One mole of mixture of CO and CO₂ requires exactly 20gm. NaOH to convert all the CO₂ into Na₂CO₃. How many more grams of NaOH would it require for conversion in to Na₂CO₃, if the mixture is completely oxidised into CO₂

$$\begin{gathered} {\mathrm{CO}}_2+2\mathrm{NaOH}\to {\mathrm{Na}}_2{\mathrm{CO}}_3+{\mathrm{H}}_2\mathrm{O} \\ 44\mathrm{g} 2\times 40\mathrm{g} \\ ? 20\mathrm{g} \\ \mathrm{Mass} \mathrm{of} {\mathrm{CO}}_2=\frac{20}{80}\times 44=11\mathrm{g} \\ {\mathrm{n}}_{{\mathrm{CO}}_2}=\frac{11}{44}=0.25\mathrm{moles} \\ {\mathrm{n}}_{\mathrm{CO}}=1-0.25=0.75\mathrm{moles} \\ \mathrm{Consider}, \\ \mathrm{CO}+\frac{1}{2}{\mathrm{O}}_2\to {\mathrm{CO}}_2 \\ 0.75\mathrm{mol} 0.75\mathrm{mol} \\ \mathrm{we} \mathrm{know} \mathrm{that}, 0.25\mathrm{moles} \mathrm{of} {\mathrm{CO}}_2 \mathrm{require} 20\mathrm{g} \mathrm{NaOH} \\ \mathrm{Mass} \mathrm{of} \mathrm{NaOH} \mathrm{required} \mathrm{for} 0.75\mathrm{moles} \mathrm{of} {\mathrm{CO}}_2=\frac{0.75}{0.25}\times 20=60\mathrm{g} \end{gathered}$$