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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» <html><body><p></p>Solution :`underset(106)(Na_(2)CO_(3))+underset(2xx36.5)(2HCl) to 2NaCl +H_(2)O+CO_(2)` <br/> Eq. mass 53 `"" `36.5 <br/> 1 g eq. `""` 1 g eq. <br/> `underset(100)(CaCO_(3))+underset(2xx36.5)(2HCl)to CaCl_(2)+H_(2)O+CO_(2)` <br/> Eq. 50 `"" `36.5 <br/> 1g eq `"" `1 g eq. <br/> Let x go `CaCO_(3)` be <a href="https://interviewquestions.tuteehub.com/tag/present-1163722" style="font-weight:bold;" target="_blank" title="Click to know more about PRESENT">PRESENT</a> in the mixture <br/> Mass of `Na_(2)CO_(3)` in the mixture =(1.03-x)g <br/> No. of equivalents of `CaCO_(3)=(x )/(50)` <br/> No. of g equivalents of `Na_(2)CO_(3)=((1.03-x))/(53)` <br/> No. of g equivalents in 20 mL. N HCl`=("Normality " xx "<a href="https://interviewquestions.tuteehub.com/tag/vol-723961" style="font-weight:bold;" target="_blank" title="Click to know more about VOL">VOL</a>.")/(1000)` <br/> `=(1xx20)/(1000)=(1)/(50)` <br/> At equivalence point, <br/> No. of g equivalents of `CaCO_(3)` + No. of g equivalents of `Na_(2)CO_(3)`=No. of <a href="https://interviewquestions.tuteehub.com/tag/gram-1010695" style="font-weight:bold;" target="_blank" title="Click to know more about GRAM">GRAM</a> equivalents of HCl <br/> `(x)/(50)+(1.03-x)/(53)=(1)/(50)` <br/> or x=0.50 <br/> `CaCO_(3)=0.50 g, % CaCO_(3)=(0.50)/(1.03)xx100=48.54` <br/> `Na_(2)CO_(3)=0.53 g, % Na_(2)CO_(3)=(0.53)/(1.03)xx100=51.46`</body></html>


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