What is the mass of the precipitate formed when an excess of barium chloride solution interacts with 200 g

What is the mass of the precipitate formed when an excess of barium chloride solution interacts with 200 g with a solution of aluminum sulfate with a mass fraction of 40.4%.

The reaction of interaction of aluminum sulfate with barium chloride is described by the following chemical reaction equation:

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

When 3 mol of barium chloride and 1 aluminum sulfate react, 3 mol of insoluble barium sulfate is synthesized.

Let’s calculate the chemical amount of a substance that is contained in 200 grams of a 40.4% solution of aluminum sulfate.

M Al2 (SO4) 3 = 27 x2 + (32 + 16 x 4) x 3 = 342 grams / mol;

N Al2 (SO4) 3 = 200 x 0.404 / 342 = 0.236 mol;

Thus, 0.236 x 3 = 0.708 mol of barium sulfate will be synthesized.

Let’s calculate its weight.

M BaSO4 = 137 + 32 + 16 x 4 = 233 grams / mol;

m BaSO4 = 233 x 0.708 = 164.96 grams;