The mass of `CaCO_(3)` be present in the mixture.
The mass of `MgCO_(3)` in the mixture=(3.68-x)g
`underset(100g)(CaCO_(3)) overset("Heat")to underset(56g)(CaO)+CO_(2)`
`underset(84g)(MgCO_(3)) overset("Heat")to underset(40g)(MgO)+CO_(2)`
`(56xxx)/(100)+(40xx(3.68-x))/(84)=1.92`
On solving x=2
Percentage of `CaCO_(3)=(2)/(3.68)xx100=54.35%`
Percentage of `MgCO_(3)=100-54.35=45.65%`