`2.5 g` of a sample containing `Na_(2)CO_(3) , NaHCO_(3)` and some non-volantile impurity on gentle heating loses `12%` of its weight. Residue is dissolved in `100 mL` water and its `10 mL` portion required `15mL 0.1M` aqueous solution of `BaCl_(2)` for complete precipitation of carbonates. Determine mass percentage of `Na_(2)CO_(3)` in the original sample?

Answer

`42.4%Na_(2)CO_(3)`

Explanation

Weight loss is due to conversion fo `NaHCO_(3)` into `Na_(2)CO_(3) : 31 g` weight is lost per mole of `NaHCO_(3)`
`implies 0.3 g wt`. loss from `(0.3)/(31)` mol of `NaHCO_(3)` producing `(0.3)/(62)` moles of `Na_(2)CO_(3)`
Total moles of carbonate `=15xx10^(-3)`
implies Moles of carbonate in original sample `=0.015 - (3)/(620)=0.01`
Mass of `Na_(2)CO_(3)` in original sample `=1.06 implies 42.4%Na_(2)CO_(3)`