# Upon interaction of 300 g of a solution of copper (2) chloride with hydrogen sulfide water,

Upon interaction of 300 g of a solution of copper (2) chloride with hydrogen sulfide water, a precipitate with an amount of 0.2 mol was formed. Calculate the mass fraction of copper (2) chloride (in%) in the original solution.

Let’s compose the reaction equation. Let’s find the quantitative ratios of substances.

СuCl2 + Н2S → СuS ↓ + 2HCl.

According to the reaction equation, there is 1 mol of CuS per 1 mol of СuCl2. Substances are in quantitative ratios 1: 1.

n (CuS) = n (CuCl2) = 0.2 mol.

Let’s find the mass of СuCl2.

M (CuCl2) = 135 g / mol.

m = n × M.

m = 135 g / mol × 0.2 mol = 27 g.

W = m (substance): m (solution) × 100%,

W = (27g: 300g) 100% = 9%.

Answer: 9%.

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