1 gm of a mixture of calcium carbonate and magnesium carbonate gave on ignition 240 mL of carbon dioxide at S.T.P. What is the percentage composition of the mixture ?

Explanation

`CaCO_(3) to CaO + CO_(2)`
`underset(84 g)(MgCO_(3)) to MgO +underset(22400 mL)(CO_(2))`
Let `CaCO_(3)` present in the mixture = x g
Now 100 g of `CaCO_(3)` give `CO_(2) = 22400` mL
x g of `CaCO_(2)` will give `CO_(2) = (22400 xx x)/100`
=224x g
Similarly, 84 g of `MgCO_(3)` give`CO_(2) = 22400` mL
(1-x) g of `MgCO_(3)` give `CO_(2) = (22400 xx (1-x))/84`
Total `CO_(2)` evolved at N.T.P. = 240 mL
(1-x) of Mg