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28 July, 02:40

3.7 grams of aluminum combine with excess copper (II) sulfate with 10 grams of copper produced. What is the percent yield? The equation is balanced below.

2Al + 3CuSO4 - -> Cu + Al2 (SO4) 3

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  1. 28 July, 03:58
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    77% percent yield

    Explanation:

    2 Al + 3 CuSO4 → Al2 (SO4) 3 + 3 Cu

    (3.7 g Al) / (26.98154 g Al/mol) x (3 mol Cu / 2 mol Al) x (63.5463 g Cu/mol) = 13.07 g Cu in theory

    (10 g) / (13.07) = 0.765 = 77% yield
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