A mixture of 2.3 g formic acid and 4.5 g ox | vedclass

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(C) Formic acid $(HCOOH)$ reacts with conc. $H_2SO_4$ to produce $CO$ and $H_2O$: $HCOOH \xrightarrow{conc. H_2SO_4} CO + H_2O$.Oxalic acid $(H_2C_2O_

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$2.8$

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(C) Formic acid $(HCOOH)$ reacts with conc. $H_2SO_4$ to produce $CO$ and $H_2O$: $HCOOH \xrightarrow{conc. H_2SO_4} CO + H_2O$.Oxalic acid $(H_2C_2O_4)$ reacts with conc. $H_2SO_4$ to produce $CO$,$CO_2$,and $H_2O$: $H_2C_2O_4 \xrightarrow{conc. H_2SO_4} CO + CO_2 + H_2O$.$KOH$ pellets absorb the acidic gas $CO_2$,leaving $CO$ as the remaining gaseous product.Moles of $HCOOH = \frac{2.3 \ g}{46 \ g/mol} = 0.05 \ mol$. This produces $0.05 \ mol$ of $CO$.Moles of $H_2C_2O_4 = \frac{4.5 \ g}{90 \ g/mol} = 0.05 \ mol$. This produces $0.05 \ mol$ of $CO$ and $0.05 \ mol$ of $CO_2$.Total moles of $CO$ remaining = $0.05 \ mol + 0.05 \ mol = 0.1 \ mol$.Mass of $CO = 0.1 \ mol 	imes 28 \ g/mol = 2.8 \ g$.

$A$ 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 of the remaining product at $STP$ is .......... $g$.

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