A calcium chloride solution was added to 200 g of a silver nitrate solution until the precipitation ceased.

A calcium chloride solution was added to 200 g of a silver nitrate solution until the precipitation ceased. The mass of the filtered and dried precipitate was 28.7 g. Calculate the mass fraction of silver nitrate in the starting material.

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

n = m: M.

M (AgCl) = 108 + 35.5 = 143.5 g / mol.

n = 28.7 g: 143 g / mol = 0.2 mol.

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

2AgNО3 + CaCl2 = 2AgCl ↓ + Ca (NO3) 2.

According to the reaction equation, 2 mol of silver chloride accounts for 2 mol of silver nitrate. Substances are in quantitative ratios 1: 1.

The amount of silver chloride substance and the amount of silver nitrate substance will be equal.

n (AgCl) = n (AgNO3) = 0.2 mol.

Find the mass of silver nitrate.

m = n × M.

M (AgNO3) = 108 + 14 + 48 = 170 g / mol.

m = 170 g / mol × 0.2 mol = 34 g.

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

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

W = (34g: 200) × 100% = 17%. 