Determine the mass of the precipitate that forms as a result of the exchange reaction

Determine the mass of the precipitate that forms as a result of the exchange reaction between the silver nitrate solution and the silver nitrate solution and the solution containing 13 35 g of aluminum chloride.

Given:
m (AlCl3) = 13.35 g
To find:
m (AgCl)
Decision:
AlCl3 + 3AgNO3 = 3AgCl + Al (NO3) 3
n (AlCl3) = m / M = 13.35 g / 133.5 g / mol = 0.1 mol
n (AlCl3): n (AgCl) = 1: 3
n (AgCl) = 0.1 mol * 3 = 0.3 mol
m (AgCl) = n * M = 0.3 mol * 143.5 g / mol = 43.05 g
Answer: 43.05 g