Question

Identify X in the equation given below. $FeC{{l}_{3}}+S{{O}_{2}}+{{H}_{2}}O\to FeC{{l}_{2}}+HCl+X$
(A)- ${{H}_{2}}S$
(B)- $S{{O}_{2}}$
(C)- ${{H}_{2}}S{{O}_{4}}$
(D)- ${{H}_{2}}S{{O}_{3}}$

Answer 1

Hint: According to the law of conservation of mass, a balanced chemical equation must have the same number of atoms of each element on the reactant and product sides. Count the number of atoms of each type and make them equal on both sides of the equation.

Complete step by step answer:
We have been given an unbalanced chemical equation with one product missing.
Let us try to write the balanced chemical equation for the given reaction to identify X.
The given chemical equation is
     \[FeC{{l}_{3}}+S{{O}_{2}}+{{H}_{2}}O\to FeC{{l}_{2}}+HCl+X\]

First we need to identify the type of atoms present in the equation. We have Fe, Cl, H, O and S atoms.
Here, ferric chloride ($FeC{{l}_{3}}$), sulphur dioxide ($S{{O}_{2}}$) and water (${{H}_{2}}O$) are reactants. Ferrous chloride ($FeC{{l}_{2}}$), hydrochloric acid (HCl) and X are the products.
Iron is getting reduced from $F{{e}^{3+}}$ in ferric chloride to $F{{e}^{2+}}$ in ferrous chloride.
The oxidation state of H, i.e. +1, remains unchanged during the reaction from ${{H}_{2}}O$ to HCl.

In the reactant side, if $FeC{{l}_{3}}$ is being reduced and ${{H}_{2}}O$ is not undergoing any change in oxidation state, it means that $S{{O}_{2}}$ is getting oxidized to X.
Now, let us balance the number of each type of atom on both sides of the equation.

Balance the number of H atoms. One molecule of water has 2 H atoms, so two molecules of HCl are required for 2 H atoms on the product side. Now the equation becomes
     \[FeC{{l}_{3}}+S{{O}_{2}}+{{H}_{2}}O\to FeC{{l}_{2}}+2HCl+X\]

 Now, the number of chlorine atoms on the reactant side cannot be made equal to 4. So by taking 2 molecules of $FeC{{l}_{3}}$, we get 6 atoms of Cl on the reactant side.
     \[2FeC{{l}_{3}}+S{{O}_{2}}+{{H}_{2}}O\to FeC{{l}_{2}}+2HCl+X\]

We can now balance the number of Fe atoms. For 2 molecules of $FeC{{l}_{3}}$ on the reactant side we need 2 molecules of$FeC{{l}_{2}}$on the product side, i.e.
     \[2FeC{{l}_{3}}+S{{O}_{2}}+{{H}_{2}}O\to 2FeC{{l}_{2}}+2HCl+X\]

So the compound X must contain one S and 3 O atoms. Balance the number of S atoms. But X cannot be $S{{O}_{3}}$ as it is not given in the options above.

From the options given above, it is clear that X has only one S atom.
However, we can change the number of O atoms by adding 4 molecules of ${{H}_{2}}O$.
This will increase the H atoms by 2 and O atoms by 1 (there were earlier 3 O atoms) on the reactant side.
           \[2FeC{{l}_{3}}+S{{O}_{2}}+2{{H}_{2}}O\to 2FeC{{l}_{2}}+2HCl+X\]

Looking at the above equation, we can say that the compound X in the equation is sulphuric acid, i.e. ${{H}_{2}}S{{O}_{4}}$.
So, the correct answer is “Option C”.

Note: We may simply see that one S atom and two O atoms missing on the product side and mistake the compound X as ${{H}_{2}}S{{O}_{3}}$, which is wrong. Carefully balance the number of atoms of each type in the equation by changing the stoichiometric coefficient of the molecules. Never change the subscripts in the molecular formulae of the reactant and product molecules to balance an equation.