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CHAPTER
9
Chemical Equation
Calculations
Section 9.1 Interpreting a Chemical Equation
2.
General Equation:
(a)
2 mol A
(b)
1LD
×
×
2A + 3B → 2C + D
2 mol C
2 mol A
3 L B
1 L D
=
2 mol C
= 3LB
4.
General Equation: 3 A + B → 2 C
(a) conservation of mass law: 1.50 g A + 1.65 g B = 3.15 g C
(b) conservation of mass law: 9.45 g C – 4.50 g A = 4.95 g B
6.
(a)
MM of P4
MM of O2
MM of P2 O3
123.88 g P4
123.88 g P4
(b)
MM of P4
MM of O2
MM of P2 O5
123.88 g P4
123.88 g P4
P4 (s) + 3 O2 (g) →
2 P2 O3 (s)
= 4(30.97 g P)
= 2(16.00 g O)
= 2(30.97 g P) + 3(16.00 g O)
+ 3(32.00 g O2 )
+ 96.00 g O2
219.88 g
→
→
→
2(109.94 g P2 O3 )
219.88 g P2 O3
219.88 g
P4 (s) + 5 O2 (g) →
2 P2 O5 (s)
= 4(30.97 g P)
= 2(16.00 g O)
= 2(30.97 g P) + 5(16.00 g O)
+ 5(32.00 g O2 )
+ 160.00 g O2
283.88 g
→
→
→
= 123.88 g/mol
= 32.00 g/mol
= 109.94 g/mol
= 123.88 g/mol
= 32.00 g/mol
= 141.94 g/mol
2(141.94 g P2 O3 )
283.88 g P2 O3
283.88 g
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Chemical Equation Calculations
57
Section 9.2 Mole–Mole Problems
8.
H2 (g) + Cl2 (g)
→
2 HCl(g)
Moles of H2 that react:
×
5.00 mol HCl
10.
1 mol H2
2 mol HCl
=
2.50 mol H2
3 Ba(s) + N2 (g) → Ba3 N2 (s)
Moles of Ba(s) that react:
3 mol Ba
0.100 mol Ba3 N2 × 1 mol Ba N = 0.300 mol Ba
3 2
Moles of N2 (g) that react:
1 mol N2
0.100 mol Ba3 N2 × 1 mol Ba N = 0.100 mol N2
3 2
12.
2 C4 H1 0(g) + 13 O2 (g) → 8 CO2 (g) + 10 H2 O(g)
Moles of C4 H1 0(g) that react:
2 mol C4 H1 0
2.50 mol H2 O × 10 mol H O = 0.500 mol C4 H1 0
2
Moles of O2 (g) that react:
13 mol O2
2.50 mol H2 O × 10 mol H O = 3.25 mol O2
2
Section 9.3 Types of Stoichiometry Problems
14.
Given
volume
Unknown
mass
Type of Stoichiometry
mass–volume problem
16.
Given
volume
Unknown
volume
Type of Stoichiometry
volume–volume problem
18.
Given
mass
Unknown
mass
Type of Stoichiometry
mass–mass problem
58
Chapter 9
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Section 9.4 Mass–Mass Problems
20.
Cu(s) + 2 AgNO3 (aq) → Cu(NO3 )2 (aq) + 2 Ag(s)
MM of AgNO3 = 169.88 g/mol
MM of Ag = 107.87 g/mol
1 mol Ag
1.00 g Ag × 107.87 g Ag
×
2 mol AgNO3
2 mol Ag
×
169.88 g AgNO3
1 mol AgNO3
= 1.57 g AgNO3
22.
2 Zn(s) + O2 (g) → 2 ZnO(s)
MM of O2 = 32.00 g/mol
MM of ZnO = 81.39 g/mol
1 mol O2
32.00 g O2
1 mol ZnO
1.00 g of ZnO g ZnO × 81.39 g ZnO × 2 mol ZnO × 1 mol O = 0.197 g O2
2
24.
2 Bi(s) + 3 Cl2 (g) → 2 BiCl3 (s)
MM of Cl2 = 70.90 g/mol
MM of BiCl3 = 315.33 g/mol
1 mol BiCl3
3 mol Cl2
70.90 g Cl2
3.52 g BiCl3 × 315.33 g BiCl × 2 mol BiCl × 1 mol Cl
= 1.19 g Cl2
3
3
2
26.
2 Co(s) + 3 HgCl2 (aq) → 2 CoCl3 (aq) + 3 Hg(l)
MM of HgCl2 = 271.49 g/mol
MM of Hg = 200.59 g/mol
3 mol HgCl2
271.49 g HgCl2
1 mol Hg
5.11 g Hg × 200.59 g Hg × 3 mol Hg
× 1 mol HgCl
= 6.92 g HgCl2
2
28.
