Download CALCULATIONS FROM CHEMICAL EQUATIONS

CHAPTER 9
CALCULATIONS FROM CHEMICAL EQUATIONS
SOLUTIONS TO REVIEW QUESTIONS
1.
A mole ratio is the ratio between the mole amounts of two atoms and/or molecules involved in a chemical
reaction.
2.
In order to convert grams to moles the molar mass of the compound under consideration needs to be
determined.
3.
The balanced equation is
Ca3 P2 þ 6 H2 O ! 3 CaðOHÞ2 þ 2 PH3
2 mol PH3
¼ 2 mol PH3
(a) Correct: ð1 mol Ca3 P2 Þ
1 mol Ca3 P2
(b)
Incorrect: 1 g Ca3P2 would produce 0.4 g PH3
1 mol
2 mol PH3
33:99 g
ð1 g Ca3 P2 Þ
¼ 0:4 g PH3
182:2 g 1 mol Ca3 P2
mol
(c)
(d)
(e)
Correct: see equation
Correct: see equation
Incorrect: 2 mol Ca3P2 requires 12 mol H2O to produce 4.0 mol PH3.
6 mol H2 O
ð2 mol Ca3 P2 Þ
1 mol Ca3 P2
¼ 12 mol H2 O
(f)
Correct: 2 mol Ca3P2 will react with 12 mol H2O (3 mol H2O are present in excess) and 6 mol
Ca(OH)2 will be formed.
3 mol CaðOHÞ2
ð2 mol Ca3 P2 Þ
¼ 6 mol CaðOHÞ2
1 mol Ca3 P2
1 mol
6 mol H2 O
18:02 g
(g) Incorrect: ð200: g Ca3 P2 Þ
¼ 119 g H2 O
182:2 g 1 mol Ca3 P2
mol
(h)
The amount of water present (100. g) is less than needed to react with 200. g Ca3P2 H2O is the
limiting reactant.
Incorrect: water is the limiting reactant.
1 mol
2 mol PH3
33:99 g
ð100: g H2 OÞ
¼ 62:9 g PH3 ðtheoreticalÞ
18:02 g 6 mol H2 O
mol
- 84 -
- Chapter 9 4.
The balanced equation is
2 CH4 þ 3 O2 þ 2 NH3 ! 2 HCN þ 6 H2 O
(a) Correct
2 mol HCN
¼ 10:7 mol HCN ðnot 12 mol HCNÞ
(b) Incorrect: ð16 mol O2 Þ
3 mol O2
(c) Correct
6 mol H2 O
¼ 36 mol H2 O ðnot 4 mol H2 OÞ
(d) Incorrect: ð12 mol HCNÞ
2 mol HCN
(e) Correct
(f) Incorrect: O2 is the limiting reactant
2 mol HCN
¼ 2 mol HCN ðnot 3 mol HCNÞ
ð3 mol O2 Þ
3 mol O2
5.
The theoretical yield of a chemical reaction is the maximum amount of product that can be produced based
on a balanced equation. The actual yield of a reaction is the actual amount of product obtained.
6.
You can calculate the percent yield of a chemical reaction by dividing the actual yield by the theoretical
yield and multiplying by one hundred.
- 85 -
- Chapter 9 -
SOLUTIONS TO EXERCISES
1. (a)
(b)
(c)
(d)
2. (a)
(b)
(c)
(d)
3. (a)
(b)
(c)
(d)
(e)
4. (a)
(b)
(c)
1 mol
ð25:0 g KNO3 Þ
¼ 0:247 mol KNO3
101:1 g
1 mol
ð56 mmol NaOHÞ
¼ 0:056 mol NaOH
1000 mmol
1 mol
2
¼ 4:4 mol ðNH4 Þ2 C2 O4
ð5:4 10 gðNH4 Þ2 C2 O4 Þ
124:1 g
The conversion is: mL sol ! g sol ! g H2SO4 ! mol H2SO4
1:727 g 0:800 g H2 SO4
1 mol
ð16:8 mL solutionÞ
¼ 0:237 mol H2 SO4
g solution
mL
98:09 g
1000 g
1 mol
ð2:10 kg NaHCO3 Þ
¼ 25:0 mol NaHCO3
kg
84:01 g
1g
1 mol
ð525 mg ZnCl2 Þ
¼ 3:85 103 mol ZnCl2
1000 mg 136:3 g
1 mol
24
ð9:8 10 molecules CO2 Þ
¼ 16 mol CO2
6:022 1023 molecules
0:789 g
1 mol
ð250 mL C2 H5 OHÞ
¼ 4:3 mol C2 H5 OH
mL
46:07 g
106:9 g
¼ 273 g FeðOHÞ3
ð2:55 mol FeðOHÞ3 Þ
mol
1000 g
ð125 kg CaCO3 Þ
¼ 1:25 105 g CaCO3
kg
17:03 g
ð10:5 mol NH3 Þ
¼ 179 g NH3
mol
1 mol
36:46 g
ð72 mmol HClÞ
¼ 2:6 g HCl
1000 mmol
mol
3:119 g
ð500:0 mL Br2 Þ
¼ 1559:5 g Br2 ¼ 1:560 103 g Br2
mL
154:8 g
ð0:00844 mol NiSO4 Þ
¼ 1:31 g NiSO4
mol
60:05 g
ð0:0600 mol HC2 H3 O2 Þ
¼ 3:60 g HC2 H3 O2
mol
514:2 g
ð0:725 mol Bi2 S3 Þ
¼ 373 g Bi2 S3
mol
- 86 -
- Chapter 9 (d)
(e)
1 mol
180:2 g
ð4:50 10 molecules C6 H12 O6 Þ
¼ 1:35 g C6 H12 O6
mol
6:022 1023 molecules
1:175 g 0:200 g K2 CrO4
ð75 mL solutionÞ
¼ 18 g K2 CrO4
g solution
mL
21
5. Larger number of molecules: 10.0 g H2O or 10.0 g H2O2
Water has a lower molar mass than hydrogen peroxide. 10.0 grams of water has a lower molar mass,
contains more moles, and therefore more molecules than 10.0 g of H2O2.
