Determine the mass of the salt formed by merging 150 g of a 13.05 barium nitrate solution and a sulfuric acid solution

Determine the mass of the salt formed by merging 150 g of a 13.05 barium nitrate solution and a sulfuric acid solution. What mass of a 10% sulfuric acid solution will this require?

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, an equal amount of insoluble salt is synthesized.

Let’s find 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 = 150 x 0.1305 / 261 = 0.075 mol;

The same amount of barium sulfate will be synthesized.

Let’s calculate its weight.

M BaSO4 = 137 + 32 + 16 x 4 = 233 grams / mol; m BaSO4 = 0.075 x 233 = 17.475 grams;

Let’s find the weight of this amount of sulfuric acid.

M H2SO4 = 2 + 32 + 16 x 4 = 98 grams / mol; m H2SO4 = 0.075 x 98 = 7.35 grams;

The weight of a 10% acid solution will be 7.35 / 0.1 = 73.5 grams;