Question

A mixture of oxalic acid (COOH-COOH) and formic acid (HCOOH) is heated with conc. sulphuric acid and the gas evolved is collected and treated with KOH solution.The volume of solution decreases by 1/6.Calculate the molar ratio of two acids in the original acid

Answer 1

Suppose number of moles of formic acid=x moles
Number of moles of oxalic acid=y moles
HCOOH
Concentration of H2SO4
CO(g)+H2O(l)..........(1)
(COOH)2
Conc.H2SO4
CO+CO2+H2O.................(II)
So number of gaseous product in equation(I)=x
Number of gaseous exchange in equation=2y
So total number of moles of gaseous product=(x+2y)
As only CO2 is only absorbed by KOH,therefore fraction of CO2=Y/(x+2y)=1/6
6y=x+2y
or 
4y=x
x/y=4/1
Hence molar ratio=4:1

Answer 2

Answer:

Suppose number of moles of formic acid=x moles

Number of moles of oxalic acid=y moles

HCOOH

Concentration of H2SO4

CO(g)+H2O(l)..........(1)

(COOH)2

Conc.H2SO4

CO+CO2+H2O.................(II)

So number of gaseous product in equation(I)=x

Number of gaseous exchange in equation=2y

So total number of moles of gaseous product=(x+2y)

As only CO2 is only absorbed by KOH,therefore fraction of CO2=Y/(x+2y)=1/6

6y=x+2y

or 

4y=x

x/y=4/1

Hence molar ratio=4:1

Explanation: