An excess of silver nitrate solution was added to a solution of aluminum chloride with a mass of 53.2 g
An excess of silver nitrate solution was added to a solution of aluminum chloride with a mass of 53.2 g and a mass fraction of 5%. Calculate the mass of the precipitate formed.
Find the mass of AlCl3 in solution.
W = m (substance): m (solution) × 100%,
m (substance) = (m (solution) × W): 100%.
m (AlCl3) = (53.2 g × 5%): 100% = 2.66 g.
Let’s find the amount of substance AlCl3 by the formula:
n = m: M.
M (AlCl3) = 133.5 g / mol.
n = 2.66 g: 133.5 g / mol = 0.02 mol.
Let’s compose the reaction equation, find the quantitative ratios of substances.
AlCl3 + 3 AgNO3 = 3 AgCl ↓ + Al (NO3) 3.
According to the reaction equation, there is 3 mol of AgCl per 1 mol of AlCl3. The substances are in quantitative ratios of 1: 3.
n (AgCl) = 3 n (AlCl3) = 0.02 × 3 = 0.06 mol.
Let’s find the mass of AgCl by the formula:
m = n × M,
М (AgCl) = 143.5 g / mol.
m = 0.06 mol × 143.5 g / mol = 8.61 g.
Answer: 8.61 g.