What is the percent yield if 4.65 g of copper is produced when 1.87 g of aluminum reacts with an excess of copper(II) sulfate?

Apr 26, 2016

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

Explanation:

The balanced equation for this would be

"2Al" + "3CuSO"_4 → "3Cu" + "Al"_2("SO"_4)_3

1.87 g of Al is $\text{0.0693 moles Al" × "3 moles of Cu"/"2 moles of Al" = "0.10396 moles of Cu}$.

This corresponds to 6.606 g of Cu, which is the theoretical yield.

"Actual"/"Theoretical" × 100 % = "% yield": 4.65/6.606 × 100 % = 70.4 %