Question

What is the balanced equation for heptane $ ({C_7}{H_{16}}) $ burning in oxygen to make carbon dioxide and water?

Answer 1

Hint :We are asked to balance a particular equation, here balance means that the same number of atoms for every one of the elements in the reactants and products sides must be maintained. This has to be done in order to maintain the conservation law. The reactants are heptane, oxygen and the products here are water and carbon dioxide.

Complete Step By Step Answer:
We need to balance equations in a systematic manner.
Write down what is given in the question:
Reactants: $ {C_7}{H_{16}},\;{O_2} $
Products: $ {H_2}O,\;C{O_2} $
So the equation will look like this initially (unbalanced):
 $ \Rightarrow {C_7}{H_{16}} + {O_2} \to C{O_2} + {H_2}O $
Normally hydrogen and oxygen vary widely while balancing equations, so we will leave them for the end and balance other elements first.
Choose to balance the numbers in complex compounds; here that would be heptane.
There is no coefficient in front of heptane but let us put $ 1 $ to make it clear;
 $ \Rightarrow 1{C_7}{H_{16}} + {O_2} \to C{O_2} + {H_2}O $
Proceeding to balance $ C $ ;
Since on the left we have only $ 1 $ heptane, we know that the value of carbon atoms is $ 7 $ , so to balance this, put $ 7 $ as coefficient of carbon for carbon dioxide since there was only $ 1 $ carbon atom on left initially.
 $ \Rightarrow 1{C_7}{H_{16}} + {O_2} \to 7C{O_2} + {H_2}O $
Now we continue to balance $ H $ ;
This can be done by putting an $ 8 $ for water on the product side since on the reactant side we do not need to make any changes to heptane and the number of atoms on $ H $ is $ 16 $ .
 $ \Rightarrow 1{C_7}{H_{16}} + {O_2} \to 7C{O_2} + 8{H_2}O $
Moving onto balance $ O $ ;
We see that the product side has more number of $ O $ atoms now so we need to bring the reactant side at par with the product side. To do this the total atoms on $ O $ on the product side is $ 22 $ , since it is a multiple of $ 2 $ , we just need to keep a $ 11 $ as coefficient of $ O $ on the reactant side to maintain the balance.
 $ \Rightarrow 1{C_7}{H_{16}} + 11{O_2} \to 7C{O_2} + 8{H_2}O $
Now we can recheck if number of all the atoms on either side match, since it does we can say;
 $ \therefore $ The overall equation when balanced looks like this : $ {C_7}{H_{16}} + 11{O_2} \to 7C{O_2} + 8{H_2}O $


Note :
In a practical situation we can keep in mind that water vapor and carbon dioxide gas are generated only when a hydrocarbon is fully burned (complete combustion). If the temperature at a time is low enough, the water vapor can easily cool down to condense and turn into liquid form. It all depends on the circumstances in which the combustion occurs. If this process of combustion comes into action outdoors, the water vapor might end up condensing rapidly. Otherwise, water can stay in its vapor form when indoor or within a confined area where it is maintained that the temperature never goes down below $ 100^\circ C $ .