Question

What volume of oxygen $\text{ (}{{\text{O}}_{\text{2}}}\text{) }$ measured at $\text{ }{{\text{0}}^{\text{0}}}\text{C }$ and 1 atm pressure, is needed to burn completely 1L, of propane gas ($\text{ }{{\text{C}}_{\text{3}}}{{\text{H}}_{\text{8}}}\text{ }$) measured under the same conditions?
A) 5 L
B) 10 L
C) 7 L
D) 6 L

Answer 1

Hint: The reaction in which the substance reacts with the oxygen gas and liberates the energy in the form of light and heat is known as the combustion reaction. The general combustion reaction of hydrocarbon is as shown below:
$\text{ }{{\text{C}}_{\text{x}}}{{\text{H}}_{\text{y}}}\text{ (g) +}\left( \text{x + }\dfrac{\text{y}}{\text{4}} \right){{\text{O}}_{\text{2}}}\text{(g)}\to \text{ xC}{{\text{O}}_{\text{2}}}\text{(g)+}\dfrac{\text{y}}{\text{4}}{{\text{H}}_{\text{2}}}\text{O(g) }$
Where x and y are the number of carbon and oxygen atoms in the hydrocarbon respectively.

Complete Solution :
We are provided the following data:
Propane gas $\text{ }{{\text{C}}_{\text{3}}}{{\text{H}}_{\text{8}}}\text{ }$ is burned in presence of oxygen gas at $\text{ }{{\text{0}}^{\text{0}}}\text{C }$ and 1 atm pressure.
We are interested to determine the volume of oxygen gas required to burn 1 litre of propane gas.
The combustion reaction of hydrocarbons with oxygen may result in carbon dioxide and water and liberates a large amount of energy. Many combustion reactions are found to occur with hydrocarbons which are made of hydrogen and carbon.

- Let's consider the combustion of the reaction of propane. Here propane is used as a fuel and it combusts in presence of oxygen to produce carbon dioxide and water. The balanced combustion reaction of propane is as shown below:
$\text{ }{{\text{C}}_{\text{3}}}{{\text{H}}_{\text{8}}}\text{(}g)\text{ + 5}{{\text{O}}_{\text{2}}}\text{(}g\text{) }\to \text{ 3C}{{\text{O}}_{\text{2}}}\text{(g) + 4}{{\text{H}}_{\text{2}}}\text{O (g) }$
We can say that 1 mole of propane reacts with the 5 moles of $\text{ }{{\text{O}}_{\text{2}}}\text{ }$ gas to give 3 moles of carbon dioxide and 4 moles of a water molecule. That is the mole ratio of propane to the oxygen gas is written as:
$\text{ }{{\text{C}}_{\text{3}}}{{\text{H}}_{\text{8}}}\text{ : }{{\text{O}}_{\text{2}}}\text{ }\Rightarrow \text{ 1 : 5 }$

According to Avogadro's law the volume of the gas is directly related to the amount of gas or the moles of gas present. Thus mathematically it can be written as:
$\text{ V }\propto \text{ n }$
Thus we can say that one mole i.e. one litre of propane requires 5 litres of oxygen gas.
So, the correct answer is “Option A”.

Note: Note that, the combustion reaction should contain the oxygen as the reactant otherwise reaction would not proceed further. The water produced is in the gaseous state rather than the liquid state because the combustion reaction is always accompanied by the high temperatures.