Determine the mass of the resulting precipitate obtained by mixing 120 g of a 30% barium

Determine the mass of the resulting precipitate obtained by mixing 120 g of a 30% barium chloride solution with 200 g of a 10% sulfuric acid solution.

When barium chloride interacts with sulfuric acid, an insoluble barium sulfate salt is synthesized.

The process is described by the following equation:

BaCl2 + H2SO4 = BaSO4 + 2HCl;

1 mole of barium chloride reacts with 1 mole of sulfuric acid. This synthesizes 1 mole of water-insoluble barium sulfate.

Let’s calculate the chemical amount of the available substance.

M BaCl2 = 137 + 35.5 x 2 = 208 grams / mol;

N BaCl2 = 120 x 0.3 / 208 = 0.173 mol;

M H2SO4 = 2 + 32 + 16 x 4 = 98 grams / mol;

N H2SO4 = 200 x 0.1 / 98 = 0.204 mol;

0.173 mol of barium chloride reacts with 0.173 mol of sulfuric acid. This synthesizes 0.173 mol of water-insoluble barium sulfate.

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

The sediment weight will be:

m BaSO4 = 0.173 x 233 = 40.31 grams;