Question

$KMn{O}_4$ reacts with $H_2{O}_2$ in an acidic medium. The number of moles of oxygen produced per mole of $KMn{O}_4$ is

• A. 2.5
• B. 5
• C. 1.25
• D. 2

Answer 1

Answer: (A)

For the reaction,
$2\stackrel{+7}{M}n{O}_4^{-}+5H_2{\stackrel{-4}{O}}_2+6H^{+}⟶$
$2M{n}^{2+}+8H_2\stackrel{-2}{O}+5O_2$
Let the number of moles of oxygen produced per mole $KMn{O}_4$ be $x$.
Number of equivalent =Number of moles
$\times$ change in oxidation state
Number of equivalent of $KMn{O}_4=1\times 5$
Number of equivalent of $H_2{O}_2=x\times 2$
$\therefore (\text{ Number of eq. }{)}_{KMn{O}_4}$
$=(\text{ Number of eq. }{)}_{H_2{O}_2}$
$1\times 5=x\times 2$
$x=\frac{5}{2}=2.5$