Question

An iron sample contains $18\%$ $F{e_3}{O_4}$. What is the amount so that it is precipitated as $F{e_2}{O_3}$, which weighs 0.40g?
A. 2.15 g
B. 1.075 g
C. 4.30 g
D. 2.01 g

Answer 1

An iron sample that contains $18\%$ $F{e_3}{O_4}$ tells us that the sample is not pure. Mass percent represents the contribution of an element to the total molar mass of the compound. It is used to find out the impure amount of $18\%$ $F{e_3}{O_4}$.

Complete step-by-step answer:
Iron (II,III) oxide reacts with oxygen at very high temperatures to produce iron(III) oxide. This reaction only takes place when iron is heated in presence of plenty of oxygen. The chemical equation for the same reaction is shown below:
$4F{e_3}{O_4}(s) + {O_2}(g) \to 6F{e_2}{O_3}(s)$
From this equation, we can say that 1 mole of $F{e_3}{O_4}$ gives 1.5 moles of $F{e_2}{O_3}$. We can calculate the molar mass of both the compounds by adding the masses of their individual constituents. Therefore, the molar mass of $F{e_3}{O_4}$ is 232 g for 1 mole of compound. But for $F{e_2}{O_3}$ , 1 mole of compound has 160 g of molar mass and for 1.5 moles, it will be 240 g.
Amount of $F{e_3}{O_4}$ required by 0.40 g of $F{e_2}{O_3}$ = $\dfrac{{232}}{{240}} \times 0.40$
= $0.3867g$
We are given that an iron sample contains $18\%$ $F{e_3}{O_4}$ , so the $18\%$ of $F{e_3}{O_4}$ is calculated as $0.3867 \times \dfrac{{100}}{{18}}g$ $= 2.15g$ of $F{e_3}{O_4}$

Hence, the correct option is (A).

Additional information: $F{e_2}{O_3}$ is also known as hematite which is grey or red in colour and $F{e_3}{O_4}$ is also known as magnetite which is black in colour. The reverse reaction can also take place but in presence of less amount of oxygen.

Note: In general, the mass percent of iron in an iron oxide of formula $F{e_3}{O_4}$ Is $72.35\%$. It is a mixed oxide where Fe is present in both +2 and +3 oxidation states whereas $F{e_2}{O_3}$ is a simple oxide in which iron has an oxidation state of +3.