Formic acid ($\text{HCOOH}$) and oxalic acid ($\text{H}_2\text{C}_2\text{O}_4$) undergo dehydration in the presence of concentrated $\text{H}_2\text{SO}_4$.

1. Dehydration of formic acid: $\text{HCOOH} \xrightarrow{\text{conc. H}_2\text{SO}_4} \text{H}_2\text{O} + \text{CO}$ Molar mass of $\text{HCOOH} = 46 \text{ g mol}^{-1}$. Moles of $\text{HCOOH} = \frac{2.3 \text{ g}}{46 \text{ g mol}^{-1}} = 0.05 \text{ mol}$. Thus, $0.05 \text{ mol}$ of $\text{CO}$ is produced.

2. Dehydration of oxalic acid: $\text{H}_2\text{C}_2\text{O}_4 \xrightarrow{\text{conc. H}_2\text{SO}_4} \text{H}_2\text{O} + \text{CO} + \text{CO}_2$ Molar mass of $\text{H}_2\text{C}_2\text{O}_4 = 90 \text{ g mol}^{-1}$ . Moles of $\text{H}_2\text{C}_2\text{O}_4 = \frac{4.5 \text{ g}}{90 \text{ g mol}^{-1}} = 0.05 \text{ mol}$. Thus, $0.05 \text{ mol}$ of $\text{CO}$ and $0.05 \text{ mol}$ of $\text{CO}_2$ are produced.

Total moles of $\text{CO}$ evolved $= 0.05 + 0.05 = 0.1 \text{ mol}$. Total moles of $\text{CO}_2$ evolved $= 0.05 \text{ mol}$.

When the gaseous mixture is passed through $\text{KOH}$ pellets, $\text{CO}_2$ is absorbed by $\text{KOH}$ to form $\text{K}_2\text{CO}_3$. The remaining gaseous product is only $\text{CO}$. Weight of the remaining product ($\text{CO}$) $= \text{Moles} \times \text{Molar mass}$ Weight of $\text{CO} = 0.1 \text{ mol} \times 28 \text{ g mol}^{-1} = 2.8 \text{ g}$.