Answer to Question #86213 in General Chemistry for Mista

Question

What is the percent yield of copper when 13.8g of Cu2O reacts with Cu2S and 14.7 grams of copper is recovered?

2Cu2O + Cu2S → 6Cu + SO2

Answer 1

2Cu₂O + Cu₂S$\to$ 6Cu + SO₂

m(Cu)=$\dfrac{m(Cu2O)*6*M(Cu)}{2*M(Cu2O}=\dfrac{13.8g*6*63.546g}{2*143.09g/mol}=18.385g$

%yield=$\dfrac{14.7g}{18.385g}*100=79.95$