Through 500 g of a 10% solution of copper (II) sulfate, 4.48 liters of hydrogen

Through 500 g of a 10% solution of copper (II) sulfate, 4.48 liters of hydrogen sulfide were passed. Determine the proportion of sulfuric acid in the resulting solution.

Given:
m solution (CuSO4) = 500 g
ω (CuSO4) = 10%
V (H2S) = 4.48 L

To find:
ω (H2SO4) -?

Decision:
1) CuSO4 + H2S => CuS ↓ + H2SO4;
2) m (CuSO4) = ω (CuSO4) * m solution (CuSO4) / 100% = 10% * 500/100% = 50 g;
3) n (CuSO4) = m (CuSO4) / M (CuSO4) = 50/160 = 0.313 mol;
4) n (H2S) = V (H2S) / Vm = 4.48 / 22.4 = 0.2 mol;
5) n (H2SO4) = n (H2S) = 0.2 mol;
6) m (H2SO4) = n (H2SO4) * M (H2SO4) = 0.2 * 98 = 19.6 g;
7) m (H2S) = n (H2S) * M (H2S) = 0.2 * 34 = 6.8 g;
8) n (CuS) = n (H2S) = 0.2 mol;
9) m (CuS) = n (CuS) * M (CuS) = 0.2 * 96 = 19.2 g;
10) m solution = m solution (CuSO4) + m (H2S) – m (CuS) = 500 + 6.8 – 19.2 = 487.6 g;
11) ω (H2SO4) = m (H2SO4) * 100% / m solution = 19.6 * 100% / 487.6 = 4.02%.

Answer: Mass fraction of H2SO4 is 4.02%.