### NEET Solved Paper 2018 Question 12

##### Question: A mixture of 2.3 g formic acid and 4.5 g oxalic acid is treated with conc. $ {H_2}S{o_4} $ . The evolved gaseous mixture is passed through KOH pellets. Weight (in g) of the remaining product at STP will be [NEET - 2018]

#### Options:

A) 2.8

B) 3.0

C) 1.4

D) 4.4

## Show Answer

#### Answer:

Correct Answer: A

#### Solution:

$ \underset{2.3g\text{ or }( \frac{1}{20}mol )}{\mathop{HCOOH}},\xrightarrow{Conc\text{.}{H_2}S{O_4}}\underset{\frac{1}{20}mol}{\mathop{\text{CO(g)}}},\text{+}{H_2}\text{O(l)} $ $ \begin{aligned} & COOH\xrightarrow{Conc\text{.}{H_2}S{O_4}}\underset{\frac{1}{20}mol}{\mathop{\text{CO(g)}}},\text{+}\underset{\frac{1}{20}mol}{\mathop{C{O_2}}},\text{(g)+}{H_2}\text{O(l)} \\ & \underset{4.5gor( \frac{1}{20}mol )}{\mathop{ \begin{aligned} & | \\ & COOH \\ \end{aligned}}}, \\ \end{aligned} $

Gaseous mixture formed is $ CO $ and $ C{O_2} $ when it is passed through KOH, only $ C{O_2} $ is absorbed. So the remaining gas is CO.

So, weight of remaining gaseous product CO is $ \frac{2}{20}\times 28=2.8g $

So, the correct option is 2.8