3.68 g of a mixture of CaCO₃ and MgCO₃ is heated to liberate 0.04 mole of CO₂. The mole % of CaCO₃ and MgCO₃ in the mixture respectively:

Let the mass of CaCO₃ = x g

Then, mass of  ${\mathrm{MgCO}}_3=(3.68-\mathrm{x})\mathrm{g}$
$\text{Moles of }{\mathrm{CaCO}}_3=\frac{\mathrm{x}}{100}$
$\text{Moles of }{\mathrm{MgCO}}_3=\frac{(3.68-\mathrm{x})}{84}$

$\frac{x}{100}+\frac{3.68-x}{84}=0.04$
$\Rightarrow x=2$
$\therefore {\mathrm{n}}_{{\mathrm{CaCO}}_3}=\frac{2}{100}=0.02 \text{and }\therefore {\mathrm{n}}_{{\mathrm{MgCO}}_3}=\frac{1.68}{84}=0.02$

$\text{Mole }\mathrm{\%} \text{of }{\mathrm{CaCO}}_3=\frac{0.02}{0.04}\times 100=50\mathrm{\%}$
$\text{Mole }\mathrm{\%} {\mathrm{MgCO}}_3=\frac{0.02}{0.04}\times 100=50\mathrm{\%}$