Question

Which of the following statements is/are correct about the following reaction?
$F{{e}_{3}}{{O}_{4}}\xrightarrow{\Delta }F{{e}_{2}}{{O}_{3}}$
(A)- The equivalent weight of $F{{e}_{3}}{{O}_{4}}$ is ${{M}_{1}}$, where ${{M}_{1}}$ is the molecular weight of $F{{e}_{3}}{{O}_{4}}$
(B)- The equivalent weight of $F{{e}_{3}}{{O}_{4}}$ is $\dfrac{{{M}_{1}}}{3}$, where ${{M}_{1}}$ is the molecular weight of $F{{e}_{3}}{{O}_{4}}$
(C)- The equivalent weight of $F{{e}_{2}}{{O}_{3}}$ is $\dfrac{3{{M}_{2}}}{2}$, where ${{M}_{2}}$ is the molecular weight of $F{{e}_{2}}{{O}_{3}}$
(D)- The equivalent weight of $F{{e}_{2}}{{O}_{3}}$ is $\dfrac{{{M}_{2}}}{2}$ , where ${{M}_{2}} $is the molecular weight of $F{{e}_{2}}{{O}_{3}}$

Answer 1

Hint: First find the oxidation state of iron in the iron oxide of the reactant side and the iron oxide of the product side. Then compare and figure out the change in the oxidation state. Check how it differs from the molar mass value of particular iron oxide.

Complete step by step solution:
-The iron compound with chemical formula $F{{e}_{3}}{{O}_{4}}$ occurs in nature as the mineral magnetite. This iron oxide is known by many names such as ferrous ferric oxide, ferroso ferric oxide, iron (II,III)oxide, magnetite, black iron oxide, lodestone and rust iron (II) diiron (III) oxide.
-The iron compound with chemical formula $F{{e}_{2}}{{O}_{3}}$ occurs in nature as the hematite mineral. This iron oxide is known by many other names such as ferric oxides, hematite, ferric iron, red iron oxide, rouge, maghemite, colcothar, and rust.
-$F{{e}_{3}}{{O}_{4}}\xrightarrow{\Delta }F{{e}_{2}}{{O}_{3}}$
In the above reaction, the ferrous ferric oxide is heated which converts it to ferric iron. The oxidation state of iron in ferrous ferric oxide ($F{{e}_{3}}{{O}_{4}}$) is +3 and +2 while in ferric oxide ($F{{e}_{2}}{{O}_{3}}$) is +4.
$\begin{align}
& F{{e}_{3}}{{O}_{4}}=FeO.F{{e}_{2}}{{O}_{3}} \\
& FeO=x-2=0\Rightarrow x=+2 \\
& F{{e}_{2}}{{O}_{3}}=2x+3(-2)=0\Rightarrow x=+3 \\
\end{align}$
$\begin{align}
& F{{e}_{2}}{{O}_{3}}\Rightarrow 2x+3(-2)=0 \\
& 2x=+6 \\
& x=\dfrac{6}{2}=3 \\
\end{align}$
Hence, there is a change in the oxidation state by +1.
So, the equivalent weight of $F{{e}_{3}}{{O}_{4}}=\dfrac{\text{Molar mass}}{1}=\text{Molar mass value}$
Therefore, the equivalent weight of $F{{e}_{3}}{{O}_{4}}$is equal to the molar mass value.

Hence, the correct answer is option A.

Note: Let us now see some of the applications and uses of $F{{e}_{3}}{{O}_{4}}$ and $F{{e}_{2}}{{O}_{3}}$-
-$F{{e}_{3}}{{O}_{4}}$ is used as a black pigment and hence is known as Mars black. It is also used as a catalyst in the Haber process and in the water-gas shift reaction. Bluing is a passivation process that produces a layer of $F{{e}_{3}}{{O}_{4}}$ on the surface of steel as a protection against rust. The nanoparticles of $F{{e}_{3}}{{O}_{4}}$ are used as contrast agents in Magnetic Resonance Imaging techniques.
-$F{{e}_{2}}{{O}_{3}}$ is the main source of iron for the steel industry which is readily attacked by acids. A very fine powder of $F{{e}_{2}}{{O}_{3}}$ is known as ‘jeweller’s rouge’ or ‘red rouge’ which is used to put the final polish on metallic jewellery and lenses. $F{{e}_{2}}{{O}_{3}}$ was the most common magnetic particle used in all types of magnetic storage and recording media. The red colour of iron(III) oxide is mainly responsible for the calamine lotion as pink colour.