How would you balance the equation for the combustion of octane:\[{{C}_{8}}{{H}_{18}}(l)+{{O}_{2}}(g)\to C{{O}_{2}}(g)+{{H}_{2}}O(l)\]?

VerifiedVerified
116.4k+ views
Hint:. The required answer to this question is dependent on the general concept of chemistry which tells about how to balance a chemical equation. This includes the fact that the number of moles of each atom in the reactant side should equal to that in the product side.

Complete step by step answer:
In the lower classes of chemistry which deals with the very basic chemistry, we have come across the concept of balancing the given chemical equation and how to interpret it.
We shall now refresh the concept so that we will approach the required answer.
- A chemical equation is the interpretation of the compound which reacts together to form new or modified compounds.
- The compounds reacting together are called reactants and the compounds which are newly formed are called as the products.
- The reactants are written on the left hand side of the reaction and products on the right hand side of the equation which is separated by an arrow that represents the reaction going from right to left side.
- Balancing of this chemical equation means that the number of moles of each atom on the reactant side should be equal to that of the product side.
Thus, in the above given reaction of combustion of octane that is given by,
\[{{C}_{8}}{{H}_{18}}(l)+{{O}_{2}}(g)\to C{{O}_{2}}(g)+{{H}_{2}}O(l)\]
Here, the total number of carbon atoms on the reactant side is 8 whereas the total number of hydrogen atoms on the reactant side is 18 and that of oxygen atom is 2.
Thus, on the product side the same number of atoms should be present and this balances the given equation.
Now, according to the above facts given, we get a balance equation as shown below,
\[2{{C}_{8}}{{H}_{18}}(l)+25{{O}_{2}}(g)\to 16C{{O}_{2}}(g)+18{{H}_{2}}O(l)\]
Thus, this above equation is the required answer.

Note: Note that while balancing the equation, try to equalise on both the reactant as well as product side accordingly in all the possible ways and not just by looking into the product side but also by looking into the reactant side.