1. For the balanced equation shown below, if 93.8 grams of PCl5 were reacted with 20.3 grams of H2O, how many grams of H3PO4 would be produced?
PCl5+4H2O=>H3PO4+5HCl
P = 31
Cl = 35.5
H = 1
O = 16
31 * 1 = 31
35.5 * 5 = 177.5
31 + 177.5 = 208.5 grams/mole
93.8/208.5 = .449 moles of PCl5
1 * 2 = 2
16 * 1 = 16
2 + 16 = 18
20.3/18 = 1.127 moles of H2O
1/5 * .449/1.127
H2O = 1.127 PCl5 = 2.245
Limiting reagent = H2O
H3PO4
1 * 3 = 3
31 * 1 = 31
16 * 4 = 64
3 + 31 + 64 = 98
18 * 4 = 72
98/72 = 1.36
1.36(20.3) = 27.63 grams of H3PO4
2. For the balanced equation shown below, if 18.3 grams of C2H5Cl were reacted with 37.3 grams of O2, how many grams of Cl2 would be produced?
4C2H5Cl+13O2=>8CO2+10H2O+2Cl2
C = 12
H = 1
Cl = 35.5
O = 16
12 * 2 = 24
1 * 5 = 5
35.5 * 1 = 35.5
24 + 5 + 35.5 = 64.5 grams/mole
18.3/64.5 = .283 moles of C2H5Cl
16 * 2 = 32
32 grams/mole
37.3/32 = 1.165 moles of O2
4/13 * .283/1.165
O2 = 4.66 C2H5Cl = 3.679
Limiting reagent = C2H5Cl
35.5 * 2 = 71
71 * 2 = 142
64.5 * 4 = 258
(142/258)*29.5=10.1 grams of Cl2
3. For the balanced equation shown below, if 23.9 grams of P4 were reacted with 20.8 grams of O2, how many grams of P4O10 would be produced?
P4+5O2=>P4O10
P = 31
O = 16
31 * 4 = 124 grams/mole
23.9/124 = .19 moles of P4
16 * 2 = 32 grams/mole
20.8/32 = .65 moles of O2
1/5 * .19/.65
O2 = .65 P4 = .95
Limiting reagent = O2
31 * 4 = 124
16 * 10 = 160
124 + 160 = 284 grams/mole
284 * 1 = 284
32 * 5 = 160
284/160 = 1.775
1.775 * 20.8 = 36.9 grams of P4O10
4. For the balanced equation shown below, if 15.3 grams of C2H3OCl were reacted with 16.0 grams of O2, how many grams of CO2 would be produced?
C2H3OCl+2O2=>2CO2+H2O+HCl
C = 12
H = 1
O = 16
Cl = 35.5
12 * 2 = 24
1 * 3 = 3
16 * 1 = 16
35.5 * 1 = 35.5
24 + 3 + 16 + 35.5 = 78.5 gram/mole
15.3/78.5 = .194 moles of C2H3OCl
16 * 2 = 32
32 grams/mole
16/32 = .5 moles of O2
1/2 * .194/.5
O2 = .5 C2H3OCl = .388
Limiting reagent = C2H3OCl
12 * 1 = 12
16 * 2 = 31
12 + 31 = 43 grams/mole
43 * 2 = 86
78.5 * 1 = 78.5
86/78.5 * 15.3 = 16.76 grams
5. For the balanced equation shown below, if 36.8 grams of CBr4 were reacted with 4.56 grams of O2, how many grams of Br2 would be produced?
CBr4+O2=>CO2+2Br2
C = 12
Br = 80
O = 16
12 * 1
80 * 4 = 320
12 + 320 = 332 grams/mole
36.8/332 = .11 moles
16 * 2 = 32 grams/mole
4.56/32 = .1425 moles
limiting reagent = CBr4
80 * 2 = 160
160 * 2 = 320
320/332 = .964
36.8 * .964 = 35.5 grams
6. For the balanced equation shown below, if 21.4 grams of C2H4 were reacted with 129 grams of O2, how many grams of H2O would be produced?
