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Answers to Finding Limiting Reagents Practice Problems
1. CCl4 + O2 --> C02 + 2Cl2
CCl4 - 29.5 grams O2 - 9.92 grams
GFW = 12 + 4 (35.5) = 154 GFW = 32
# moles = 29.5/154 # moles = 9.92/32
# moles = .19 # moles = .31
Since the ratio is 1:1, and .31 is not 1:1 of .19 but rather 2 times as much as it, CCl3 is the limiting reagent.
2. 2C2H3O2Cl + 3O2 --> 4CO2 + 2H2O + HCl
C2H3O2Cl - 47.8 grams O2 - 43.5 grams
GFW = 2 (12) + 3 (1) + 2 (16) + 35.5 = 94.5 GFW = 32
# moles = 47.8/94.5 # moles = 43.5/32
# moles = .51 # moles = 1.4
Since the ratio is 2:3 and .51 is not 2/3 as much as 1.4, C2H3O2Cl is the limiting reagent.
3. 4C2H3Br3 + 11O2 --> 8CO2 + 6H20 + 6Br2
C2H3Br3 - 39.8 grams O2 - 25.1 grams
GFW = 2 (12) + 3 (1) + 3 (80) = 267 GFW = 32
# moles = 39.8/267 # moles = 25.1/32
# moles = .149 # moles = .784
Since the ratio is 4:11 and .149 is not 4/11 as much as .748, C2H3Br3 is the limiting reagent.
4. 4C2H3Br3 + 11O2 --> 8CO2 + 6H2O + 6Br2
C2H3Br3 - 39.8 grams O2 - 25.1 grams
GFW = 2 (12) + 3 (1) + 3 (80) = 267 GFW= 32
# moles = grams/GFW # moles = grams/GFW
# moles = 39.3/267 # moles = 25.1/32
# moles = .149 # moles = .784
Since the ratio is 4:11, and .149 is not 4/11 of .784, C2H3Br3 is the limiting reagent.
5. SiO2 + 3C --> 2SiC + CO2
SiO2 - 3.91 grams C - 3.63 grams
GFW = 28 + 2 (16) = 60 GFW = 12
# moles = 3.91/60 # moles = 3.63/12
# moles = .07 # moles = .3025
Since the ratio is 1:3, and .07 is not 1/2 of .3025, SiO2 is the limiting reagent.
6. 2C6H4Cl2 + 13O2 --> 4CO2 + 6HO4 + ClO2
C6H4Cl2 - 5.08 grams O2 - 4.07 grams
GFW = 6 (12) + 4 (1) + 2 (35.5) = 147 GFW = 32
# moles = 5.08/147 # moles = 4.07/32
# moles = .035 # moles = .127
Since the ratio is 2:13, and .127 is not 13/2 as much as .035, O2 is the limiting reagent.
7. 2Zns + 3O2 --> ZnO + SO2
ZnS - 21 grams O2 - 13.6 grams
GFW = 65.4 + 32 = 97.4 GFW = 32
# moles = 21/97.4 # moles = 13.6/32
# moles = .216 # moles = .425
Since the ratio is 2:3 and .216 is not 2/3 as much as .425, ZnS is the limiting reagent.
8. 3HCl + Na3PO4 --> H3NaPO2 + CLNa2O
HCl - 19.4 grams Na3PO4 - 54.6 grams
GFW = 1 + 35.5 = 36.5 GFW = 3 (23) + 31 + 4 (16) = 164
# moles = 19.4/36.5 # moles = 54.6/164
# moles = .532 # moles = .333
Since the ratio is 3:1 and .532 is not 3 times as much as .333, HCl is the limiting reagent.
9. 4C2H3O2F + 7O2 --> CO2 + 2HO2 + O4 + F2O
C2H3O2F - 9.99 grams O2 - 9.06 grams
GFW = 2 (12) + 3 + 2 (16) + 19 = 78 GFW = 32
# moles = 9.99/78 # moles = 9.06/32
# moles = .13 # moles = .283
Since the ratio is 4:7 and .13 is not 4/7 as much as .283, C2H3O2F is the limiting reagent.
