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Q. For decolourisation of $1$ mole of acidified $KMnO_4$ the moles of $H_2O_2$ required are

AIIMSAIIMS 1980Redox Reactions

Solution:

$H _{2} O _{2}$ reduces acidified $KMnO _{4}$ to colourless $MnSO _{4}$.
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$2\, mol$ of $KMnO _{4}$ oxidise $5 \,mol$ of $H _{2} O _{2}$
$\therefore 1 \,mol$ of $KMnO _{4}$ will oxidise $\frac{5}{2} mol$ of $H _{2} O _{2}$.