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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............. .

Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :`underset(underset(46 g)("Formic acid"))(HCOOH) overset(conc.H_(2)SO_(4))rarrunderset(28g)(CO)+ H_(2)O` <br/> Now 46 g of COOH evolve CO = 28 g <br/> `:. 2.3 g` of HCOOH will evolve `CO = (28)/(46) <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 2.3` <br/> = 1.4 g <br/>`{:(""COOH),("|" overset(conc.H_(2)SO_(4))rarr CO+CO_(2)+H_(2)O),(COOH""28 g),("Oxalic acid"),(""90g):}` <br/> When the <a href="https://interviewquestions.tuteehub.com/tag/gaseous-467815" style="font-weight:bold;" target="_blank" title="Click to know more about GASEOUS">GASEOUS</a> mixture of `(CO+CO_(2)` is passed through KOH pellets, `CO_(2)` is <a href="https://interviewquestions.tuteehub.com/tag/absorbed-846166" style="font-weight:bold;" target="_blank" title="Click to know more about ABSORBED">ABSORBED</a> while CO phase out <br/> `2KOH + CO_(2) rarr K_(2)CO_(3) + H_(2)O` <br/> Now, 90g of oxalic acid evolve `CO = 28 g` <br/> `:. 4.5 g` of oxalic acid will evolve CO <br/> `= (28)/(90) xx 4.5 = 1.4 g` <br/> Total amount of CO <a href="https://interviewquestions.tuteehub.com/tag/evolved-2623419" style="font-weight:bold;" target="_blank" title="Click to know more about EVOLVED">EVOLVED</a> = 1.4 + 1.4 = 2.8 g</body></html>


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