`6.2g` of a sample containing `Na_2CO_3, NaHCO_3` and non-volatile inert impurity on gentle heating loses 5% of its mass due to reaction `2NaHCO_3 rarr Na_2 CO_3 + H_2 O + CO_2` . Residue is dissolved in water and formed `100mL` solution and its `10mL` portion requires `7.5 mL` of `0.2M` aqueous solution of `BaCI_2` for complete precipitation of carbonates. Determine mass (in gram) 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)`