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A mixture of HCOOH and H(2)C(2)O(4) is h...

A mixture of `HCOOH` and `H_(2)C_(2)O_(4)` is heated with conc. `H_(2)SO_(4)`. The gas produced is collected and on treating with `KOH` solution the volume of the gas decreases by `1//6th`. Calculate molar ratio of two acids in original mixure.

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Learn that Conc. `H_(2)SO_(4)` is a dehydrating agent, so, from both HCOOH and `H_(2)C_(2)O_(4)` we must remove water. That is `H_(2)C_(2)O_(4) overset("Conc "H_(2)SO_(4))rarr CO+CO_(2)+H_(2)O and HCOOH overset("Conc "H_(2)SO_(4))rarr CO+H_(2)O.`
Say, the total volume of gases produced was 36 litre. The volume of `CO_(2)` must be 6 litre as `CO_(2)` would dissolve in `NaOH` base. This also means that volume of CO produced from oxalic acid would be 6 litre. So, volume of CO produced from `HCOOH` would be `36 - 12 = 24` litre.
Moles of `H_(2)C_(2)O_(4)` should proportionately be equal to 6 and that of `HCOOH` must be 24. Ratio of moles of `HCOOH` and `H_(2)C_(2)O_(4)` will be `(24)/(6)=4.`
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