An excess of barium nitrate solution was added to a solution containing 49 grams

An excess of barium nitrate solution was added to a solution containing 49 grams of sulfuric acid. Find the mass of the precipitated sediment.

Barium nitrate reacts with sulfuric acid. This results in the formation of water-insoluble barium sulfate, which precipitates. This reaction is described by the following chemical equation.

Ba (NO3) 2 + H2SO4 = BaSO4 + 2HNO3;

Barium nitrate reacts with sulfuric acid in equivalent 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 = 49/98 = 0.5 mol;

Barium sulfate will be synthesized in the same molar amount (sulfuric acid is taken in excess).

Let’s find its weight.

M BaSO4 = 137 + 32 + 16 x 4 = 233 grams / mol; m BaSO4 = 0.5 x 233 = 116.5 grams;