A mixture of `CaCO_(3)` and `MgCO_(3)` weighing `1.84 g` on heating left a residue weighing `0.96 g`. Calculate the percentage of each in the mixture.

Explanation

In this problem, first try to find, what is the residue.
`{:(CaCO_3toCaO+CO_2),(MgCO_3toMgO+CO_2):}]`
`CO_(2)` (carbon diozixde) being a gas escapes, so residue includes `MgO` and `CaO`.
Let the sample contains `x g` of `CaCO_(3)`.
`implies` Mass of `MgCO_(3) = (1.84 - x) g`
moles of `CaCO_(3) = (x)/(100)`
and mole of `MgCO_(3) = (1.84 - x)/(84)`
From stiochiometry, we have:
1 mol of `CaCO_(3) -= 1 mol of CaO`
`(x)/(100) mol -= (x)/(100) mol of CaO`
`-= (x)/(100) xx 56 g of CaO`
`1 mol of MgCO_(3) -= 1 mol MgO`
`(1.84 - x)/(84) mol of MgCO -= (1.84 - x)/(84) mol of MgO`
`-= ((1.84 - x))/(84) xx 40 g of MgO`
`implies` Total of `MgO` and `CaO` (i.e., residue) `= 0.96 g`
`implies (56 x)/(100) + (40)(84) (1.84 - x) = 0.96 g`
Solving for `x`, we get `x = 1.0 g`
So, we have `1.0 g` of `CaCO_(3)` and
`(1.84 - 1.0) =- 0.84 g` of `MgCO_(3)` in the sample
`% CaCO_(3) = (1.0)/(1.84) xx 100 = 54.34%`
`% Mg CO_(3) = (100 - 54.34)% = 45.66%`