6. Larger number of molecules: 25.0 g HCl or 85 g C6H12O6
1 mol
6:022 1023 molecules
ð25:0 g HClÞ
¼ 4:13 1023 molecules HCl
36:46 g
mol
1 mol
6:022 1023 molecules
ð85:0 g C6 H12 O6 Þ
¼ 2:84 1023 molecules C6 H12 O6
180:2 g
mol
HCl contains more molecules
7. Mole Ratios
12 CO 2 þ 11 H 2O ! C12 H22 O11 þ 12 O2
(a)
(b)
(c)
12 mol CO2
11 mol H2 O
11 mol H2 O
1 mol C12 H22 O11
(d)
12 mol O2
12 mol CO2
(f)
12 mol O2
1 mol C12 H22 O11
6 mol O2
1 mol C4 H9 OH
5 mol H2 O
6 mol O2
(d)
1 mol C4 H9 OH
4 mol CO2
5 mol H2 O
1 mol C4 H9 OH
4 mol CO2
5 mol H2 O
(f)
(e)
1 mol C12 H22 O11
12 mol CO2
11 mol H2 O
12 mol O2
8. Mole ratios
C4 H9 OH þ 6 O 2 ! 4 CO 2 þ 5 H 2O
(a)
(b)
(c)
(e)
9. CO2 þ 4 H 2 ! CH 4 þ 2 H2O
2 mol H2 O
¼ 50: mol H2 O
(a) ð25 mol CO2 Þ
1 mol CO2
1 mol CH4
¼ 6:0 mol CH4
(b) ð12 mol H2 OÞ
2 mol H2 O
- 87 -
4 mol CO2
6 mol O2
- Chapter 9 10. The balanced equation is H2 SO4 þ 2 NaOH ! Na2 SO4 þ 2 H 2O
2 mol NaOH
¼ 34 mol NaOH
(a) ð17 mol H2 SO4 Þ
1 mol H2 SO4
1 mol Na2 SO4
(b) ð21 mol NaOHÞ
¼ 11 mol Na2 SO4
2 mol NaOH
11. The balanced equation is
MnO2 ðsÞ þ 4 HClðaqÞ ! Cl2 ðgÞ þ MnCl2 ðaqÞ þ 2 H2 OðlÞ
4 mol HCl
¼ 4:20 mol HCl
(a) ð1:05 mol MnO2 Þ
1 mol MnO2
1 mol MnCl2
(b) ð1:25 mol H2 OÞ
¼ 0:625 mol MnCl2
2 mol H2 O
12. Al4 C3 þ 12 H2 O ! 4 AlðOHÞ3 þ 3 CH 4
1 mol
12 mol H2 O
¼ 8:33 mol H2 O
(a) ð100: g Al4 C3 Þ
144:0 g 1 mol Al4 C3
4 mol AlðOH3 Þ
¼ 0:800 mol AlðOHÞ3
(b) ð0:600 mol CH4 Þ
3 mol CH4
13. Grams of CaCl2
CaCO3 þ 2 HCl ! CaCl 2 þ H2 O þ CO 2
The conversion is: g CaCO3 ! mol CaCO3 ! mol CaCl2 ! g CaCl2
1 mol
1 mol CaCl2
111:0 g
ð50:0 g CaCO3 Þ
¼ 55:4 g CaCl2
100:1 g 1 mol CaCO3
mol
14. Grams of AlBr3
2 Al þ 6 HBr ! 2 AlBr3 þ 3 H 2
The conversion is: g Al ! mol Al ! mol AlBr3 ! g AlBr3
1 mol
2 mol AlBr3
266:7 g
ð25:2 g AlÞ
¼ 249 g AlBr3
2 mol Al
26:98 g
mol
15. The balanced equation is Fe2 O3 þ 3C ! 2Fe þ 3CO
The conversion is: kg Fe2 O3 ! kmol Fe2 O3 ! kmol Fe ! kg Fe
55:85 kg
2 kmol Fe
1 kmol
ð125 kg Fe2 O3 Þ
¼ 87:4 kg Fe
kmol
159:7 kg 1kmol Fe2 O3
- 88 -
- Chapter 9 16. The balanced equation is 3 Fe þ 4 H2 O ! Fe3O4 þ 4 H 2
Calculate the grams of both H2O and Fe to produce 375 g Fe3O4
1 mol
4 mol H2 O
18:02 g
ð375 g Fe3 O4 Þ
¼ 117 g H2 O
231:6 g 1 mol Fe3 O4
mol
1 mol
3 mol Fe
55:85 g
ð375 g Fe3 O4 Þ
¼ 271 g Fe
231:6 g 1 mol Fe3 O4
mol
17. The balanced equation is: 2 C 12H 4Cl 6 þ 23 O 2 þ 2 H2O ! 24 CO2 þ 12 HCl
2 mol H2 O
¼ 0:870 mol H2 O
(a) ð10:0 mol O2 Þ
23 mol O2
12 mol HCl 36:46 g
(b) ð15:2 mol H2 OÞ
¼ 3:33 103 g HCl
2 mol H2 O
mol
1 mol
24 mol CO2
(c) ð76:5 g HClÞ
¼ 4:20 mol CO2
12 mol HCl
36:46 g
1 mol
2 mol C12 H4 Cl6
360:9 g
(d) ð100:25 g CO2 Þ
¼ 68:51g C12 H4 Cl6
24 mol CO2
44:01 g
mol
1000 g
1 mol
12 mol HCl
36:46 g
(e) ð2:5 kg C12 H4 Cl6 Þ
¼ 1:5 103 g HCl
1 kg
360:9 g 2 mol C12 H4 Cl6
mol
18. 4 HgS þ 4 CaO ! 4 Hg þ 3 CaS þ CaSO4
3 mol CaS
(a) ð2:5 mol CaOÞ
¼ 1:9 mol CaS
4 mol CaO
4 mol Hg
200:6 g
(b) ð9:75 mol CaSO4 Þ
¼ 7:82 103 g Hg
1 mol CaSO4
mol
1 mol
4mol CaO
(c) ð97:25 g HgSÞ
¼ 0:4179 mol CaO
232:7 g 4mol HgS
1 mol
4mol Hg
200:6 g
(d) ð87:6 g HgSÞ
¼ 75:5 g Hg
232:7 g 4mol HgS
mol
1000g
1 mol
3mol CaS 72:15 g
¼ 2:50 103 g CaS
(e) ð9:25 kg HgÞ
1kg
200:6 g
4mol Hg
1 mol
19.
