Calculate the mass of the precipitate formed by the interaction of 200 g of a 7.2% sodium sulfate

Calculate the mass of the precipitate formed by the interaction of 200 g of a 7.2% sodium sulfate solution with an excess of barium chloride solution.

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 the substance, which is contained in 200 grams of 7.2% sodium sulfate solution.
M Na2SO4 = 23 x 2 + 32 + 16 x 4 = 142 grams / mol;
N Na 2SO4 = 200 x 0.072 / 142 = 0.101 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 0.101 = 23.533 grams;