To a solution of hydrochloric acid weighing 36.5 g with a mass fraction of hcl 10% was added copper (II) oxide weighing 8 g

| August 6, 2021 | education

To a solution of hydrochloric acid weighing 36.5 g with a mass fraction of hcl 10% was added copper (II) oxide weighing 8 g Calculate the amount of substance and the mass of the resulting salt

Find the mass of hydrochloric acid in the solution.

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

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

m (HCl) = (36.5 g × 10%): 100% = 3.65 g.

Find the amount of HCl by the formula:

n = m: M.

М (HCl) = 36.5 g / mol.

n = 3.65 g: 36.5 g / mol = 0.1 mol (deficiency).

Let’s find the amount of substance CuO:

M (CuO) = 80 g / mol.

n = 8 g: 80 g / mol = 0.1 mol.

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

CuO + 2HCl = CuCl2 + H2O.

According to the reaction equation, there is 1 mol of CuCl2 per 2 mol of HCl. The substances are in quantitative ratios of 2: 1. The amount of CuCl2 is 2 times less than the amount of HCl.

n (CuCl2) = ½ n (HCl) = 0.1: 2 = 0.05 mol.

Let’s find the mass of CuCl2 by the formula:

m = n × M,

M (CuCl2) = 135 g / mol.

m = 0.05 mol × 135 g / mol = 6.75 g.

Answer: 6.75 g.



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