What is the mass of the precipitate obtained by the interaction of an excess of sodium sulfate

What is the mass of the precipitate obtained by the interaction of an excess of sodium sulfate solution with 74.6 g of barium chloride solution with a mass fraction of a solute of 15%?

The reaction between sodium sulfate and barium chloride is described by the following chemical reaction equation:

BaCl2 + Na2SO4 = BaSO4 + 2NaCl;

When 1 mol of barium chloride and 1 mol of sodium sulfate interact, 1 mol of insoluble barium sulfate is synthesized.

Let’s calculate the chemical amount of a substance that is contained in 74.6 grams of a 15% barium chloride solution.

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

N BaCl2 = 74.6 x 0.15 / 208 = 0.054 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 0.054 = 12.58 grams;