Question

$0.7$ g sample of iron ore was dissolved in acid. Iron was reduced to $+2$ state and it required $50$ mL of $\frac{M}{50}$ $KMn{O}_4$ solution for titration. The percentage of $Fe$ and $F{e}_3{O}_4$ in the ore respectively, are :

• A. $40\%$, $55.24\%$
• B. $55.24\%$, $40\%$
• C. $8\%$ , $11\%$
• D. $11\%$, $8\%$

Answer 1

Answer: (A)

$40\%$, $55.24\%$

Ore of iron is $F{e}_3{O}_4$ which is mixture of $FeO+F{e}_2{O}_3$.

In $F{e}_2{O}_3$, $Fe$ is in $+3$ state which is reduced to $+2$ state.
$F{e}^{3+}+e^{-}\to F{e}^{2+}$ (In $FeO,Fe$ is in $+2$ state)
$Fe$ in $+2$ state is equivalent of $Mn{O}_4^{⊝}$.

$\therefore Mn{O}_4^{⊝}+5F{e}^{2+}+H^{⨁}\to M{n}^{2+}+5F{e}^{3+}+H_{2}O$

Milliequivalents of $KMn{O}_4=50\times \frac{1}{50}\times 5$ $=5$ meq.

$=5$ meq. of $F{e}^{2+}$
$=5\times {10}^{-3}\times 56$ g of $F{e}^{2+}$ or $Fe$
$=0.280$ g

$\%$ of $Fe=\frac{0.280}{0.7}\times 100=40\%$

Again $3Fe=F{e}_3{O}_4$

$3\times 56$ g $=(3\times 56+64)$ g of $F{e}_3{O}_4$
$168$ g $\equiv 232$ g of $F{e}_3{O}_4$

$\%$ of $F{e}_3{O}_4=\frac{40\times 232}{168}=55.24\%$