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Example 1. Sn + HF → SnF 2 + H 2 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 → SnF 2 + H 2. 245 g SnF 2. 121 g SnF 2. Finishing. - PowerPoint PPT Presentation
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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?