18.4g of mixture CaCO₃ and MgCO₃ on heating gives 4.0g of magnesium oxide. The volume of carbon dioxide produced at STP in this process is:

The correct answer is: 4.48 L

MgCO₃ + CaCO₃  MgO + CaO + 2CO₂

$1$ mole of magnesium carbonate (molecular weight $84$ g/mol) gives $1$ mole of magnesium oxide (molecular weight is $40$ g/mol).
Thus, $4.0$ g of magnesium oxide is obtained from $8.4$ g ($0.1$ mol) of magnesium carbonate.
Thus, in $18.4$ g of mixture, $18.4-8.4=10$ g ($0.1$ mol) of calcium carbonate (molecular weight $=100$ g/mole) is present.
A mixture of $1$ mol of magnesium carbonate and $1$ mole of calcium carbonate gives $2$ moles of carbon dioxide.
Hence, a mixture of $0.1$ mole of magnesium carbonate and $0.1$ mole of calcium carbonate will give $0.2$ moles of carbon dioxide.
At STP, $1$ mole of carbon dioxide occupies $22.4$ L.
Hence, $0.2$ mole of carbon dioxide will occupy $0.2\times 22.4=4.48$ L at STP.