The decomposition of a certain mass of CaCO₃ gave 11.2 dm³ of CO₂ gas at STP. The mass of KOH required to completely neutralize the gas is

Weight of 11.2 dm³ of CO₂ gas at STP is 44/2 = 22 g
$\underset{56\mathrm{g}}{\mathrm{KOH}}+\underset{44\mathrm{g}}{{\mathrm{CO}}_2}\to {\mathrm{KHCO}}_3$
KOH required for complete neutralization of 22 g CO₂ $\mathrm{is} \frac{56}{44}\mathrm{x}22 = 28 \mathrm{g}$