What mass of barium sulfate is formed by the interaction of 9.8 g of sulfuric acid with barium hydroxide?

The reaction of interaction of sulfuric acid with barium hydroxide is described by the following chemical reaction equation:

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

In the interaction of 1 mol of barium hydroxide and 1 mol of sulfuric acid, 1 mol of insoluble barium sulfate is synthesized.

Let’s calculate the chemical amount of a substance that is contained in 9.8 grams of sulfuric acid.

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

N H2SO4 = 9.8 / 98 = 0.1 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 = 233 x 0.1 = 23.3 grams;