A mixture of 2.3 g formic acid and 4.5 g oxalic acid is treated with conc. H2SO4. The evolved gaseou

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Question

A mixture of 2.3 g formic acid and 4.5 g oxalic acid is treated with conc. H2SO4. The evolved gaseous mixture is passed through KOH pellets. Weight (in g) of the remaining product at STP will be

Accepted Answer

Correct Answer: D — 2.8 gStep-by-step Solution:When treated with conc. H2SO4 (a dehydrating agent):Formic acid (HCOOH, M = 46 g/mol) undergoes dehydration:HCOOH → CO + H2OOxalic acid (H2C2O4, M = 90 g/mol) undergoes dehydration:H2C2O4 → CO + CO2 + H2OMoles of formic acid = 2.3 / 46 = 0.05 mol → produces 0.05 mol CO.Moles of oxalic acid = 4.5 / 90 = 0.05 mol → produces 0.05 mol CO + 0.05 mol CO2.Total gases evolved: CO = 0.05 + 0.05 = 0.10 mol; CO2 = 0.05 mol; H2O (absorbed by H2SO4).When passed through KOH pellets, CO2 is absorbed: CO2 + 2KOH → K2CO3 + H2O.The remaining gas is only CO.Mass of CO = 0.10 mol × 28 g/mol = 2.8 g.Why other options are incorrect:Option C (4.4 g) would correspond to the mass of CO2 alone (0.1 mol × 44 = 4.4 g), incorrectly assuming CO2 is the remaining gas.Options A and B do not correspond to any valid product calculation from this reaction.Key Concept: Conc. H2SO4 dehydrates formic acid and oxalic acid to give CO and CO2 respectively. KOH absorbs acidic gases (CO2, SO2) but not CO. Molar mass: CO = 28 g/mol, CO2 = 44 g/mol.

A mixture of 2.3 g formic acid and 4.5 g oxalic acid is treated with conc. H2SO4. The evolved gaseous mixture is passed through KOH pellets. Weight (in g) of the remaining product at STP will be

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A mixture of 2.3 g formic acid and 4.5 g oxalic acid is treated with conc. H2SO4. The evolved gaseous mixture is passed through KOH pellets. Weight (in g) of the remaining product at STP will be

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