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
