Calculate the mass of silver chloride precipitated by the interaction of 380 silver nitrate containing 10%

Calculate the mass of silver chloride precipitated by the interaction of 380 silver nitrate containing 10% impurity with hydrochloric acid.

Since in technical silver nitrate, according to the condition of the problem, there are 10% impurities, the mass of pure silver nitrate is: 380 × 0.9 = 342 g.

The equation for the reaction of silver nitrate with hydrochloric acid:

Ag NO3 + H Cl = Ag Cl + HNO3.

From the reaction equation it follows that from 1 mole of silver nitrate, 1 mole of silver chloride is obtained.

Molar masses:

M (Ag NO3) = 108 + 14 + 3 × 16 = 170 g / mol;

M (Ag Cl) = 108 + 35.5 = 143.5 g / mol.

Let’s solve the proportion: 170 – 143.5; 342 – x, whence x = 342 × 143.5 / 170 = 288.7 g Ag Cl.