# An excess of silver nitrate solution was added to a solution of aluminum chloride with a mass of 53.2 g

**An excess of silver nitrate solution was added to a solution of aluminum chloride with a mass of 53.2 g and a mass fraction of 5%. Calculate the mass of the precipitate formed.**

Find the mass of AlCl3 in solution.

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

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

m (AlCl3) = (53.2 g × 5%): 100% = 2.66 g.

Let’s find the amount of substance AlCl3 by the formula:

n = m: M.

M (AlCl3) = 133.5 g / mol.

n = 2.66 g: 133.5 g / mol = 0.02 mol.

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

AlCl3 + 3 AgNO3 = 3 AgCl ↓ + Al (NO3) 3.

According to the reaction equation, there is 3 mol of AgCl per 1 mol of AlCl3. The substances are in quantitative ratios of 1: 3.

n (AgCl) = 3 n (AlCl3) = 0.02 × 3 = 0.06 mol.

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

m = n × M,

М (AgCl) = 143.5 g / mol.

m = 0.06 mol × 143.5 g / mol = 8.61 g.

Answer: 8.61 g.