# Find the mass of the precipitate formed by the action of 300 g of sulfuric acid per 150 g of barium chloride.

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. In this case, the same amount of insoluble salt is synthesized.

Let’s determine the chemical amount of sulfuric acid.

M H2SO4 = 2 + 32 + 16 x 4 = 98 grams / mol; N H2SO4 = 300/98 = 3.061 mol;

Find the chemical amount of barium chloride.

M BaCl2 = 137 + 35.5 x 2 = 208 grams / mol; N BaCl2 = 150/208 = 0.72 mol;

The same amount of barium sulfate will be synthesized (sulfuric acid is taken in excess).

Let’s determine its weight.

M BaSO4 = 137 + 32 + 16 x 4 = 233 grams / mol; m BaSO4 = 0.72 x 233 = 167.76 grams;

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