For decolourisation of 1 mole of KMnO_ 4 , the moles of H_ 2 O_ 2...

The reaction for decolourisation of 1 mole of $\mathrm{KMnO}_{4}$ by $\mathrm{H}_{2} \mathrm{O}_{2}$ are given as follow:

2 $\mathrm{KMnO}_{4}+$3 $\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{K}_{2} \mathrm{SO}_{4}+$2 $\mathrm{MnSO}_{4}$

+3 $\mathrm{H}_{2} \mathrm{O}+$5[$\mathrm{O}]$

$5 \mathrm{H}_{2} \mathrm{O}_{2}+5[\mathrm{O}] \longrightarrow 5 \mathrm{H}_{2} \mathrm{O}+5 \mathrm{O}_{2}$
The overall reaction is

2 $\mathrm{KMnO}_{4}+$3 $\mathrm{H}_{2} \mathrm{SO}_{4}+$5 $\mathrm{H}_{2} \mathrm{O}_{2} \longrightarrow \mathrm{K}_{2} \mathrm{SO}_{4}+$2 $\mathrm{MnSO}_{4}$

+8 $\mathrm{H}_{2} \mathrm{O}+$5 $\mathrm{O}_{2}$

As 2 moles of $\mathrm{KMnO}_{4}$ require $\mathrm{H}_{2} \mathrm{O}_{2}=5$ moles $\therefore 1$ mole of $\mathrm{KMnO}_{4}$ require $\mathrm{H}_{2} \mathrm{O}_{2}=\frac{5}{2}=2.5$ moles