Question

What volume of propane is burnt for every $100\,c{m^3}$ of oxygen used in the following reaction.
${C_3}{H_8}\,\, + 5{O_2} \to 3C{O_2}\,\, + 4{H_2}O$

Answer 1

This question is simply solved by using the unitary method. By looking at the chemical reaction, we can say that one mole of propane reacts with five moles of oxygen to produce carbon dioxide and water. Also, we have given the volume of oxygen in the question.

Complete answer:
Here, in this question we are given the volume of oxygen equals to $100\,c{m^3}$
Chemical equation-
${C_3}{H_8}\,\, + 5{O_2} \to 3C{O_2}\,\, + 4{H_2}O$
Also, one mole of propane reacts with five mole of oxygen. We know that mole is just a measurement of the amount of substance. It can be used for the mass of the substance or volume of the substance. In this question, we are talking about the volume so, mole can be related to the volume of the substance. Hence, we can say that one volume of propane reacts with five volumes of oxygen.
From the equation,
$5$ V of ${O_2}$ required = $1$ V of propane
Or, $5\,c{m^3}$ of ${O_2}$ required = $1\,c{m^3}$of propane
$100\,c{m^3}$ of ${O_2}$ will require = $\dfrac{{100}}{5}$= $20\,c{m^3}$ of propane
Therefore, $20\,c{m^3}$ volume of propane is burnt for every $100\,c{m^3}$ of oxygen.

Note:
This question is based on Avogadro’s law. According to this law, Volume is directly proportional to the number of moles (amount) of the gases if the temperature and pressure remains constant. This law is applicable for the ideal gases. This relation is derived from kinetic theory of gases assuming that the gases should behave ideally. This is also applicable for real gases but at low pressure and high temperature.