As a result of the interaction of solutions of silver nitrate and potassium chloride, taken in excess

As a result of the interaction of solutions of silver nitrate and potassium chloride, taken in excess, a precipitate with a mass of 2.87 g formed. Calculate the mass of the initial solution of silver nitrate with a mass fraction of 17% taken for the reaction.

1.Let’s find the amount of the substance of the precipitated precipitate – silver chloride by the formula:

n = m: M.

M (AgCl) = 108 + 35 = 143 g / mol.

n = 2.87 g: 143 g / mol = 0.02 mol.

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

KCl + AgNO3 = AgCl ↓ + KNO3.

According to the reaction equation, there is 1 mole of silver chloride for 1 mole of silver nitrate. Substances are in quantitative ratios 1: 1.

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

n (AgNO3) = n (AgCl) = 0.02 mol.

3.Let’s find the mass of AgNO3.

m = n M.

M (Ag NO3) = 108 + 14 + 48 = 170 g / mol.

m = 0.02 mol × 170 g / mol = 3.4 g.

4. Find the mass of the silver nitrate solution.

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

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

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

m (solution) = (3.4 g × 17%): 100% = 0.578 g.

Answer: m (solution) = 0.578 g.