Question

One mole of a mixture of $\mathrm{CO}$ and ${\mathrm{CO}}_2$ requires exactly $20\mathrm{g}$ of $\mathrm{NaOH}$ in solution for complete conversion of all the ${\mathrm{CO}}_2$ into ${\mathrm{Na}}_2{\mathrm{CO}}_3$​. How many more grams of $\mathrm{NaOH}$ would it require for conversion into ${\mathrm{Na}}_2{\mathrm{CO}}_3$. If the mixture (one mole) is completely oxidised to ${\mathrm{CO}}_2$?

Answer 1

Step 1 Number of moles of $\mathrm{NaOH}$:

The Sodium hydroxide reacts with Carbon dioxide,

$\underset{\mathrm{Sodium}\mathrm{hydroxide}}{2\mathrm{NaOH}\left(\mathrm{s}\right)}+\underset{\mathrm{Carbon}\mathrm{dioxide}}{{\mathrm{CO}}_2\left(\mathrm{g}\right)}\to \underset{\mathrm{Sodium}\mathrm{carbonate}}{{\mathrm{Na}}_2{\mathrm{CO}}_3\left(\mathrm{aq}\right)}+\underset{\mathrm{Water}}{{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}$

Two moles of $\mathrm{NaOH}$ will react with one mole of ${\mathrm{CO}}_2$.

The given mass of $\mathrm{NaOH}$ is $20\mathrm{g}$

so, the number of moles of $\mathrm{NaOH}$ $=\frac{\mathrm{Given}\mathrm{mass}}{\mathrm{Molecular}\mathrm{mass}}=\frac{20}{40}=0.5\mathrm{g}$

Step 2 Mass of ${\mathrm{Na}}_2{\mathrm{CO}}_3$:

0.5 moles $\mathrm{NaOH}$ will react with 0.25 mole of ${\mathrm{CO}}_2$to give ${\mathrm{Na}}_2{\mathrm{CO}}_3$

so, 1 mole $\mathrm{NaOH}$ will react with 0.5 mole of ${\mathrm{CO}}_2$ to give ${\mathrm{Na}}_2{\mathrm{CO}}_3$

Therefore,2 mole $\mathrm{NaOH}$ will react with 1 mole of ${\mathrm{CO}}_2$ to give ${\mathrm{Na}}_2{\mathrm{CO}}_3$

That is, the mass of $\mathrm{NaOH}$ required to completely oxidize one mole of ${\mathrm{CO}}_2$ to ${\mathrm{Na}}_2{\mathrm{CO}}_3$ $=\mathrm{Number}\mathrm{of}\mathrm{moles}\times \mathrm{Molecular}\mathrm{mass}=2\times 40=80\mathrm{g}$

Hence, the mass of $\mathrm{NaOH}$ required to completely oxidize one mole of ${\mathrm{CO}}_2$ to ${\mathrm{Na}}_2{\mathrm{CO}}_3$ is $80\mathrm{g}$