10. 2CH4S + 7O2 --> CO2 + OH + S7O2
CH4S - 53.4 grams O2 - 197 grams
GFW = 12 + 4 (1) + 32 = 48 GFW = 32
# moles = 53.4/48 # moles = 197/32
# moles = 1.11 # moles = 6.2
Since the ratio is 2:7 and 1.11 is not 2/7 as much as 6.2, CH4S is the limiting reagent.
Answers to Using Limiting Reagents Problems (Note - LR is limiting reagent)
1. 3HCl + Na3PO4 --> H3PO4 + 3NaCl
HCl - 48 grams Na3PO4 - 48.3 grams
GFW = 1 + 35.5 = 36.5 GFW = 3 (23) + 31 + 4 (16) = 164
# moles = 48/36.5 # moles = 48.3/164
# moles = 1.32 # moles = .295 (LR)
.295 x 3/1 = .885 moles
# moles = g/GFW
GFW of NaCl (product in problem) = 23 + 35.5 = 58.5
.885 = g/58.5
g = 51.77
51.77 grams of NaCl would be produced.
2. CF4 + O2 --> CO2 + 2F2
CF4 - 38.2 grams O2 - 9.65 grams
GFW = 12 + 4 (19) = 88 GFW = 32
# moles = 38.2/88 # moles = 9.65/32
# moles = .434 # moles = .302 (LR)
.302 x 1/1 = .302
GFW of CO2 = 12 + 32 = 44
.302 = g/44
g = 13.288
13.288 grams of CO2 would be produced.
3. CaH2 + 2H2O --> Ca(OH)2 + 2H2
CaH2 - 76.5 grams H2O - 120 grams
GFW = 40 + 2 (1) = 42 GFW = 2 (1) + 16 = 18
# moles = 76.5/42 # moles = 120/18
# moles = 1.82 (LR) # moles = 6.67
1.82 x 1/1 = 1.82
GFW of Ca(OH)2 = 40 + 2 (16) + 2 (1) = 74
1.82 = g/74
g= 134.7
134.7 grams of Ca(OH)2 would be produced.
4. Fe3O4 + 4H2 --> 3Fe + 4H2O
Fe3O4 - 30.2 grams H2 - 1.82 grams
GFW = 3 (56) + 4 (16) = 232 GFW = 2 (1) = 2
# moles = 30.2/232 # moles = 1.82/2
# moles = .130 (LR) # moles = .91
.130 x 4/1 = .52
GFW of H2O = 2 (1) + 16 = 18
.52 = g/18
g = 9.36
9.36 grams of H2O would be produced.
5. 4C5N + 11O2 --> 4CO2 + 10H2O + 4NO
CH5N - 53.6 grams O2 - 87.9 grams
GFW = 12 + 5 (1) + 14 = 31 GFW = 32
# moles = 53.6/31 # moles = 87.9/32
# moles = 1.73 # moles = 2.75 (LR)
2.75 x 4/11 = 1
GFW of CO2 = 12 + 2 (16) = 44
# moles = grams/GFW
1 = g/44
44 grams of CO2 would have been produced.
6. 4C2H3Br3 + 11O2 --> 8CO2 + 6H2O = 6Br2
C2H3Br2 - 39.8 grams O2 = 25.1 grams
GFW = 2 (12) + 3 (1) + 3 (80) = 267 GFW = 32
# moles = 39.8/267 # moles = 25.1/32
# moles = .149 (LR) # moles = .784
.149 x 6/4 = .2235
GFW of H2O = 2 (1) + 16 = 18
.2235 = g/18
g = 4.023
4.023 grams of H2O would have been produced.
7. 3Ca(OH)2 + 2H3PO4 --> Ca3(PO4)2 + 6OH
Ca(OH)2 - 19.7 grams H3PO4 - 25.2 grams
GFW = 40 + 2 (1) + 2 (16) = 74 GFW = 3 (1) + 31 + 4 (16) = 98
# moles = 19.7/74 # moles = 25.2/98
# moles = .266 (LR) # moles = .257
.266 x 1/3 = .078
GFW of Ca3(PO4)2 = 3 (40) + 2 (31) + 8 (16) = 310
.078 = g/310
g= 24.18
24.18 grams of Ca3(PO4)2 would have been produced.
