Calculate the mass of barium sulfate precipitated when the solutions are drained, one of which contains 416 g

Calculate the mass of barium sulfate precipitated when the solutions are drained, one of which contains 416 g of barium chloride, and the other 200 g of sulfuric acid.

Barium chloride reacts with sulfuric acid. At the same time, a water-insoluble barium sulfate salt is synthesized, which precipitates. The reaction is described by the following chemical reaction equation.

BaCl2 + H2SO4 = BaSO4 + 2HCl;

Barium chloride reacts with sulfuric acid in equal molar amounts. This produces the same amount of insoluble barium sulfate.

Let’s calculate the chemical amount of sulfuric acid.

M H2SO4 = 2 + 32 + 16 x 4 = 98 grams / mol; N H2SO4 = 200/98 = 2.041 mol;

Find the chemical amount of barium chloride.

M BaCl2 = 137 + 35.5 x 2 = 208 grams / mol; N BaCl2 = 416/208 = 2 mol;

The same amount of barium sulfate will be synthesized (sulfuric acid is taken in excess).

Let’s calculate its weight.

M BaSO4 = 137 + 32 + 16 x 4 = 233 grams / mol; m BaSO4 = 2 x 233 = 466 grams;