Example 1

Sn + HF → SnF2 + H2

If you react 62.5 g of hydrofluoric acid with 92.0 g of tin and you produce 109 g of tin (II) fluoride, what was your percent yield?

Answer Sn + 2 HF → SnF2 + H2

62.5 g HF 1 mol HF 1 mol SnF2 156.7 g SnF2

20.008 g HF

2 mol HF 1 mol SnF2

92.0 g Sn 1 mol Sn 1 mol SnF2 156.7 g SnF2

118.7 g Sn 1 mol Sn 1 mol SnF2

245 g SnF2

121 g SnF2

Finishing

Percent yield = actual yield/ theoretical x 100

109 g / 121.45… g =

89.7 %

Example 2 NH3 + O2 NO + H2O

How much NO can be produced from 52 g of NH3 and 79 g of O2 if you have a 84% yield?

4 4 65

52 g NH3 1 mol NH3 4 mol NO 30.01 g NO

17.034 g NH3 4 mol NH3 1 mol NO

79 g O2 1 mol O2 4 mol NO 30.01 g NO

32 g O2 5 mol O2 1 mol NO

=92 g NO

=59 g NO

Finishing

Don’t round the number until you report your final answer

Actual yield /59.26975 g = .84 Actual yield = 50. g NO

Question 3 HCl + C4H10 C2H6 + Cl2

You collect 43.5 g of chlorine gas from 84.9 g of HCl and 143 g of C4H10. What was your percent yield?

2 2

84.9 g HCl 1 mol HCl 1 mol Cl2 70.9 g Cl2

36.458 g HCl 2 mol HCl 1 mol Cl2

=82.6 g Cl2

143 g C4H10 1 mol C4H10 1 mol Cl2 70.9 g Cl2

58.12 g C4H10 1 mol C4H10 1 mol Cl2

=174 g Cl2

Finishing

Actual yield / theoretical yield = percent yield

43.5 g / 82.5526…g = 52.7%

Homework

NiS + O2 NiO + SO2

How many grams of sulfur dioxide gas can be made from 84.9 g of Nickel (II) sulfide and 143 g of oxygen assuming a 68.7% yield?