(a)
Hydrogen
Oxygen
Hydrogen is the limiting reactant.
!
- 89 -
- Chapter 9 (b)
Hydrogen
Bromine
Bromine is the limiting reactant.
!
20.
(a)
Lithium
Iodine
No limiting reactant.
!
(b)
Silver
Chlorine
Silver is the limiting reactant.
!
21.
(a)
Potassium
Chlorine
Potassium is the limiting reactant.
!
(b)
Aluminum
Oxygen
Oxygen is the limiting reactant.
!
22.
(a)
Nitrogen
Oxygen
Oxygen is the limiting reactant.
!
- 90 -
- Chapter 9 (a)
Iron
Hydrogen
Oxygen
Water is the limiting reactant.
!
23. (a)
KOH
16:0 g
þ
HNO3
12:0 g
!
KNO3
þ
H2 O
Choose one of the products and calculate its mass that would be produced from each given
reactant. Using KNO3 as the product:
1 mol
1 mol KNO3
101:1 g
ð16:0 g KOHÞ
¼ 28:8 g KNO3
1 mol KOH
56:10 g
mol
1 mol
1 mol KNO3 101:1 g
ð12:0 g HNO3 Þ
¼ 19:3 g KNO3
1 mol KOH
63:02 g
mol
Since HNO3 produces less KNO3, it is the limiting reactant and KOH is in excess.
(b)
2 NaOH
10:0 g
þ
H2 SO4
10:0 g
!
Na2 SO4
þ 2 H2O
Choose one of the products and calculate its mass that would be produced from each given
reactant. Using H2O as the product:
1 mol
2 mol H2 O
18:02 g
ð10:0 g NaOHÞ
¼ 4:51 g H2 O
40:00 g 2 mol NaOH
mol
1 mol
2 mol H2 O
18:02 g
ð10:0 g H2 SO4 Þ
¼ 3:67 g H2 O
98:09 g 1 mol H2 SO4
mol
Since H2SO4 produces less H2O, it is the limiting reactant and NaOH is in excess.
24. (a)
2 BiðNO 3Þ3
50:0 g
þ
3 H 2S
6:00 g
!
Bi2 S3
þ
6 HNO3
Choose one of the products and calculate its mass that would be produced from each given
reactant. Using Bi2S3 as the product:
1 mol
1 mol Bi2 S3
514:2 g
¼ 32:5 g Bi2 S3
ð50:0 g BiðNO3 Þ3 Þ
395:0 g 2 mol BiðNO3 Þ3
mol
1 mol
1 mol Bi2 S3
514:2 g
ð6:00 g H2 SÞ
¼ 30:2 g Bi2 S3
3 mol H2 S
34:09 g
mol
Since H2S produces less Bi2S3, it is the limiting reactant and Bi(NO3)3 is in excess.
- 91 -
- Chapter 9 (b)
3 Fe
40 :0 g
þ
4 H2 O
16:0 g
!
þ
Fe3 O4
4 H2
Choose one of the products and calculate its mass that would be produced from each given
reactant. Using H2 as the product:
1 mol
4 mol H2
2:016 g
ð40:0 g FeÞ
¼ 1:93 g H2
3 mol Fe
55:85 g
mol
1 mol
4 mol H2
2:016 g
ð16:0 g H2 OÞ
¼ 1:79 g H2
18:02 g 4 mol H2 O
mol
Since H2O produces less H2, it is the limiting reactant and Fe is in excess.
25. Limiting reactant calculations
2 AlðOHÞ3 þ 3 H2 SO4 ! Al 2ðSO 4Þ3 þ 6 H2 O
(a)
Reaction between 35.0 g Al(OH)3 and 35.0 g H2SO4.
Convert each amount to moles of Al2(SO4)3.
1 mol Al2 ðSO4 Þ3
1 mol
¼ 0:224 mol Al2 ðSO4 Þ3
ð35:0 g AlðOHÞ3 Þ
2 mol AlðOHÞ3
78:00 g
1 mol Al2 ðSO4 Þ3
1 mol
¼ 0:119 mol Al2 ðSO4 Þ3
ð35:0 g H2 SO4 Þ
3 mol H2 SO4
98:09 g
H2SO4 is the limiting reactant. The yield is 0.119 mol Al2(SO4)3
(b)
Reaction between 45.0 g H2SO4 and 25.0 g Al(OH)3.
Calculate the grams of Al2(SO4)3 from each reactant
1 mol Al2 ðSO4 Þ3
1 mol
342:2 g
ð45:0 g H2 SO4 Þ
¼ 52:3 g Al2 ðSO4 Þ3
3 mol H2 SO4
98:09 g
mol
1 mol
1 mol Al2 ðSO4 Þ3
342:2 g
25:0 g AlðOHÞ3
¼ 54:8 g Al2 ðSO4 Þ3
2 mol AlðOHÞ3
78:09 g
mol
H2SO4 is the limiting reactant. The yield is 52.3 g Al2 (SO4)3
Al(OH)3 is the excess reactant.
