Question

When iron pyrite ($Fe{S_2}$) is roasted then it gives $F{e_2}{O_3}$ and $S{O_2}$. In this reaction, equivalent weight of $Fe{S_2}$ will be ( M = Molecular weight of $Fe{S_2}$)
(A) $\dfrac{M}{1}$
(B) $\dfrac{M}{5}$
(C) $\dfrac{M}{{10}}$
(D) $\dfrac{M}{{11}}$

Answer 1

This question is based on the topic of equivalent weight. Equivalent weight is the mass of one equivalent i.e. the mass of a given substance which will combine with or displace a fixed quantity of another substance. The equivalent weight of an element is the mass which displaces or combines 1.008 gram of hydrogen or 8.0 grams of oxygen or 35.5 grams of chlorine.

Complete step by step answer:
First of all, let's see the reaction of iron pyrite with oxygen.
$4Fe{S_2} + 11{O_2} \to 2F{e_2}{O_3} + 8S{O_2}$
It is a type of combustion reaction. A combustion reaction is a high-temperature exothermic reaction involving fuel and an oxidant to proceed. In the current case, iron pyrite is the fuel and atmospheric oxygen acts as an oxidant. This combustion reaction generates iron (III) oxides. Typically, the pyrite ore contains up to 85% iron pyrite and 15 % inert materials which reacts with excess air (up to 200% excess) to generate sulfur dioxide ($S{O_2}$). The waste containing iron oxide is further removed. Based on the enthalpy of the reaction, 5656 KJ of heat is generated per kg of ore. This calculation assumes that the reaction is maintained at a constant temperature. The release of energy makes it an exothermic reaction and the given reaction. It is also a redox reaction which can be defined as a chemical reaction in which electrons are transferred between two reactants participating in it. This transfer of electrons can be identified by observing the changes in the oxidation states of the given reacting species.
Since, It’s a redox reaction. Therefore, number of e- gained = number of e- lost
1 mole of $Fe{S_2}$ = ($\dfrac{11}{4}$ mole of ${O_2}$)
Each of the oxygen atoms takes 2 electrons for reduction, therefore from one ${O_2}$ molecule, 4 electrons are absorbed for reduction.
Therefore, number of moles of electrons gained by ${O_2}$ will be $\dfrac{11}{4} \times 4 = 11$
Therefore, number of moles of electron lost by 1 mole of $Fe{S_2}$ = 11

Therefore the equivalent weight of $Fe{S_2}$ = $\dfrac{M}{{11}}$.

Note: Always remember that a reaction which includes both oxidation and reduction is known as a redox reaction. If in any reaction, any of them (oxidation part or reduction part) is missing them that would not be called a redox reaction.