\(2.3\; \mathrm{g}\) ફોર્મિક અસિડ અને \(4.5 \;\mathrm{g}\) ઓક્ઝેલિક એસિડ ના મિશ્રણનું સાંદ્ર \(\mathrm{H}_{2} \mathrm{SO}_{4}\) સાથે પ્રક્રિયા કરવામાં આવે છે,ઉત્પન્ન થયેલા વાયુમય મિશ્રણ \(KOH\)ની ગોળીઓ માંથી પસાર થાય છે,તો \(STP\) એ વધેલ નીપજનું વજન .........\(g\)

\(\mathrm{HCOOH} \xrightarrow[dehydrating Agent]{H_2SO_4} \)\( \mathrm{CO}+\mathrm{H}_{2} \mathrm{O}\left(\begin{array}{l}{\mathrm{H}_{2} \mathrm{O}\text { abosrbed }} \\ {\mathrm{by} \mathrm{H}_{2} \mathrm{SO}_{4}}\end{array}\right)\)
(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
\(=\frac{1}{10} \times 28\)
\(={2.8 \mathrm{g}}\)