Question

In the chemical reaction between stoichiometric quantities of $KMn{O}_4$ and $KI$ in weakly basic solution, what is the number of moles of $I_2$ released for $4$ moles of $KMn{O}_4$ consumed?

Answer 1

Answer: 6

In slightly alkaline medium, $Mn{O}_4^{-}$reduces to $Mn{O}_2$ while $I$ is oxidized to $I_2$.
$Mn{O}_4^{-}+2H_{2}O+3e^{-}⟶Mn{O}_2+4O{H}^{-})\times 2$
$2I^{-}⟶I_2+2e^{-})\times 3$
$2Mn{O}_4^{-}+6I^{-}+4H_{2}O⟶2Mn{O}_2+3I_2+8O{H}^{-}$
So, $2mol$ of $KMn{O}_4$ can produce $3mole$ of $I_2$
$\therefore 4mol$ of $KMn{O}_4$ can produce $6$ mole of $I_2$