A 4.0 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 loss in weight is due to removal of `CO_(2) " and " H_(2)O` which escape out on heating.
Wt. of `Na_(2)CO_(3)` in the product ` = 2.00 - 0.248 = 1.752 `g
Let wt. `NaHCO_(3) = (2.00 -x) g`
Since `Na_(2)CO(3)` in the products contains x g of unchanged reactant `Na_(2)CO_(3) ` and rest produced from `NaHCO_(3)`.
The wt. of `Na_(2)CO_(3)` produced by `NaHCO_(3) = (1.752 - x)`
`underset((2.0 -x)) (NaHCO_(3)) to underset((1.752 -x))(Na_(2)CO_(3) + )(H_(2)O + CO_(2))uparrow `
Appiying POAC for Na atom.
` 1 xx ` moies of `NaHCO_(3) = 2 xx ` moies of `Na_(2)CO_(3) rArr ((2 -x))/84 = 2 xx ((1 .752 -x))/106`
` :. x = 1.328 `g
` :. % " of " Na_(2)CO_(3) = (1.328)/2 xx 100 = 66.4%`