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NEET generalGeneralMedium

Question

A mixture of 2.3 g formic acid and 4.5 g oxalic acid is treated with conc. H2SO4H_2SO_4. The evolved gaseous mixture is passed through KOH pellets. Weight (in g) of the remaining product at STP will be

1

2.8

2

3.0

3

1.4

4

4.4

Step-by-Step Solution

HCOOH Conc.H2SO4\xrightarrow{Conc. H_2SO_4} CO(g) + H2OH_2O(l). Moles of HCOOH = 2.3/46 = 1/20 mol. COOH-COOH Conc.H2SO4\xrightarrow{Conc. H_2SO_4} CO(g) + CO2CO_2(g) + H2OH_2O(l). Moles of oxalic acid = 4.5/90 = 1/20 mol. Gaseous mixture is CO and CO2CO_2. KOH absorbs CO2CO_2. Remaining gas is CO. Total moles of CO = 1/20 + 1/20 = 2/20 mol. Weight = (2/20) * 28 = 2.8 g.

Exam Context & Concepts Covered

This question aligns with the NEET general syllabus, specifically targeting concepts from General. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

generalmixtureformicoxalictreatedevolved

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