2 Na3 PO4 (aq) + 3 Ca(OH)2 (aq) → Ca3 (PO4 )2 (s) + 6 NaOH(aq)
MM of Ca(OH)2 = 74.10 g/mol
MM of Ca3 (PO4 )2 = 310.18 g/mol
1 mol Ca3 (PO4 )2
3 mol Ca(OH)2
74.10 g Ca(OH)2
2.39 g Ca3 (PO4 )2 × 310.18 g Ca (PO ) × 1 mol Ca (PO ) × 1 mol Ca(OH)
3
4 2
3
4 2
2
= 1.71 g Ca(OH)2
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Chemical Equation Calculations
59
Section 9.5 Mass–Volume Problems
30.
Fe2 (CO3 )3 (s) → Fe2 O3 (s) + 3 CO2 (g)
MM of Fe2 (CO3 )3 = 291.73 g/mol
1 mol Fe2 (CO3 )3
3 mol CO2
22.4 L CO2
5.00 g Fe2 (CO3 )3 × 291.73 g Fe (CO ) × 1 mol Fe (CO ) × 1 mol CO
2
3 3
2
3 3
2
×
32.
1000 mL CO2
1 L CO2
= 1150 mL CO2
2 HgO(s) → 2 Hg(l) + O2 (g)
MM of HgO = 216.59 g/mol
1 mol HgO
2.50 g HgO × 216.59 g HgO
×
1 mol O2
22.4 L O2
2 mol HgO × 1 mol O2
×
1000 mL O2
1 L O2
= 129 mL O2
34.
MnCl2 (s) + H2 SO4 (aq) → MnSO4 (aq) + 2 HCl(g)
MM of MnCl2 = 125.84 g/mol
1 mol MnCl2
125.84 g MnCl2
1 L HCl
1 mol HCl
49.5 mL HCl × 1000 mL HCl × 22.4 L HCl × 2 mol HCl × 1 mol MnCl
2
= 0.139 g MnCl2
36.
2 H2 O2 (l) → 2 H2 O(l) + O2 (g)
MM of H2 O2 = 34.02 g/mol
1 L O2
1 mol O2
55.0 mL O2 × 1000 mL O × 22.4 L O
2
2
×
2 mol H2 O2
34.02 g H2 O2
1 mol O2 × 1 mol H2 O2
= 0.167 g H2 O2
60
Chapter 9
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Section 9.6 Volume–Volume Problems
38.
H2 (g) + I2 (g) → 2 HI(g)
2 mL HI
125 mL H2 × 1 mL H = 2.50 × 102 mL HI
2
40.
3 H2 (g) + N2 (g) → 2 NH3 (g)
3 mL H2
45.0 mL NH3 × 2 mL NH = 67.5 mL H2
3
42.
2 CO(g) + O2 (g) → 2 CO2 (g)
10.0 L CO ×
44.
2 L CO2
2 L CO = 10.0 L CO2
2 Cl2 (g) + 3 O2 (g) → 2 Cl2 O3 (g)
3 L O2
1.75 L Cl2 O3 × 2 L Cl O
2 3
2.63 L O2
46.
1000 mL O2
1 L O2
= 2630 mL O2
2 SO2 (g) + O2 (g) → 2 SO3 (g)
25.0 L O2
48.
×
= 2.63 L O2
×
2 L SO2
1 L O2 = 50.0 L SO2
2 N2 (g) + 5 O2 (g) → 2 N2 O5 (g)
500.0 cm3 N2 O5 ×
5 cm3 O2
3
3
2 cm3 N2 O5 = 1.250 × 10 cm O2
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Chemical Equation Calculations
61
Section 9.7 Limiting Reactant Concept
50.
2 NO(g) + O2 (g) → 2 NO2 (g)
1.00 mol NO
2.00 mol O2
×
×
2 mol NO2
2 mol NO
= 1.00 mol NO2
2 mol NO2
1 mol O2
= 4.00 mol NO2
The limiting reactant is NO because it produces less product. This reaction
produces 1.00 mol NO2 .
52.
N2 (g) + O2 (g) → 2 NO(g)
×
0.125 mol N2
0.125 mol O2
×
2 mol NO
1 mol N2
= 0.250 mol NO
2 mol NO
1 mol O2
= 0.250 mol NO
Neither N2 or O2 is a limiting reactant because each produces the same amount
of product. This reaction produces 0.250 mol NO.
54.
