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

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