An excess of lead nitrate solution was added to 89 g of a solution of aluminum chloride with a mass

An excess of lead nitrate solution was added to 89 g of a solution of aluminum chloride with a mass fraction of 15.0%. Determine the mass of the precipitate.

Let’s find the mass of aluminum chloride in solution.

The mass fraction of a substance is calculated by the formula:

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

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

m (substance) = (89g × 15%): 100% = 13.35 g.

Let’s find the amount of aluminum chloride substance by the formula:

n = m: M.

M (AlCl3) = 27 + 35.5 × 3 = 133.5 g / mol.

n = 13.35 g: 133.5 g / mol = 0.1 mol.

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

2AlCl3 + 3Pb (NO3) 2 = 3 PbCl2 ↓ + 2Al (NO3) 3

According to the reaction equation, for 2 mol of aluminum chloride there is 3 mol of lead chloride. Substances are in quantitative ratios 2: 3 = 1: 1.5.

The amount of PbCl2 will be 1.5 times greater than the amount of aluminum chloride.

n (PbCl2) = 1.5 n (AlCl3) = 0.1 × 1.5 = 0.15 mol.

Let’s find the mass of PbCl2.

m = n × M.

M (PbCl2) = 207 + 71 = 278 g / mol.

m = 278 g / mol × 0.15 mol = 41.7 g.

Answer: 41.7 g.