2 H2 (g) + O2 (g) → 2 H2 O(l)
5.00 mol H2
×
2 mol H2 O
2 mol H2
= 5.00 mol H2 O
1.50 mol O2
×
2 mol H2 O
1 mol O2
= 3.00 mol H2 O
The limiting reactant is O2 because it produces less water. This reaction produces
3.00 mol H2 O.
62
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56.
2 C2 H6 (g) + 7 O2 (g) → 4 CO2 (g) + 6 H2 O(g)
6 mol H2 O
2 mol C2 H6
= 3.00 mol H2 O
6 mol H2 O
7 mol O2
= 2.57 mol H2 O
1.00 mol C2 H6 ×
3.00 mol O2
×
The limiting reactant is O2 because it produces less water. This reaction produces
2.57 mol H2 O.
58.
2 Co(s) + 3 S(s) →
(a)
Co2 S3 (s)
1.00 mol Co
×
1.00 mol S
×
1 mol Co2 S3
2 mol Co
1 mol Co2 S3
3 mol S
= 0.500 mol Co2 S3
= 0.333 mol Co2 S3
The limiting reactant is S because it produces less product. This reaction produces
0.333 mol Co2 S3 .
After reaction:
mol of Co
= 1.00 – 2(0.333)
mol of S
= 1.00 – 3(0.333)
mol of Co2 S3 = 0.00 + 0.333
(b)
= 0.333 mol
= 0.000 mol
= 0.333 mol
2.00 mol Co
×
1 mol Co2 S3
2 mol Co
= 1.00 mol Co2 S3
3.00 mol S
×
1 mol Co2 S3
3 mol S
= 1.00 mol Co2 S3
Neither Co or S is a limiting reactant because each produces the same amount of
product. This reaction produces 1.00 mol CoS.
After reaction:
mol of Co
=
mol of S
=
mol of Co2 S3 =
2.00 – 2.00
3.00 – 3.00
0.00 + 1.00
= 0.00 mol
= 0.00 mol
= 1.00 mol
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Chemical Equation Calculations
63
Section 9.8 Limiting Reactant Problems
60.
FeO(l) + Mg(l) → Fe(l) + MgO(s)
1 mol FeO
10.0 g FeO × 71.85 g FeO
×
1 mol Mg
40.0 g Mg × 24.31 g Mg
×
1 mol Fe
55.85 g Fe
×
1 mol FeO
1 mol Fe
1 mol Fe
1 mol Mg
= 7.77 g Fe
55.85 g Fe
× 1 mol Fe
= 91.9 g Fe
Therefore, FeO is the limiting reactant and will be consumed before Mg. This
reaction produces 7.77 g Fe.
62.
Fe2 O3 (l) + 2 Al(l) → 2 Fe(l) + Al2 O3 (s)
1 mol Fe2 O3
37.5 g Fe2 O3 × 159.70 g Fe O
2 3
175 g Al
1 mol Al
× 26.98 g Al
×
×
2 mol Fe
1 mol Fe2 O3
2 mol Fe
2 mol Al
×
×
55.85 g Fe
1 mol Fe
55.85 g Fe
1 mol Fe
= 26.2 g Fe
= 362 g Fe
Therefore, Fe2 O3 is the limiting reactant because it will be consumed before Al.
This reaction produces 26.2 g Fe.
64.
Mg(OH)2 (s) + H2 SO4 (l) → MgSO4 (s) + 2 H2 O(l)
1 mol Mg(OH)2
0.605 g Mg(OH)2 × 58.33 g Mg(OH)
2
1 mol MgSO4
× 1 mol Mg(OH)
2
×
120.38 g MgSO4
1 mol MgSO4
= 1.25 g MgSO4
1.00 g H2 SO4
1 mol H2 SO4
× 98.09 g H SO
2
4
×
1 mol MgSO4
1 mol H2 SO4
×
120.38 g MgSO4
1 mol MgSO4
= 1.23 g MgSO4
Therefore, H2 SO4 is the limiting reactant because it will be consumed before
Mg(OH)2 . This reaction produces 1.23 g MgSO4 .
64
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66.
2 Al(OH)3 (s) + 3 H2 SO4 (l) → Al2 (SO4 )3 (aq) + 6 H2 O(l)
×
3.00 g Al(OH)3
1 mol Al(OH)3
78.01 g Al(OH)3
×
6 mol H2 O
2 mol Al(OH)3
×
18.02 g H2 O
1 mol H2 O
×
18.02 g H2 O
1 mol H2 O
= 2.08 g H2 O
×
1.00 g H2 SO4
1 mol H2 SO4
98.09 g H2 SO4
×
6 mol H2 O
3 mol H2 SO4
= 0.367 g H2 O
Therefore, H2 SO4 is the limiting reactant because it will be consumed before
Al(OH)3 . This reaction produces 0.367 g H2 O.
