Question

$3$ mole of $FeS{O}_4$ are oxidized by $a$ mole of $KMn{O}_4$ in acid medium where as $3$ moles of $Fe{C}_2{O}_4$ are oxidized by $b$ mole of $KMn{O}_4$ in acid medium. The ratio of $a$ and $b$ is:

• A. $1/3$
• B. $1/2$
• C. $1/4$
• D. $1/5$

Answer 1

Answer: (A)

$1/3$

$nf$ for $FeS{O}_4=1$ as $Fe⟶F{e}^{+3}$

$nf$ for $KMn{O}_4$ in acidic medium is $=5$

$M{n}^{+7}⟶M{n}^{+2}$

$\Rightarrow n\times nf\left(FeS{O}_4\right)=n\times nf\left(KMn{O}_4\right)$

$\Rightarrow 3\times 1=a\times 5$

$\Rightarrow a=\frac{6}{5}$

Similarly,

for $Fe{C}_2{O}_4$, $F{e}^{+2}⟶F{e}^{+3}$, $nf=1$

$\Rightarrow C_2{O}_4^{2-}⟶C{O}_2$

$\Rightarrow 2C^{+3}⟶C^{+4}$ $nf=2$

Total $nf=1+2=3$

$\Rightarrow n\times nf\left(Fe{C}_2{O}_4\right)=n\times nf\left(KMn{O}_4\right)$

$\Rightarrow 3\times 3=b\times 5$

$\Rightarrow b=\frac{9}{5}$

$\therefore \frac{a}{b}=\frac{3}{5}\times \frac{5}{9}=\frac{1}{3}$ .