Study Guide Percent Yield - Mr. Fischer



Study Guide Percent Yield

To solve these you must first solve for the amount of the product that is produced. This is a three-step problem and the answer is the calculated yield. To get the percent yield you must divide the actual yield, by the theoretical yield times 100.

1. What is the percent yield when 5.22g of O2 reacts with H2 to produce 3.00 grams of water.

2H2 + O2 ( 2H2O

1) 5.22 g x 1 mole O2 = 0.163125 mole O2

32.00 g

2) 0.1631 mole O2 x 2 moles H2O = 0.32625 moles H2O

1 mole O2

3) 0.32625 moles H2O x 18 g = 5.8725 grams H2O

1 mole H2O theoretical yield

Percent Yield = Actual yield x 100

Calculated yield

3.00 grams H2O x 100 = 51.1 %

5.88 grams H2O

2. What is the percent yield when 4.68g of Fe reacts with S to produce 2.88g of FeS.

Fe + S ( FeS

1) 4.86 g x 1 mole Fe = 0.083795 mole O2

55.85 g

2) 0.083795 mole Fe x 1 moles FeS = 0.083795 moles FeS

1 moleFe

3) 0.083795 moles FeS x 87.92 g = 7.37 grams FeS

1 mole FeS theoretical yield

2.88 grams FeS x 100 = 39.1%

7.32 grams FeS

3. What is the percent yield when 5.87g of Mg(OH)2 reacts with HCl to form 8.84g of MgCl2 and water.

Mg(OH)2 + 2HCl ( MgCl2 + 2H2O

1) 5.87 g x 1 mole = 0.1006688 mole Mg(OH)2

58.31g

2) 0.1006688 mole Mg(OH)2 x 1 moles MgCl2 = 0.1006688 moles MgCl2

1 mole Mg(OH)2

3) 0.1006688 moles MgCl2 x 92.21 g = 9.58467 grams MgCl2

1 mole MgCl2 theoretical yield

8.84 grams MgCl2 x 100 = 92.2%

9.58 grams MgCl2

4. What is the percent yield when 6.25g of AgNO3 reacts with NaCl to form NaNO3 and 4.12g of AgCl.

AgNO3 + NaCl ( NaNO3 + AgCl

1) 6.25 g x 1 mole AgNO3 = 0.036786mole AgNO3

169.9g

2) 0.036786 mole x 1 moles AgCl = 0.036786 moles AgCl

1 mole AgNO3

3) 0.036786 moles x 143.35 g = 5.27 grams AgCl

1 mole AgCl theoretical yield

4.12grams AgCl x 100 = 78.2%

5.27 grams AgCl

5. What is the percent yield when 7.81g of HCl reacts with NaOH to produce 5.24g of NaCl and H2O.

HCl + NaOH ( NaCl + H2O

1) 7.81 g x 1 mole HCl = 0.214266 mole HCl

36.45g

2) 0.214266 mole HCl x 1 moles NaCl = 0.214266 moles NaCl

1 mole HCl

3) 0.214266 moles x 58.45 g = 12.52 grams NaCl

1 mole NaCl theoretical yield

5.24 grams NaCl x 100 = 41.9%

12.52 grams NaCl

6. What is the percent yield when 6.33g of H2SO4 reacts with NaOH to produce 5.92g of Na2SO4 and water.

H2SO4 + 2NaOH ( Na2SO4 + 2H2O

1) 6.33 g x 1 mole H2SO4 = 0.0645457 mole H2SO4

98.07g

2) 0.0645457 mole H2SO4 x 1 moles Na2SO4 = 0.0645457 moles Na2SO4

1 mole H2SO4

3) 0.0645457 moles Na2SO4 x 142.07 g = 9.17 grams Na2SO4

1 mole Na2SO4 theoretical yield

5.92 grams Na2SO4 x 100 = 64.6%

9.17grams Na2SO4

7. What is the percent yield when 43.25g of CaC2 reacts with water to produce 33.71g of Ca(OH)2 and C2H2.

CaC2 + 2H2O ( Ca(OH)2 + C2H2

1) 43.25 g x 1 mole CaC2 = 0.6749375 mole CaC2

64.08g

2) 0.6749375 mole CaC2 x 1 moles Ca(OH)2 = 0.6749375 moles Ca(OH)2

1 mole CaC2

3) 0.6749375 moles Ca(OH)2 x 74.08 g = 50.00 grams Ca(OH)2

1 mole Ca(OH)2 theoretical yield

33.71 grams Ca(OH)2 x 100 = 67.42 %

50.00grams Ca(OH)2

8. What is the percent yield when 65.14g of CaCl2 reacts with Na2CO3 to produce 52.68g of Na2CO3 and NaCl.

CaCl2 + Na2CO3 ( CaCO3 + 2NaCl

1) 65.14 g x 1 mole CaCl2 = 0.58695 mole CaCl2

110.98g

2) 0.58695 mole CaCl2 x 1 moles CaCO3 = 0.58695 moles CaCO3

1 mole CaCl2

3) 0.58695 moles CaCO3 x 100.08 g = 58.74 grams CaCO3

1 mole CaCO3 theoretical yield

52.68 grams CaCO3 x 100 = 89.7% grams

58.74grams CaCO3

9. What is the percent yield when 4.687g of SF4 reacts with I2O5 to produce 6.281g of IF5 and SO2.

5SF4 + 2I2O5 ( 4IF5 + 5SO2

1) 4.687 g x 1 mole SF4 = 0.04337 mole SF4

108.07g

2) 0.04337 mole SF4 x 4 moles IF5 = 0.034696 moles IF5

5 mole SF4

3) 0.034696 moles IF5 x 221.9 g = 7.699 grams IF5

1 mole IF5 theoretical yield

6.281 grams IF5 x 100 = 81.58%

7.699 grams IF5

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