A `2.0 g` sample of a mixture containing sodium carbonate, sodium bicarbonate and sodium sulphate is gently heated till the evolution of `CO_(2)` ceases. The volume of `CO_(2)` at `750mm Hg` pressure and at `298 K` is measured to be `123.9 mL`. A `1.5 g` of the same sample requires `150 mL` of `(M//10) HCl` for complete neutralisation. Calculate the percentage composition of the components of the mixture.

Explanation

`CO_(2)` is evolved due to following reaction :
`2NaHCO_(3) rarr Na_(2)CO_(3) + H_(2)O + CO_(2)`
Moles of `CO_(2)` produced `= (pV)/(RT)`
`=(750)/(760) xx (123.9)/(1000) xx (1)/(0.082 xx 298)`
`= 5 xx 10^(-3)`
`rArr` Moles of `NaHCO_(3)` in `2g` sample `= 2 xx 5 xx 10^(-3) = 0.01`
`rArr` millimol of `NaHCO_(3)` in `1.5g` sample
`= (0.01)/(2) xx 1.5 xx 1000 = 7.5`
Let the `1.5g` sample contain x millimol `Na_(2)CO_(3)` then
`2x + 7.5 =` millimol of `HCl = 15`
`rArr x = 3.75`
`rArr` Mass of `NaHCO_(3) = (7.5 xx 84)/(1000) = 0.63 g`
Mass of `Na_(2)CO_(3) = (3.75 xx 106)/(1000) =0.3975g`
`rArr %` mass of `Na_(2)CO_(3) = (0.3975)/(1.5) xx 100`
`= 26.5%`