Question

How do you balance $C{{H}_{3}}OH+{{O}_{2}}\to C{{O}_{2}}+{{H}_{2}}O$ ?

Answer 1

As we know that in a balanced reaction, basically the stoichiometric coefficient of a reactant is found to be equal to that of the product. In order to balance the equation, we must make sure that both sides of the reaction that is reactant as well as product side have equal numbers of atoms. We must try to balance the oxygen atoms first, and then the other atoms later.

Complete answer:
- We will balance the equation $C{{H}_{3}}OH+{{O}_{2}}\to C{{O}_{2}}+{{H}_{2}}O$. Let us discuss about the steps to balance this chemical equation:
- As we are being provided with the following equation:
$C{{H}_{3}}OH+{{O}_{2}}\to C{{O}_{2}}+{{H}_{2}}O$
-In order to balance the equation, we will first balance the atoms other than that of H and O. So, we will first balance the C atoms. So, we can write the equation as:
$2C{{H}_{3}}OH+{{O}_{2}}\to 2C{{O}_{2}}+{{H}_{2}}O$
- Now, we will balance the H atoms, as we can see here that the number of hydrogen atoms is less than that of the number of oxygen atoms. Hence, we will write chemical equation as:
$2C{{H}_{3}}OH+{{O}_{2}}\to 2C{{O}_{2}}+4{{H}_{2}}O$
- Now, we will balance the remaining O atoms, and can write the balanced chemical equation as:
$2C{{H}_{3}}OH+3{{O}_{2}}\to 2C{{O}_{2}}+{{H}_{2}}O$
- Hence, we can conclude that we can balance the given equation $C{{H}_{3}}OH+{{O}_{2}}\to C{{O}_{2}}+{{H}_{2}}O$ as $2C{{H}_{3}}OH+3{{O}_{2}}\to 2C{{O}_{2}}+{{H}_{2}}O$

Note:
- As we know that we cannot change the molecular formula of the compound, when we balance the equation. This is due to the fact that whenever we add a number in the place of a stoichiometric coefficient, at that time we do not change the molecular formula of the compound.