# Iron sulfide 2 weighing 17.6 g was treated with an excess of hydrogen chloride solution.

Iron sulfide 2 weighing 17.6 g was treated with an excess of hydrogen chloride solution. The released gas was passed through 160 ml of a 25% solution of copper sulfate 2 (density 1.2). Calculate the mass fraction in% of copper sulfate 2 in the final solution.

Given:
m (FeS) = 17.6 g
V solution (CuSO4) = 160 ml
ω (CuSO4) = 25%
ρ solution (CuSO4) = 1.2 g / ml

To find:
ω2 (CuSO4) -?

Decision:
1) FeS + 2HCl => FeCl2 + H2S ↑;
H2S + CuSO4 => CuS ↓ + H2SO4;
2) n (FeS) = m / M = 17.6 / 88 = 0.2 mol;
3) n (H2S) = n (FeS) = 0.2 mol;
4) m solution (CuSO4) = ρ solution * V solution = 1.2 * 160 = 192 g;
5) m (CuSO4) = ω * m solution / 100% = 25% * 192/100% = 48 g;
6) n (CuSO4) = m / M = 48/160 = 0.3 mol;
7) n react. (CuSO4) = n (H2S) = 0.2 mol;
8) n rest. (CuSO4) = n (CuSO4) – n reag. (CuSO4) = 0.3 – 0.2 = 0.1 mol;
9) m rest. (CuSO4) = n rest. * M = 0.1 * 160 = 16 g;
10) m (H2S) = n * M = 0.2 * 34 = 6.8 g;
11) n (CuS) = n (H2S) = 0.2 mol;
12) m (CuS) = n * M = 0.2 * 96 = 19.2 g;
13) m2 solution = m (H2S) + m solution (CuSO4) – m (CuS) = 6.8 + 192 – 19.2 = 179.6 g;
14) ω2 (CuSO4) = m rest. (CuSO4) * 100% / m2 solution = 16 * 100% / 179.6 = 8.9%.

Answer: The mass fraction of CuSO4 is 8.9%.

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