Question

How do you balance the equation:
${{C}_{4}}{{H}_{10}}+{{O}_{2}}\to C{{O}_{2}}+{{H}_{2}}O$

Answer 1

Hint: This reaction is the combustion of butane because the butane is reacted with oxygen to form carbon dioxide and water. To balance the equation, first balance the carbon atoms and then balance the hydrogen and oxygen atoms.

Complete answer:
The reaction given in the question contains butane and oxygen as the reactant and carbon dioxide and water as the products. So, this reaction is the combustion of butane. Now, for balancing this reaction first we have to balance the elements in the reactions other than hydrogen and oxygen atoms. In this reaction only carbon is there other than hydrogen and oxygen. First, we have to balance the number of carbon atoms. The reaction is:
${{C}_{4}}{{H}_{10}}+{{O}_{2}}\to C{{O}_{2}}+{{H}_{2}}O$
There are 4 carbon atoms on the reactant side and there is only one carbon atom on the product side. So, we have to multiply the carbon dioxide by 4. The reaction is:
${{C}_{4}}{{H}_{10}}+{{O}_{2}}\to 4C{{O}_{2}}+{{H}_{2}}O$
Now, we have to balance hydrogen and oxygen atoms in the reaction. There are 10 hydrogen atoms on the reactant side and only 2 hydrogen atoms are there on the product side. So, we have to multiply the water molecule by 5. The reaction is:
${{C}_{4}}{{H}_{10}}+{{O}_{2}}\to 4C{{O}_{2}}+5{{H}_{2}}O$
There are 2 oxygen atoms in the reactant side and there are 8 + 5 = 13 oxygen atoms on the product side. So, 13 is an odd number, we have to multiply the oxygen atoms in the reactant side with $\dfrac{13}{2}$, we get:
${{C}_{4}}{{H}_{10}}+\dfrac{13}{2}{{O}_{2}}\to 4C{{O}_{2}}+5{{H}_{2}}O$
Now, if we want to remove the fraction then we have to multiply the whole equation with, we get:
$2{{C}_{4}}{{H}_{10}}+13{{O}_{2}}\to 8C{{O}_{2}}+10{{H}_{2}}O$
Now, all the elements are balanced, so the reaction is balanced.

Note:
There is a general equation for the combustion of alkane:
${{C}_{n}}{{H}_{2n+2}}+\left( \dfrac{3n+1}{2} \right){{O}_{2}}\to nC{{O}_{2}}+(n+1){{H}_{2}}O$
So, by putting the value of n in this equation we can directly get the balanced equation. In the question the reactant has 4 carbon atoms so, the n is 4:
${{C}_{4}}{{H}_{10}}+\left( \dfrac{3(4)+1}{2} \right){{O}_{2}}\to 4C{{O}_{2}}+(4+1){{H}_{2}}O$
${{C}_{4}}{{H}_{10}}+\dfrac{13}{2}{{O}_{2}}\to 4C{{O}_{2}}+5{{H}_{2}}O$