Problem One
For the balanced equation CCl4+2HF => CCl2F2+2HCl,
if 39.1 grams of CCl4 were reacted with 18.0 grams of HF, how many grams of CCl2F2 would be produced?
Step One: You must find the limiting reagent
C * 1 = 12 * 1 = 12 H * 1 = 1 * 1 = 1
Cl * 4 = 35 * 4 = 140 F * 1 = 19 * 1 = 19
12 + 140 = 152 grams/mole 1 + 19 = 20 grams/mole
39.1 grams = 0.257 moles 18 grams = 0.9 moles
152 g/m 20 g/m
CCl4 = Limiting Reagent
Step Two: Compare the molar ratio of the limiting reagent to the product
0.257 moles * (1 mole of CCl2F2 / 1 moles of CCl4) = 0.257 moles
Step Three: Convert back into grams
C * 1 = 12 * 1 = 12
Cl * 2 = 35 * 2 = 70
F * 2 = 19 * 2 = 38
12 + 70 + 38 = 120 grams/mole
0.257 moles * (120 g/m) = 30.84 grams
Problem Two
For the balanced equation 2NH3+CO2=>CO(NH2)2+H2O,
if 77.5 grams of NH3 were reacted with 59.2 grams of CO2, how many grams of CO(NH2)2 would be produced?
Step One: You must find the limiting reagent
N * 1 = 14 * 1 = 14 C * 1 = 12 * 1 = 12
H * 3 = 1 * 3 = 3 O * 2 = 16 * 2 = 32
14 + 3 = 17 grams/mole 12 + 32 = 44 grams/mole
77.5 grams = 4.55 moles 59.2 grams = 1.34 moles
17 g/m 44 g/m
CO2 = Limiting Reagent
Step Two: Compare the molar ratio of the limiting reagent to the product
1.34 grams/mole * (1 mole of CO(NH2)2 / 1 mole of CO2) = 1.34 moles
Step Three: Convert back into grams
C * 1 = 12 * 1 = 12
O * 1 = 16 * 1 = 16
N * 2 = 14 * 2 = 28
H * 4 = 1 * 4 = 4
12 + 16 + 28 + 4 = 60 grams/mole
1.34 moles * (60g/m) = 80.4 grams
Problem Three
3. For the balanced equation 4C2H3F3+7O2=>8CO+6H2O+6F2,
if 25.5 grams of C2H3F3 were reacted with 11.1 grams of O2, how many grams of H2O would be produced?
Step One: Find the limiting reagent
C * 2 = 12 * 2 = 24 O * 2 = 16 * 2 = 32 grams/mole
H * 3 = 1 * 3 = 3
F * 3 = 19 * 3 = 57
24 + 3 + 57 = 84 grams/mole
25.5 grams = 0.303 moles 11.1 grams = 0.346 moles
84 g/m 32 g/m
O2 = Limiting reagent
Step Two: Compare the molar ratio of the limiting reagent to the product
0.346 moles * (6 moles of H20 / 7 moles of O2) = 0.296 moles
Step Three: Convert back to moles
0.296 moles * (18 g/m) = 5.33 grams
Problem Four
4. For the balanced equation Fe3O4+4H2=>3Fe+4H2O,
if 24.4 grams of Fe3O4 were reacted with 1.08 grams of H2, how many grams of Fe would be produced?
Step One: Find the limiting reagent
Fe * 3 = 56 * 3 = 168 H * 2 = 1 * 2 = 2 grams/mole
O * 4 = 16 * 4 = 64
168 + 64 = 232 grams/mole
24.4 grams = 0.105 moles 1.08 grams = 0.54 moles
232 g/m 2 g/m
Fe3O4 = Limiting reagent
Step Two: Compare the molar ratio of the limiting reagents to the product
0.105 moles * (3 moles of Fe / 1 mole of Fe3O4) = 0.31 moles
Step Three: Convert back into grams
0.31 moles * (56 g/m) = 17.66 grams
Problem Five
5. For the balanced equation C2H5NSCl+4O2=>2CO+2H2O+NO+SO3+HC,
if 20.3 grams of C2H5NSCl were reacted with 29.9 grams of O2, how many grams of SO3 would be produced?
