Question

The composition of a sample of wustite is ${\text{F}}{{\text{e}}_{{\text{0}}{\text{.92}}}}{\text{O}}$. What percentage of the iron is present in the form of Fe(III)?
A. $5$ %
B. $7.08$ %
C. $16$ %
D. $23.6$ %

Answer 1

Hint:We have the total amount of Fe so, we will assume that one type of Fe is x. Then we can determine the amount of second type Fe. Then we will substitute the total positive charge equal to the total negative charge to determine the value of X. we will multiply the X by a hundred to get the amount of iron present as Fe (III).

Complete step-by-step solution:It is given that the composition of a sample of wustite is ${\text{F}}{{\text{e}}_{{\text{0}}{\text{.92}}}}{\text{O}}$. The amount of Fe is $0.92$. Out of $0.92$, some amount is of Fe(III) and some amount is of Fe(II).
We assume that the amount of Fe (III) is X so, the amount of Fe (II) will be $0.92 - {\text{X}}$.
Now, the given compound is neutral. So, total positive charge is equal to total negative charge.
Here, we have two type of positive charge total charge will be sum of charge of Fe (II) and Fe (II) charge so,Fe (III) = $\,3 \times {\text{X}}$
Fe (II) = $\,2 \times \left( {0.92 - {\text{X}}} \right)$
${{\text{O}}^{2 - }}$ = $\,2 \times 1.0$
 On putting the positive charge equal to negative charge we get,
$\left( {{\text{3X}}\,} \right){\text{ + }}\,\,2 \times \left( {0.92 - {\text{X}}} \right)\,\, = \,2 \times 1.0$
$\Rightarrow {\text{3X}}\,\,{\text{ + }}\,\,18.4 - 2{\text{X}}\,\, = \,2.0$
$\Rightarrow {X} = 2.0 - 1.84$
$\therefore {X} = 0.16$
So, the amount of Fe (III) is $0.16$.
Now we will multiply the amount of Fe (III) with hundred to convert it into percent.
\[0.16\, \times \,100\, = \,16\]%
So, the percentage of the iron present in the form of Fe(III) is $16$%.

Therefore, option (C) $16$%, is correct.

Note:The given compound is neutral. So, the total positive charge is equal to the total negative charge. This is known as electroneutrality. According to electroneutrality, the total positive charge of a compound will always be equal to the total negative charge. This type of metal oxides in which a metal is present in both oxidation states, II and III are known as spinal. In spinal Metal(II) ions are present in tetrahedral voids and Metal (III) ions are present in octahedral voids. The oxide ions form the FCC unit cell.