An excess of silver nitrate solution was added to a solution of hydrochloric acid with a volume of 50 ml

An excess of silver nitrate solution was added to a solution of hydrochloric acid with a volume of 50 ml and a density of 1.1 g / ml, the mass fraction of HCl in which is 20%. Determine the mass of the sediment formed.

Find the mass of the hydrochloric acid solution.

m = Vp

m = 50 cm3 × 1.1 g / ml = 55 g.

Find the mass of HCl in solution.

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

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

m (HCl) = (55 g × 20%): 100% = 11 g.

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

n = m: M.

M (HCl) = 36.5 g / mol.

n = 11 g: 36.5 g / mol = 0.3 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) = 0.3 mol.

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

m = n × M,

М (AgCl) = 143.5 g / mol.

m = 0.3 mol × 143.5 g / mol = 43.05 g.

Answer: 43.05.