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.

Explanation

`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`