`1.42 ` g of a mixture of `caCO_(3) and MgCO_(3) ` was dissolved in 200 mL of `0.2 `N HCl solution which was then diluted to 250 mL . 10 mL of this solution was neutralised by 12 mL of (N/30) `Na_(2)CO_(3)` .Find out the percentage of each in the mixture . ( Ca = 40 , Mg = 24 , C = 12 , O = 16)

Explanation

Let the wt. of `CaCO_(3)` be x g .
` :. ` wt . Of `MgCO_(3) = (1.42- x ) g `
` :. ` eq of `CaCO_(3) = x/50` and eq. of `MgCO_(3) = (1.42 - x)/42 " "…(Eqn.4)`
`( " eq. wt of " CaCO_(3) = 100/2 = 50 " eq. wt of " MgCO_(3) = 84/2 = 42)`
Total m.e of `CaCO_(3) and MgCO_(3) = x/50 xx 1000 + (1.42 - x)/42 xx 1000 ...(Eqn.3)`
m.e of HCl = `0.2 xx 200 = 40 `
From the given question it is clear that m.e of HCl is greater than those of `caCO_(3) and MgCO_(3)`
` :. ` m.e of excess HCl = m.e of HCl - m.e of `CaCO_(3) and MgCO_(3)`
` = 40 - { 1000 (x/50 + (1.42 -x)/42 )}`
` :. ` the m.e of the resulting solution does not change on dilution .
` :. ` normality of excess HCl in the diluted resulting solution
` = (m.e)/250 = (4 - {1000(x/50+(1.42-x)/42)})/250`
` :. ` m.e of 10 mL of the resulting solution
` (4 - {1000(x/50+(1.42-x)/42)})/250xx10`
m.e of `na_(2)CO_(3)` solution ` = 1/30 xx12 ` ...(Eqn.1)
` :. (40 - {1000(x/50+(1.42-x)/42)})/250 xx 10 = 1/ 30 xx 12 " " ( Eqn. 2)`
` :. x = 1 `
` :. % of CaCO_(3) = 1/(1.42) xx 100 = 70.4 % `
% of `MgCO_(3) = 100 - 70.4 = 29 .6 % `