(c)
Reaction between 2.5 mol Al(OH)3 and 5.5 mol H2SO4.
Convert each amount to moles of product.
1 mol Al2 ðSO4 Þ3
2:5 mol AlðOHÞ3
¼ 1:3 mol Al2 ðSO4 Þ3
2 mol AlðOHÞ3
1 mol Al2 ðSO4 Þ3
¼ 1:8 mol Al2 ðSO4 Þ3
ð5:5 mol H2 SO4 Þ
3 mol H2 SO4
Al(OH)3 is the limiting reactant. 1.3 mol Al2 (SO4)3 produced
6 mol H2 O
¼ 7:5 mol H2 S produced
2:5 mol AlðOHÞ3
2 mol AlðOHÞ3
3 mol H2 SO4
¼ 3:8 mol H2 SO4 produced
2:5 mol AlðOHÞ3
2 mol AlðOHÞ3
- 92 -
- Chapter 9 5:5 mol H2 SO4 3:8 mol H2 SO4 ¼ 1:7 mol H2 SO4 unreacted
When the reaction is complete, 1.3 mol Al2(SO4)3, 7.5 mol H2O, and 1.7 mol H2SO4 will be in the
container.
26. The balanced equation is P4 þ 6 Cl2 ! 4 PCl3
(a) Reaction between 20.5 g P4 and 20.5 g Cl2
Convert each amount to moles of PCl3
1 mol
4 mol PCl3
¼ 0:662 mol PCl3
ð20:5 g P4 Þ
123:9 g
1 mol P 4
1 mol
4 mol PCl3
¼ 0:193 mol PCl3
ð20:5 g Cl2 Þ
6 mol Cl2
70:90 g
(b)
(c)
Cl2 is the limiting reactant. The yield is 0.193 mol PCl3.
Reaction between 55 g Cl2 and 25 g P4.
Calculate the grams of PCl3 from each reactant.
1 mol
4 mol PCl3
137:3 g
ð55 g Cl2 Þ
¼ 71 g PCl3
6 mol Cl2
70:90 g
mol
1 mol
4 mol PCl3
137:3 g
ð25 g P4 Þ
¼ 1:1 102 g PCl3
1 mol P4
123:9 g
mol
Cl2 is the limiting reactant. The yield is 71 g PCl3.
P4 is the excess reactant.
Reaction between 15 mol P4 and 35 mol Cl2.
Convert each amount to moles of product.
4 mol PCl3
¼ 60: mol PCl3
ð15 mol P4 Þ
1 mol P4
4 mol PCl3
¼ 23 mol PCl3 produced
ð35 mol Cl2 Þ
6 mol Cl2
Cl2 is the limiting reactant.
1 mol P4
¼ 5:8 mol P4 reacted
ð35 mol Cl2 Þ
6 mol Cl2
15 mol P4 5:8 mol P4 ¼ 9 mol P4 left over
When the reaction is complete, 23 mol PCl3 and 9 mol P4 will be the container.
27. X 8 þ 12 O2 ! 8 XO3
The conversion is: g O2 ! mol O2 ! mol X8
1 mol
1 mol X 8
¼ 0:3125 mol X 8
ð120:0 g O2 Þ
80:0 g X 8 ¼ 0:3125 mol X 8
32:00 g 12 mol O2
80:0 g
¼ 256 g= mol X 8
0:3125 mol
g
256
mol ¼ 32:0 g
molar mass X ¼
8
mol
Using the periodic table we find that the element with 32.0 g/mol is sulfur.
- 93 -
- Chapter 9 28. X þ 2 HCl ! XCl 2 þ H 2
The conversion is: g H 2 ! mol H 2 ! mol X
1 mol
1 mol X
¼ 1:20 mol X
ð2:42 g H2 Þ
2:016 g 1 mol H2
78:5 g X ¼ 1:20 mol X
78:5 g
¼ 65:4 g=mol
1:20 mol
Using the periodic table we find that the element with atomic mass 65.4 is zinc.
29. Limiting reactant calculation and percent yield
Fe2 O3 þ 3 CO ! 2 Fe þ 3 CO 2
1000 g
1 mol
2 mol Fe
55:85 g
ð15:0 kg Fe2 O3 Þ
¼ 1:05 104 g Fe
kg
159:7 g 1 mol Fe2 O3
mol
1000 g
1 mol
2 mol Fe
55:85 g
ð9:0 kg COÞ
¼ 1:20 104 g Fe
kg
28:01 g 3 mol CO
mol
The limiting reactant is Fe2O3. 1.05 104 g Fe is the theoretical yield.
1 kg
ð1:05 10 gÞ
¼ 10:5 kg Fe
1000 g
actual yield
8:5 kg
Percent yield ¼
ð100Þ ¼
ð100Þ ¼ 81% yield of Fe
theoretical yield
10:5 kg
4
30.
