1.03 g mixture of sodium carbonate and calcium carbone require 20 mL N HCl for complete neutralisation. Calculate the percentage of sodium carbonate and calcium carbonate in the given mixture.

1.03 g mixture of sodium carbonate and calcium carbone require 20 mL N

1.	1.03 g mixture of sodium carbonate and calcium carbone require 20 mL N HCl for complete neutralisation. Calculate the percentage of sodium carbonate and calcium carbonate in the given mixture.
Answer» Solution :`underset(106)(Na_(2)CO_(3))+underset(2xx36.5)(2HCl) to 2NaCl +H_(2)O+CO_(2)` Eq. mass 53 `"" `36.5 1 g eq. `""` 1 g eq. `underset(100)(CaCO_(3))+underset(2xx36.5)(2HCl)to CaCl_(2)+H_(2)O+CO_(2)` Eq. 50 `"" `36.5 1g eq `"" `1 g eq. Let x go `CaCO_(3)` be PRESENT in the mixture Mass of `Na_(2)CO_(3)` in the mixture =(1.03-x)g No. of equivalents of `CaCO_(3)=(x )/(50)` No. of g equivalents of `Na_(2)CO_(3)=((1.03-x))/(53)` No. of g equivalents in 20 mL. N HCl`=("Normality " xx "VOL.")/(1000)` `=(1xx20)/(1000)=(1)/(50)` At equivalence point, No. of g equivalents of `CaCO_(3)` + No. of g equivalents of `Na_(2)CO_(3)`=No. of GRAM equivalents of HCl `(x)/(50)+(1.03-x)/(53)=(1)/(50)` or x=0.50 `CaCO_(3)=0.50 g, % CaCO_(3)=(0.50)/(1.03)xx100=48.54` `Na_(2)CO_(3)=0.53 g, % Na_(2)CO_(3)=(0.53)/(1.03)xx100=51.46`