Zinc weighing 16.25 g was dissolved in an excess of concentrated sulfuric acid. The released hydrogen sulfide

Zinc weighing 16.25 g was dissolved in an excess of concentrated sulfuric acid. The released hydrogen sulfide, the yield of which was 80%, was passed through an excess of copper (II) sulfate solution. Determine the mass of the precipitated sediment.

Given:
m (Zn) = 16.25 g
η (H2S) = 80%

To find:
m (draft) -?

Decision:
1) 4Zn + 5H2SO4 => 4ZnSO4 + 4H2O + H2S ↑;
H2S + CuSO4 => H2SO4 + CuS ↓;
2) M (Zn) = Ar (Zn) = 65 g / mol;
M (CuS) = Mr (CuS) = Ar (Cu) * N (Cu) + Ar (S) * N (S) = 64 * 1 + 32 * 1 = 96 g / mol;
3) n (Zn) = m (Zn) / M (Zn) = 16.25 / 65 = 0.25 mol;
4) n theory. (H2S) = n (Zn) / 4 = 0.25 / 4 = 0.0625 mol;
5) n practical (H2S) = η (H2S) * n theory. (H2S) / 100% = 80% * 0.0625 / 100% = 0.05 mol;
6) n (CuS) = n practical (H2S) = 0.05 mol;
7) m (CuS) = n (CuS) * M (CuS) = 0.05 * 96 = 4.8 g.

Answer: The mass of CuS is 4.8 g.