A `2.0 g` of mixture of `Na_(2)CO_(3)` and `NaHCO_(3)` loses `0.248 g` when heated to `300^(@)C`, the temperature at which `NaHCO_(3)` decomposes to `Na_(2)CO_(3)`, `CO_(2)` and `H_(2) O`. What is the percentage of `Na_(2) CO_(3)` in mixture?

Explanation

The balanced chemical equation is:
`underset(168)(2NaHCO_(3)) to Na_(2)CO_(3) + underset(18 g)(H_(2)O) + underset(44 g)(CO_(2))`
`CO_(2)` and `H_(2)O` will escape as gases at `300^(@)` C. Therefore, loss in weight would correspond to the weight of `H_(2)O` and `CO_(2)`.
Total mass of C02 + water lost by heating 168 g of `NaHCO_(3)` = (18 + 44) = 62 g
62 g of weight is lost by heating `NaHCO_(3)`
`=168/62 xx 0.248 = 0.672`
2 g of sample contains `NaHCO_(3) =0.672 g`
Weight of `Na_(2)CO_(3)` in sample `=2 xx 0.672 = 1.344`g
`therefore` Percentage of `Na_(2)CO_(3) = (1.344)/2 xx 100 = 67.2%`