Determine the yield of the reaction product if the oxidation of 102.4 g of copper

Determine the yield of the reaction product if the oxidation of 102.4 g of copper with an excess of concentrated sulfuric acid yielded 230.4 g of copper (II) sulfate.

1. Let’s write down the reaction equation:

Cu + 2H2SO4 = CuSO4 + SO2 + 2H2O;

2.Calculate the chemical amount of copper:

n (Cu) = m (Cu): M (Cu) = 102.4: 64 = 1.6 mol;

3.determine the theoretical amount of copper sulfate:

ntheor (CuSO4) = n (Cu) = 1.6 mol;

4.Calculate the practical chemical amount of the obtained sulfate:

npractice (CuSO4) = m (CuSO4): M (CuSO4);

M (CuSO4) = 64 + 32 + 4 * 16 = 160 g / mol;

npract (CuSO4) = 230.4: 160 = 1.44 mol;

5.Let’s find the yield of the reaction product:

ν (CuSO4) = npract (CuSO4): ntheor (CuSO4) = 1.44: 1.6 = 0.9 or 90%.

Answer: 90%.