An excess of silver nitrate solution was added to a solution of aluminum chloride with a mass of 399 g and a mass fraction of 10%

An excess of silver nitrate solution was added to a solution of aluminum chloride with a mass of 399 g and a mass fraction of 10%. Calculate the mass of the precipitate formed.

Given: mp-pa (AlCl3) = 399g wp-pa (AlCl3) = 10% = 0.1 + excess AgNO3 solution
m (AgCl) -?
Solution: AlCl3 + 3AgNO3 —> Al (NO3) 3 + 3AgCl n (AlCl3) = 399g * 0.1 / 133.5g / mol = 0.3 mol n (AlCl3) – 0.3 mol 3n (AgCl) – x Therefore n (AgCl) = 0.9 mol m (AgCl) = 0.9 mol * 143.5 g / mol = 129.15 g
Answer: 129.15g