What mass of copper (II) sulfide is formed by passing 4.48 L of hydrogen sulfide through a 10%

What mass of copper (II) sulfide is formed by passing 4.48 L of hydrogen sulfide through a 10% solution of copper (II) sulfate with a volume of 363.6 ml, if the density of the solution is 1.1 g / ml?

Given:
V (H2S) = 4.48 L
ω (CuSO4) = 10%
V solution (CuSO4) = 363.6 ml
ρ solution (CuSO4) = 1.1 g / ml

Find:
m (CuS) -?

1) H2S + CuSO4 => CuS ↓ + H2SO4;
2) m solution (CuSO4) = ρ solution * V solution = 1.1 * 363.6 = 399.96 g;
3) m (CuSO4) = ω * m solution / 100% = 10% * 399.96 / 100% = 39.996 g;
4) M (CuSO4) = Mr (CuSO4) = Ar (Cu) * N (Cu) + Ar (S) * N (S) + Ar (O) * N (O) = 64 * 1 + 32 * 1 + 16 * 4 = 160 g / mol;
5) n (CuSO4) = m / M = 39.996 / 160 = 0.25 mol;
6) n (CuS) = n (CuSO4) = 0.25 mol;
7) M (CuS) = Mr (CuS) = Ar (Cu) * N (Cu) + Ar (S) * N (S) = 64 * 1 + 32 * 1 = 96 g / mol;
8) m (CuS) = n * M = 0.25 * 96 = 24 g.

Answer: The mass of CuS is 24 g.