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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
Ca 3P2 + 6 H 2O ¡ 3 Ca1OH22 + 2 PH 3
(a)
Correct: 11 mol Ca3P22 ¢
(b)
Incorrect: 1 g Ca 3P2 would produce 0.4 g PH 3
(1 g Ca 3P2)a
2 mol PH3
≤ = 2 mol PH3
1 mol Ca3P2
33.99 g
2 mol PH 3
1 mol
b¢
b = 0.4 g PH 3
≤a
182.2 g 1 mol Ca 3P2
mol
(c)
Correct: see equation
(d)
Correct: see equation
(e)
Incorrect: 2 mol Ca 3P2 requires 12 mol H 2O to produce 4.0 mol PH 3 .
12 mol Ca 3P22 ¢
(f)
6 mol H 2O
≤ = 12 mol H 2O
1 mol Ca 3P2
Correct: 2 mol Ca 3P2 will react with 12 mol H 2O (3 mol H 2O are present in excess)
and 6 mol Ca(OH)2 will be formed.
12 mol Ca 3P22 ¢
3 mol Ca(OH)2
≤ = 6 mol Ca(OH)2
1 mol Ca 3P2
18.02 g
6 mol H 2O
1 mol
b¢
b = 119 g H 2O
≤a
182.2 g 1 mol Ca 3P2
mol
The amount of water present (100. g) is less than needed to react with
200. g Ca 3P2 . H 2O is the limiting reactant.
(g)
Incorrect: (200. g Ca 3P2)a
(h)
Incorrect: water is the limiting reactant.
1100. g H 2O2a
33.99 g
2 mol PH 3
1 mol
b¢
b = 62.9 g PH 3
≤a
18.02 g 6 mol H 2O
mol
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- Chapter 9 4.
The balanced equation is
2 CH 4 + 3 O2 + 2 NH 3 ¡ 2 HCN + 6 H 2O
(a) Correct
2 mol HCN
(b) Incorrect: 116 mol O22 ¢
≤ = 10.7 mol HCN (not 12 mol HCN)
3 mol O2
(c) Correct
6 mol H 2O
(d) Incorrect: 112 mol HCN2 ¢
≤ = 36 mol H 2O (not 4 mol H 2O)
2 mol HCN
(e) Correct
(f)
Incorrect: O2 is the limiting reactant
13 mol O22 ¢
2 mol HCN
≤ = 2 mol HCN (not 3 mol HCN)
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.
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Page 97
CHAPTER 9
SOLUTIONS TO EXERCISES
1.
2.
3.
(a)
125.0 g KNO32a
1 mol
b = 0.247 mol KNO3
101.1 g
(b)
156 mmol NaOH2a
(c)
15.4 * 102 g (NH 4)2C2O42a
(d)
The conversion is: mL sol ¡ g sol ¡ g H 2SO4 ¡ mol H 2SO4
1 mol
b = 0.056 mol NaOH
1000 mmol
1 mol
b = 4.4 mol (NH 4)2C2O4
124.1 g
116.8 mL solution2a
1.727 g 0.800 g H 2SO4
1 mol
b¢
b = 0.237 mol H 2SO4
≤a
mL
g solution
98.09 g
(a)
12.10 kg NaHCO32a
1000 g
1 mol
ba
b = 25.0 mol NaHCO3
kg
84.01 g
(b)
1525 mg ZnCl 22a
(c)
19.8 * 1024 molecules CO22 ¢
(d)
1250 mL C2H 5OH2a
(a)
12.55 mol Fe(OH)32a
(b)
1125 kg CaCO32a
(c)
110.5 mol NH 32a
(d)
172 mmol HCl2a
36.46 g
1 mol
ba
b = 2.6 g HCl
1000 mmol
mol
(e)
1500.0 mL Br22a
3.119 g
b = 1559.5 g Br2 = 1.560 * 103 g Br2
mL
1g
1 mol
ba
b = 3.85 * 10-3 mol ZnCl 2
1000 mg 136.3 g
1 mol
≤ = 16 mol CO2
6.022 * 1023 molecules
0.789 g
1 mol
ba
b = 4.3 mol C2H 5OH
mL
46.07 g
106.9 g
b = 273 g Fe(OH)3
mol
1000 g
b = 1.25 * 105 g CaCO3
kg
17.03 g
b = 179 g NH 3
mol
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- Chapter 9 -
4.
154.8 g
b = 1.31 g NiSO4
mol
(a)
(0.00844 mol NiSO4)a
(b)
(0.0600 mol HC2H 3O2)a
(c)
(0.725 mol Bi 2S3)a
(d)
(4.50 * 1021 molecules C6H 12O6) ¢
60.05 g
b = 3.60 g HC2H 3O2
mol
514.2 g
b = 373 g Bi 2S3
mol
180.2 g
1 mol
b
≤a
23
mol
6.022 * 10 molecules
= 1.35g C6H 12O6
(e)
5.
(75 mL solution2a
1.175 g 0.200 g K 2CrO4
ba
b = 18 g K 2CrO4
mL
g solution
Larger number of molecules: 10.0 g H 2O or 10.0 g H 2O2
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 H 2O2 .
6.
