Question

How would you balance the following equation:
${\text{Fe + }}{{\text{O}}_{\text{2}}} \to {\text{F}}{{\text{e}}_{\text{2}}}{{\text{O}}_{\text{3}}}$

Answer 1

Hint: For balancing a chemical equation, we have to maintain equal molecularity of each atom present on the reactant side as well as on the product side of the given chemical reaction.

Complete step by step answer:
Given that, Iron reacts with oxygen to form Iron (III) Oxide and chemical reaction for this is shown as follow:
${\text{Fe + }}{{\text{O}}_{\text{2}}} \to {\text{F}}{{\text{e}}_{\text{2}}}{{\text{O}}_{\text{3}}}$
For balancing the above given equation we have to consider following points in mind:
- First we calculate the number of each atom on the left hand side as well as on the right hand side of the given chemical equation.
- So on the left hand side or reactant side only one iron (${\text{Fe}}$) atom and on the right hand side or product side of the chemical reaction two iron (${\text{Fe}}$) atoms are present, which is not correct.
- On the left hand side two atoms of oxygen (${\text{O}}$) are present but on the right hand side three oxygen atoms are present which is not correct.
So, we will balance the above given reaction by the following manner so that each atom will have the same molecularity on the reactant side as well as on the product of the reaction. And balanced equation is shown as:
     ${\text{4Fe + 3}}{{\text{O}}_{\text{2}}} \to 2{\text{F}}{{\text{e}}_{\text{2}}}{{\text{O}}_{\text{3}}}$
- In the above balanced equation four iron atoms (${\text{Fe}}$) and six oxygen (${\text{O}}$) atoms are present on the left hand side as well as on the right hand side of the chemical reaction.

Note: Here some of you may think order of reaction and molecularity of reaction are the same things but that will be true only for the elementary (single step reaction) reactions only.