8. 4C2H3Br + 11O2 --> 8CO2 + 6H2O + 2Br2
CaH3Br - 41.8 grams O2 - 24.9 grams
GFW = 2 (12) + 3 (1) + 80 = 107 GFW = 32
# moles = 41.8/107 # moles = 24/9/32
# moles = .391 # moles = .778 (LR)
.778 x 6/11 = .424
GFW of H2O = 2 (1) + 16 = 18
.424 = g/18
g = 7.632
7.632 grams of H2O would have been produced.
9. 4C2H5Cl2 + 13O2 --> 8CO2 + 10H2O + 4Cl2
C2H5Cl2 - 57.5 grams O2 - 38.6 grams
GFW = 2 (12) + 5 (1) + 2 (35.5) = 100 GFW = 32
# moles = 57.5/100 # moles = 38.6/32
# moles = .575 # moles = 1.21 (LR)
1.21 x 4/13 = .372
GFW of Cl2 = 2 (35.5) = 71
.372 = g/71
g = 26.4
26.4 grams of Cl2 would have been produced.
10. 4FeS2 + 11O2 --> 2Fe2O3 + 8SO2
FeS2 - 41 grams O2 - 54 grams
GFW = 56 + 2 (32) = 120 GFW = 32
# moles = 41/120 # moles = 54/32
# moles = .342 (LR) # moles = 1.69
.342 x 2/4 = .171
GFW of Fe2O3 = 2 (56) + 3 (16) = 160
.171 = g/160
g = 27.4
27.4 grams of Fe2O3 would have been produced.
Answers to Theoretical Yield Problems
1. For the balanced equation shown below, if the reaction of 35.9 grams of CH3Cl produces 6.92 grams of H2O, what is the percent yield?
4CH3Cl + 7O2 --> 4CO2 +6H2O + 2Cl2
GFW of CH3Cl = 12 + 3 (1) + 35.5 = 50.5
35.9/50.5 = .711 mol
.711 x 6/4 = 1.07
GFW of the product (H2O) = 2 (1) + 16 = 18
18 x 1.07 = 19.3 g (theoretical yield)
Percentage Yield = (actual yield/theoretical yield) x 100
Percentage Yield = (6.92/19.3) x 100
Percentage Yield = 36 %
2. For the balanced equation shown below, if the reaction of 32.7 grams of CH2Cl2 produces 2.21 grams of H2O, what is the percent yield?
2CH2Cl2 + 3O2 --> 2CO2 + 2H2O + 2Cl2
GFW of CH2Cl2 = 12 + 2 (1) + 2 (35.5) = 85
32.7/85 = .385 moles
.385 x 2/2 = .385
GFW of the product H2O = 18
18 x .385 = 6.93 (theoretical yield)
% Yield = (actual yield/theoretical yield) x 100
% Yield = (2.21/6.93) x 100
% Yield = 32 %
3. For the balanced equation shown below, if the reaction of 23.1 grams of O2 produces 69.8 grams of CO, what is the percent yield?
2C2H3O2Br + O2 --> 4CO + 2H2O + 2HBr
GFW of O2 = 32
23.1/32 = .722 moles
.722 x 4/1 = 2.888
GFW of the product CO = 12 + 16 = 28
28 x 2.888 = 80.86 (theoretical yield)
% Yield = (69.8/80.86) x 100
% Yield = 86.32 %
4. For the balanced equation shown below, if the reaction fo 94.9 grams of C3H6 produces 152 grams of CO, what is the percent yield?
C3H6 + 3O2 --> 3CO + 2H2O
GFW of C3H6 = 3 (12) + 6 (1) = 42
94.9/42 = 2.26 moles
2.26 x 3/1 = 6.78
GFW of the product CO = 28
28 x 6.79 = 189.84 (theoretical yield)
% Yield = (152/189.84) x 100
% Yield = 80.1 %
5. For the balanced equation shown below, if the reaction of 63.5 grams of H2O produces 178 grams of Fe3O4, what is the percent yield?
