What mass of precipitate is formed when 100 ml of 1 M BaCl₂ reacts with 100 ml of 1 M Na₂SO₄?
- 23.3 g
- 20.8 g
- 11.6 g
- 46.6 g
📝 Detailed explanation
BaCl₂ + Na₂SO₄ → BaSO₄ + 2NaCl.n(BaCl₂) = 0.1 mol, n(Na₂SO₄) = 0.1 mol.
n(BaSO₄) = 0.1 mol.
M(BaSO₄) = 137 + 32 + 64 = 233 g/mol.
Mass = 0.1 * 233 = 23.3 g.