cl

1 mol NH

= 0.0588 mol NH4Cl

CaO(s) + CO2(g)  CaCO3(s)

molar mass CaO = 56.08 g

1.00 g CaO 

1 mol

56.08 g

= 0.0178 mol CaO

0.0178 mol CaO 

1 mol CaCO

1 mol CaO

= 0.0178 mol CaCO3

4Na(s) + O2(g)  2Na2O(s)

molar mass Na = 22.99 g

1.00 g Na 

1 mol

22.99 g

= 0.0435 mol Na

192

Chapter 9: Chemical Quantities

0.0435 mol Na 

2 mol Na O

4 mol Na

= 0.0217 mol Na2O

2P(s) + 3Cl2(g)  2PCl3(l)

molar mass P = 30.97 g

1.00 g P 

1 mol

30.97 g

= 0.0323 mol P

0.0323 mol P 

2 mol PCl

2 mol P

= 0.0323 mol PCl3

79.

molar mass CuSO4 = 159.6 g

4.21 g CuSO4 

1 mol

159.6 g

= 0.0264 mol CuSO4

molar mass Ba(NO3)2 = 261.3 g

7.94 g Ba(NO3)2 

1 mol

261.3 g

= 0.0304 mol Ba(NO3)2

molar mass water = 18.02 g; 1.24 mg = 0.00124 g

0.00124 g 

1 mol

18.02 g

= 6.88  10–5 mol H2O

molar mass W = 183.9 g

9.79 g W 

1 mol

183.9 g

= 5.32  10–2 mol W

molar mass S = 32.07 g; 1.45 lb = 1.45(454) = 658 g

658 g S 

1 mol

32.07 g

= 20.5 mol S

molar mass C2H5OH = 46.07 g

4.65 g C2H5OH 

1 mol

46.07 g

= 0.101 mol C2H5OH

molar mass C = 12.01 g

12.01 g C 

1 mol

12.01 g

= 1.00 mol C

80.

molar mass HNO3 = 63.0 g

5.0 mol HNO3 

63.0 g

1 mol

= 3.2  102 g HNO3

molar mass Hg = 200.6 g

193

Chapter 9: Chemical Quantities

0.000305 mol Hg 

200.6 g

1 mol

= 0.0612 g Hg

molar mass K2CrO4 = 194.2 g

2.31  10–5 mol K2CrO4 

194.2 g

1 mol

= 4.49  10–3 g K2CrO4

molar mass AlCl3 = 133.3 g

10.5 mol AlCl3 

133.3 g

1 mol

= 1.40  103 g AlCl3

molar mass SF6 = 146.1 g

4.9  104 mol SF6 

146.1 g

1 mol

= 7.2  106 g SF6

molar mass NH3 = 17.01 g

125 mol NH3 

17.01 g

1 mol

= 2.13  103 g NH3

molar mass Na2O2 = 77.98 g

0.01205 mol Na2O2 

77.98 g

1 mol

= 0.9397 g Na2O2

81.

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

BaCl2(aq) + H2SO4(aq)  BaSO4(s) + 2HCl(aq)

0.145 mol BaCl2 

1 mol H SO

1 mol BaCl

= 0.145 mol H2SO4

AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq)

0.145 mol AgNO3 

1 mol NaCl

1 mol AgNO

= 0.145 mol NaCl

Pb(NO3)2(aq) + Na2CO3(aq)  PbCO3(s) + 2NaNO3(aq)

0.145 mol Pb(NO3)2 

3 2

1 mol Na CO

1 mol Pb(NO )

= 0.145 mol Na2CO3

C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)

0.145 mol C3H8 

5 mol O

1 mol C H

= 0.725 mol O2

82.

2SO2(g) + O2(g)  2SO3(g)

molar masses: SO2, 64.07 g; SO3, 80.07 g; 150 kg = 1.5  105 g

1.5  105 g SO2 

1 mol

64.07 g

= 2.34  103 mol SO2

194

Chapter 9: Chemical Quantities

2.34  103 mol SO2 

2 mol SO
2 mol SO

= 2.34  103 mol SO3

2.34  103 mol SO3 

80.07 g

1 mol

= 1.9  105 g SO3 = 1.9  102 kg SO3

83.

