3.68g of mixture of CaCO_(3) and MgCO_(3) , on reaction with 1000 mL of N/10 HCl solution produced1.76" g of " CO_(2) . Calculate the percentage of each in the mixture .

3.68g of mixture of CaCO_(3) and MgCO_(3) , on reaction with 1000 mL o

1.	3.68g of mixture of CaCO_(3) and MgCO_(3) , on reaction with 1000 mL of N/10 HCl solution produced1.76" g of " CO_(2) . Calculate the percentage of each in the mixture .
Answer» Solution :`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 MOLEOF `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 HCLSOLUTION . Andso 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 % `