Question

What is the percent yield if 4.65 grams of copper is produced when 1.897 grams of aluminum reacts with an excess of copper (II) sulfate?
2Al + 3CuSO4 (aq) —> Al2(SO4)3 (aq) + 3Cu

Answer 1

The percentage yield if 4.65 g of copper is produced will be 70.4 %.

Explanation:

The balanced equation is given as

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

1.87 g of Al =  0.0693 moles of Al *  3 moles of Cu / 2 moles of Al

                  = 0.10396 moles of Cu.

This corresponds to 6.606 g of Cu which refers to theoretical yield.

% yield = (actual / theoretical) * 100

            = (4.65 / 6.606) * 100

% yield = 70.4 %.