A mixture of 2.3 g formic acid 4.5 g oxalic acid is treated with cone. H 2 SO 4 .The evolved gaseous mixture is passed through KOH pellets. Weight (in g) of the remaining product at STP will be:
Text Solution
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No. of moles of HCOOH are:
n = 0.05 moles
No. of moles of H 2 C 2 O 4 are:
n = 0.05 moles
HCOOH 
CO (g] + H 2 O (1)
Initial 0.05 mol 0 0
Final 0.05 mol 0.05mol
H 2 C 2 O 4 
CO + CO 2 + H 2 O (1)
Initial 0.05mol 0 0
Final 0 0.05 mol 0.05mol
CO 2 is absorbed by KOH and the remaining product is CO.
Therefore total no, of moles of CO formed in both the reactions is,
0.05+ 0,05 = 0,1 mole
Mass of CO formed = 0.1 x 28 = 2.8 gm
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