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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.
Solution
| 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%
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