Find the mass fraction of the yield of barium sulfate if, upon the interaction of 17.1 g

Find the mass fraction of the yield of barium sulfate if, upon the interaction of 17.1 g of barium hydroxide with sulfuric acid, 20 g of barium sulfate were obtained

Barium hydroxide 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 (OH) 2 + H2SO4 = BaSO4 + 2H2O;

Barium hydroxide reacts with sulfuric acid in equivalent molar amounts. In this case, an equal amount of insoluble salt is synthesized.

Let’s calculate the chemical amount of barium hydroxide.

M Ba (OH) 2 = 137 + 16 x 2 + 2 = 171 grams / mol; N Ba (OH) 2 = 17.1 / 171 = 0.1 mol;

Barium sulfate will be synthesized in the same molar amount.

Let’s calculate its weight.

M BaSO4 = 137 + 32 + 16 x 4 = 233 grams / mol; m BaSO4 = 0.1 x 233 = 23.3 grams;

The reaction yield will be 20 / 23.3 = 0.858 = 85.8%;