When 15 g of a mixture of NaCl and `Na_(2)CO_(3)` is heated with dilute HCl, 2.5 g of `CO_(3)` is evolved at NTP. Calculate percentage composition of the original mixture.

Explanation

Following reactions take place :
`NaCl + HCl rarr` No reaction
`Na_(2)CO_(3)+2HCl rarr 2NaCl+H_(2)O+CO_(2)`
Let 'X' g of `Na_(2)CO_(3)` is there in the mixture.
Moles of `Na_(2)CO_(3)=(x)/(106)`
From stoichiometry,
1 mole of `Na_(2)CO_(3)-=` 1 mole of `CO_(2)`
`rArr (x)/(106)=(2.5)/(44)`
`rArr x = (2.5)/(44)xx 106`
`rArr x = 6.022 g`
`therefore % Na_(2)CO_(3) = 40.15`
Hence, % NaCl in the mixture = 100 - 40.15 = 59.85 .