2ZnS(s) + 3O2(g)  2ZnO(s) + 2SO2(g)

molar masses: ZnS, 97.45 g; SO2, 64.07 g; 1.0  102 kg = 1.0  105 g

1.0  105 g ZnS 

1 mol

97.45 g

= 1.026  103 mol ZnS

1.026  103 mol ZnS 

2 mol SO

2 mol ZnS

= 1.026  103 mol SO2

1.026  103 mol SO2 

64.07 g

1 mol

= 6.6  104 g SO2 = 66 kg SO2

84.

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

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

3.25 g Na2O2 

1 mol

77.98 g

= 0.0417 mol Na2O2

0.0417 mol Na2O2 

1 mol O

2 mol Na O

= 0.0209 mol O2

0.0209 mol O2 

32.00 g

1 mol

= 0.667 g O2

85.

Cu(s) + 2AgNO3(aq)  Cu(NO3)2(aq) + 2Ag(s)

millimolar masses: Cu, 63.55 mg; AgNO3, 169.9 mg

1.95 mg AgNO3 

1 mmol

169.9 mg

= 0.01148 mmol AgNO3

0.01148 mmol AgNO3 

1 mmol Cu

2 mmol AgNO

= 0.005740 mmol Cu

0.005740 mmol Cu 

63.55 g

1 mol

= 0.365 mg Cu

86.

Zn(s) + 2HCl(aq)  ZnCl2(aq) + H2(g)

molar masses: Zn, 65.38 g; H2, 2.016 g

2.50 g Zn 

1 mol

65.38 g

= 0.03824 mol Zn

0.03824 mol Zn 

1 mol H

1 mol Zn

= 0.03824 mol H2

195

Chapter 9: Chemical Quantities

0.03824 mol H2 

2.016 g

1 mol

= 0.0771 g H2

87.

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

26.04 g

= 5.760 mol C2H2

5.760 mol C2H2 

5 mol O

2 mol C H

= 14.40 mol O2

14.40 mol O2 

32.00 g

1 mol

= 4.6  102 g O2

88.

2Na(s) + Br2(l)  2NaBr(s)

molar masses: Na, 22.99 g; Br2, 159.8 g; NaBr, 102.9 g

5.0 g Na 

1 mol

22.99 g

= 0.2175 mol Na

5.0 g Br2 

1 mol

159.8 g

= 0.03129 mol Br2

Intuitively, we would suspect that Br2 is the limiting reactant, because there is much less
Br2 than Na on a mole basis. To prove that Br2 is the limiting reactant, the following
calculation is needed:

0.03129 mol Br2 

2 mol Na

1 mol Br

= 0.06258 mol Na.

Clearly, there is more Na than this present, so Br2 limits the reaction extent and the
amount of NaBr formed.

0.03129 mol Br2 

2 mol NaBr

1 mol Br

= 0.06258 mol NaBr

0.06258 mol NaBr 

102.9 g

1 mol

= 6.4 g NaBr

Zn(s) + CuSO4(aq)  ZnSO4(aq) + Cu(s)

molar masses: Zn, 65.38 g; Cu, 63.55 g; ZnSO4, 161.5 g; CuSO4, 159.6 g

5.0 g Zn 

1 mol

65.38 g

= 0.07648 mol Zn

5.0 g CuSO4 

1 mol

159.6 g

= 0.03132 mol CuSO4

As the coefficients of Zn and CuSO4 are the same in the balanced chemical equation, an
equal number of moles of Zn and CuSO4 would be needed for complete reaction. There is
less CuSO4 present, so CuSO4 must be the limiting reactant.

