What mass of barium sulfate is formed by the interaction of a solution containing 2.4 grams of barium

What mass of barium sulfate is formed by the interaction of a solution containing 2.4 grams of barium chloride with sodium sulfate.

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 2.4 grams of sodium sulfate.

M Na2SO4 = 23 x 2 + 32 + 16 x 4 = 142 grams / mol;

N Na2SO4 = 2.4 / 142 = 0.017 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 0.017 = 3.961 grams;