3Fe + 4H2O --> Fe3O4 + 4H2
GFW of H2O = 18
63.5/18 = 3.53 moles
3.53 x 1/4 = .8825
GFW of the product Fe3O4 = 3 (56) + 4 (16) = 232
232 x .8825 = 204.7 (theoretical yield)
% Yield = (178/204.7) x 100
% Yield = 87%
6. For the balanced equation shown below, if the reaction of 59.5 grams of H2O produces 5.84 grams of H2, what is the percent yield?
CaH2 + 2H2O --> Ca(OH)2 + 2H2
GFW of H2O = 18
59.5/18 = 3.31 moles
3.31 x 2/2 = 3.31
GFW of the product H2 = 2
2 x 3.31 = 6.62 (theoretical yield)
% Yield = (5.84/6.62) x 100
% Yield = 88.2%
7. For the balanced equation shown below, if the reaction of 0.0228 grams of C3H6O produces 0.0381 grams of CO2, what is the percent yield?
C3H6O + 4O2 --> 3CO2 + 3H2O
GFW of C3H6O = 3 (12) + 6 + 16 = 58
0.0228/58 = 3.93 x 10E-4
3.93 x 10E-4 x 3/1 = .001
GFW of product CO2 = 12 + 32 = 44
44 x .001 = .044 (theoretical yield)
% Yield = (.0381/.044) x 100
% Yield = 86.6%
8. For the balanced equation shown below, if the reaction of 63.6 grams of C2H5NSCl produces 24.4 grams of CO, what is the percent yield?
4C2H5NSCl + 15O2 --> 8CO + 10H2O + 4NO + 4SO2 + 2Cl2
GFW of C2H5NSCl = 2 (12) + 5 + 14 +32 + 35.5 = 110.5
63.6/110.5 = .576 moles
.576 x 8/4 = 1.152
GFW of product CO = 12 + 16 = 28
28 x 1.152 = 32.256 (theoretical yield)
% Yield = (24.4/32.256) x 100
% Yield = 75.64%
9. For the balanced equation shown below, if the reaction of 22.5 grams of O2 produces 2.49 grams of H2O, what is the percent yield?
4C2H3Cl + 11O2 --> 8CO2 + 6H2O + 2Cl2
GFW of O2 = 32
22.5/32 = .703 moles
.703 x 6/11 = .383
GFW of product H2O = 18
18 x .383 = 6.9 (theoretical yield)
% Yield = (2.49/6.9) x 100
% Yield = 36.1%
10. For the balanced equation shown below, if the reaction of 7.19 grams of H3PO3 produces 5.19 grams of H3PO4, what is the percent yield?
4H3PO3 --> 3H3PO4 + PH3
GFW of H3PO3 = 3 (1) + 31 + 3 (16) = 82
7.19/82 = .088 moles
.088 x 3/4 = .066
GFW of product H3PO4 = 3 + 31 + 4 (16) = 98
98 x .066 = 6.47 (theoretical yield)
% Yield = (5.19/6.47) x 100
% Yield = 80.22%
Answers to Percentage Yield Problems
1. For the balanced equation shown below, if the reaction of 14.7 grams of C4H8S produces an 87.1% yield, how many grams of H2O would be produced?
2C4H8S + 15O2 --> 8CO2 + 8H2O + 2SO3
GFW of C4H8S = 88
14.7/88 = .17 moles
.17 x 8/2 = .68
GFW of the the product H2O = 18
18 x .68 = 12.24 g (theoretical yield)
% Yield = Actual yield/theoretical yield
% Yield = (87.1/100) x 12.24 = 10.45
10.45 grams of H2O would be produced.
2. For the balanced equation shown below, if the reaction of 56.5 grams of H2S produces a 45% yield, how many grams of HCl would be produced?
2BiCl3 + 3H2S --> Bi2S3 + 6HCl
GFW of H2S = 2 + 32 = 34
56.5/34 = 1.66 moles
1.66 x 6/3 = 3.32
GFW of the product HCl = 1 + 35.5 = 36.5
36.5 x 3.32 = 121.18 g (theoreticial yield)
% Yield = (45/100) x 121.18 = 54.53
54.53 grams of HCl would be produced.
3. For the balanced equation shown below, if the reaction of 24.1 grams of H2O produces a 36.3% yield, how many grams of HNO3 would be produced?
