To a solution containing 32 g of copper sulfate (2), 24 g of iron filings were added. How much copper will be released in this case?

Given:
m (CuSO4) = 32 g
m (Fe) = 24 g

To find:
m (Cu) -?

Decision:
1) CuSO4 + Fe => Cu ↓ + FeSO4;
2) M (CuSO4) = Mr (CuSO4) = Ar (Cu) + Ar (S) + Ar (O) * 4 = 64 + 32 + 16 * 4 = 160 g / mol;
M (Fe) = Mr (Fe) = Ar (Fe) = 56 g / mol;
M (Cu) = Mr (Cu) = Ar (Cu) = 64 g / mol;
3) n (CuSO4) = m (CuSO4) / M (CuSO4) = 32/160 = 0.2 mol;
4) n (Fe) = m (Fe) / M (Fe) = 24/56 = 0.4 mol;
5) The amount of substance CuSO4 in a deficiency:
n (Cu) = n (CuSO4) = 0.2 mol;
6) m (Cu) = n (Cu) * M (Cu) = 0.2 * 64 = 12.8 g.

Answer: The mass of Cu is 12.8 g.