Larger number of molecules: 25.0 g HCl or 85 g C6H 12O6
125.0 g HCl2a
1 mol
6.022 * 1023 molecules
ba
b = 4.13 * 1023 molecules HCl
36.46 g
mol
185.0 g C6H 12O62a
1 mol
6.022 * 1023 molecules
ba
b
180.2 g
mol
= 2.84 * 1023 molecules C6H 12O6
HCl contains more molecules
7.
Mole ratios
2 C3H 7OH + 9 O2 ¡ 6 CO2 + 8 H 2O
(a)
6 mol CO2
2 mol C3H 7OH
(d)
8 mol H 2O
2 mol C3H 7OH
(b)
2 mol C3H 7OH
9 mol O2
(e)
6 mol CO2
8 mol H 2O
(c)
9 mol O2
6 mol CO2
(f)
8 mol H 2O
9 mol O2
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- Chapter 9 8.
9.
Mole ratios
3 CaCl 2 + 2 H 3PO4 ¡ Ca 3(PO4)2 + 6 HCl
(a)
3 mol CaCl 2
1 mol Ca 3(PO4)2
(d)
1 mol Ca 3(PO4)2
2 mol H 3PO4
(b)
6 mol HCl
2 mol H 3PO4
(e)
6 mol HCl
1 mol Ca 3(PO4)2
(c)
3 mol CaCl 2
2 mol H 3PO4
(f)
2 mol H 3PO4
6 mol HCl
C2H 5OH + 3 O2 ¡ 2 CO2 + 3 H 2O
17.75 mol C2H 5OH2 ¢
10.
The balanced equation is 4 HCl + O2 ¡ 2 Cl2 + 2 H2O
15.60 mol HCl2 ¢
11.
12.
2 mol CO2
≤ = 15.5 mol CO2
1 mol C2H 5OH
2 mol Cl2
≤ = 2.80 mol Cl2
4 mol HCl
The balanced equation is
MnO2(s) + 4 HCl(aq) ¡ Cl2(g) + MnCl2(aq) + 2 H2O(l)
(a)
11.05 mol MnO22 ¢
(b)
11.25 mol H2O2 ¢
4 mol HCl
≤ = 4.20 mol HCl
1 mol MnO2
1 mol MnCl2
≤ = 0.625 mol MnCl2s
2 mol H2O
Al 4C3 + 12 H 2O ¡ 4 Al(OH)3 + 3 CH 4
(a)
1100. g Al 4C32a
12 mol H 2O
1 mol
b¢
≤ = 8.33 mol H 2O
144.0 g 1 mol Al 4C3
(b)
10.600 mol CH 42 ¢
4 mol Al(OH)3
≤ = 0.800 mol Al(OH)3
3 mol CH 4
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- Chapter 9 13.
Grams of NaOH
Ca(OH)2 + Na 2CO3 ¡ 2 NaOH + CaCO3
The conversion is: g Ca(OH)2 ¡ mol Ca(OH)2 ¡ mol NaOH ¡ g NaOH
1500 g Ca(OH)22a
14.
40.00 g
1 mol
2 mol NaOH
ba
ba
b = 5 * 102 g NaOH
74.10 g 1 mol Ca(OH)2
mol
Grams of Zn 3(PO4)2
3 Zn + 2 H 3PO4 ¡ Zn 3(PO4)2 + 3 H 2
The conversion is: g Zn ¡ mol Zn ¡ mol Zn 3(PO4)2 ¡ g Zn 3(PO4)2
110.0 g Zn2a
1 mol Zn 3(PO4)2 386.1 g
1 mol
b¢
b = 19.7 g Zn 3(PO4)2
≤a
65.39 g
3 mol Zn
mol
15.
The balanced equation is Fe 2O3 + 3 C ¡ 2 Fe + 3 CO
The conversion is: kg Fe 2O3 ¡ kmol Fe 2O3 ¡ kmol Fe ¡ kg Fe
55.85 kg
1 kmol
2 kmol Fe
1125 kg Fe 2O32a
ba
ba
b = 87.4 kg Fe
159.7 kg 1 kmol Fe 2O3
kmol
16.
The balanced equation is 3 Fe + 4 H 2O ¡ Fe 3O4 + 4 H 2
Calculate the grams of both H 2O and Fe to produce 375 g Fe 3O4
17.
1375 g Fe 3O42 ¢
18.02 g
4 mol H 2O
1 mol
b = 117 g H 2O
≤a
≤a
231.6 g 1 mol Fe 3O4
mol
1375 g Fe 3O42a
55.85 g
3 mol Fe
1 mol
b¢
b = 271 g Fe
≤a
231.6 g 1 mol Fe 3O4
mol
The balanced equation is 2 C2H 6 + 7 O2 ¡ 4 CO2 + 6 H 2O
(a)
115.0 mol C2H 62 ¢
(b)
18.00 g H 2O2a
(c)
175.0 g C2H 62a
(d)
12.75 mol of H2O2 ¢
7 mol O2
≤ = 52.5 mol O2
2 mol C2H 6
44.01 g
4 mol CO2
1 mol
b¢
b = 13.0 g CO2
≤a
18.02 g 6 mol H 2O
mol
44.01 g
4 mol CO2
1 mol
b¢
b = 2.20 * 102 g CO2
≤a
30.07 g 2 mol C2H 6
mol
44.01 g
4 mol CO2
b = 80.7 g CO2
≤a
6 mol H2O
mol
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- Chapter 9 -
18.
