An excess of hydrochloric acid solution was added to 500 g of a solution of silver (I)

An excess of hydrochloric acid solution was added to 500 g of a solution of silver (I) nitrate with a mass fraction of 0.0592. Determine the mass of the precipitate formed in grams.

Let’s write the reaction equation:
AgNO3 + HCl = AgCl ↓ + HNO3

Let’s find the amount of silver nitrate:
m (AgNO3) = m (p-pa) * ω (AgNO3) = 500 * 0.0592 = 29.6 g.

n (AgNO3) = m (AgNO3) / Mr (AgNO3) = 29.6 / 170 = 0.17 mol;

Let’s find the mass of silver chloride:
n (AgCl) = n (AgNO3) = 0.17 mol;

m (AgCl) = n (AgCl) * Mr (AgCl) = 0.17 * 143.5 = 24.395 g.

Answer: 24.395 g.