## Problem One

1. For the balanced equation C3H6+3O2=>3CO+3H2O,

if the reaction of 21.3 grams of O2 produces a 47.3% yield, how many grams of CO would be produced ?

Step One: Figure out how many moles of the limiting reagent you have

O * 2 = 16 * 2 = 32 grams/mole

21.3 grams = 0.66 moles

32 g/m

Step Two: Determine how many moles of product would be formed

0.66 moles * (3 moles of CO2 / 3 moles of O2) = 0.66 moles

Step Three: Multiply your result by the molar mass of the product

0.66 moles * (28 g/m) = 18.48

Step Four: Solve for the actual yield

47.3 % = (actual yield) * 100

18.48 grams

Actual yield = 8.74 grams

## Problem Two

2. For the balanced equation 2C6H6+15O2=>12CO2+6H2O,

if the reaction of 9.86 grams of C6H6 produces 4.52 grams of H2O, what is the percent yield?

Step One: Figure out how many moles of the limiting reagent you have

C * 6 = 12 * 6 = 72

H * 6 = 1 * 6 = 6

72 + 6 = 78 grams/mole

9.86 grams = 0.126 moles

78 g/m

Step Two: Determine how many moles of product would be formed

0.126 moles * (6 moles of H2O / 2 moles of C6H6) = 0.379 moles

Step Three: Multiply your result by the molar mass of the product

0.379 moles * (18 g/m) = 6.826 grams

Step Four: Solve for the percent yield

percent yield = 4.52 grams * 100

6.826 grams

percent yield = 66.21%

## Problem Three

3. For the balanced equation 2C2H3O2F+3O2=>4CO2+2H2O+2HF,

if the reaction of 24.1 grams of O2 produces a 55.7% yield, how many grams of HF would be produced ?

Step One: Figure out how many moles of the limiting reagent you have

O * 2 = 16 * 2 = 32 grams/mole

24.1 grams = 0.753 moles

32 g/m

Step Two: Determine how many moles of product would be formed

0.753 moles * (2 moles of HF / 3 moles of O2) = 0.502

Step Three: Multiply your results by the molar mass of the product

H * 1 = 1 * 1 = 1

F * 1 = 19 * 1 = 19

1 + 19 = 20 grams/mole

0.502 moles * (20 g/m) = 10.041 grams

Step Four: Solve for the actual yield

55.7 yield = (actual yield) * 100

10.041 grams

Actual Yield = 5.59%

## Problem Four

4. For the balanced equation 2C4H8O+11O2=>8CO2+8H2O,

if the reaction of 54.7 grams of O2 produces 16.0 grams of H2O, what is the percent yield?

Step One: Figure out how many moles of the limiting reagent you have

O * 2 = 16 * 2 = 32 g/m

54.7grams = 1.709 moles

32g/m

Step Two: Determine how many moles of product will be formed

1.709 moles * (8 moles of H2O / 11 moles of O2) = 1.243 moles

Step Three: Multiply your results by the molar mass of the product

1.243 moles * (18 g/m) = 22.377 grams

Step Four: Solve for percent yield

percent yield = 16 grams * 100

22.377 grams

percent yield = 71.5%

## Problem Five

5. For the balanced equation Fe3O4+4H2=>3Fe+4H2O,

if the reaction of 24.8 grams of H2 produces a 63.7% yield, how many grams of H2O would be produced ?

Step One: Figure out how many moles of the limiting reagent you have

H * 2 = 1 * 2 = 2

24.8 grams = 12.4 moles

2 g/m

Step Two: Determine how many moles of product will be formed

12.4 moles * (4 moles of H2O / 4 moles of H2) = 12.4 moles

Step Three: Multiply your results by the molar mass of the product

12.4 moles * (18 g/m) = 223.2 grams

Step Four: Solve for actual yield

63.7 % = (actual yield) * 100

223.2 grams

actual yield = 142.17%

## Problem Six

6. For the balanced equation 2NH3+CO2=>CO(NH2)2+H2O,

if the reaction of 81.9 grams of CO2 produces 26.7 grams of H2O, what is the percent yield?