CuðsÞ þ 2 Ag NO3ðaqÞ ! 2 AgðsÞ þ CuðNO 3Þ2 ðaqÞ
1 mol
2 mol Ag 107:9 g
ð125 g CuÞ
¼ 424 g Ag
63:55 g 1 mol Cu
mol
1 mol
2 mol Ag
107:9 g
ð275 g AgNO3 Þ
¼ 175 g Ag
169:9 g 2 mol AgNO3
mol
The limiting reactant is AgNO3. 175 g Ag is the theoretical yield
Percent yield ¼
31. (a)
actual yield
135 g
ð100Þ ¼
ð100Þ ¼ 77:1% yield of Ag
theoretical yield
175 g
Limiting reactant problem
3 Cu þ 8 HNO3 ! 3 CuðNO3 Þ2 þ 4 H2 O þ 2 NO
1 mol 3 mol CuðNO3 Þ2
187:6
ð27:5 g CuÞ
¼ 81:2 g CuðNO3 Þ2
3 mol Cu
63:55
1 mol
1 mol
3 mol CuðNO3 Þ2
187:6 g
ð125 g HNO3 Þ
¼ 140 g CuðNO3 Þ2
8 mol HNO3
63:02 g
1 mol
- 94 -
- Chapter 9 (b)
(c)
The excess reactant is HNO3
The percent yield is 87.3%. The actual yield is
ð0:873Þð81:2Þ ¼ 70:9 g CuðNO3 Þ2 actual yield
32. Fe2 O3(s) þ 6 HCl(aq) ! 2 FeCl3(aq) þ 3H2O(l)
(a)
(b)
Limiting reactant problem.
1 mol
2 mol FeCl3
ð35 g Fe2 O3 Þ
¼ 0:44 mol FeCl3
159:7 g 1 mol Fe2 O3
1 mol
2 mol FeCl3
¼ 0:32 mol FeCl3
ð35 g HClÞ
6 mol HCl
36:46 g
HCl is the limiting reactant.
1 mol
3 mol H2 O
ð35 g HClÞ
¼ 0:48 mol H2 O
36:46 g
6 mol HCl
0.32 mol FeCl3 and 0.48 mol H2O are the theoretical yields.
Fe2O3 is the excess reactant.
1 mol
1 mol Fe2 O3
159:7 g
ð35 g HClÞ
¼ 26 g Fe2 O3 reacted
6 mol HCl
36:46 g
mol
35 g Fe2 O3 26 g Fe2 O3 ¼ 9 g Fe2 O3 excess
(c)
162:2 g
¼ 48 g FeCl3
The percent yield is 92.5%. The actual yield is ð0:925Þð0:32 mol FeCl3 Þ
mol
33. No. There are not enough screwdrivers, wrenches or pliers. 2400 screwdrivers, 3600 wrenches and
1200 pliers are needed for 600 tool sets.
34. The balanced equation is C6 H12 O6 ! 2 C2H5OH þ 2 CO 2
453:6 g
1 mol
2 mol C2 H5 OH 46:07 g
1 mL
1L
¼ 169 L C2 H5 OH
ð575 lb C6 H12 O6 Þ
lb
180:2 g 1 mol C6 H12 O6
mol
0:789 g 1000 mL
35. Consider the reaction A ! 2B and assume that you have 1 gram of A. This does not guarantee that you will
produce 1 gram of B because A and B have different molar masses. One gram of A does not contain the same
number of molecules as 1 gram of B. However, 1 mole of A does have the same number of molecules as one
mole of B. (Remember, 1 mole ¼ 6.022 1023 molecules always.) If you determine the number of moles in
one gram of A and multiply by 2 to get the number of moles of B . . . then from that you can determine the
grams of B using its molar mass. Equations are written in in terms of moles not grams.
36. 4 KO2 þ 2 H2O þ 4 CO2 ! 4 KHCO3 þ 3 O2
(a)
0:85 g CO2
1 mol
4 mol KO2
0:019 mol KO2
¼
min
min
44:01 g 4 mol CO2
0:019 mol KO2
ð10:0minÞ ¼ 0:19 mol KO2
min
- 95 -
- Chapter 9 -
(b)
g CO2
mol CO2
mol O2
g O2
g O2
!
!
!
!
min
min
min
min
hr
0:85 g CO2
1 mol
3 mol O2
32:00 g 60 min
28 g O2
¼
min
hr
44:01 g 4 mol CO2
1:0 mol
1:0 hr
The conversion is:
H SO
! 12 C þ11 H O
2
37. C12 H22 O11
(a)
4
2
453:6 g
1 mol
12 mol C
12:01 g
ð2:0 lb C12 H22 O11 Þ
¼ 3:8 102 g C
lb
342:3 g 1 mol C12 H22 O11
mol
453:6 g
1 mol
11 mol H2 O
18:02 g
ð2:0 lb C12 H22 O11 Þ
¼ 5:3 102 g H2 O
lb
342:3 g 1 mol C12 H22 O11
mol
From 2:0 lb C12 H22 O11 , 3:8 102 g C and 5:3 102 g H2 O are yielded.
(b)
ð25:2 g C12 H22 O11 Þ
1 mol C12 H22 O11
11 mol H2 O
18:02 g
1 mL
¼ 14:7 mL H2 O
1 mol C12 H22 O11
mol
0:994 g
342:3 g
38. 2 CH3OH þ 3 O2 ! 2 CO2 þ 4 H2O
The conversion is:
mL CH3 OH ! g CH3 OH ! mol CH3 OH ! mol O2 ! g O2
0:72 g
1 mol
3 mol O2
32:00 g
ð60:0 mL CH3 OHÞ
¼ 65 g O2
mL
32:04 g 2 mol CH3 OH
mol
39. The balanced equation is 7 H2O2 þ N2H4 ! 2 HNO3 þ 8 H2O
The conversion is:
(a)
(b)
(c)
(d)
1000 g
1 mol
2 mol HNO3
63:02 g
ð75 kg N2 H4 Þ
¼ 2:9 105 g HNO3
1 mol N2 H4
1 kg
32:05 g
mol
1000 mL 1:41 g
1 mol
8 mol H2 O
18:02 g
ð250 L H2 O2 Þ
¼ 2:1 105 g H2 O
1L
1 mL
34:02 g 7 mol H2 O2
mol
1 mol
1 mol N2 H4
32:02 g
ð725 g H2 O2 Þ
¼ 97:6 g N2 H4
34:02 g 7 mol H2 O2
mol
Reaction between 750 g of N2H2 and 125 g of H2O2.
