A solution of sulfuric acid (196 g, mass fraction of acid 10%) was added to a solution of barium chloride

A solution of sulfuric acid (196 g, mass fraction of acid 10%) was added to a solution of barium chloride (solution mass 125 g, mass fraction of salt 10%) and calculate the mass of the precipitate formed as a result of the reaction.

Barium chloride reacts with sulfuric acid. In this case, a water-insoluble barium sulfate salt is formed, which turns into an insoluble precipitate. The reaction is described by the following chemical reaction equation.

BaCl2 + H2SO4 = BaSO4 + 2HCl;

Barium chloride reacts with sulfuric acid in equal molar amounts. In this case, the same amount of insoluble salt is synthesized.

Let’s calculate the chemical amount of sulfuric acid.

M H2SO4 = 2 + 32 + 16 x 4 = 98 grams / mol; N H2SO4 = 196 x 0.1 / 98 = 0.2 mol;

Determine the chemical amount of barium chloride.

M BaCl2 = 137 + 35.5 x 2 = 208 grams / mol; N BaCl2 = 125 x 0.1 / 208 = 0.06 mol;

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

Let’s determine its weight.

M BaSO4 = 137 + 32 + 16 x 4 = 233 grams / mol; m BaSO4 = 0.06 x 233 = 13.98 grams;