19.
(e)
125.0 mol C2H62 ¢
(f)
1125 g H2O2 ¢
32.00 g
7 mol O2
b = 2.80 * 103 g O2
≤a
2 mol C2H6
mol O2
30.07 g
2 mol C2H6
1 mol
b = 69.5 g C2H6
≤¢
≤ a
18.02 g
6 mol H2O
mol
4 FeS2 + 11 O2 ¡ 2 Fe 2O3 + 8 SO2
(a)
11.00 mol FeS22a
2 mol Fe 2O3
b = 0.500 mol Fe 2O3
4 mol FeS2
(b)
14.50 mol FeS22 ¢
11 mol O2
≤ = 12.4 mol O2
4 mol FeS2
(c)
11.55 mol Fe 2O32 ¢
8 mol SO2
≤ = 6.20 mol SO2
2 mol Fe 2O3
(d)
10.512 mol FeS22 ¢
64.07 g
8 mol SO2
b = 65.6 g SO2
≤a
4 mol FeS2
mol
(e)
140.6 g SO22a
(f)
1221 g Fe 2O32a
(a)
11 mol O2
1 mol
b¢
≤ = 0.871 mol O2
64.07 g 8 mol SO2
120.0 g
4 mol FeS2
1 mol
b¢
b = 332 g FeS2
≤a
159.7 g 2 mol Fe 2O3
mol
Hydrogen
Oxygen
Hydrogen is the limiting reactant.
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- Chapter 9 -
20.
21.
(b)
Hydrogen
Bromine
Bromine is the limiting reactant.
(a)
Lithium
Iodine
No limiting reactant.
(b)
Silver
Chlorine
Silver is the limiting reactant.
(a)
Potassium
Chlorine
Potassium is the limiting reactant.
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- Chapter 9 -
22.
23.
(b)
Aluminum
Oxygen
Oxygen is the limiting reactant.
(a)
Nitrogen
Oxygen
Oxygen is the limiting reactant.
(b)
Iron
Hydrogen
Oxygen
Water is the limiting reactant.
(a)
HNO3
KOH
+
KNO3
+
H 2O
¡
16.0 g
12.0 g
Choose one of the products and calculate its mass that would be produced from
each given reactant. Using KNO3 as the product:
116.0 g KOH2 ¢
1 mol KNO3 101.1 g
1 mol
b = 28.8 g KNO3
≤a
≤a
56.10 g
1 mol KOH
mol
112.0 g HNO32a
1 mol KNO3 101.1 g
1 mol
b¢
b = 19.3 g KNO3
≤a
63.02 g
1 mol KOH
mol
Since HNO3 produces less KNO3 , it is the limiting reactant and KOH is in excess.
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- Chapter 9 (b)
2 NaOH
10.0 g
H 2SO4
10.0 g
+
¡
Na 2SO4
2 H 2O
+
Choose one of the products and calculate its mass that would be produced from
each given reactant. Using H 2O as the product:
110.0 g NaOH2a
18.02 g
2 mol H 2O
1 mol
b¢
b = 4.51 g H 2O
≤a
40.00 g 2 mol NaOH
mol
110.0 g H 2SO42a
18.02 g
2 mol H 2O
1 mol
b¢
b = 3.67 g H 2O
≤a
98.09 g 1 mol H 2SO4
mol
Since H 2SO4 produces less H 2O, it is the limiting reactant and NaOH is in excess.
24.
(a)
2 Bi(NO3)3
50.0 g
3 H 2S
6.00 g
+
Bi 2S3
¡
+
6 HNO3
Choose one of the products and calculate its mass that would be produced from
each given reactant. Using Bi 2S3 as the product:
150.0 g Bi(NO3)3)a
(6.00 g H 2S)a
514.2 g
1 mol Bi2S3
1 mol
ba
ba
b = 32.5 g Bi2S3
395.0 g 2 mol Bi(NO3)3
mol
1 mol Bi 2S3 514.2 g
1 mol
b¢
b = 30.2 g Bi 2S3
≤a
34.09 g
3 mol H 2S
mol
Since H 2S produces less Bi 2S3 , it is the limiting reactant and Bi(NO3)3 is in excess.
(b)
3 Fe
40.0 g
4 H 2O
16.0 g
+
¡
Fe 3O4
+
4 H2
Choose one of the products and calculate its mass that would be produced from
each given reactant. Using H 2 as the product:
(40.0 g Fe)a
4 mol H 2 2.016 g
1 mol
b¢
b = 1.93 g H 2
≤a
55.85 g
3 mol Fe
mol
116.0 g H 2O2a
2.016 g
4 mol H 2
1 mol
b¢
b = 1.79 g H 2
≤a
18.02 g 4 mol H 2O
mol
Since H 2O produces less H 2 , it is the limiting reactant and Fe is in excess.
25.
