A 2 - g sample containing `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 the given mixture ? (Na = 23, C = 12 , O = 16 and H = 1 )

Explanation

On heating `Na_(2) CO_(3)and NaHCO_(3), Na_(2)CO_(3)` remains unchanged while `NaHCO_(3)` changes into `Na_(2) CO_(3) , CO_(2) and H_(2) O` . The loss in weight is due to remove of `CO_(2) and H_(2)O` which escape out on heating .
`:.` wt. of `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)` only = (1.752- x) g
Now, we have ,
` {:(NaHCO_(3) to Na_(2)CO_(3) + (H_(2)O + CO_(2) )uarr),("(2 . 0 - x) g (1 . 752 - x) g "):}`
Applying POAC for Na atoms,
moles of Na in ` NaHCO_(3) ` = moles of Na in `Na_(2)CO_(3)`
` 1 xx ` moles of `NaHCO_(3) = 2xx ` moles of `Na_(2) CO_(3)`
`(2. 0 - x)/( 84) = 2 xx (1 . 752 - x)/( 106) [{:(NaHCO_(3) = 84 ),(Na_(2) CO_(3) = 106):}]`
`x = (82)/( 62) = 1.328 g`
`:. % of Na_(2) CO_(3) = (1.328)/( 2 . 0) xx 100 = 664 %`