# A solution of silver nitrate was added to hydrochloric acid weighing 365 grams with a mass fraction of 10%.

**A solution of silver nitrate was added to hydrochloric acid weighing 365 grams with a mass fraction of 10%. Determine the mass of the resulting residue.**

Find the mass of HCl in solution.

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

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

m (HCl) = (365 g × 10%): 100% = 36.5 g.

Let’s find the amount of HCl by the formula:

n = m: M.

M (HCl) = 36.5 g / mol.

n = 36.5 g: 36.5 g / mol = 1 mol.

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

HCl + AgNO3 = AgCl ↓ + HNO3.

According to the reaction equation, there is 1 mol of AgCl per 1 mol of HCl. Substances are in quantitative ratios 1: 1.

n (AgCl) = n (HCl) = 1 mol.

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

m = n × M,

М (AgCl) = 143.5 g / mol.

m = 1 mol × 143.5 g / mol = 143.5 g.

Answer: 143.5 g.