Limiting reactant calculations
C3H 8 + 5 O2 ¡ 3 CO2 + 4 H 2O
(a)
Reaction between 20.0 g C3H 8 and 20.0 g O2
Convert each amount to moles of CO2
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- Chapter 9 -
120.0 g C3H82a
120.0 g O22a
3 mol CO2
1 mol
b¢
≤ = 1.36 moles CO2
44.09 g 1 mol C3H8
3 mol CO2
1 mol
b¢
≤ = 0.375 moles CO2
32.00 g
5 mol O2
O2 is the limiting reactant. The yield is 0.375 moles CO2 .
(b)
Reaction between 20.0 g C3H 8 and 80.0 g O2
Convert each amount to moles of CO2
120.0 g C3H82a
180.0 g O22a
3 mol CO2
1 mol
b¢
≤ = 1.36 moles CO2
44.09 g 1 mol C3H8
3 mol CO2
1 mol
b¢
≤ = 1.50 moles CO2
32.00 g
5 mol O2
C3H 8 is the limiting reactant. The yield is 1.36 moles CO2 .
(c)
Reaction between 2.0 mol C3H 8 and 14.0 mol O2
According to the equation, 2 mol C3H 8 will react with 10 mol O2 . Therefore, C3H 8
is the limiting reactant and 4.0 mol O2 will remain unreacted.
12.0 mol C3H 82 ¢
(2.0 mol C3H 8) ¢
3 mol CO2
≤ = 6.0 mol CO2 produced
1 mol C3H 8
4 mol H 2O
≤ = 8.0 mol H 2O produced
1 mol C3H 8
When the reaction is completed, 6.0 mol CO2 , 8.0 H 2O, and 4.0 mol O2 will be in
the container.
26.
The balanced equation is 2 C3H6 + 9 O2 ¡ 6 CO2 + 6 H2O
(a) Reaction between 15.0 g C3H6 and 15.0 g O2.
Convert each amount to moles of H2O
115.0 g C3H62 ¢
115.0 g O22 ¢
6 mol H2O
1 mol
≤¢
≤ = 1.07 mol H2O
42.08 g 2 mol C3H6
6 mol H2O
1 mol
≤¢
≤ = 0.313 mol H2O
32.00 g
2 mol O2
The O2 is the limiting reactant. The yield is 0.313 mol H2O.
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- Chapter 9 (b)
Reaction between 12.0 g of C3H6 and 25.0 g of O2.
Convert each amount to moles of H2O
112.0 g C3H62 ¢
125.0 g O22 ¢
6 mol H2O
1 mol
≤¢
≤ = 0.856 mol H2O
42.08 g 2 mol C3H6
6 mol H2O
1 mol
≤¢
≤ = 0.521 mol H2O
32.00 g
9 mol O2
O2 is the limiting reactant. The yield is 0.521 mol H2O.
(c)
Reaction between 5.0 mol of C3H6 and 15.0 mol of O2.
Convert each to moles of CO2
(5.0 mol C3H6) ¢
6 mol CO2
≤ = 15 mol CO2
2 mol C3H6
6 mol CO2
≤ = 10 mol CO2
9 mol O2
Since O2 is the limiting reactant. C3H6 will be left unreacted.
(15.0 mol O2) ¢
27.
X8 + 12 O2 ¡ 8 XO3
The conversion is: g O2 ¡ mol O2 ¡ mol X8
(120.0 g O22a
1 mol X8
1 mol
b¢
≤ = 0.3125 mol X8
32.00 g 12 mol O2
80.0 g X8 = 0.3125 mol X8
80.0 g
= 256 g>mol X8
0.3125 mol
g
256
g
mol
= 32.0
molar mass X =
8
mol
Using the periodic table we find that the element with 32.0 g>mol is sulfur.
28.
X + 2 HCl ¡ XCl 2 + H 2
The conversion is: g H 2 ¡ mol H 2 ¡ mol X
12.42 g H 22a
1 mol
1 mol X
b¢
≤ = 1.20 mol X
2.016 g 1 mol H 2
78.5 g X = 1.20 molX
78.5 g
= 65.4 g>mol
1.20 mol
Using the periodic table we find that the element with atomic mass 65.4 is zinc.
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- Chapter 9 29.
Limiting reactant calculation and percentage yield
2 Al + 3 Br2 ¡ 2 AlBr3
Reaction between 25.0 g Al and 100. g Br2
Calculate the grams of AlBr3 from each reactant.
125.0 g Al2a
2 mol AlBr3 266.7 g
1 mol
b¢
b = 247 g AlBr3
≤a
26.98 g
2 mol Al
mol
(100. g Br2)a
2 mol AlBr3 266.7 g
1 mol
b¢
b = 111 g AlBr3
≤a
159.8 g
3 mol Br2
mol
Br2 is limiting; 111 g AlBr3 is the theoretical yield of product.
Percent yield = a
30.
64.2 g
actual yield
b11002 = a
b 11002 = 57.8%
theoretical yield
111 g
Percent yield calculation
Fe(s) + CuSO4(aq) ¡ Cu(s) + FeSO4(aq)
1400. g CuSO42a
% yield = a
31.
