Question

$0.804$ g sample of iron ore was dissolved in acid. Iron was oxidised to $+2$ state and it required $47.2$ mL of $0.112NKMn{O}_4$ solution for titration. Calculate $\%$ of $F{e}_3{O}_4$ in ore (as nearest integer).

Answer 1

The redox changes are as follows:
$Fe⟶F{e}^{2+}+2e$
$F{e}^{2+}⟶F{e}^{3+}+e$
$5e+M{n}^{7+}⟶M{n}^{2+}$
$\therefore$ Meq. of $F{e}^{2+}$ = Meq. of $KMn{O}_4$
$\frac{w}{\left(\frac{56}{1}\right)}\times =47.2\times 0.112$
$\therefore w_F{e}^{2+}=0.296g$
$\therefore$ Purity of $Fe=\frac{0.296\times 100}{0.804}=36.82\%$
Now, $F{e}_3{O}_4⟶3Fe$
$\because 3\times 56gFe$ obtained by $232gF{e}_3{O}_4$
$\therefore 0.296gFe$ is obtained by $\frac{232\times 0.296}{56\times 3}=0.409gF{e}_3{O}_4$
$\therefore \%$ of $F{e}_3{O}_4=\frac{0.409}{0.804}\times 100=50.84\%$
So, nearest integer value is $51$.