Question

10 g of a mixture of sodium chloride and anhydrous sodium sulphate is dissolved in water. An excess of barium chloride solution is added and 6.99 g of barium sulphate is precipitated according to the equation given below:

\[\ce{Na2SO4 + BaCl2 -> BaSO4 + 2NaCl}\]

Calculate the percentage of sodium sulphate in the original mixture.

Answer 1

Na2SO4	+	BaCl2	→	BaSO4	+	2NaCl
[2(Na) + S + 4(O)]				6.99 g
[(2 × 23) + 32 + (4 × 16)]				[137 + 32 + 64]
142 g				233 g

Now, 233 g of BaSO₄ is produced by Na₂SO₄ = 142 g

6.99 g BaSO₄ will be produced by = `(6.99 × 142)/233`

= 4.26 g Na₂SO₄

∴ Na₂SO₄ in original mixture = `(4.26 × 100)/10`

= 42.6%