Step One: Figure out how many moles of the limiting reagent you have

C * 1 = 12 * 1 = 12

O * 2 = 16 * 2 = 32

12 + 32 = 44 grams/mole

81.9 grams = 1.86 moles

44 g/m

Step Two: Determine how many moles of product will be formed

1.86 moles * (1 mole of H2O / 1 mole of CO2) = 1.86 moles

Step Three: Multiply your results by the molar mass of the product

1.86 moles * (18 g/m) = 33.5 grams

Step Four: Solve for percent yield

percent yield = 26.7 grams * 100

33.5 grams

percent yield = 79.69%

## Problem Seven

7. For the balanced equation K2O+H2O=>2KOH,

if the reaction of 27.5 grams of K2O produces a 69.5% yield, how many grams of KOH would be produced ?

Step One: Figure out how many moles of the limiting reagent you have

K * 2 = 40 * 2 = 78

O * 1 = 16 * 1 = 16

78 + 16 = 94 grams/mole

27.5 grams = 0.292 moles

94 g/m

Step Two: Determine how many moles of product will be formed

0.286 moles * (2 moles of KOH / 1 mole of K2O) = 0.585 moles

Step Three: Multiply your results by the molar mass of the product

0.585 moles * (56 g/m) = 532.76 grams

Step Four: Solve for actual yield

69.5 % = (actual yield) * 100

32.76 grams

actual yield = 22.77 grams

## Problem Eight

8. For the balanced equation C6H5N2Cl=>C6H5Cl+N2,

if the reaction of 63.8 grams of C6H5N2Cl produces 8.35 grams of N2, what is the percent yield?

Step One: Figure out how many moles of the limiting reagent you have

C * 6 = 12 * 6 = 72

H * 5 = 1 * 5 = 5

N * 2 = 14 * 2 = 28

Cl * 1 = 35 * 1 = 35

72 + 5 + 28 + 35 = 140 grams/mole

63.8 grams = 0.455 moles

140 g/m

Step Two: Determine how many moles of product will be formed

0.455 moles * (1 mole of N2 / 1 mole of C6H5N2Cl) = 0.455 moles

Step Three: Multiply your results by the molar mass of the product

0.455 moles * (28 g/m) = 12.76 grams

Step Four: Solve for percent yield

Percent yield = 8.35 grams * 100

12.76 grams

Percent yield = 65.43%

## Problem Nine

9. For the balanced equation 2C4H10O2+11O2=>8CO2+10H2O,

if the reaction of 94.4 grams of O2 produces a 72.3% yield, how many grams of H2O would be produced ?

Step One: Figure out how many moles of the limiting reagent you have

O * 2 = 16 * 2 = 32 grams/mole

94.4 grams = 2.95 moles

32 g/m

Step Two: Determine how many moles of product will be formed

2.95 moles * (10 moles of H2O / 11 moles of O2) = 2.68 moles

Step Three: Multiply your results by the molar mass of the product

2.68 moles * (18 g/m) = 48.27 grams

Step Four: Solve for actual yield

72.3 % = (actual yield) * 100

48.27 grams

actual yield = 34.9 %

## Problem Ten

10. For the balanced equation 2C6H4Cl2+13O2=>12CO2+2H2O+4HCl,

if the reaction of 52.8 grams of O2 produces 31.7 grams of CO2, what is the percent yield?

Step One: Figure out how many moles of the limiting reagent you have

O * 2 = 16 * 2 = 32 grams/mole

52.8 grams = 1.65 moles

32 g/m

Step Two: Determine how many moles of product will be formed

1.65 moles * (12 moles of CO2 / 13 moles of O2) = 1.52 moles

Step Three: Multiply your results by the molar mass of the product

1.52 moles * (44 g/m) = 67.01 grams

Step Four: Solve for percent yield

percent yield = 31.7 grams * 100

67.01 grams

percent yield = 47.3 %