2 text{ mol} of FeSO_4 are oxidized by X text | vedclass

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Question

(A) In acidic medium,the $n$-factor of $KMnO_4$ is $5$. For $FeSO_4$: The $n$-factor is $1$ $(Fe^{2+} \rightarrow Fe^{3+} + e^-)$. Equivalents of $FeS

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$1 : 3$

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(A) In acidic medium,the $n$-factor of $KMnO_4$ is $5$. For $FeSO_4$: The $n$-factor is $1$ $(Fe^{2+} ightarrow Fe^{3+} + e^-)$. Equivalents of $FeSO_4 = 	ext{Equivalents of } KMnO_4$. $2 	imes 1 = X 	imes 5 \Rightarrow X = 2/5$. For $FeC_2O_4$: The $n$-factor is $3$ ($Fe^{2+} ightarrow Fe^{3+} + e^-$ and $C_2O_4^{2-} ightarrow 2CO_2 + 2e^-$). Equivalents of $FeC_2O_4 = 	ext{Equivalents of } KMnO_4$. $2 	imes 3 = Y 	imes 5 \Rightarrow Y = 6/5$. Ratio $X : Y = (2/5) : (6/5) = 2 : 6 = 1 : 3$.

$2 \text{ mol}$ of $FeSO_4$ are oxidized by $X \text{ mol}$ of $KMnO_4$ whereas $2 \text{ mol}$ of $FeC_2O_4$ are oxidized by $Y \text{ mol}$ of $KMnO_4$. The ratio of $X$ and $Y$ is:

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