# 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;