Step One: Find the limiting reagent
C * 2 = 12 * 2 - 24 O * 2 = 16 * 2 = 32 grams/mole
H * 5 = 1 * 5 = 5
N * 1 = 14 * 1 = 14
S * 1 = 32 * 1 = 32
Cl * 1 = 35 * 1 = 35
24 + 5 + 14 + 32 + 35 = 110 grams/mole
20.3grams = 0.184 moles 29.9 grams = 0.934 moles
110g/m 32g/m
C2H5NSCl = Limiting reagent
Step Two: Compare the molar ratio of the limiting reagents to the product
0.184 moles * (1 mole of SO3 / 1 mole of C2H5NSCl) = 0.184 moles
Step Three: Convert back into grams
S * 1 = 32 * 1 = 32
O * 3 = 16 * 3 = 48
32 * 48 = 80 grams/mole
0.184 moles * (80 g/m) = 14.7 grams
Problem Six
6. For the balanced equation 2C4H8O+11O2=>8CO2+8H2O,
if 6.30 grams of C4H8O were reacted with 10.8 grams of O2,how many grams of CO2 would be produced?
Step One: Find the limiting reagent
C * 4 = 12 * 4 = 48 O * 2 = 16 * 2 = 32 grams/mole
H * 8 = 1 * 8 = 8
O * 1 = 16 * 1 = 16
48 + 8 + 16 = 72 grams/mole
6.30 grams = 0.087 moles 10.8 grams = 0.337 moles
72 g/m 32 g/m
O2 = Limiting reagent
Step Two: Compare the molar ratio of the limiting reagent to the product
0.337 moles (8 moles of CO2 / 11 moles of O2) = 0.245 moles
Step Three: Convert back into grams
0.245 moles * (44 g/m) = 10.8 grams
Problem Seven
7. For the balanced equation 4C2H3Br3+7O2=>8CO+6H2O+6Br2,
if 19.3 grams of C2H3Br3 were reacted with 2.81 grams of O2, how many grams of H2O would be produced?
Step One: Find the limiting reagent
C * 2 = 12 * 2 = 24 O * 2 = 16 * 2 = 32 grams/mole
H * 3 = 1 * 3 = 3
Br * 3 = 79 * 3 = 237
24 + 3 + 237 = 264 grams/mole
19.3 grams = 0.073 moles 2.81 grams = 0.087 moles
264 g/m 32 g/m
O2 = Limiting reagent
Step Two: Compare the molar ratio of the limiting reagent to the product
0.087 moles * (6 moles of H2O / 7 moles of O2) = 0.075 moles
Step Three: Convert back into grams
0.075 * (18 g/m) = 1.35 grams
Problem Eight
8. For the balanced equation 4C2H5N+15O2=>8CO2+10H2O+4NO,
if 28.1 grams of C2H5N were reacted with 49.9 grams of O2, how many grams of NO would be produced?
Step One: Find the limiting reagent
C * 2 = 12 * 2 = 24 O * 2 = 16 * 2 = 32 grams/mole
H * 5 = 1 * 5 = 5
N * 1 = 14 * 1 = 14
24 + 5 + 14 = 43 grams/mole
28.1 grams = 0.65 moles 49.9 grams = 1.55 moles
43 g/m 32 g/m
O2 = Limiting reagent
Step Two: Compare the molar ratio of the limiting reagent to the product
1.55 moles * (4 moles of NO / 15 moles of O2) = 0.415 moles
Step Three: Convert back into grams
0.415 moles * (30 g/m) = 12.47 grams
Problem Nine
9. For the balanced equation 4C6H5F+29O2=>24CO2+10H2O+2F,
if 2.58 grams of C6H5F were reacted with 4.88 grams of O2, how many grams of H2O would be produced?
Step One: Find the limiting reagent
C * 6 = 12 * 6 = 72 O * 2 = 16 * 2 = 32 grams/mole
H * 5 = 1 * 5 = 5
F * 1 = 19 * 1 = 19
72 + 5 + 19 = 96 grams/mole
2.58 grams = 0.026 moles 4.88 grams = 0.152 moles
96 g/m 32 g/m
O2 = Limiting reagent
Step Two: Compare the molar ratio of the limiting reagent to the product
0.152 moles * (10 moles of H2O / 29 grams of O2) = 0.052 moles
Step Three: Convert back into grams
0.052 moles * (18 g/m) = 0.946 grams
Problem Ten
10. For the balanced equation C6H5Cl+7O2=>6CO2+2H2O+HCl,
if 68.4 grams of C6H5Cl were reacted with 223 grams of O2, how many grams of H2O would be produced?
Step One: Finding the limiting reagent
C * 6 = 12 * 6 = 72 O * 2 = 16 * 2 = 32 grams/mole
H * 5 = 1 * 5 = 5
Cl * 1 = 35 * 1 = 35
72 + 5 + 35 = 112 grams/mole
68.8 grams = 0.614 moles 223 grams = 6.96 moles
112 g/m 32 g/m
C6H5Cl = Limiting reagent
Step Two: Compare the molar ratio of the limiting reagent to the product
0.614 moles * (2 moles of H2O / 1 mole of C6H5Cl) = 1.22 moles
Step Three: Convert back into grams
1.22 moles * (18 g/m) = 22.1 grams
