Question

A solution of sodium sulphate containing 0.2 moles of sulphate ions is treated with a solution of barium chloride containing 14.2g of chloride ions. Total weight of precipitate formed is (atomic mass of : Na=23u, S=32u, O=16u, Ba=137u, Cl=35.5u)
1)28.6g
2)11.7g
3)70.0g
4)46.6g

Answer 1

answer : option (4) 46.6 g

explanation : no of moles of Na2SO4 = 0.2 mol

In one mole of BaCl2 two moles of Cl– ions.

no of moles of chloride ions = 14.2g/35.5g = 0.4 mol

so, number of moles of BaCl2 = 0.4/2 = 0.2 mol

Na2SO4 + BaCl2 ⇔2NaCl + BaSO4 (↓)

one mole of Na2SO4 reacts with one mole of BaCl2 and produces one mole of BaSO4 (Precipitation).

⇒0.2 mol of Na2SO4 reacts with 0.2 mol of BaCl2 and produces 0.2 mol of BaSO4 (precipitation).

weight of BaSO4 = 0.2 × (137 + 32 + 64)

= 0.2 × 233

= 46.6 g