Cu2S + O2à2Cu + SO2 If the reaction of .760 Kg of copper sulfide(I) with excess...
Cu2S + O2à2Cu + SO2
If the reaction of .760 Kg of copper sulfide(I) with excess oxygen produces 0.310 kg of copper metal, what is the percentage yield?
- 40.8%
- 39.9%
- 204%
- 51.1%
- 102%
Homework Answers
Answer
51.1%
Explanation
Cu2S + O2 -------> 2Cu + SO2
Stochiometrically, 1mole of Cu2S gives 2moles of Cu
given moles of Cu2O = 760g/159.16= 4.775mol
moles of Cu can be maximum obtained by 4.775moles of Cu2O = 2 × 4.775mol = 9.55mol
mass of Cu can be maximum obtained = 9.55mol × 63.55g/mol = 606.9g = 0.6069kg
Theoretical yield = 0.6069kg
percentage yield = ( 0.310kg/0.6069kg) × 100 = 51.1%
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Cu2S + O2à2Cu + SO2
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