`3.68 ` g of mixture of `CaCO_(3) and MgCO_(3)` , on reaction with 1000 mL of N/10 HCl solution produced `1.76 " g of " CO_(2)` . Calculate the percentage of each in the mixture .

Explanation

`CaCO_(3) +2HCI to CaCl_(2) +H_(2)O + CO_(2)`
`MgCO_(3) +2HCl to MgCl_(2) +H_(2)O + CO_(2)`
From the equation , 2 eq. of HCl produces 1 mole of `CO_(2)`
` :. `1 eq. HCl produces 1/2 molee of `CO_(2)`
` :. " eq. wt of " CO_(2) = 44/2 = 22 `
` :. " eq. of " CO_(2) = (1.76 )/22 = 0.08 " " ...(Eqn . 4i) `
From the given problem it is clear that the number of equivalents of the mixture of `caCO_(3) and MgCO_(3)` is lesser than that of HCl solution . And so the `CO_(2)` is produced by the complete consuumption of `CaCO_(3) and MgCO_(3)` .
` :. " eq. of " CaCO_(3) + " eq. of " MgCO_(3) = " eq. of " CO_(2)`
`x/50 +(3.68 - x)/42 = 0.08 ( " x = amount of " CaCO_(3))`
x = 2.0 g
` :. % of CaCO_(3) = 2/(3.68) xx 100 = 54 .34 % `
% of `Mg CO_(3) = 100 - 54. 34 = 45 . 66 % `