KMnO_4 reacts with ferrous ammonium sulphate | vedclass

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(D) From the balanced chemical equation:$MnO_4^- + 5Fe^{2+} + 8H^+ 	o Mn^{2+} + 5Fe^{3+} + 4H_2O$$1 \ 	ext{mole of } KMnO_4 	ext{ reacts with } 5 \

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$50 \ mL$ of $0.1 \ M$ $FeSO_4$

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(D) From the balanced chemical equation:$MnO_4^- + 5Fe^{2+} + 8H^+ 	o Mn^{2+} + 5Fe^{3+} + 4H_2O$$1 \ 	ext{mole of } KMnO_4 	ext{ reacts with } 5 \ 	ext{moles of } Fe^{2+}$.Using the stoichiometry: $n_{Fe^{2+}} = 5 	imes n_{KMnO_4}$$M_1 V_1 = 0.1 \ M 	imes 10 \ mL = 1 \ 	ext{mmol of } KMnO_4$.Therefore,moles of $Fe^{2+}$ required = $5 	imes 1 \ 	ext{mmol} = 5 \ 	ext{mmol}$.For $0.1 \ M$ $FeSO_4$ solution,$V = \frac{	ext{moles}}{	ext{Molarity}} = \frac{5 \ 	ext{mmol}}{0.1 \ M} = 50 \ mL$.Thus,$10 \ mL$ of $0.1 \ M$ $KMnO_4$ is equivalent to $50 \ mL$ of $0.1 \ M$ $FeSO_4$.

$KMnO_4$ reacts with ferrous ammonium sulphate according to the equation
$MnO_4^- + 5Fe^{2+} + 8H^+ \to Mn^{2+} + 5Fe^{3+} + 4H_2O$
Here $10 \ mL$ of $0.1 \ M$ $KMnO_4$ is equivalent to:

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$A$ solution of $10 \ mL$ of $\frac{M}{10} \ FeSO_4$ was titrated with $KMnO_4$ solution in an acidic medium. The amount of $KMnO_4$ used will be:

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$KMnO_4$ reacts with ferrous ammonium sulphate according to the equation$MnO_4^- + 5Fe^{2+} + 8H^+ \to Mn^{2+} + 5Fe^{3+} + 4H_2O$Here $10 \ mL$ of $0.1 \ M$ $KMnO_4$ is equivalent to: