NEET · Chemistry · STD 11 - 1. Some basic concept of chemistry
A mixture of \(2.3\; \mathrm{g}\) formic acid and \(4.5 \;\mathrm{g}\) oxallic 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 \(STP\) ..........\(g\)
- A \(1.4\)
- B \(3\)
- C \(2.8\)
- D \(4.4\)
Answer & Solution
Correct Answer
(C) \(2.8\)
Step-by-step Solution
Detailed explanation
\(\mathrm{HCOOH} \xrightarrow[dehydrating Agent]{H_2SO_4}\)\(\mathrm{CO}+\mathrm{H}_{2} \mathrm{O}\left (\begin{array}{l}{\mathrm{H}_{2} \mathrm{O}\text { abosrbed }} \\ {\mathrm{by} \mathrm{H}_{2} \mathrm{SO}_{4}}\end{array}\right)\)
(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
\(=\frac{1}{10} \times 28\)
\(={2.8 \mathrm{g}}\)
(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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