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The composition of a sample of Wustite is $Fe_{0..93}O_{1.00}$.What percentage of the iron is present in the form of $Fe(III)$?

$\begin{array}{1 1}(a)\;15.05\%& (b)\;14.04\% \\(c)\;15.00\%&(d)\;14\%\end{array} $

1 Answer

We know that in pure iron oxide(FeO),iron and oxygen are present in the ratio 1: 1
In wustite $(Fe_{0.93}O_{1.00})$,some of the $Fe^{2+}$ ions are missing and the number of $Fe^{2+}$ ions present in 0.93 instead of 1.
From here we find the number of missing $Fe^{2+}$ ions
$\Rightarrow 1.0-0.93=0.07$
Since each $Fe^{2+}$ ion carries two units of positive charge so the total positiv charge missing
$\Rightarrow 0.07\times 2=0.14$
For maintenance of electrical neutrality this much (ie 0.14) positive charge has to be compensated by the presence of $Fe^{3+}$ ions.
If one $Fe^{3+}$ ion replaces one $Fe^{2+}$ ion then there is an increase of one unit positive charge.So to compensate 0.14 unit positive charge we require 0.14 $Fe^{3+}$ ions to replace $Fe^{2+}$ ion.
So,$100Fe^{2+}$ ions have $Fe^{3+}$ ions=$\large\frac{0.14}{0.93}$$\times 100=15.05\%$
Hence (a) is the correct option.
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