$18.4{ g}^{ }$ of a mixture of ${CaCO}_3$ and ${MgCO}_3$ on heating gives $4.0 g$ of magnesium oxide. The volume of ${CO}_2$ produced at $STP$ in this process is

The chemical reaction is as follows:

$Mg{CO}_3+{CaCO}_3⟶MgO+CaO+2{CO}_2$

From the given reaction, 1 mole of $MgCO{CO}_3$gives 1 mole of $MgO$

4grams$\mathrm{of} \mathrm{MgO} \mathrm{will} \mathrm{be} \mathrm{obtained} \mathrm{from}$ $\frac{84}{40}\times 4=8.4\mathrm{g} \mathrm{MgO}$

Therefore the weight of ${CaCO}_3$ will be = 18.4-8.4=10g

Using the formula: $moles=\frac{given mass}{molecular mass}$

Moles of ${CaCO}_3$=$\frac{10}{100}=0.1mol$

Therefore 0.1mole ${CaCO}_3+0.1moleof$$MgC{O}_3$$=0.2$ moles of carbon dioxide

We know that 1 mole =22.4 L

Therefore 0.2 mole=4.48 L

Hence the correct option is (B).