A mixture of 2.3 g formic acid and 4.5 g oxalic acid is treated with conc.${\mathrm{H}}_2{\mathrm{SO}}_4$. The evolved gaseous mixture is passed through KOH pellets. Weight of the remaining product at STP will be ____ grams.

$$\begin{gathered} \underset{\mathrm{n}=\frac{2.3}{46}=0.05}{\mathrm{HCOOH}} \stackrel{{\mathrm{H}}_2{\mathrm{SO}}_4}{\to }\underset{0.05\mathrm{moles}}{ \mathrm{CO}}+{\mathrm{H}}_2\mathrm{O} \\ \underset{\mathrm{n}=\frac{4.5}{90}=0.05}{{\left(\mathrm{COOH}\right)}_2} \stackrel{{\mathrm{H}}_2{\mathrm{SO}}_4}{\to } \underset{0.05}{\mathrm{CO}}+\underset{0.05}{{\mathrm{CO}}_2}+{\mathrm{H}}_2\mathrm{O} \end{gathered}$$

Gaseous mixture formed is CO and CO₂. When it is passed through KOH, only CO₂ is absorbed. So the remaining gas is CO. 

So, the no. of moles of gaseous product CO= 0.05+0.05=0.1mole

Weight of CO=$0.1\times 28=2.8\mathrm{g}$

Hence, the correct answer 2.8 g.