KMnO_4 reacts with oxalic acid according to t | vedclass

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(B) From the balanced chemical equation,$2 \ 	ext{moles}$ of $MnO_4^-$ react with $5 \ 	ext{moles}$ of $C_2O_4^{2-}$.The number of moles of $KMnO_4$

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$50 \ mL$ of $0.1 \ M$ $C_2H_2O_4$

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(B) From the balanced chemical equation,$2 \ 	ext{moles}$ of $MnO_4^-$ react with $5 \ 	ext{moles}$ of $C_2O_4^{2-}$.The number of moles of $KMnO_4$ used is $n = M 	imes V = 0.1 \ 	ext{mol/L} 	imes 0.020 \ 	ext{L} = 0.002 \ 	ext{moles}$.According to the stoichiometry,$2 \ 	ext{moles}$ of $KMnO_4$ require $5 \ 	ext{moles}$ of oxalic acid $(C_2H_2O_4)$.Therefore,$0.002 \ 	ext{moles}$ of $KMnO_4$ require $(5/2) 	imes 0.002 = 0.005 \ 	ext{moles}$ of $C_2H_2O_4$.Checking the options:For option $B$: $n = 0.1 \ 	ext{M} 	imes 0.050 \ 	ext{L} = 0.005 \ 	ext{moles}$.Thus,$50 \ 	ext{mL}$ of $0.1 \ 	ext{M}$ $C_2H_2O_4$ is equivalent.

List-$I$	List-$II$ (At $STP$)
$(A)$ $10 \ g \ CaCO_3 \xrightarrow{\Delta} \text{decomposition}$	$(i)$ $0.224 \ L \ CO_2$
$(B)$ $1.06 \ g \ Na_2CO_3 \xrightarrow{\text{Excess } HCl}$	$(ii)$ $4.48 \ L \ CO_2$
$(C)$ $2.4 \ g \ C \xrightarrow{\text{Excess } O_2} \text{combustion}$	$(iii)$ $0.448 \ L \ CO_2$
$(D)$ $0.56 \ g \ CO \xrightarrow{\text{Excess } O_2} \text{combustion}$	$(iv)$ $2.24 \ L \ CO_2$
	$(v)$ $22.4 \ L \ CO_2$

$KMnO_4$ reacts with oxalic acid according to the equation:
$2MnO_4^- + 5C_2O_4^{2-} + 16H^+ \to 2Mn^{2+} + 10CO_2 + 8H_2O$
Here,$20 \ mL$ of $0.1 \ M$ $KMnO_4$ is equivalent to:

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$KMnO_4$ reacts with oxalic acid according to the equation:$2MnO_4^- + 5C_2O_4^{2-} + 16H^+ \to 2Mn^{2+} + 10CO_2 + 8H_2O$Here,$20 \ mL$ of $0.1 \ M$ $KMnO_4$ is equivalent to: