Question

A mixture of 2.3g of formic acid and 4.5g of 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
(A) 1.4
(B) 3.0
(C) 4.4
(D) 2.8

Answer 1

The answer to this question includes the fact that when both acids are treated with another acid they release carbon monoxide gas and only one releases carbon dioxide gas and this is absorbed by $KOH$ pellets and the answer is only the mass of carbon monoxide.

Complete step by step solution:
We have learnt in our previous chapters of chemistry that the number of moles of the products formed is given by the ratio of their given weight to that of their molecular weight.
Thus, the reaction of formic acid with conc.${{H}_{2}}S{{O}_{4}}$ is as shown below,
$HCOOH\xrightarrow{{{H}_{2}}S{{O}_{4}}}{{H}_{2}}O+C{{O}_{(g)}}\uparrow$
Similarly, for the oxalic acid having formula ${{C}_{2}}{{H}_{2}}{{O}_{4}}$ on treatment with conc.${{H}_{2}}S{{O}_{4}}$ will be
${{C}_{2}}{{H}_{2}}{{O}_{4}}\xrightarrow{{{H}_{2}}S{{O}_{4}}}C{{O}_{2}}(g)+CO(g)+{{H}_{2}}O$
Now, 2.3 g of formic acid have $n=\dfrac{2.3}{12+2(16)+1}$
Therefore, $n=\dfrac{2.3}{46}=0.05moles$
Thus, 0.05 moles of formic acid produces 0.5 moles of water and carbon monoxide.
Similarly, 4.5 g of oxalic acid have $n=\dfrac{4.5}{2(12)+2(1)+4(16)}$
Thus, $n=\dfrac{4.5}{90=0.05moles}$
Therefore, 0.05 moles of oxalic acid produces 0.5 moles of carbon dioxide, 0.05 moles of carbon monoxide and 0.05 moles of water.
Now, in a mixture if these two gases total number of moles of carbon dioxide produces = 0.05 moles and that of carbon monoxide is 0.5+0.5= 0.1 moles .
Now, when we pass these gases into $KOH$ pellets, it absorbs carbon dioxide gas.
Therefore the remaining product is 0.1 moles of carbon monoxide.
Thus, mass of $CO$ produced is $0.1\times [12+16]$
$=0.1\times 28=2.8g$

Thus, the correct answer is option (D) 2.8g.

Note: While calculating the mass of remaining product, one should be careful about the fact that which gas is absorbed by the pellets of bases whether $KOH$ or $NaOH$ and both absorbs the $C{{O}_{2}}$ gas present in air. Otherwise it may lead to wrong answers.