KMnO_4 reacts with oxalic acid according to t | vedclass

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(B) From the balanced chemical equation: $2 \ mol$ of $MnO_4^-$ reacts with $5 \ mol$ of $C_2O_4^{2-}$.Using the stoichiometry relation: $\frac{n_{MnO

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$50 \ mL$ of $0.1 \ M$ $H_2C_2O_4$

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(B) From the balanced chemical equation: $2 \ mol$ of $MnO_4^-$ reacts with $5 \ mol$ of $C_2O_4^{2-}$.Using the stoichiometry relation: $\frac{n_{MnO_4^-}}{2} = \frac{n_{C_2O_4^{2-}}}{5}$.Given: $V_{MnO_4^-} = 20 \ mL$,$M_{MnO_4^-} = 0.1 \ M$.$n_{MnO_4^-} = M 	imes V = 0.1 	imes 20 = 2 \ mmol$.Therefore,$n_{C_2O_4^{2-}} = \frac{5}{2} 	imes n_{MnO_4^-} = 2.5 	imes 2 = 5 \ mmol$.Checking option $(B)$: $50 \ mL$ of $0.1 \ M$ $H_2C_2O_4$ gives $n = 50 	imes 0.1 = 5 \ mmol$. This matches the requirement.

$KMnO_4$ 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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