What is the mass of sediment obtained by the reaction of 40 g of barium chloride and 100 g of sulfuric acid.

The equation for the reaction of sulfuric acid with barium chloride:
BaCI₂ + H₂SO₄ = BaSO₄ ↓ + 2HCI.
During the reaction, barium sulfate precipitates, the mass of which must be determined.
Determine the number of moles of starting materials:
n (BaCI₂) = m (BaCI₂) ÷ Mr (BaCI₂) = 40 ÷ 208 = 0.19 mol.
n (H₂SO₄) = m (H₂SO₄) ÷ Mr (H₂SO₄) = 100 ÷ 98 = 1.02 mol.
Sulfuric acid will be in excess, and barium chloride will be deficient, because 1 mole of substances enters into the reaction.
Hence n (BaSO₄) = n (BaCI₂) = 0.19 mol.
We find the mass of the sediment:
m (BaSO₄) = n (BaSO₄) • Mr (BaSO₄) = 0.19 • 233 = 44.3 g.