Question

The percentage of iron present as $Fe(III)$ in $F{e}_{0.93}{O}_{1.0}$ is

• A. 8.3 %
• B. 9.6 %
• C. 11.5 %
• D. 17.7 %

Answer 1

Answer: (C)

Let $x$ atoms of $F{e}^{3+}$ ions are present. This means $xF{e}^{3+}$ ions have been replaced by $F{e}^{2+}$ ions. No. of $F{e}^{2+}$ ions $=0.93-x$ For electrical neutrality,

$2(0.93-x)+3x=2$

$1.86+x=2$ or $x=0.14$

Fraction of $F{e}^{3+}=0.14,F{e}^{2+}=0.93-0.14=0.79$

Formula : $F{e}_{0.79}^{2+}F{e}_{0.14}^{3+}{O}_{1.0}^{2-}$

Total molar mass $=0.93\times 56+1\times 16=68.08$

$\%$ of $Fe(II)=\frac{0.14\times 56}{68.08}\times 100=11.5\%$