Convert each amount to grams of H2O.
1 mol
8 mol H2 O
18:02 g
ð750 g N2 H4 Þ
¼ 3:4 103 g H2 O
32:05 g 1 mol N2 H4
mol
1 mol
8 mol H2 O
18:02 g
ð125 g H2 O2 Þ
¼ 75:7 g H2 O
34:02 g 7 mol H2 O2
mol
75.7 g H2O can be produced.
- 96 -
- Chapter 9 (e)
Since H2O2 is the limiting reactant, N2H4 is in excess.
1 mol
1 mol N2 H4 32:05 g
ð125 g H 2O 2Þ
¼ 16:8 g N2 H4 reacted
34:02 g 7 mol H2 O2
mol
750 g N2H4 given 16.8 g N2 H4 used ¼ 730 g N2H4 remaining
40. The balanced equation is
16 HCl þ 2 KMnO4 ! 5 Cl2 þ 2 KCl þ 2 MnCl2 þ 8 H2O
(a)
(b)
(c)
(d)
Reaction between 25 g KMnO4 and 85 g HCl. Convert each to moles of MnCl2.
1 mol KMnO4
2 mol MnCl2
¼ 0:16 mol MnCl2
ð25 g KMnO4 Þ
158:04 g KMnO4
2 mol KMnO4
1 mol
2 mol MnCl2
ð85 g HClÞ
¼ 0:29 mol MnCl2
16 mol HCl
36:46 g
KMnO4 is the limiting reactant; 0.16 mol MnCl2 produced.
1 mol
8 mol H2 O 18:02 g
ð75 g KClÞ
¼ 73 g H2 O
74:55 g
2 mol KCl
mol
1 mol
5 mol Cl2
70:90 g
ð150 g HClÞ
¼ 91 g Cl2
36:46 g 16 mol HCl
mol
75 g
Theoretical yield is 91 g Cl2; Percent yield:
ð100Þ ¼ 82% yield
91 g
Reaction between 25 g HC1 and 25 g KMnO4. Convert each amount to grams of CL2.
1 mol
5 mol Cl2
70:90 g
¼ 15 g Cl2
ð25 g HClÞ
36:46 g 16 mol HCl
mol
1 mol
5 mol Cl2
70:90 g
ð25 g KMnO4 Þ
¼ 28 g Cl2
158:04 g 2 mol KMnO4
mol
HC1 is the limiting; KMnO4 is in excess; 15 g Cl2 will be produced.
(e)
Calculate the mass of unreacted KMnO4:
1 mol
2 mol KMnO4
158:04 g
ð25 g HClÞ
¼ 14 g KMnO4 will react.
16 mol HCl
36:46 g
mol
Unreacted KMnO4 ¼ 25 g 14 g ¼ 11 g KMnO4 remain unreacted.
- 97 -
- Chapter 9 41. The balanced equation is
4 Ag þ 2 H2S þ O2 ! 2 Ag2S þ 2 H2O
1 mol
2 mol Ag2 S 247:9 g
¼ 1:3 g Ag2 S
(a) ð1:1 g AgÞ
107:09 g
4 mol Ag
mol
1 mol
2 mol Ag2 S 247:9 g
ð0:14 g H2 SÞ
¼ 1:0 g Ag2 S
34:09 g
2 mol H2 S
mol
1 mol
2 mol Ag2 S 247:9 g
ð0:080 g O2 Þ
¼ 1:2 g Ag2 S
32:00 g
1 mol O2
mol
(b)
H2S is limiting; 1.0 g Ag2S forms.
1 mol
2 mol H2 S 34:09 g
ð1:1 g AgÞ
¼ 0:17 g H2 S reacts
107:9 g
4 mol Ag
mol
0.17 g 0.14 g ¼ 0.03 grams more H2S needed to completely react Ag.
42. Equation: CaO þ H2O ! Ca(OH)2
1 mol 1 mol CaðOHÞ2
74:10 g
ð35:55 g CaOÞ
¼ 46:97 g CaðOHÞ2
1 mol CaO
56:08
1 mol
43. (a)
(b)
CuðNO3 Þ2 ðaqÞ þ 2 NaOHðaqÞ ! CuðOHÞ2 ðsÞ þ 2 NaNO 3ðaqÞ
1 mol
1 mol CuðOHÞ2
97:57 g
15:25 g CuðNO3 Þ2
¼ 7:931 g CuðOHÞ2
187:6 g 1 mol CuðNO3 Þ2
mol
1 mol
1 mol CuðOHÞ2
97:57 g
ð12:75 g NaOHÞ
¼ 15:55 g CuðOHÞ2
2 mol NaOH
40:00 g
mol
The theoretical yield is 7.931 g Cu(OH)2
actual yield
5:23 g
The percent yield ¼
ð100Þ ¼
ð100Þ ¼ 65:9 of yield of Cu(OH)2
theoretical yield
7:931 g
44. C2 H5 OH þ 3 O2 ! 2 CO2 þ 3 H2 O
2 mol CO2
¼ 5:0 mol CO2
(a) ð2:5 mol C2 H5 OHÞ
1 mol C2 H5 OH
2 mol CO2
¼ 5:0 mol CO2
ð7:5 mol O2 Þ
3 mol O2
Neither reactant is limiting.
3 mol H2 O
ð2:5 mol C2 H5 OHÞ
¼ 7:5 mol H2 O
1 mol C2 H5 OH
When the reaction is complete, there will be 5.0 mol CO2 and 7.5 mol H2O.