63.55 g
1 mol
1 mol Cu
b¢
b = 159 g Cu (theoretical yield)
≤a
159.6 g 1 mol CuSO4
mol
151 g
actual yield
b 11002 = a
b11002 = 95.0% yield of Cu
theoretical yield
159 g
The balanced equation is 3 C + 2 SO2 ¡ CS2 + 2 CO2
Calculate the g C needed to produce 950 g CS2 taking into account that the yield of CS2
is 86.0%. First calculate the theoretical yield of CS2 .
950 g CS2
= 1.1 * 103 g CS2 1theoretical yield2
0.860
Now calculate the grams of coke needed to produce 1.1 * 103 g CS2 .
11.1 * 103 g CS22a
32.
12.01 g
1 mol
3 mol C
b¢
b = 5.2 * 102 g C
≤a
76.15 g 1 mol CS2
mol
The balanced equation is CaC2 + 2 H 2O ¡ C2H 2 + Ca1OH22
First calculate the grams of pure CaC2 in the sample from the amount of C2H 2 produced.
10.540 mol C2H22 ¢
1 mol CaC2 64.10 g CaC2
b = 34.6 g of pure CaC2 in the sample
≤a
1 mol C2H2
mol CaC2
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- Chapter 9 -
Now calculate the percent CaC2 in the impure sample.
34.6 g CaC2
≤ 11002 = 77.8% CaC2 in the impure sample
44.5 g sample
No. There are not enough screwdrivers, wrenches or pliers. 2400 screwdrivers,
3600 wrenches and 1200 pliers are needed for 600 tool sets.
¢
33.
34.
A subscript is used to indicate the number of atoms in a formula. It cannot be changed
without changing the identity of the substance. Coefficients are used only to balance atoms in
chemical equations. They may be changed as needed to achieve a balanced equation.
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 = 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 Á then from
that you can determine the grams of B using its molar mass. Equations are written in
terms of moles not grams.
36.
4 KO2 + 2 H 2O + 4 CO2 ¡ 4 KHCO3 + 3 O2
(a)
(b)
¢
0.85 g CO2
4 mol KO2
0.019 mol KO2
1 mol
b¢
≤a
≤ =
min
44.01 g
4 mol CO2
min
¢
0.019 mol KO2
b110.0 min2 = 0.19 mol KO2
min
The conversion is:
¢
37.
(a)
g CO2
g O2
g O2
mol CO2
mol O2
¡
¡
¡
¡
min
min
min
min
hr
0.85 g CO2
28 g O2
32.00 g 60.0 min
3 mol O2
1 mol
b¢
ba
b =
≤a
≤a
min
44.01 g 4 mol CO2
mol
1.0 hr
hr
1750 g C6H 12O62a
2 mol C2H 5OH 46.07 g
1 mol
b¢
b = 380 g C2H 5OH
≤a
180.2 g 1 mol C6H 12O6
mol
1750 g C6H 12O62a
44.01 g
2 mol CO2
1 mol
b¢
b = 370 g CO2
≤a
180.2 g 1 mol C6H 12O6
mol
Alternate Solution: 750 g C2H6O6 - 380 g
C2H5OH = 370 g CO2 by the conservation of mass method.
(b)
1380 g C2H 5OH2a
1 mL
b = 480 mL C2H 5OH
0.79 g
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- Chapter 9 38.
2 CH 3OH + 3 O2 ¡ 2 CO2 + 4 H 2O
The conversion is:
mL CH 3OH ¡ g CH 3OH ¡ mol CH 3OH ¡ mol O2 ¡ g O2
160.0 mL CH 3OH2a
39.
0.72 g
32.00 g
3 mol O2
1 mol
b¢
b = 65 g O2
≤¢
≤a
mL
32.04 g 2 mol CH 3OH
mol
The balanced equation is 7 H2O2 + N2 H4 ¡ 2 HNO3 + 8 H2O
The conversion is:
(a)
175 kg N2H42a
1000 g
2 mol HNO3 63.02 g
1 mol
b¢
b = 2.9 * 105 g HNO3
≤¢
≤a
1 kg
32.05 g
1 mol N2H4
mol
(b)
1250 L H2O22a
18.02 g
8 mol H2O
1 mol
1000 mL 1.41 g
b¢
ba
b
≤¢
≤a
1L
1 mL
34.02 g 7 mol H2O2
mol
= 2.1 * 105 g H2O
(c)
1725 g H2O22a
1 mol N2H4 32.05 g
1 mol
b¢
b = 97.6 g N2H4
≤a
34.02 g 7 mol H2O2
mol
(d)
Reaction between 750 g of N2H2 and 125 g of H2O2.
Convert each amount to grams of H2O.
1750 g N2H42a
18.02 g
8 mol H2O
1 mol
b¢
b = 3.4 * 103 g H2O
≤a
32.05 g 1 mol N2H4
mol
1125 g H2O22a
18.02 g
8 mol H2O
1 mol
b¢
b = 75.7 g H2O
≤a
34.02 g 7 mol H2O2
mol
75.7 g H2O can be produced.
(e)
Since H2O2 is the limiting reactant, N2H4 is in excess.
1125 g H2O22a
1 mol N2H4 32.05 g
1 mol
b¢
b = 16.8 g N2H4 reacted
≤a
34.02 g 7 mol H2O2
mol
750 g N2H4 given - 16.8 g N2 H4 used = 730 g N2 H4 remaining
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Page 110
- Chapter 9 40.
