`2.0` g of mixture of carconate , bicarbonate and chloride of sodium , on heating , produced 56 mL of `CO_(2)` at NTP . `1.6` g of the same mixture required 25 mL of N HCl solution for neutralisation . Calculate the percentage of `Na_(2)CO_(3), NaHCO_(3)` and NaCl in the mixture from the given data .

Explanation

On heating the given mixture , only `NaHCO_(3)` decomposes as
`2NaHCO_(3) to Na_(2)CO_(3) +H_(2)O + CO_(2)`
` :. ` eq. of `NaHCO_(3)` = eq. of `CO_(2) = 56/11200 " "….(Eqn . 4ii)`
( 1 eq. of `CO_(2)` occupies 11200 mL at NTP )
` :. wt of NaHCo_(3) = 56/1120 xx 84 = 0.42 g `
( eq. wt . of `NaHCO_(3) = 84`)
` :. % of NaHCO_(3) = (0.42)/2 xx 100 = 21 % `
Now , if x is the weight of NaCl in `1.6` g of the mixture
then wt. of `NaHCO_(3) = 0.336 " g " (i.e ., 21 % " of " 1.6 g)`
and wt. of `Na_(2)CO_(3) = 1.6 - 0.336 - x = (1.264 - x) g `
Since `Na_(2)CO_(3)` are neutralised by HCl solution as :
`Na_(2)CO_(3) +2HCl to 2NaCl +H_(2)O +CO_(2)`
`NaHCO_(3) +HCl to NaCl +H_(2)O + CO_(2)`
m.e of `Na_(2)CO_(3)` + m.e of `NaHCO_(3)` = m.e of HCl
or eq. of `Na_(2)CO_(3) xx 1000 " eq. of " NaHCO_(3) xx 100 ` = m.e of HCl ...(Eqn.3)
`(1.264 -x)/53 xx1000 +(0.336)/84 xx 1000 = 1 xx 25 `
` x = 0.151 g `
` :. " " % of NaCl = (0.151)/(1.6) xx 100 = 9.42 % `
and % of `Na_(2)CO_(3) = 100 - (21 +9.42) = 69.58 % `
Thus , `{:{(Na_(2)CO_(3)=69.58%),(NaHCO_(3)=21.00%),(NaCl=9.42%):}` .