A mixture of 2.3 g formic acid and 4.5 g oxalic acid is treated with conc. `H_(2)SO_(4)`. The evolved gaseous mixture is passed through KOH pellets. Weight (in g) of the remaining product at STP will be

Options

A. 1.4
B. `3.0`
C. 2.8
D. 4.4

Answer

C

Explanation

`n_(HCOOH) = (2.3 g)/(46 g mol^(-1)) = 0.05 mol`
`n_((COOH)_(2)) = (4.5 g)/(90 g mol^(-1)) = 0.5 mol`
`underset(0.05 mol)(COOH)_(2) rarr underset(0.05 mol)(CO) + underset(0.05 mol)(H_(2) O)`
`underset(0.05 mol)((CHOOH)_(2)) rarr underset(0.05 mol)(CO) + underset(0.05 mol)(CO_(2)) + underset(0.05 mol)(H_(2) O)`
Since `CO_(2) (g)` is absorbed by `KOH` pellets we are left with only `CO (g)` at `STP` as `H_(2) O` will be formed.
`n_(CO) = 0.05 + 0.05 = 0.1`
`m_(CO) = (n_(CO)) ("molar mass"_(CO))`
`= (0.1 mol) (25 g mol^(-))`
`= 2.8 g`