Question

The composition of a sample of wustite (ore of Fe) is $F{e}_{0.93}$ $O_{1.00}$. What percentage of the iron is present in the form of Fe(III)?

• A. $14.05\%$
• B. $16.05\%$
• C. $13.05\%$
• D. $15.05\%$

Answer 1

The correct option is D

$15.05\%$

The give compound is electrically neutral. If there were only ${Fe}^{2+}$, electrical neutrality is not possible since 1.00 quantity of ${Fe}^{2+}$ is required.
$\therefore$ 0.93 Fe constitutes ${Fe}^{2+}$and ${Fe}^{3+}$
Let ${Fe}^{3+}$ be x in quantity and ${Fe}^{2+}$ be \(0.93-x)\ in quantity.
Oxygen is in -2 state. For neutrality,

and 3 (x) +(0.93-x) x 2 = 2
x=0.14 and 0.93-0.14=0.79

$\therefore$ Percentage of ${Fe}^{3+}$ = $\frac{0.14}{0.93}$ $\times$ 100 = 15.05 %