Question

How would you balance ${{H}_{2}}{{O}_{2}}\to {{H}_{2}}O+{{O}_{2}}$?

Answer 1

Hint In order to balance the equation multiply the coefficient of each term of the reactant and product to make sure that the number of atoms of every element on both the sides remain same throughout.

Complete step by step solution:
In order to answer our question, we need to observe the equation closely. Now, we know that mass is constant. That means that in any chemical equation, for a given mass of reactants, the mass of the product will be the same. In our case, we have the reactant as ${{H}_{2}}O$. So, the combined mass of these two gases will be the mass of the products i.e ${{H}_{2}}O$ and ${{O}_{2}}$ formed. There is one more rule in chemical equations that no new atom or substance can be formed for a given set of reactants, in our question both the reactants and products contain the hydrogen as well as the oxygen atom. In no way some other atom like nitrogen or helium is formed anywhere.
Now, using the above two laws of combination, we can solve our question. On observing the equation, we find 1 molecule of hydrogen peroxide is giving 1 molecule of water and 1 molecule of oxygen. Now the molecular formula of diatomic atoms and the molecule of water cannot change so we have to manipulate the coefficients to get the equation balanced. Now, we can see that only one molecule of hydrogen peroxide is available in reactants. So, we multiply the product ${{H}_{2}}O$ with 2. Now, we have 4 oxygen atoms and 4 hydrogen atoms in the product. In order to balance the hydrogen on the reactant side we multiply it by 2 so that total atoms become 4.
Finally, we obtain 4 hydrogen and 4 oxygen atoms both on reactant and product side and

Hence, the reaction is balanced. The full balanced equation is given by:
\[2{{H}_{2}}{{O}_{2}}\to 2{{H}_{2}}O+{{O}_{2}}\]

NOTE: When an equation is balanced, it also gives us an idea about the number of moles formed. In our case, 2 moles of hydrogen peroxide decomposes to give 1 mole of oxygen and 2 moles of water.