The mass of the precipitate that precipitates upon the interaction of 7.3 g of hydrochloric acid with 17 g of silver nitrate.

Let’s implement the solution:

According to the condition of the problem, we write down the process:
HCl + AgNO3 = AgCl + HNO3 – ion exchange, a precipitate of silver chloride was formed;

Let’s make the calculations:
M (HCl) = 36.5 g / mol;

M (AgNO3) = 169.8 g / mol;

M (AgCl) = 143.3 g / mol.

Let’s determine the amount of starting materials:
Y (HCl) = m / M = 7.3 / 36.5 = 0.2 mol (substance in excess);

Y (AgNO3) = m / M = 17 / 169.8 = 0.1 mol (deficient substance);

Y (AgNO3) = 0.1 mol since the amount of these substances is 1 mol;

Calculations are carried out for the substance in deficiency.

We find the mass of the sediment:
m (AgCl) = Y * M = 0.1 * 143.3 = 14.33 g

Answer: obtained silver chloride weighing 14.33 g