C2H4+3O2=>2CO2+2H2O
C = 12
H = 1
O = 16
12 * 2 = 24
1 * 4 = 4
24 + 4 = 28 grams/mole
21.4/ 28 = .764 moles of C2H4
16 * 2 = 32 grams/mole
129/32 = 4.03 moles of O2
1/3 * .764/4.03
.764 * 3 = 2.29
O2 = 4.03 C2H4 = 2.29
Limiting reagent = C2H4
1 * 2 = 2
16 * 1 = 16
2 + 16 = 18 grams/mole
18 * 2 = 36
36/28 = 1.285
1.285 * 21.4 = 27.5 grams
7. For the balanced equation shown below, if 13.2 grams of Fe were reacted with 31.5 grams of Cl2, how many grams of FeCl3 would be produced?
2Fe+3Cl2=>2FeCl3
Fe = 56
Cl = 35.5
56 grams/mole
13.2/56 = .24 moles of Fe
35.5 * 2 = 71 grams/mole
31.5/71 = .443 moles of Cl2
2/3 * .24/.443
.443 * 2 = .886 = Cl2 .24 * 3 = .72 = Fe
Limiting reagent = Fe
56 * 1 = 56
35.5 * 3 = 106.5
56 + 106.5 = 162.5
162.5 * 2 = 325
56 * 2 = 112
325/112 = 2.9
13.2 * 2.9 = 38.28 grams
8. For the balanced equation shown below, if 41.4 grams of Al were reacted with 190 grams of Cr2O3, how many grams of Cr would be produced?
2Al+Cr2O3=>Al2O3+2Cr
Al = 27
Cr = 52
O = 16
27 grams/mole
41.4/27 = 1.53 moles of Al
52 * 2 = 104
16 * 3 = 48
104 + 48 = 152 grams/mole
190/152 = 1.25 moles of Cr2O3
2/1 * 1.53/1.25
1.53 = Al 2 * 1.25 = 2.5 = Cr2O3
Limiting reagent = Al
52 * 2 = 104
27 *2 = 54
104/54 = 1.93
1.93 * 41.4 = 79.9 grams
9. For the balanced equation shown below, if 59.3 grams of CH2S were reacted with 80.6 grams of O2, how many grams of SO3 would be produced?
CH2S+3O2=>CO2+H2O+SO3
C = 12
H = 1
S = 32
O = 16
12 *1 = 12
1 * 2 = 2
32 * 1 = 32
12 + 2 + 32 = 46 grams/mole
59.3/46 = 1.29 moles of CH2S
16 * 2 = 32 grams/mole
80.6/32 = 2.51 moles of O2
1/3 * 1.29/2.51
O2 = 2.51 CH2S = 1.29 * 3 = 3.87
Limiting reagent = O2
32 * 1 = 32
16 * 3 = 48
32 + 48 = 80
32 * 3 = 96
80/96 = .83
.83 * 80.6 = 66.649 grams
10. For the balanced equation shown below, if 11.0 grams of CH3COF were reacted with 4.97 grams of H2O, how many grams of CH3COOH would be produced?
CH3COF+H2O=>CH3COOH+HF
C = 12
H = 1
O = 16
F = 19
12 * 1 = 12
1 * 3 = 3
12 * 1 = 12
16 * 1 = 16
19 * 1 = 19
12 + 3 + 12 + 16 + 19 = 62 grams/mole
11/46 = .177 moles of CH3COF
1 * 2 = 2
16 * 1 = 16
2 + 16 = 18 grams/mole
4.97/18 = .276 moles of H2O
limiting reagent = CH3COF
12 * 1 = 12
1 * 3 = 12
12 * 1 = 12
16 * 1 = 16
16 * 1 = 16
1 * 1 = 1
3 + 12 + 12 +16 + 16 + 1 = 69 grams/mole
60/62 = .967
.967 * 11 = 10.6 grams