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 $4$ mol of $KMn{O}_4$ consumed?

• A. 6
• B. 6.0
• C. 6.00

Answer 1

Answer: (A), (B), (C)

The ionic form of reaction of $KMn{O}_4$ and KI in weakly basic solution is given as:

$2\stackrel{7}{Mn}{O}_4^{-}+6{\stackrel{-1}{I}}^{-}+4H_{2}O\to 2\stackrel{+4}{Mn}{O}_2+3\stackrel{0}{I_2}+8O{H}^{-}$
Here, oxidation state of $Mn$ decreases from $+7$ in $KMn{O}_4$ to $+4$ in $Mn{O}_2$
n- factor of $KMn{O}_4=\text{change in oxidation number per formula unit}=3$
oxidation state of $I$ increases from $-1$ in $I^{-}$ to $0$ in $I_2$
now, n- factor of $I_2=\text{change in oxidation number per}{I}_2\text{molecule}=2$
$\text{Since, n-factor}\times \text{Number of mol(n)}=\text{number of g-equivalents}$
and at completion of reaction,
g-equivalents of $KMn{O}_4$ consumed = g-equivalents of $I_2$ formed
So, $3\times molofKMn{O}_4=2\times molof{I}_2$
$3\times 4=2\times molof{I}_2$
Amount of $I_2=6$mol