Determine the mass of the precipitate formed when interacting with an excess of copper chloride solution

| February 24, 2021 | education

Determine the mass of the precipitate formed when interacting with an excess of copper chloride solution if the solution contains 1.7 g of hydrogen sulfide.

Let’s find the amount of the substance hydrogen sulfide H2S by the formula:

n = m: M.

M (H2S) = 34 g / mol.

n = 1.7 g: 34 g / mol = 0.25 mol.

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

CuCl2 + H2S = CuS ↓ + 2HCl

According to the reaction equation, 1 mol of hydrogen sulfide accounts for 1 mol of copper sulfide (precipitate). The substances are in quantitative ratios of 1: 1.

The amount of substance will be equal.

n (H2S) = n (CuS) = 0.25 mol.

Let’s find the mass of sulfur sulfide.

m = n × M,

M (CuS) = 64 + 32 = 96 g / mol.

m = 0.25 mol × 96 g / mol = 24 g.

Answer: 24 g.



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