A mixture of `NaCl` and `Na_(2)CO` is given On heating `12 g` of the mixture with dilute `HCl, 2.24 g` of `CO_(2)` is removed. Calculate the amounts of each in the mixture.

Explanation

Reactions: `NaCl + HCl rarr` no reaction
`Na_(2) CO_(3) + 2 HCl rarr 2 NaCl + CO_(2) + H_(2) O`
Let `x g` of `Na_(2)CO_(3)` in the mixture.
`:. (x)/(106) mol of Na_(2) CO_(3) = (x)/(106) mol of CO_(2)`
`:. (x)/(106) = (2.24)/(44)` or `x = 5.4 g Na_(2) CO_(3)`
Weight of `NaCl` in mixture `= 12 - 5.4 = 6.6 g`