N2O5 + H2O --> 2HNO3
GFW of H2O = 18
24.1/18 = 1.34 moles
1.34 x 2/1 = 2.68
GFW of the product HNO3 = 1 + 14 + 3 (16) = 63
63 x 2.68 = 168.64 g (theoretical yield)
% Yield = (36.3/100) x 168.64 = 61.22
61.22 grams of HNO3 would be produced.
4. For the balanced equation shown below, if the reaction of 40.1 grams of HCl produces a 66.9% yield, how many grams of CaCl2 would be produced?
CaO + 2HCl --> CaCl2 + H2O
GFW of HCl = 1 + 35.5 = 36.5
40.1/36.5 = 1.1 moles
1.1 x 1/2 = .55
GFW of the product CaCl2 = 40 + 2 (35.5) = 111
111 x .55 = 61.05 g (theoretical yield)
% Yield = (66.9/100) x 61.05 = 40.84
40.84 grams of CaCl2 would be produced.
5. For the balanced equation shown below, if the reaction of 73.1 grams of O2 produces a 46.2% yield, how many grams of CO2 would be produced?
2C6H4Cl2 + 13O2 --> 12CO2 + 2H2O + 4HCl
GFW of O2 = 32
73.1/32 = 2.3 moles
2.3 x 12/13 = 2.12
GFW of the product CO2 = 12 + 2 (16) = 44
44 x 2.12 = 93.28 g (theoretical yield)
% Yield = (46.2/100) x 93.28 = 43.1
43.1 grams of CO2 would be produced.
6. For the balanced equation shown below, if the reaction of 48.8 grams of O2 produces a 49.2% yield, how many grams of H2O would be produced?
C4H10S + 8O2 --> 4CO2 + 5H2O + SO3
GFW of O2 = 32
48.8/32 = 1.525 moles
1.525 x 5/8 = .953
GFW of the product H2O = 2 + 16 = 18
18 x .953 = 17.154 (theoretical yield)
% Yield = (49.2/100) x 17.154 = 8.44
8.44 grams of H2O would be produced.
7. For the balanced equation shown below, if the reaction of 34.0 grams of C2H3F produces a 36.3% yield, how many grams of H2O would be produced?
2C2H3F + 3O2 --> 4CO + 2H2O + 2HF
GFW of C2H3F = 2 (12) + 3 + 19 = 46
34/46 = .74 moles
.74 x 2/2 = .74
GFW of the product H2O = 18
18 x .74 = 13.32 (theoretical yield)
% yield = (36.3/100) x 13.32 = 4.84
4.84 grams of H2O would be produced.
8. For the balanced equation shown below, if the reaction of 18.1 grams of S produces a 50.1% yield, how many grams of Cu2S would be produced?
2Cu + S --> Cu2S
GFW of S = 32
18.1/32 = .566 moles
.566 x 1/1 = .566
GFW of the product Cu2S = 2 (63.5) + 32 = 159
159 x .566 = 90 (theoretical yield)
% yield = (50.1/100) x 90 = 45.1
45.1 grams of Cu2S would be produced.
9. For the balanced equation shown below, if the reaction of 10.7 grams of C2H3OF produces a 54.0% yield, how many grams of HF would be produced?
C2H3OF + 2O2 --> 2CO2 + H2O + HF
GFW of C2H3OF = 2 (12) + 3 + 16 +19 = 62
10.7/62 = .173 moles
.173 x 1/1 = .173
GFW of the product HF = 1 + 19 = 20
20 x .173 = 3.46 (theoretical yield)
% Yield = (54.0/100) x 3.46 = 1.87
1.87 grams of HF would be produced.
10. For the balanced equation shown below, if the reaction of 99.6 grams of FeS2 produces a 68.1% yield, how many grams of Fe2O3 would be produced?
4FeS2 + 11O2 --> 2Fe2O3 + 8SO2
GFW of FeS2 = 56 + 2 (32) = 120
99.6/120 =.83 moles
.83 x 2/4 = .415
GFW of the product Fe2O3 = 2 (56) + 3 (16) = 160
160 x .415 = 66.4 (theoretical yield)
% Yield = (68.1/100) x 66.4 = 45.2
45.2 grams of Fe2O3 would be produced.