- 98 -
- Chapter 9 (b)
1 mol
2 mol CO2
44:01 g
ð225 g C2 H5 OHÞ
¼ 430 g CO2
46:07 g 1 mol C2 H5 OH
mol
1 mol
3 mol H2 O
18:02 g
ð225 g C2 H5 OHÞ
¼ 264 g H2 O
46:07 g 1 mol C2 H5 OH
mol
45. The balanced equation is Zn þ 2 HC1 ! ZnCl2 þ H2
180.0 g Zn 35 g Zn ¼ 145 g Zn reacted with HCl
1 mol
1 mol H2
2:016 g
(a) ð145 g ZnÞ
¼ 4:47 g H2 produced
65:39 g 1 mol Zn
mol
1 mol
2 mol HCl 36:46 g
(b) ð145 g ZnÞ
¼ 162 g HCl reacted
65:39 g
1 mol Zn
mol
1 mol
2 mol HCl 36:46 g
(c) ð180:0 g ZnÞ
¼ 201 g HCl reacts
65:39 g
1 mol Zn
mol
201 g 162 g ¼ 39 g more HCl needed to react with the 180.0 g Zn
46.
FeðsÞ þ CuSO4 ðaqÞ ! CuðsÞ þ FeSO4 ðaqÞ
2:0 mol
3:0 mol
(a)
(b)
2.0 mol Fe react with 2.0 mol CuSO4 to yield 2.0 mol Cu and 2.0 mol FeSO4. 1.0 mol CuSO4 is
unreacted. At the completion of the reaction, there will be 2.0 mol Cu, 2.0 mol FeSO4, and 1.0 mol
CuSO4.
Determine which reactant is limiting and then calculate the g FeSO4 produced from that reactant.
1 mol
1 mol Cu 63:55 g
ð20:0 g FeÞ
¼ 22:8 g Cu
55:85 g
1 mol Fe
mol
1 mol
1 mol Cu
63:55 g
ð40:0 g CuSO4 Þ
¼ 15:9 g Cu
159:6 g 1 mol CuSO4
mol
Since CuSO4 produces less Cu, it is the limiting reactant. Determine the mass of FeSO4. produced
from 40.0 g CuSO4.
1 mol
1 mol FeSO4
151:9 g
¼ 38:1 g FeSO4 produced
ð40:0 g CuSO4 Þ
159:6 g 1 mol FeSO4
mol
Calculate the mass of unreacted Fe.
1 mol
1 mol Fe
55:85 g
ð40:0 g CuSO4 Þ
¼ 14:0 g Fe will react
159:6 g 1 mol FeSO4
mol
Unreacted Fe ¼ 20.0 g 14.0 g ¼ 6.0 g. Therefore, at the completion of the reaction,
15.9 g Cu, 38.1 g FeSO4, 6.0 g Fe, and no CuSO4 remain.
- 99 -
- Chapter 9 47. Limiting reactant calculation
CO(g) þ 2 H2(g) ! CH3OH(l)
Reaction between 40.0 g CO and 10.0 g H2: determine the limiting reactant by calculating the amount of
CH3OH that would be formed from each reactant.
1 mol
1 mol CH3 OH 32:04 g
¼ 45:8 g CH3 OH
28:01 g
1 mol CO
mol
1 mol
1 mol CH3 OH 32:04 g
ð10:0 g H2 Þ
¼ 79:5 g CH3 OH
2:016 g
2 mol H2
mol
ð40:0 g COÞ
CO is limiting; H2 is in excess; 45.8 g CH3OH will be produced.
Calculate the mass of unreacted H2:
ð40:0 g COÞ
1 mol
28:01 g
2 mol H2
1 mol CO
2:016 g
¼ 5:76 g H2 react
mol
10.0 g H2 5.76 g H2 ¼ 4.2g H2 remain unreacted
48. The balanced equation is C6H12O6 ! 2 C2H5OH þ 2 CO2
(a) First calculate the theoretical yield.
1 mol
ð750 g C6 H12 O6 Þ
180:2 g
2 mol C2 H5 OH
1 mol C6 H12 O6
46:07 g
mol
¼ 3:8 102 g C2 H5 OH ðtheoretical yieldÞ
Then take 84.6% of the theoretical yield to obtain the actual yield.
actual yield ¼
ðtheoretical yieldÞð84:6Þ ð3:8 102 g C2 H5 OHÞð84:6Þ
¼
100
100
¼ 3:2 102 g C2 H5 OH
(b)
475 g C2H5OH represents 84.6% of the theoretical yield. Calculate the theoretical yield.
theoretical yield ¼
475 g
¼ 561 g C2 H5 OH
0:846
Now calculate the g C6H12O6 needed to produce 561 g C2H5OH.
1 mol
1 mol C6 H12 O6
180:2 g
ð561 g C2 H5 OHÞ
¼ 1:10 103 g C6 H12 O6
46:07 g 2 mol C2 H5 OH
mol
49. The balanced equations are:
CaCl2(aq) þ 2 AgNO3(aq) ! Ca(NO3)2(aq) þ 2 AgCl(s)
MgCl2(aq) þ 2 AgNO3(aq ) ! Mg(NO3)2(aq) þ 2 AgCl(s)
- 100 -
- Chapter 9 1 mol of each salt will produce the same amount (2 mol) of AgCl. MgCl2 has a higher percentage of Cl
than CaCl2 because Mg has a lower atomic mass than Ca. Therefore, on an equal mass basis, MgCl2 will
produce more AgCl than will CaCl2.
Calculations show that 1.00 g MgCl2 produces 3.01 g AgCl, and 1.00 g CaCl2 produces 2.56 g AgCl.
