A mixture of 2.3 g formic acid (HCOOH) and 4.5 g oxalic acid ((COOH)2) is treated with conc. H2SO4. The evolved gaseous mixture is passed through KOH pellets. The weight (in g) of the remaining gas at STP will be

Correct answer: (A) (A) 2.8. Answer: (A) 2.8 g. Solution: Conc. H2SO4 dehydrates the acids: HCOOH ->[conc. H2SO4] CO + H2O and (COOH)2 ->[conc. H2SO4] CO + CO2 + H2O. Moles: HCOOH = 2.3/46 = 1/20 mol giving 1/20 mol CO; (COOH)2 = 4.5/90 = 1/20 mol giving 1/20 mol CO and 1/20 mol CO2. KOH absorbs the acidic CO2, leaving CO: total = 2/20 mol, so mass = (2/20)× 28 = 2.8 g.

NEET 2018 · ChemistryDerivative reactivityPrevious Year Question

A mixture of 2.3 g formic acid ( $HCOOH$ ) and 4.5 g oxalic acid ( $(COOH) X_{2}$ ) is treated with conc. $H X_{2} SO X_{4}$ . The evolved gaseous mixture is passed through $KOH$ pellets. The weight (in g) of the remaining gas at STP will be

Answer: (A) (A) 2.8. \textbf{Answer:} (A) 2.8 g.

A.(A) 2.8✓
B.(B) 3.0
C.(C) 1.4
D.(D) 4.4

Correct Answer

(A) (A) 2.8

Solution & Explanation

\textbf{Answer:} (A) 2.8 g. \textbf{Solution:} Conc. $H X_{2} SO X_{4}$ dehydrates the acids: $HCOOH conc \cdot H X_{2} SO X_{4} CO + H X_{2} O$ and $(COOH) X_{2} conc \cdot H X_{2} SO X_{4} CO + CO X_{2} + H X_{2} O$ . Moles: $HCOOH = 2.3/46 = 1/20$ mol giving $1/20$ mol $CO$ ; $(COOH) X_{2} = 4.5/90 = 1/20$ mol giving $1/20$ mol $CO$ and $1/20$ mol $CO X_{2}$ . $KOH$ absorbs the acidic $CO X_{2}$ , leaving $CO$ : total $= 2/20$ mol, so mass $= (2/20) \times 28 = 2.8$ g.

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A mixture of 2.3 g formic acid (HCOOH) and 4.5 g oxalic acid ((COOH)X2​) is treated with conc. HX2​SOX4​. The evolved gaseous mixture is passed through KOH pellets. The weight (in g) of the remaining gas at STP will be

Solution & Explanation

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A mixture of 2.3 g formic acid ( $HCOOH$ ) and 4.5 g oxalic acid ( $(COOH) X_{2}$ ) is treated with conc. $H X_{2} SO X_{4}$ . The evolved gaseous mixture is passed through $KOH$ pellets. The weight (in g) of the remaining gas at STP will be