The reaction of 19.6 g of copper (II) hydroxide with hydrochloric acid gave 134 g of a solution of copper (II) chloride.

The reaction of 19.6 g of copper (II) hydroxide with hydrochloric acid gave 134 g of a solution of copper (II) chloride. Calculate the mass fraction of salt in the resulting solution.

1) Let’s compose the reaction equation:
Cu (OH) 2 + 2HCl = CuCl2 + 2H2O
2) Calculate the amount of the substance of the reacted copper (II) hydroxide:
n (Cu (OH) 2) = m (Cu (OH) 2) / M (Cu (OH) 2) = 19.6 / 98 = 0.2 mol.
3) Determine the amount of substance, the mass of the formed copper (II) chloride and its mass fraction in the resulting solution:
n (CuCl2) = n (Cu (OH) 2) = 0.2 mol
m (CuCl2) = n (CuCl2) ∙ M (CuCl2) = 0.2 ∙ 135 = 27 g.
ω (CuCl2) = (m (CuCl2) / m (solution)) ∙ 100% = (27/134) ∙ 100 = 20.15%
Answer: 20.15%.