When oxalic acid reacts with conc. $H_{2} {SO}_{4}$ two gases produced are neutral and acidic in nature respectively. Potassium hydroxide absorbs one of the two gases. The product formed during this absorption and the gas which gets absorbed are respectively :

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$K_{2} {CO}_{3} and {CO}_{2}$

${KHCO}_{3} and CO$

${KHCO}_{3} and {CO}_{2}$

$K_{2} {CO}_{3} and CO$

answer is A.

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Detailed Solution

Re-action can be given as,

$(COOH)_{2} \overset{conc. H_{2} {SO}_{4}, heat}{\to} CO + {CO}_{2} + H_{2} O$

While ${CO}_{2}$ is naturally acidic, $CO$ is neutral.

$2 KOH + {CO}_{2} \to K_{2} {CO}_{3} + H_{2} O$

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