The balanced equation is
16 HCl + 2 KMnO4 ¡ 5 Cl2 + 2 KCl + 2 MnCl2 + 8 H2O
(a)
Reaction between 25 g KMnO4 and 85 g HCl. Convert each to moles of MnCl2.
125 g KMnO42a
1 mol KMnO4
2 mol MnCl2
ba
b = 0.16 mol MnCl2
158.04 g KMnO4 2 mol KMnO4
2 mol MnCl2
1 mol
ba
b = 0.29 mol MnCl2
36.46 g
16 mol HCl
KMnO4 is the limiting reactant; 0.16 mol MnCl2 produced.
185 g HCl2a
(b)
175 g KCl2a
(c)
1150 g HCl2a
8 mol H2O 18.02 g
1 mol
ba
ba
b = 73 g H2O
74.55 g
2 mol KCl
mol
70.90 g
5 mol Cl2
1 mol
ba
ba
b = 91 g Cl2
36.46 g 16 mol HCl
mol
75 g
Theoretical yield is 91 g Cl2; Percent yield: a 91 g b(100) = 82% yield
(d)
Reaction between 25 g HCl and 25 g KMnO4. Convert each amount to grams of CL2.
125 g HCl2a
70.90 g
5 mol Cl2
1 mol
ba
ba
b = 15 g Cl2
36.46 g 16 mol HCl
mol
125 g KMnO42a
70.90 g
5 mol Cl2
1 mol
ba
ba
b = 28 g Cl2
158.04 g 2 mol KMnO4
mol
HCl is the limiting; KMnO4 is in excess; 15 g Cl2 will be produced.
(e)
Calculate the mass of unreacted KMnO4:
125 g HCl2a
2 mol KMnO4 158.04 g
1 mol
ba
ba
b = 14 g KMnO4 will react.
36.46 g
16 mol HCl
mol
Unreacted KMnO4 = 25 g -14 g = 11 g KMnO4 remain unreacted.
41.
The balanced equation is
4Ag + 2 H2S + O2 ¡ 2 Ag2S + 2 H2O
(a)
11.1 g Ag2a
2 mol Ag2S 247.9 g
1 mol
ba
ba
b = 1.3 g Ag2S
107.9 g
4 mol Ag
mol
10.14 g H2S2a
2 mol Ag2S 247.9 g
1 mol
ba
ba
b = 1.0 g Ag2S
34.09 g
2 mol H2S
mol
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- Chapter 9 -
10.080 g O22a
(b)
2 mol Ag2S 247.9 g
1 mol
ba
ba
b = 1.2 g Ag2S
32.00 g
1 mol O2
mol
H2S is limiting 1.0 g Ag2S forms.
2 mol H2S 34.09 g
1 mol
ba
ba
b = 0.17 g H2S reacts
11.1 g Ag2a
107.9 g
4 mol Ag
mol
0.17 g - 0.14 g = 0.03 grams more H2S needed to completely react Ag.
42.
Mass of the beaker
26.500 g beaker + Ca(OH)2
26.095 g beaker + CaO
0.405 g H 2O
absorbed
10.405 g H 2O2a
1 mol
b = 2.25 * 10-2 mol H 2O absorbed
18.02 g
Since the reaction is a 1:1 mole, the amount of CaO in the beaker is 2.25 * 10-2 mol.
Convert to grams.
12.25 * 10-2 mol CaO2a
26.095 g
-1.26 g
24.835 g
43.
56.08 g CaO
b = 1.26 g CaO in the beaker.
mol
beaker + CaO
CaO
mass of the beaker
Pb(NO3)2(aq) + 2 KI(aq) ¡ PbI 2(s) + 2 KNO3(aq)
The solid is lead (II) iodide, PbI 2 .
(b) Double displacement reaction.
(c) Calculate the moles of each reactant.
1 mol
b = 0.045 mol Pb(NO3)2
[15 g Pb(NO3)2] + a
331.2 g
(a)
115 g KI2a
1 mol
b = 0.090 mol KI
166.0 g
Stoichiometric quantities of reactants are used.
Theoretical yield of PbI 2 is 0.045 mol.
Actual yield: 16.68 g PbI 22a
Percent yield: a
1 mol
b = 0.0145 mol PbI 2
461.1 g
0.0145 mol
b11002 = 32% yield
0.045 mol
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Page 112
- Chapter 9 44.
Composition of a mixture of KNO3 and KCl.
In the mixture only KCl reacts with AgNO3 .
KCl(aq) + AgNO3(aq) ¡ AgCl(s) + KNO3(aq)
14.33 g AgCl2a
74.55 g KCl
b = 2.25 g KCl in the mixture
143.7 g AgCl
10.00 g mixture - 2.25 g KCl = 7.75 g KNO3
45.
a
2.25 g KCl
b11002 = 22.5% KCl
10.00 g mixture
a
7.75 g KNO3
b11002 = 77.5% KNO3
10.00 g mixture
The balanced equation is Zn + 2 HCl ¡ ZnCl 2 + H 2
180.0 g Zn - 35 g Zn = 145 g Zn reacted with HCl
46.
