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

Out of `Na_(2)CO_(3) , NaHCO_(3)` and `Na_(2)SO_(4)` only `NaHCO_(3)` decompose on heating to give `CO_(2)` gas, according to the equation `= ( PV)/( RT) = ( 750 xx 123.9)/( 760 xx 1000 xx 0.082 xx 298) = 5 xx 10^(-3)`
`:.` moles of `NaHCO_(3) = 2 xx 5 xx 10^(-3) = 0.01`
Equivalents of HCl used `=(150 xx(1)/(10))/( 1000) = 1.5 xx 10^(-2)`
Equivalents of `NaHCO_(3)` in 1.5 g `= 0.01 xx (1.5)/( 2) = 7.5 xx 10^(-3)`
`:.` Equivalents of `Na_(2)CO_(3)`
`= 1.5 xx 10^(-2) - 7.5 xx 10^(-3) = 7.5 xx 10^(-3)`
Moles of `Na_(2)CO_(3) = ( 7.5 xx 10^(-3))/( 2)`
( when `Na_(2)CO_(3) ` reacts with HCl it gives `NaCl, CO_(2) ` and `H_(2)O` . No atom undergoes changes in oxidation state . `:.` 'n' factor of `Na_(2) CO_(3) = 2) = 3.75 xx 10^(-3)`
Mass of `NaHCO_(3) ` in `1.5g = 7.5 xx 10^(-3) xx 84 = 0.63 g `
Mass of `Na_(2)CO_(3) ` in 1.5 g `= 3.75 xx 10^(-3) xx 106 = 0.3975 g `
`:.` mass of `Na_(2) SO_(4) = 1.5- 0.63- 0.3975 = 0.4725 g `
Percentage of `NaHCO_(3) = ( 0.63 ) /( 1.5 ) x 100 = 42%`
Percentage of `Na_(2) CO_(3) = ( 0.3975) /(1.5) xx 100 = 26.5%`
Percentage of `Na_(2) SO_(4) = ( 0.4725)/( 1.5) xx 100 = 31.5%`