Question

When non-stoichiometric compound $F{e}_{0.95}O$ is heated in the presence of oxygen, it converts into $F{e}_2{O}_3$, then which of the following statements are correct?

• A. Equivalent weight of $F{e}_{0.95}O$ is $\frac{M}{0.5}$ where $M$ is molecular weight of $F{e}_{0.95}O$
• B. The number of moles of $F{e}^{3+}$ and $F{e}^{2+}$ in $1mole$ of $F{e}_{0.95}O$ are $0.1$ and $0.85$ respectively
• C. The number of moles of $F{e}^{3+}$ and $F{e}^{2+}$ in $1moles$ of $F{e}_{0.95}O$ are $0.85$ and $0.10$ respectively
• D. The $\%$ composition of $F{e}^{2+}$ and $F{e}^{3+}$ in the non-stoichiometric compound is $89.47\%$ and $10.53\%$ respectively

Answer 1

Answer: (B), (D)

The $\%$ composition of $F{e}^{2+}$ and $F{e}^{3+}$ in the non-stoichiometric compound is $89.47\%$ and $10.53\%$ respectively

$F{e}_{0.95}O+O_2\to F{e}_2{O}_3$
$\text{Let x be the fraction of}F{e}^{3+}\text{in the compound then}F{e}^{2+}=(0.95-x)$
$x\times 3+(0.95-x)\times 2-2=0$
$x=0.1$
$\%ofF{e}^{2+}=\frac{0.85}{0.95}\times 100=89.47\%$