50. The balanced equation is Li2O þ H2O ! 2 LiOH
The conversion is: g H2O ! mol H2O ! mol Li2O ! g Li2O ! kg Li2O
2500 g H2 O
1 mol
1 mol Li2 O 29:88 g
1 kg
4:1 kg Li2 O
¼
astronaut day 18:02 g
1 mol H2 O
mol
1000 g
astronaut day
4:1 kg Li2 O
ð30 daysÞð3 astronautsÞ ¼ 3:7 102 kg Li2 O
astronaut day
51. The balanced equation is
H2SO4 þ 2 NaCl ! Na2SO4 þ 2 HCl
First calculate the g HCl to be produced
1000 mL
1:20 g
ð20:0 L HCl solutionÞ
ð0:420Þ ¼ 1:01 104 g HCl
1L
1:00 mL
Then calculate the g H2SO4 required to produce the HCl
1 mol
ð1:01 10 g HClÞ
36:46 g
4
1 mol H2 SO4
2 mol HCl
¼ 1:36 104 g H2 SO4
Finally, calculate the kg H2SO4 (96%)
1:00 g H2 SO4 solution
1 kg
ð1:36 10 g H2 SO4 Þ
¼ 14 kg concentrated H2 SO4
0:96 g H2 SO4
1000 g
4
52. Percent yield of H2SO4
1 mol
ð100:0 g SÞ
¼ 3:118 mol S to start with
32:07 g
3.118 mol S ! 3.118 mol SO2 10% ¼ 2.806 mol SO2
0:3118
2:8062
2.806 mol SO2 ! 2.806 mol SO3 10% ¼ 2.525 mol SO3
0:2806
2:5254
- 101 -
- Chapter 9 2.525 mol SO3 ! 2.525 mol H2SO4 10% ¼ 2.273 mol H2SO4
0:2525
2:2725
98:09 g
ð2:273 mol H2 SO4 Þ
¼ 223:0 g H2 SO4 formed
mol
1 mol H2 SO4
ð3:118 mol SÞ
¼ 3:118 mol H2 SO4 ðtheoretical yieldÞ
1 mol S
2:273 mol H2 SO4
ð100Þ ¼ 72:90% yield
3:118 mol H2 SO4
Alternate Solution:
Calculation of yield. There are three chemical steps to the formation of H2SO4. Each step has a 10% loss of yield.
Step 1:
Step 2:
Step 3:
100% yield 10% ¼
90.00% yield
90.00% yield 10% ¼ 81.00% yield
81.00% yield 10% ¼ 72.90% yield
Now calculate the grams of product. One mole of sulfur will yield a maximum of 1 mol H2SO4. Therefore
3.118 mol S will give a maximum of 3.118 mol H2SO4.
1 mol H2 SO4
98:09 g
ð3:118 mol SÞ
ð0:7290Þ ¼ 223:0 g H2 SO4 yield
1 mol S
mol
53. According to the equations, the moles of CO2 come from both reactions and the moles H2O come from
only the first reaction.
So the mol NaHCO3 ¼ 2 mol H2O ¼ 2 0.0357 mol ¼ 0.0714 mol NaHCO3
84:01 g
ð0:0714 mol NaHCO3 Þ
¼ 6:00 g NaHCO3 in the sample
mol
(10.00 g NaHCO3 þ Na2CO3) 6.00 g NaHCO3 ¼ 4.00 g Na2CO3 in the sample
6:00 g NaHCO3
ð100Þ ¼ 60:0% NaHCO3
10:00 g
4:00 g Na2 CO3
ð100Þ ¼ 40:0% Na2 CO3
10:00 g
54.
(a)
Hydrogen
Carbon
Oxygen
There is no limiting reactant
= CH4
= O2
= H2O
= CO2
- 102 -
- Chapter 9 55. The balanced equation is 2 KClO3 ! KCl3 þ 3 O2
12.82 g mixture 9.45 g residue ¼ 3.37 g O2 lost by heating
Because the O2 lost came only from KClO3, we can use it to calculate the amount of KClO3 in the mixture
The conversion is: g O2 ! mol O2 ! mol KClO3 ! g KClO3
1 mol
1 mol KClO3
122:6 g
ð3:37 g O2 Þ
¼ 8:61 g KClO3 in the mixture
3 mol O2
32:00 mol
mol
8:61 g KClO3
ð100Þ ¼ 67:2% KClO3
12:82 g sample
56. The balanced equation is
Al(OH)3(s) þ 3 HCl(aq) ! AlCl3(aq) þ 3 H2O(l)
The conversion is: L HC1 ! g HCl ! mol HCl ! mol A1(OH)3 ! g A1(OH)3
2:5 L
day
3:0 g HCl
L
1 mol
1 mol AlðOHÞ3
78:00 g
¼ 5:3 g AlðOHÞ3 =day
3 mol HCl
36:46 g
mol
Now calculate the number of 400. mg tablets that can be made from 5.3 g A1(OH)3
5:3 g AlðOHÞ3
day
1000 mg
g
1 tablet
¼ 13 tablets=day
400: mg
57. 4 P þ 5 O2 ! P4O10
P4O10 þ 6 H2O ! 4 H3PO4
In the first reaction:
1 mol
ð20:0 g PÞ
¼ 0:646 mol P
30:97 g
1 mol
ð30:0 g O2 Þ
¼ 0:938 mol O2
32:00 g
This is a ratio of
0:646 mol P
3:44 mol P
¼
0:938 mol O2 5:00 mol O2
Therefore, P is the limiting reactant and the P4O10 produced is:
1 mol P4 O10
ð0:646 mol PÞ
¼ 0:162 mol P4 O10
4 mol P
- 103 -
- Chapter 9 In the second reaction:
1 mol
ð15:0 g H2 OÞ
¼ 0:832 mol H2 O
18:02 g
and we have 0.162 mol P4O10. The ratio of
H2 O 0:832 mol 5:14 mol
is
¼
P4 O10 0:162 mol 1:00 mol
Therefore, H2O is the limiting reactant and the H3PO4 produced is:
4 mol H3 PO4
97:99 g
ð0:832 mol H2 OÞ
¼ 54:4 g H3 PO4
6 mol H2 O2
mol
- 104 -