68.
N2 (g) + 2 O2 (g) → 2 NO2 (g)
95.0 mL N2
×
2 mL NO2
1 mL N2
= 190 mL NO2
45.0 mL O2
×
2 mL NO2
2 mL O2
= 45.0 mL NO2
(1.90 × 102 mL NO2 )
Therefore, O2 is the limiting reactant because it will be consumed before N2 . This
reaction produces 45.0 mL NO2 .
70.
2 N2 (g) + 3 O2 (g) → 2 N2 O3 (g)
45.0 mL N2
×
70.0 mL O2
×
2 mL N2 O3
2 mL N2
2 mL N2 O3
3 mL O2
= 45.0 mL N2 O3
= 46.7 mL N2 O3
Therefore, N2 is the limiting reactant because it will be consumed before O2 . This
reaction produces 45.0 mL N2 O3 .
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72.
2 SO2 (g) + O2 (g) → 2 SO3 (g)
1.25 L SO2
×
2 L SO3
2 L SO2
= 1.25 L SO3
3.00 L O2
×
2 L SO3
1 L O2
= 6.00 L SO3
Therefore, SO2 is the limiting reactant because it will be consumed before O2 .
This reaction produces 1.25 L SO3 .
74.
4 HCl(g) + O2 (g) → 2 Cl2 (g) + 2 H2 O(g)
10.0 L HCl
×
2 L Cl2
4 L HCl
50.0 L O2
×
2 L Cl2
1 L O2
=
5.00 L Cl2
(1.00 × 102 mL Cl2 )
= 100 L Cl2
Therefore, HCl is the limiting reactant because it will be consumed before O2 .
This reaction produces 5.00 L Cl2 .
Section 9.9 Percent Yield
76.
Actual yield: 3.50 g acetone; theoretical yield: 3.67 g acetone
3.50 g Percent yield: 3.67 g × 100% = 95.4%
78.
Actual yield: 0.725 g K2 CO3 ; theoretical yield: 0.690 g K2 CO3
Percent yield:
66
Chapter 9
0.725 g 0.690 g × 100%
= 105%
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General Exercises
80.
The units associated with molar volume are liters per mole (L/mol).
82.
(NH4 )2 Cr2 O7 (s) → Cr2 O3 (s) + 4 H2 O(l) + N2 (g)
MM of (NH4 )2 Cr2 O7 =
1.54 g (NH4 )2 Cr2 O7
×
252.10 g/mol
1 mol (NH4 )2 Cr2 O7
252.10 g (NH4 )2 Cr2 O7
22.4 L N2
× 1 mol N
2
84.
×
1 mol N2
1 mol (NH4 )2 Cr2 O7
= 0.137 L N2 (137 mL N2 )
3 MnO2 (s) + 4 Al(s) → 3 Mn(s) + 2 Al2 O3 (s)
MM of Mn = 54.94 g/mol
MM of Al2 O3 = 101.96 g/mol
1.00 kg Mn
1000 g Mn
1 kg Mn ×
×
×
86.
1 mol Mn
54.94 g Mn ×
101.96 g Al2 O3
1 mol Al2 O3
2 mol Al2 O3
3 mol Mn
= 1240 g Al2 O3
Sb2 S3 (s) + 6 HCl(aq) → 2 SbCl3 (aq) + 3 H2 S(g)
MM of Sb2 S3 = 339.71 g/mol
MM of SbCl3 = 228.10 g/mol
3.00 g Sb2 S3
×
1 mol Sb2 S3
339.71 g Sb2 S3
×
2 mol SbCl3
1 mol Sb2 S3
×
228.10 g SbCl3
1 mol SbCl3
= 4.03 g SbCl3
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Chemical Equation Calculations
67
Challenge Exercises
88.
C 3 H8 (g) + 5 O2 (g) → 3 CO2 (g) + 4 H2 O(g)
MM of C3 H8 =
10.0 g C3 H8
×
44.11 g/mol
1 mol C3 H8
44.11 g C3 H8
×
3 mol CO2
1 mol C3 H8
×
22.4 L CO2
1 mol CO2
= 15.2 L CO2 at STP
90.
2 H2 O(l) → 2 H2 (g) + O2 (g)
MM of H2 O = 18.02 g/mol
100.0 mL H2 O ×
1.00 g H2 O
1 mL H2 O
×
1 mol H2 O
1 mol O2
×
18.02 g H2 O
2 mol H2 O ×
= 62.2 L O2 at STP
Online Exercises
92.
68
China produced the most sulfuric acid in 2000; about 24 million tons.
Chapter 9
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22.4 L O2
1 mol O2