Question

Balance the following equation stepwise: ${{H}_{2}}{{S}_{2}}{{O}_{7}}(I)+{{H}_{2}}O(I)\to {{H}_{2}}S{{O}_{4}}(I)$

Answer 1

Hint: To solve this, follow the law of conservation of mass. Identify the number of each of the elements on both the sides. Balance the side which has a lower number of atoms by multiplying it with a number.

Complete step by step answer:
We know that balancing a reaction equation means that the number of elements will be the same on the reactant as well as the product side. For this question we have to follow conservation of mass.
We know that the law of conservation of mass states that for a closed system, where transfer of mass and energy takes place, the mass of the system is always constant at any time as the mass of the system cannot change. Thus, mass can neither be added nor removed but only transfers from one form to another.
In the question, the equation given to us is:
     ${{H}_{2}}{{S}_{2}}{{O}_{7}}(I)+{{H}_{2}}O(I)\to {{H}_{2}}S{{O}_{4}}(I)$

Now, to balance this equation let us start by identifying the elements on the reactant i.e. the left hand side. We know that the elements will be the same on the product side too as they are formed from these reactants only.
In the reactant side, we have hydrogen, sulphur and oxygen.
Now let us find the number of atoms of these elements on the reactant side.
On this side we have 2 atoms of hydrogen in ${{H}_{2}}{{S}_{2}}{{O}_{7}}$ and 2 in ${{H}_{2}}O$. Therefore, the total number of hydrogen atoms is 4.
We have 2 atoms of sulphur and 7 oxygen atoms in ${{H}_{2}}{{S}_{2}}{{O}_{7}}$ and 1 in ${{H}_{2}}O$. So the total number of oxygen atoms in the reactant side is 8.
Now, let us see the product side. We have obtained only one product and it has 2 atoms of hydrogen, 1 atom of sulphur and 4 atoms of oxygen.
Now, let us compare the number of atoms in the reactant and the products-

Element	Number of atoms before reaction	Number of atoms after reaction
Hydrogen	4	2
Oxygen	8	4
Sulphur	2	1

We can understand from the above table that if we multiply the product with 2, we will get the left hand side equal to the right hand side and thus the equation will be balanced.
Therefore, we can write that the balanced equation will be-
     ${{H}_{2}}{{S}_{2}}{{O}_{7}}(I)+{{H}_{2}}O(I)\to 2{{H}_{2}}S{{O}_{4}}(I)$
And this is the required answer.

Note: If we had the equation in ionic form instead of the elemental form, we would have to balance mass as well as the charge on both the sides. We have to follow the law of conservation of mass as well as charge while writing a balanced equation. In the above equation, there was no overall charge on any of the reactant or product species and thus are neutral so we just followed the conservation of mass.