1.84 g of a mixture of `CaCO_(3) and MgCO_(3)` is strongly heated till no further loss in mass takes place. The residue is found to weigh 0.96 g. Calculate the percentage of each component in the mixture.

Explanation

Both `CaCO_(3) and MgCO_(3)` upon heating lose `CO_(2)` and the mass of residue is that of CaO and MgO. Let us calculate the same from the mixture and quate with the given mass `(0.96g)`
Let the mass of `CaCO_(3)` in the mixture = xg
`:.` Mass of `MgCO_(3)` in the mixture `(1.84 - x)g`
Step I. Calculation of mass of residue (CaO) from x g of `CaCO_(3)`
The chemical equation for the reaction is :
`{:(CaCO_(3),overset("Heat")(rarr),CaO+CO_(2)),(40+12+48,,40+16),(=100,,=56g):}`
100 g of `CaCO_(3)` upon heating from CaO (residue) = 56 g
`:.` x g of `CaCO_(3)` upon heating form CaO (residue) `= (56)/(100) xx xg`
Step II. Calculation of mass of residue (MgO) from `(1-x)` of `MgCO_(3)`
`{:(MgCO_(3),overset("Heat")(rarr),MgO+CO_(2)),(24+12+48,,24+16),(=84g,,=40g):}`
84 g of `MgCO_(3)` upon heating from MgO (residue) = 40 g
`(1-x)` of `MgCO_(3)` upon heating from MgO (residue) `= (40)/(84) xx(1-x)g`
Step III. Calculation of percentage composition of the mixture
Total mass of the residue `= [(56)/(100)xxx+(40)/(84)(1-x)]g`
But the mass of residue actually formed = 0.96 g
Equating the two, we get
`(56x)/(100)+(40(1-x))/(84)=0.96`
`4704x+7360-4000x=0.96xx100xx84=8064`
`704x=8064-7360=704,x=(704)/(704)=1`
Mass of `CaCO_(3)` in the mixture = 1g
Mass of `MgCO_(3)` in the mixture `= 1.84 - 1 = 0.84`g
Percentage of `CaCO_(3)` in the mixture `= (1)/(1.84) xx 100 = 54.35 %`
Percentage of `MgCO_(3)` in the mixture `= (0.84)/(1.84) xx 100 = 45.65 %`