Question

$KMnO_4$ reacts with $H_2O_2$ in an acidic medium. The number of moles of oxygen produced per mole of $KMnO_4$ is

• A. 2.5
• B. 5
• C. 1.25
• D. 2

Answer 1

Answer: (A)

For the reaction,
$2 \overset{+7}{ M }nO _{4}^{-}+5 H _{2} \overset{-4}{ O }_{2}+6 H ^{+} \longrightarrow $
$2 Mn ^{2+}+8 H _{2}\overset{-2}{O}+5 O _{2}$
Let the number of moles of oxygen produced per mole $KMnO _{4}$ be $x$.
Number of equivalent =Number of moles
$\times$ change in oxidation state
Number of equivalent of $KMnO _{4}=1 \times 5$
Number of equivalent of $H _{2} O _{2}=x \times 2$
$\therefore (\text { Number of eq. })_{ KMnO _{4}}$
$=(\text { Number of eq. })_{ H _{2} O _{2}}$
$1 \times 5 =x \times 2$
$x =\frac{5}{2}=2.5$