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

What mass of barium sulfate is formed by the interaction of a solution containing barium chloride m = 62.4 g with an excess of sulfuric acid.

1. We draw up a chemical equation, not forgetting to put down the coefficients:

H2SO4 + BaCl2 = BaSO4 + 2HCl.

2. Find the molar mass of BaCl2:

M (BaCl2) = 62.4 / 208.346

n (BaCl2) = 62.4 / M (BaCl2) = 62.4 / 208.346 = 0.3 mol.

3. The amount of ВаCl2 is equal to the amount of BaSO4:

n (BaCl2) = n (BaSO4) = 0.3 mol.

4. Find the mass of BaSO4:

m (BaSO4) = 0.3 * 233.34 = 70.002g.

Answer: m (BaSO4) = 70.002g.