(a)
1145 g Zn2a
2.016 g
1 mol H2
1 mol
ba
b a
b = 4.47 g H2 produced
65.39 g 1 mol Zn
mol
(b)
1145 g Zn2a
1 mol
2 mol HCl 36.46 g
ba
ba
b = 162 g HCl reacted
65.39 g
1 mol Zn
mol
(c)
1180.0 g Zn2a
1 mol
2 mol HCl 36.46 g
ba
ba
b = 201 g HCl reacts
65.39 g
1 mol Zn
mol
201 g - 162 g = 39 g more HCl needed to react wih the 180.0 g Zn
Fe(s)
2.0 mol
(a)
(b)
+
CuSO4(aq)
3.0 mol
¡
Cu(s)
+
FeSO4(aq)
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 Cu 63.55 g
1 mol
120.0 g Fe2a
ba
ba
b = 22.8 g Cu
55.85 g
1 mol Fe
mol
63.55 g
1 mol Cu
1 mol
ba
b = 15.9 g Cu
ba
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 .
140.0 g CuSO42a
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- Chapter 9 -
140.0 g CuSO42a
151.9 g
1 mol FeSO4
1 mol
b¢
b = 38.1 g FeSO4 produced
≤a
159.6 g 1 mol CuSO4
mol
Calculate the mass of unreacted Fe.
140.0 g CuSO42a
55.85 g
1 mol
1 mol Fe
b¢
b = 14.0 g Fe will react
≤a
159.6 g 1 mol CuSO4
mol
Unreacted Fe = 20.0 g - 14.0 g = 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.
47.
Limiting reactant calculation
CO(g) + 2 H 2(g) ¡ CH 3OH(l)
Reaction between 40.0 g CO and 10.0 g H 2 : determine the limiting reactant by
calculating the amount of CH 3OH that would be formed from each reactant.
140.0 g CO2a
110.0 g H 22a
1 mol CH 3OH 32.04 g
1 mol
b¢
b = 45.8 g CH 3OH
≤a
28.01 g
1 mol CO
mol
1 mol CH 3OH 32.04 g
1 mol
b¢
b = 79.5 g CH 3OH
≤a
2.016 g
2 mol H 2
mol
CO is limiting; H 2 is in excess; 45.8 g CH 3OH will be produced.
Calculate the mass of unreacted H 2 :
140.0 g CO2a
2.016 g
2 mol H 2
1 mol
b¢
b = 5.76 g H 2 react
≤a
28.01 g 1 mol CO
mol
10.0 g H 2 - 5.76 g H 2 = 4.2 g H 2 remain unreacted
48.
The balanced equation is C6H 12O6 ¡ 2 C2H 5OH + 2 CO2
(a)
First calculate the theoretical yield.
1750 g C6H 12O62a
2 mol C2H 5OH 46.07 g
1 mol
b¢
b
≤a
180.2 g 1 mol C6H 12O6
mol
= 3.8 * 102g C2H 5OH (theoretical yield)
Then take 84.6% of the theoretical yield to obtain the actual yield.
actual yield =
1theoretical yield2184.62
13.8 * 102 g C2H 5OH2184.62
=
100
100
= 3.2 * 102 g C2H 5OH
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- Chapter 9 (b)
475 g C2H 5OH represents 84.6% of the theoretical yield. Calculate the
theoretical yield.
475 g
= 561 g C2H 5OH
theoretical yield =
0.846
Now calculate the g C6H 12O6 needed to produce 561 g C2H 5OH.
1561 g C2H 5OH2a
49.
1 mol C6H 12O6 180.2 g
1 mol
b¢
b = 1.10 * 103 g C6H 12O6
≤a
46.07 g 2 mol C2H 5OH
mol
The balanced equations are:
CaCl 2(aq) + 2 AgNO3(aq) ¡ Ca(NO3)2(aq) + 2 AgCl(s)
MgCl 2(aq) + 2 AgNO3(aq) ¡ Mg1NO322(aq) + 2 AgCl(s)
1 mol of each salt will produce the same amount (2 mol) of AgCl. MgCl 2 has a higher
percentage of Cl than CaCl 2 because Mg has a lower atomic mass than Ca. Therefore, on
an equal mass basis, MgCl 2 will produce more AgCl than will CaCl 2 .
Calculations show that 1.00 g MgCl 2 produces 3.01 g AgCl, and 1.00 g CaCl 2 produces
2.56 g AgCl.
50.
51.
The balanced equation is Li 2O + H 2O ¡ 2 LiOH
The conversion is: g H 2O ¡ mol H 2O ¡ mol Li 2O ¡ g Li 2O ¡ kg Li 2O
¢
2500 g H 2O
1 kg
4.1 kg Li 2O
1 mol Li 2O 29.88 g
1 mol
b¢
ba
b =
≤a
≤a
astronaut day 18.02 g
1 mol H 2O
mol
1000 g
astronaut day
¢
4.1 kg Li 2O
≤ 130 days213 astronauts2 = 3.7 * 102 kg Li 2O
astronaut day
The balanced equation is
H 2SO4 + 2 NaCl ¡ Na 2SO4 + 2 HCl
First calculate the g HCl to be produced
120.0 L HCl solution2a
1.20 g
1000 mL
ba
b 10.4202 = 1.01 * 104 g HCl
1L
1.00 mL
Then calculate the g H 2SO4 required to produce the HCl
11.01 * 104 g HCl2a
1 mol H 2SO4 98.09 g
1 mol
b¢
b = 1.36 * 104 g H 2SO4
≤a
36.46 g
2 mol HCl
1 mol
Finally, calculate the kg H 2SO4 (96%)
11.36 * 104 g H 2SO42 ¢
1.00 g H 2SO4 solution
1 kg
b = 14 kg concentrated H 2SO4
≤a
0.96 g H 2SO4
1000 g
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- Chapter 9 52.
