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

Suppose `Na_(2)CO_(3)=xg, NaHCO_(3)=yg`
Then `Na_(2)SO_(4)=2-(x+y)g`
On heating only `NaHCO_(3)` will decompose to give `CO_(2)` as follows :
`underset(2(84)g)(2NaHCO_(3))rarrNa_(2)CO_(3)+H_(2)O+underset("22400 ml at STP")(CO_(2))`
`"y g NaHCO"_(3)" will give CO"_(2)=(22400)/(168)xx"y ml at STP"`
Actual `CO_(2)` produced at STP may be calculated as follows :
`(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2)),(760xxV_(1))/(273)=(750xx123.9)/(298),V_(1)=112.0ml`
Hence, `(22400)/(160)y=112 or y=0.84g`
`"1.5 g of the mixture requires M/10 HCl = 150 ml"`
`therefore " 2.0 g of the mixture will require M/10 HCl"=(150)/(1.5)xx2.0="200 ml = 0.02 mole HCl"`
`underset(106g)(Na_(2)CO_(3))+underset("2 moles")(2HCl)rarr2NaCl+H_(2)O+CO_(2)`
`underset(84g)(NaHCO_(3))+underset("1 mole")(HCl)rarr NaCl+H_(2)O+CO_(2)`
`xg Na_(2)CO_(3)" required HCl"=(2)/(106)xx"x moles"`
`"0.84 g "NaHCO_(3)" require HCl"=(1)/(84)xx"0.84 mole = 0.01 mole"`
`"Hence, "(2x)/(106)+0.01=0.02 or x=0.53g`
`%" of "Na_(2)CO_(3)=(0.53)/(2)xx100=26.5%`
`%" of "NaHCO_(3)=(0.84)/(2)xx100=42.0%`
`%" of "Na_(2)SO_(4)=100-(26.5+42.0)=31.5%`