196

Chapter 9: Chemical Quantities

0.03132 mol CuSO4 

1 mol ZnSO

1 mol CuSO

= 0.03132 mol ZnSO4

0.03132 mol ZnSO4 

161.5 g

1 mol

= 5.1 g ZnSO4

0.03132 mol CuSO4 

1 mol Cu

1 mol CuSO

= 0.03132 mol Cu

0.03132 mol Cu 

63.55 g

1 mol

= 2.0 g Cu

NH4Cl(aq) + NaOH(aq)  NH3(g) + H2O(l) + NaCl(aq)

molar masses: NH4Cl, 53.49 g; NaOH, 40.00 g; NH3, 17.03 g; H2O, 18.02 g;
NaCl, 58.44 g

5.0 g NH4Cl 

1 mol

53.49 g

= 0.09348 mol NH4Cl

5.0 g NaOH 

1 mol

40.00 g

= 0.1250 mol NaOH

As the coefficients of NH4Cl and NaOH are both one in the balanced chemical equation
for the reaction, an equal number of moles of NH4Cl and NaOH would be needed for
complete reaction. There is less NH4Cl present, so NH4Cl must be the limiting reactant.

As the coefficients of the products in the balanced chemical equation are also all one, if
0.09348 mol of NH4Cl (the limiting reactant) reacts completely, then 0.09348 mol of
each product will be formed.

0.09348 mol NH3 

17.03 g

1 mol

= 1.6 g NH3

0.09348 mol H2O 

18.02 g

1 mol

= 1.7 g H2O

0.09348 mol NaCl 

58.44 g

1 mol

= 5.5 g NaCl

Fe2O3(s) + 3CO(g)  2Fe(s) + 3CO2(g)

molar masses: Fe2O3, 159.7 g; CO, 28.01 g; Fe, 55.85 g; CO2, 44.01 g

5.0 g Fe2O3 

1 mol

159.7 g

= 0.03131 mol Fe2O3

5.0 g CO 

1 mol

28.01 g

= 0.1785 mol CO

Because there is considerably less Fe2O3 than CO on a mole basis, let’s see if Fe2O3 is the
limiting reactant.

0.03131 mol Fe2O3 

3 mol CO

1 mol Fe O

= 0.09393 mol CO

197

Chapter 9: Chemical Quantities

As there is 0.1785 mol of CO present, but we have determined that only 0.09393 mol CO
would be needed to react with all the Fe2O3 present, then Fe2O3 must be the limiting
reactant. CO is present in excess.

0.03131 mol Fe2O3 

2 mol Fe

55.85 g Fe

1 mol Fe O

1 mol Fe



= 3.5 g Fe

0.03131 mol Fe2O3 

3 mol CO

44.01 g CO

1 mol Fe O

1 mol CO



= 4.1 g CO2

89.

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 C2H5OH 

1 mol

46.07 g

= 0.5427 mol C2H5OH

25.0 g O2 

1 mol

32.00 g

= 0.7813 mol O2

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

0.5427 mol C2H5OH 

3 mol O

1 mol C H OH

= 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) because 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 CO

44.01 g CO

3 mol O

1 mol CO



= 22.9 g CO2

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

28.02 g

= 0.8922 mol N2

25.0 g O2 

1 mol

32.00 g

= 0.7813 mol O2

As 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. There is less O2 present on a mole basis, so O2 must be the limiting reactant.

0.7813 mol O2 

2 mol NO 30.01 g NO

1 mol O2

1 mol NO



= 46.9 g NO

2NaClO2(aq) + Cl2(g)  2ClO2(g) + 2NaCl(aq)

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

198

Chapter 9: Chemical Quantities

25.0 g NaClO2 

1 mol

90.44 g

= 0.2764 mol NaClO2

25.0 g Cl2 

1 mol

70.90 g

= 0.3526 mol Cl2

See if NaClO2 is the limiting reactant.

0.2764 mol NaClO2 

1 mol Cl

2 mol NaClO

= 0.1382 mol Cl2

As 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

58.44 g NaCl

2 mol NaClO

1 mol NaCl



= 16.2 g NaCl

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

2.016 g

= 12.40 mol H2

25.0 g N2 

1 mol

28.02 g

= 0.8922 mol N2

See if N2 is the limiting reactant.

0.8922 mol N2 

3mol H

1 mol N

= 2.677 mol H2

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

0.8922 mol N2 

2 mol NH

17.03 g NH

1 mol N

1 mol NH



= 30.4 g NH3

90.

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

32.05 g

= 0.624 mol N2H4

20.0 g O2 

1 mol

32.00 g

= 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 N

28.02 g N

1 mol N H

1 mol N



= 17.5 g N2

199

Chapter 9: Chemical Quantities

0.624 mol N2H4 

2 mol H O

18.02 g H O

1 mol N H

1 mol H O



= 22.5 g H2O

91.

