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A mixture of 2.3 g formic acid and 4.5 g oxalic acid is treated with conc. H_2SO_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>4.4<br/>1.4<br/>2.8<br/>3</p>Solution :2.3 gm formic <a href="https://interviewquestions.tuteehub.com/tag/acid-847491" style="font-weight:bold;" target="_blank" title="Click to know more about ACID">ACID</a> (HCOOH) <br/>` = (2.3)/(46) = 1/20 ` mol HCOOH <br/>4.5 gram oxalic acid `(H_2 C_2O_4)<br/>` = (4.5)/(90) = 1/20 ` mol `H_2C_2O_4`<br/>(i) `<a href="https://interviewquestions.tuteehub.com/tag/underset-3243992" style="font-weight:bold;" target="_blank" title="Click to know more about UNDERSET">UNDERSET</a>(1/20 "mol")(HCOOH) overset(<a href="https://interviewquestions.tuteehub.com/tag/conc-927968" style="font-weight:bold;" target="_blank" title="Click to know more about CONC">CONC</a>. H_2SO_4)(to) underset(1/20 "mol")(CO + H_2O)`<br/>(ii) `H_2C_2 O_4 overset(conc. H_2SO_4)(to) CO + CO_2 + H_2O` <br/>`CO_2`will be absorbed in KOH anc CO isadditional product.<br/>Total <a href="https://interviewquestions.tuteehub.com/tag/moles-1100459" style="font-weight:bold;" target="_blank" title="Click to know more about MOLES">MOLES</a> of additional `CO_2 = 1/20 + 1/20 = 1/10`mol<br/>` = 1/10 "mol" = 1/10 xx <a href="https://interviewquestions.tuteehub.com/tag/28-299271" style="font-weight:bold;" target="_blank" title="Click to know more about 28">28</a> = 2.8 g`</body></html>


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