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A student started with 1.636 grams of copper (II) oxide and produced 3.500 g of copper...

A student started with 1.636 grams of copper (II) oxide and produced 3.500 g of copper (II) sulfate. What is the percent yield? The product is a hydrate. Use appropriate significant figures. Do not put a percent sign in the answer box. Useful information: chemical equation: CuO(s) + H2SO4 (aq) ---> CuSO4 (aq) +H2O(l) Formula weight of hydrated copper (II) sulfate = 249.6 g/mol

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Answer #1

CuO(s) + H2SO4 (aq) ---> CuSO4 (aq) +H2O(l)

1 mole of CuO react with H2So4 to gives 1 mole of CuSo4 5H2O

79.5g of CuO react with H2SO4 to gives 249.6g of CuSo4.5H2O

1.636g of CuO react with H2So4 to gives = 249.6*1.636/79.5   = 5.14g of CuSo4.5H2O

percentage yield = actual yield*100/theoritical yield

                              = 3.5*100/5.14   = 68.1% >>>>answer

                               = 3.5

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