Percent yield of H 2SO4
1100.0 g S2a
1 mol
b = 3.118 mol S to start with
32.07 g
3.118 mol S ¡ 3.118 mol SO2 - 10% = 2.806 mol SO2
-0.3118
2.8062
2.806 mol SO2 ¡ 2.806 mol SO3 - 10% = 2.525 mol SO3
-0.2806
2.5254
2.525 mol SO3 ¡ 2.525 mol H 2SO4 - 10% = 2.273 mol H 2SO4
-0.2525
2.2725
98.09 g
12.273 mol H 2SO42a
b = 223.0 g H 2SO4 formed
mol
1 mol H 2SO4
13.118 mol S2a
b = 3.118 mol H 2SO4 (theoretical yield)
1 mol S
¢
2.273 mol H 2SO4
≤ 11002 = 72.90% yield
3.118 mol H 2SO4
Alternate Solution:
Calculation of yield. There are three chemical steps to the formation of H 2SO4 . Each step
has a 10% loss of yield.
Step 1:
Step 2:
Step 3:
100% yield - 10% = 90.00% yield
90.00% yield - 10% = 81.00% yield
81.00% yield - 10% = 72.90% yield
Now calculate the grams of product. One mole of sulfur will yield a maximum of 1 mol
H 2SO4 . Therefore 3.118 mol S will give a maximum of 3.118 mol H 2SO4 .
1 mol H 2SO4 98.09 g
13.118 mol S2a
ba
b 10.72902 = 223.0 g H 2SO4 yield
1 mol S
mol
53.
According to the equations, the moles of CO2 come from both reactions and the moles
H 2O come from only the first reaction.
So the mol NaHCO3 = 2 * mol H 2O = 2 * 0.0357 mol = 0.0714 mol NaHCO3
84.01 g
b = 6.00 g NaHCO3 in the sample
10.0714 mol NaHCO32a
mol
110.00 g NaHCO3 + Na 2CO32 - 6.00 g NaHCO3 = 4.00 g Na 2CO3 in the sample
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- Chapter 9 -
a
6.00 g NaHCO3
b11002 = 60.0% NaHCO3
10.00 g
4.00 g Na2CO3
a
b11002 = 40.0% Na2CO3
10.00 g
54.
The balanced equation is 2 KClO3 ¡ 2 KCl + 3 O2
12.82 g mixture - 9.45 g residue = 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
13.37 g O22a
a
55.
2 mol KClO3 122.6 g
1 mol
ba
ba
b = 8.61 g KClO3 in the mixture
32.00 g
3 mol O2
mol
8.61 g KClO3
b11002 = 67.2% KClO3
12.82 g sample
The balanced equation is
Al(OH)3(s) + 3 HCl(aq) ¡ AlCl 3(aq) + 3 H 2O(l)
The conversion is: L HCl ¡ g HCl ¡ mol HCl ¡ mol Al(OH)3 : g Al(OH)3
a
1 mol Al1OH23 78.00 g
2.5 L 3.0 g HCl
1 mol
ba
ba
b¢
b = 5.3 g Al1OH23>day
≤a
day
L
36.46 g
3 mol HCl
mol
Now calculate the number of 400. mg tablets that can be made from 5.3 g Al(OH)3
¢
56.
5.3 g Al(OH)3 1000 mg
1 tablet
b = 13 tablets>day
ba
≤a
g
day
400. mg
4 P + 5 O2 ¡ P4O10
P4O10 + 6 H 2O ¡ 4 H 3PO4
In the first reaction:
1 mol
120.0 g P2a
b = 0.646 mol P
30.97 g
1 mol
b = 0.938 mol O2
130.0 g O22a
32.00 g
3.44 mol P
0.646 mol P
=
This is a ratio of
0.938 mol O2
5.00 mol O2
Therefore, P is the limiting reactant and the P4O10 produced is:
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- Chapter 9 -
10.646 mol P2 ¢
1 mol P4O10
≤ = 0.162 mol P4O10
4 mol P
In the second reaction:
115.0 g H 2O2a
1 mol
b = 0.832 mol H 2O
18.02 g
and we have 0.162 mol P4O10 . The ratio of
H 2O
0.832 mol
5.14 mol
is
=
P4O10
0.162 mol
1.00 mol
Therefore, H 2O is the limiting reactant and the H 3PO4 produced is:
10.832 mol H 2O2 ¢
4 mol H 3PO4 97.99 g
b = 54.4 g H 3PO4
≤a
6 mol H 2O
mol
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