68 grams of aluminum sulfate reacts with 52 grams of barium chloride. What is the mass of the precipitate formed?

First, we write down the reaction equations.

Al2 (SO4) 3 + 3BaCl2 = 2AlCl3 + 3BaSO4.

Determine which compound is in excess.

M (Al2 (SO4) 3) = 342 g / mol.

M (BaCl2) = 208 g / mol.

X g of barium sulfate – 52 g of barium chloride.

342 g / mol of barium sulfate – 3 * 208 g / mol of barium chloride.

X = 342 * 52 / (3 * 208) = 28.5 g.

We find the mass of the precipitate in terms of barium chloride.

M (BaSO4) = 233 g / mol.

52 g of barium chloride – in g of barium sulfate.

3 * 208 g / mol – 3 * 233 g / mol.

Y = 3 * 233 * 52 / (3 * 208) = 58, 25 g.

Answer: the mass of the sediment is 58.25 g.