# Calculate the mass of the precipitate formed by the interaction of potassium hydroxide

**Calculate the mass of the precipitate formed by the interaction of potassium hydroxide and 108 grams of a 10% solution of copper chloride 2**

Given:

m (solution) = 108 g,

W = 10%.

m (Сu (OH) 2) -?

1.Let’s find the mass of copper chloride in the solution.

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

hence m (substance) = (m (solution) × w): 100%.

m (substance) = (108 g × 10%): 100% = 10.8 g.

2.Let’s find the amount of copper chloride substance by the formula:

n = m: M.

M (CuCl2) = 64 + 70 = 134 g / mol.

n = 10.8 g: 134 g / mol = 0.08 mol.

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

CuCl2 + 2KOH = Cu (OH) 2 ↓ + 2KCl.

According to the reaction equation, 1 mol of copper chloride accounts for 1 mol of copper hydroxide. Substances are in quantitative ratios 1: 1.

The amount of substance will be the same.

n (Cu (OH) 2) = n (CuCl2) = 0.08 mol.

4.Let’s find the mass of copper hydroxide by the formula:

m = n × M,

M (Cu (OH) 2) = 64 + 2 (1 + 16) = 98 g / mol.

m = 0.08 mol × 98 g / mol = 7.84 g.

Answer: 7.84 g.