(moles)$_{1}=\frac{2.3}{46}=\frac{1}{20} \quad \quad 0 \quad \quad \quad 0$
(moles)$_f 0\quad \quad \quad \quad \quad \quad \frac{1}{20}\quad \quad \quad \frac{1}{20}$
$\quad \quad \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4} \stackrel{\mathrm{H}_{2} \mathrm{SO}_{4}}{\longrightarrow} \mathrm{CO}+\mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O}$
$\left[\mathrm{H}_{2} \mathrm{O} \text { absorbed by } \mathrm{H}_{2} \mathrm{SO}_{4}\right]$
(moles)$_1\quad \frac{4.5}{90}=\frac{1}{20} 0 \quad \quad \quad 0 \quad \quad 0$
(moles)$_f\quad 0\quad \quad \quad \ \frac{1}{20} \quad \quad \frac{1}{20} \quad \quad \frac{1}{20}$
$\mathrm{CO}_{2}$ is absorbed by KOH. So the remaning product is only CO. moles of CO formed from both reactions
$=\frac{1}{20}+\frac{1}{20}=\frac{1}{10}$
Left mass of $\mathrm{CO}=$ moles $\times$ molar mass
$=\frac{1}{10} \times 28$
$={2.8 \mathrm{g}}$
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