A mixture of `H_2C_2O_4` and `HCOOH` is heated with conc `H_2SO_4`. The gas produced is collected and on treatment with KOH solution, the volume of the gas decreases by `(1)/(6)` calculate the molar ratio of the two acids in the original mixture.

Explanation

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.`