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) cases. The volume of CO_(2) at 750 mm Hg pressure and at 298 K is measured to be 123.9 ml. A 1.5 g of the same sample required 150 ml of M/10 HCl for complete neutralization. Calculate the percentage composition of the mixture.

A 2.0 g sample of a mixture containing sodium carbonate, sodium bicarb

1.	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) cases. The volume of CO_(2) at 750 mm Hg pressure and at 298 K is measured to be 123.9 ml. A 1.5 g of the same sample required 150 ml of M/10 HCl for complete neutralization. Calculate the percentage composition of the mixture.
Answer» SOLUTION :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%`