A mixture of 2.3 g formic acid and 4.5 g oxalic acid is treated with conc. H_2SO_4. The evolved gaseous mixture is passed through KOH pellets. Weight (in g) of the remaining product at STP will be

1. 1.4

2. 2.8

3. 3.0

4. 4.4

Doubtly Asked 30 Oct 2026 NEET UG 2018 Practice test neet Save WhatsApp Report

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Answer: 2. 2.8

Conc. \(\mathrm{H_2SO_4}\) dehydrates both acids. KOH absorbs \(\mathrm{CO_2}\), so only CO remains.

\(\mathrm{HCOOH\to CO+H_2O}\): 2.3 g / 46 = 0.05 mol, gives 0.05 mol CO.
\(\mathrm{(COOH)_2\to CO+CO_2+H_2O}\): 4.5 g / 90 = 0.05 mol, gives 0.05 mol CO and 0.05 mol \(\mathrm{CO_2}\).

Total CO = 0.10 mol, mass = 0.10 x 28 = 2.8 g.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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