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

A mixture of 2.3 g formic acid and 4.5 g oxalic acid is treated with c

1.	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» SOLUTION :`underset(underset(46 g)("Formic acid"))(HCOOH) overset(conc.H_(2)SO_(4))rarrunderset(28g)(CO)+ H_(2)O` Now 46 g of COOH evolve CO = 28 g `:. 2.3 g` of HCOOH will evolve `CO = (28)/(46) XX 2.3` = 1.4 g `{:(""COOH),("\|" overset(conc.H_(2)SO_(4))rarr CO+CO_(2)+H_(2)O),(COOH""28 g),("Oxalic acid"),(""90g):}` When the GASEOUS mixture of `(CO+CO_(2)` is passed through KOH pellets, `CO_(2)` is ABSORBED while CO phase out `2KOH + CO_(2) rarr K_(2)CO_(3) + H_(2)O` Now, 90g of oxalic acid evolve `CO = 28 g` `:. 4.5 g` of oxalic acid will evolve CO `= (28)/(90) xx 4.5 = 1.4 g` Total amount of CO EVOLVED = 1.4 + 1.4 = 2.8 g