The mass of the precipitate formed as a result of the interaction of 5.2 g of barium chloride

The mass of the precipitate formed as a result of the interaction of 5.2 g of barium chloride with a solution of sulfuric acid, taken in excess, is

Barium chloride 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.

BaCl2 + H2SO4 = BaSO4 + 2HCl;

Barium chloride reacts with sulfuric acid in equal molar amounts. This produces the same amount of insoluble barium sulfate.

Let’s calculate the chemical amount of barium chloride.

M BaCl2 = 137 + 35.5 x 2 = 208 grams / mol; N BaCl2 = 5.2 / 208 = 0.025 mol;

The same amount of barium sulfate will be synthesized.

Let’s find its weight.

M BaSO4 = 137 + 32 + 16 x 4 = 233 grams / mol; m BaSO4 = 0.025 x 233 = 8.825 grams;