When an excess of barium nitrate solution was added to a solution of sulfuric acid with a mass fraction of 10%

When an excess of barium nitrate solution was added to a solution of sulfuric acid with a mass fraction of 10%, a precipitate weighing 11.65 g was formed. Determine the mass of the initial sulfuric acid solution.

Barium nitrate 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.

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

Barium nitrate reacts with sulfuric acid in equal molar amounts. This precipitates the same amount of insoluble salt.

Find the chemical amount of barium sulfate.

M BaSO4 = 137 + 32 + 16 x 4 = 233 grams / mol; N BaSO4 = 11.65 / 233 = 0.05 mol;

The same amount of sulfuric acid will be used.

Let’s calculate its weight.

M H2SO4 = 2 + 32 + 16 x 4 = 98 grams / mol; m H2SO4 = 0.05 x 98 = 4.9 grams;

The weight of 10% sulfuric acid will be 4.9 / 0.1 = 49 grams;