To a solution containing sulfuric acid weighing 19.6 g, a solution containing barium nitrate weighing 26.1 g

To a solution containing sulfuric acid weighing 19.6 g, a solution containing barium nitrate weighing 26.1 g was added, the substance taken in excess was determined, and the amount and weight of the unreacted residue was calculated.

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. 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 = 19.6 / 98 = 0.2 mol;

Let’s calculate the chemical amount of barium nitrate.

M Ba (NO3) 2 = 137 + (14 + 16 x 3) x 2 = 261 grams / mol; N Ba (NO3) 2 = 26.1 / 261 = 0.1 mol;

Sulfuric acid is taken in excess. 0.1 mol of acid will not react.

Its weight will be:

M H2SO4 = 2 + 32 + 16 x 4 = 98 grams / mol; m H2SO4 = 0.1 x 98 = 9.8 grams;