# 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%.

Answer: 17%.