Question

How do you calculate the volume of oxygen required for complete combustion of $0.25{\text{ d}}{{\text{m}}^3}$ of methane at S.T.P?

Answer 1

Since we have to find the volume of oxygen at STP we will convert the given volume of methane into litres. We will write the balanced combustion reaction for the methane. Then we can find the volume of oxygen by finding the number of moles of oxygen produced on combustion of one mole of methane.

Complete Answer:
The combustion reaction of methane can be represented as:
$C{H_4}{\text{ + }}{O_2}{\text{ }} \to {\text{ }}C{O_2}{\text{ }} + {\text{ }}{H_2}O$
We have to balance the combustion reaction for finding the number of moles of products. Thus the balanced combustion will be represented as:
$C{H_4}{\text{ + 2}}{O_2}{\text{ }} \to {\text{ }}C{O_2}{\text{ }} + {\text{ 2}}{H_2}O$
At STP we know that one mole of gas is equivalent to $22.4{\text{ L}}$ of gas. We can also say that one mole of gas will account for $22.4{\text{ L}}$ of volume. Firstly we will convert given volume of methane into litre by using the below conversion:
$1{\text{ d}}{{\text{m}}^3} = {\text{ 1 L}}$
Thus the volume of given methane is $0.25{\text{ L}}$. Now from the above equation we can say that:
$\Rightarrow$ Number of moles of oxygen produced on combustion of one mole of methane $= {\text{ 2 moles}}$
$\Rightarrow$ Volume of oxygen produce on combustion of $22.4{\text{ L}}$ methane $= {\text{ }}\left( {{\text{2 }} \times {\text{ 22}}{\text{.4}}} \right){\text{ L}}$
$\Rightarrow$ Volume of oxygen produce on combustion of $1{\text{ L}}$ methane $= {\text{ }}\dfrac{{\left( {{\text{2 }} \times {\text{ 22}}{\text{.4}}} \right){\text{ L}}}}{{{\text{22}}{\text{.4 L}}}}$
$\Rightarrow$ Volume of oxygen produce on combustion of $0.25{\text{ L}}$ methane $= {\text{ }}\dfrac{{\left( {{\text{2 }} \times {\text{ 22}}{\text{.4}}} \right){\text{ L}}}}{{{\text{22}}{\text{.4 L}}}} \times 0.25{\text{ L}}$
$\Rightarrow$ Volume of oxygen produce on combustion of $0.25{\text{ L}}$ methane $= {\text{ 0}}{\text{.5 L}}$
Therefore we can say that the volume of oxygen required for complete combustion of $0.25{\text{ L}}$ methane is ${\text{0}}{\text{.5 L}}$.

Note:
It must be noted that the reaction must be balanced. If it is not balanced then the answer would be different and wrong. At STP the volume $22.4{\text{ L}}$ is equivalent to one mole of gas. Thus we can represent moles in terms of volume as well. Here we have used a unitary method to solve the problem.