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

2.0g of mixture of carconate , bicarbonate and chloride of sodium , on

1.	2.0g of mixture of carconate , bicarbonate and chloride of sodium , on heating , produced 56 mL of CO_(2) at NTP . 1.6g of the same mixture required 25 mL of N HCl solution for neutralisation . Calculate the percentage ofNa_(2)CO_(3), NaHCO_(3) and NaCl in the mixture from the given data .
Answer» Solution :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 ofNaHCo_(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%):}` .