92.

12.5 g theory 

40 g actual

100 g theory

= 5.0 g

93.

C5H12(l) + 8O2(g)  5CO2(g) + 6H2O(l)

molar masses: C5H12, 72.146 g; H2O, 18.016 g

20.4 g C5H12 

= 0.283 mol C5H12

0.283 mol C5H12 

= 1.70 mol H2O

1.70 mol H2O 

= 30.6 g H2O

94.

The balanced equation is: 2NaNO3  2NaNO2 + O2

0.2339 g NaNO2 

= 0.003390 mol NaNO2

0.003390 mol NaNO2 

= 0.003390 mol NaNO3

0.003390 mol NaNO3 

= 0.28815 g NaNO3

% NaNO3 =

 100 = 68.12%

200

Chapter 9: Chemical Quantities

95.

The balanced equation is: 2LiOH(s) + CO2(g)  Li2CO3(s) + H2O(l)

67.4 g LiOH 

= 2.81 mol LiOH

2.81 mol LiOH 

= 1.41 mol Li2CO3

1.41 mol Li2CO3 

= 104 g Li2CO3

96.

The balanced equation is: Fe2O3(s) + 2Al(s)  2Fe(l) + Al2O3(s)

25.69 g Fe 

= 0.4600 mol Fe

0.4600 mol Fe 

= 0.2300 mol Fe2O3

0.2300 mol Fe2O3 

= 36.73 g Fe2O3

0.4600 mol Fe 

= 0.4600 mol Al

0.4600 mol Al 

= 12.41 g Al

0.4600 mol Fe 

= 0.2300 mol Al2O3

0.2300 mol Al2O3 

= 23.45 g Al2O3

97.

2H2S(g) + 3O2(g)  2SO2(g) + 2H2O(g)

O2 is the limiting reactant (only 2.0 mol H2S needed).

3.0 mol O2 

= 2.0 mol SO2

98.

The balanced equation is: 2NH3(g) + 2Na(s)  2NaNH2(s) + H2(g)

32.8 g NH3 

= 1.93 mol NH3

16.6 g Na 

= 0.722 mol Na

Na is the limiting reactant (only 0.722 mol NH3 needed).

201

Chapter 9: Chemical Quantities

0.722 mol Na 



= 28.2 g NaNH2

0.722 mol Na 



= 0.728 g H2

99.

The balanced equation is: SO2(g) + 2NaOH(s)  Na2SO3(s) + H2O(l)

38.3 g SO2 

= 0.598 mol SO2

32.8 g NaOH 

= 0.820 mol NaOH

NaOH is the limiting reactant (only 0.410 mol SO2 needed).

0.820 mol NaOH 



= 51.7 g Na2SO3

0.820 mol NaOH 



= 7.39 g H2O

100.

2C3H6(g) + 2NH3(g) + 3O2(g)  2C3H3N(g) + 6H2O(g)

5.23  102 g C3H6 

= 12.4 mol C3H6

12.4 mol C3H6 



= 658 g C3H3N

5.00  102 g NH3 

= 29.4 mol NH3

29.4 mol NH3 



= 1560 g C3H3N

1.00  103 g O2 

= 31.25 mol O2

31.25 mol O2 



= 1110 g C3H3N

Thus, C3H6 is the limiting reactant and 658 g C3H3N is produced (660. g C3H3N if
decimals are carried through).

12.4 mol C3H6 



= 670. g H2O (672 g H2O if decimals

are carried through)

C3H6: 0 g (limiting reactant and is used up)

202

Chapter 9: Chemical Quantities

NH3:

12.4 mol C3H6 

= 12.4 mol NH3 used up

29.4 mol NH3 – 12.4 mol NH3 = 17.0 mol NH3 leftover

17.0 mol NH3 

= 290. g NH3 (289 g NH3 if decimals are carried

through)

O2:

12.4 mol C3H6 

= 18.6 mol O2 used up

31.25 mol O2 – 18.6 mol O2 = 12.65 mol O2 leftover

12.65 mol O2 

= 405 g O2

203