Solveeit Logo

Question

Question: What is the balanced equation for the combustion of pentane gas (\({{C}_{5}}{{H}_{12}}\))?...

What is the balanced equation for the combustion of pentane gas (C5H12{{C}_{5}}{{H}_{12}})?

Explanation

Solution

Combustion means the compound is heated and there is the release of energy. The combustion of pentane gas means the reactants will be pentane and oxygen so, the products will be carbon dioxide gas and water.

Complete answer:
A balanced equation means the numbers of atoms on the reactant side are equal to the number of atoms on the product side. Combustion means the compounds are heated and there is the release of energy and for the combustion oxygen is an important element used.
Pentane is a compound of group alkane in which there are 5 carbon atoms, so its formula is C5H12{{C}_{5}}{{H}_{12}}. The combustion of pentane gas means the reactants will be pentane and oxygen so, the products will be carbon dioxide gas and water. The unbalanced reaction of combustion is given below:
C5H12+O2CO2+H2O{{C}_{5}}{{H}_{12}}+{{O}_{2}}\to C{{O}_{2}}+{{H}_{2}}O
Now, we have to balance the reaction, the number of carbon atoms on the reactant side is 5 and the number of carbon atoms on the product side is 1, so we can multiply the carbon dioxide by 5. The reaction will be:
C5H12+O25CO2+H2O{{C}_{5}}{{H}_{12}}+{{O}_{2}}\to 5C{{O}_{2}}+{{H}_{2}}O
There are 12 hydrogen atoms on the reactant side while there are 2 hydrogen atoms on the product side. So, we can multiply the water molecule by 6. The reaction will be:
C5H12+O25CO2+6H2O{{C}_{5}}{{H}_{12}}+{{O}_{2}}\to 5C{{O}_{2}}+6{{H}_{2}}O
There are 2 oxygen atoms on the reactant side while there are 16 oxygen atoms on the product side. So, we can multiply the oxygen molecule by 8. The reaction will be:
C5H12+8O25CO2+6H2O{{C}_{5}}{{H}_{12}}+8{{O}_{2}}\to 5C{{O}_{2}}+6{{H}_{2}}O
This is the balanced reaction of the combustion of pentane gas.

Note:
We can also find the direct method to find the balanced equation for the combustion of any alkane. The reaction is:
CnH2n+2+(3n+12)O2nCO2+(n+1)H2O{{C}_{n}}{{H}_{2n+2}}+\left( \dfrac{3n+1}{2} \right){{O}_{2}}\to nC{{O}_{2}}+(n+1){{H}_{2}}O
For pentane, the value of n is 5.