Determine the mass of the precipitate formed by the interaction of 50 g of 18% barium hydroxide

Determine the mass of the precipitate formed by the interaction of 50 g of 18% barium hydroxide solution with sulfuric acid.

Barium hydroxide reacts with sulfuric acid. This forms a water-insoluble barium sulfate salt, which precipitates. The reaction is described by the following chemical reaction equation.

Ba (OH) 2 + H2SO4 = BaSO4 + 2H2O;

Barium hydroxide reacts with sulfuric acid in equal molar amounts. In this case, the same amount of insoluble barium sulfate is synthesized.

Find the chemical amount of barium hydroxide.

M Ba (OH) 2 = 137 + 16 x 2 + 2 = 171 grams / mol; N Ba (OH) 2 = 50 x 0.18 / 171 = 0.053 mol;

The same amount of barium sulfate will be synthesized.

Let’s determine its weight.

M BaSO4 = 137 + 32 + 16 x 4 = 233 grams / mol; m BaSO4 = 0.053 x 233 = 12.35 grams;