# A mixture of 220 g of iron (II) sulfide and 77.6 g of zinc sulfide was treated with an excess of hydrochloric acid.

A mixture of 220 g of iron (II) sulfide and 77.6 g of zinc sulfide was treated with an excess of hydrochloric acid. The evolved gas was passed through a solution of copper (II) sulfate. Calculate the volume (l) of a 10% copper sulfate solution (density 1.1 g / ml) consumed to absorb the resulting gas.

FeS + 2HCl ex. = FeCl2 + H2S ↑
ZnS + 2HCl ex. = ZnCl2 + H2S ↑
H2S + CuSO4 = CuS ↓ + H2SO4
n (FeSO4) = 220 / (56 + 32) = 220/88 = 2.5 mol
n (ZnS) = 77.6 / (65.5 + 32) = 77.6 / 97.5 = 0.8 mol
⇒ Σn (H2S) = 2.5 + 0.8 = 3.3 mol
⇒ n = (ω x ρ x V) / Mr
⇔ V = (3.3 x (63.5 + 32 + 64)) / (0.1 x 1.1) = 4 386.25 or 4.4 l.
Answer: the volume of copper sulfate is 4.4 liters.

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