Question

\[A{l_{(s)}} + {O_{2(g)}} \to A{l_2}{O_{3(s)}}\]
The above reaction is balanced so that its coefficients are the lowest whole – numbers. Find the coefficient of \[{O_2}\]
A.6
B.4
C.2
D.3
E.1

Answer 1

Hint: The chemical equation that has been given is not balanced. In order to find the value for the coefficient of \[{O_2}\] , we must first balance the equation, and then we can derive the corresponding value.

Complete step by step answer:
Before we move forward with the solution of this question, let us understand a few basic important concepts.
To balance a given chemical equation, we must follow the following steps:
First of all, we must write the given unbalanced chemical equation:
 \[A{l_{(s)}} + {O_{2(g)}} \to A{l_2}{O_{3(s)}}\]
Identify and classify the reactants and the products of the reaction.
Reactants: 1. Aluminium = Al
           2. Oxygen = \[{O_2}\]
Product: 1. Alumina = \[A{l_2}{O_3}\]
Write down the number of atoms of the given elements present in the molecules of the elements and compounds on each side:
Reactants: 1. Aluminium = Al = 1
           2. Oxygen = O= 2
Product: 1. Alumina = \[A{l_2}{O_3}\]
        Al=2
        C=0
Now, we must ensure that the number of atoms of a given element present in both the sides is equal. So, we must now cross multiply to achieve equivalent figures on both the sides.
Reactants: 1. Aluminium = \[Al = 1\left( 2 \right) = 2\]
           2. Oxygen = \[O = 2\left( 3 \right) = 6\]
Product: 1. Alumina = \[A{l_2}{O_3}\] -
         \[\begin{array}{*{20}{l}}
  {Al = 2\left( 1 \right) = 2\;} \\
  {O = 3\left( 2 \right) = 6}
\end{array}\]
In this step, we must place the coefficients before the molecules, to get balance out the number of atoms:
 \[(2)A{l_{(s)}} + (3){O_{2(g)}} \to (2)A{l_2}{O_{3(s)}}\]
Reactants: 1. Aluminium = \[Al = 1\left( 2 \right) = 2\]
           2. Oxygen = \[O = 2\left( 3 \right) = 6\]
Product: 1. Alumina = \[A{l_2}{O_3}\]-
        \[\begin{array}{*{20}{l}}
  {Al = 4\left( 1 \right) = 4\;} \\
  {O = 3\left( 2 \right) = 6}
\end{array}\]
We can see that the number of oxygen atoms are balance, but the number of aluminium atoms is not balance. Hence, let us again change the coefficients a bit to get the correct result
 \[(4)A{l_{(s)}} + (3){O_{2(g)}} \to (2)A{l_2}{O_{3(s)}}\]
Reactants: 1. Aluminium = \[Al = 1\left( 4 \right) = 4\]
           2. Oxygen = \[O = 2\left( 3 \right) = 6\]
Product: 1. Alumina = \[A{l_2}{O_3}\]-
        \[\begin{array}{*{20}{l}}
  {Al = 4\left( 1 \right) = 4\;} \\
  {O = 3\left( 2 \right) = 6}
\end{array}\]
Hence, the given chemical equation is balanced. From this, the coefficient of \[{O_2}\] can be found to be equal to 3.

Hence, Option D is the correct option.

Note:
The Law of Conservation of Mass states that the mass of the reactants must balance the mass of the products. To balance a chemical equation, the atoms of both the elements and molecules on the reactant side (left side) and product side (right side) must be equal to each other.