The weight of ${\mathrm{KMnO}}_4$ required to completely oxidise 0.25 moles of  in acid medium is .............. (molecular weight of ${\mathrm{KMnO}}_4=158)$

In acid medium $\stackrel{+7}{{\mathrm{KMnO}}_4}$ redises to give ${\mathrm{Mn}}^{+2} \mathrm{salt} \stackrel{+2}{{\mathrm{FeSO}}_4}$ oxidizes to give ${\mathrm{Fe}}^{+3}$

${\mathrm{KMnO}}_4+\stackrel{+2}{5{\mathrm{FeSO}}_4}\stackrel{{\mathrm{H}}^{+}}{\to }{\mathrm{Mn}}^{+2}{\mathrm{SO}}_4+{\mathrm{Fe}}_4^{+2}{\left({\mathrm{SO}}_4\right)}_3$

Balanced equation is

$2{\mathrm{KMnO}}_{4\left(\mathrm{aq}\right)}+10 {\mathrm{FeSO}}_{4\left(\mathrm{aq}\right)}+8{\mathrm{H}}_2{{\mathrm{SO}}_4}_{\left(\mathrm{aq}\right)}\to 2{\mathrm{MnSO}}_{4\left(\mathrm{aq}\right)}+5{\mathrm{Fe}}_2{\left({\mathrm{SO}}_4\right)}_{39\mathrm{aq})}={\mathrm{K}}_2{\mathrm{SO}}_{4\left(\mathrm{aq}\right)}+8{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{l}\right)}$

$1 \mathrm{mole} {\mathrm{KMnO}}_4=5 \mathrm{mole} {\mathrm{FeSO}}_4$

1 mole ${\mathrm{FeSO}}_4=\frac{1}{5}\mathrm{mole} {\mathrm{KMnO}}_4$

0.25 mole = $\frac{1}{5}\times 0.25=\frac{1}{20} mole KMn{O}_4$

$=\frac{158.1}{20}=7.9 \mathrm{grams}$