Question

Barium chloride was thoroughly reacted with sulfuric acid. The weight of sulfuric acid needed to completely consume barium chloride was 20 kilograms. What was the total weight of barium sulfate being produced in this reaction?

Answer 1

The reaction is:

BaCl₂ + H₂SO₄ => BaSO₄ + 2HCl

As we see from the equation, 1 mole of sulfuric acid produces 1 mole of barium sulfate. Find the total weight of barium sulfate:

$m(BaSO_4) = m(H_2SO_4)*M(BaSO_4)/M(H_2SO_4)$ , where M -- molar mass.

m(BaSO4) = 20000 * 233.4 / 98 = 47632 (g) ~ 47.6 (kg)

Answer: 47.6 kg is the total weight of barium sulfate being produced in this reaction.