Question

A gas mixture contains acetylene and carbon dioxide. $20$ litres of this mixture requires $20$ litres of oxygen for complete combustion. If all gases are measured under similar conditions of temperature and pressure, the percentage of acetylene in the mixture is:
A.$50\% $
B.$40\% $
C.$60\% $
D.$75\% $

Answer 1

Hint: One mole of compound has $22.4L$ and molar mass of the compound is equal to the sum of the molar masses of all the atoms in the compound. First balance the chemical reaction and then apply the mole concept.

Complete step by step solution:
First of all let us read about balanced chemical reactions.
Balanced chemical reaction: It is defined as the reaction in which the number of atoms on both the sides i.e. on the reactant side as well as on the product side is same.
Combustion: It is defined as the reaction in which one mole of a substance burns in air i.e. oxygen to form oxides.
Now in the question we are given with the mixture of acetylene and carbon dioxide. And we know that carbon dioxide does not give combustion reactions. So the only compound which will give combustion is acetylene.
The balanced chemical reaction when acetylene reacts with oxygen to form carbon dioxide is as follows: ${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{2}}}{\text{ + }}\dfrac{{\text{5}}}{{\text{2}}}{{\text{O}}_{\text{2}}} \to {\text{2C}}{{\text{O}}_{\text{2}}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{O}}$. Here the number of carbon atoms on both sides is equal i.e. two, number of hydrogen atoms on both the sides is equal i.e. two and the number of oxygen atoms on both the sides is equal i.e. five.
Now from the balanced chemical reaction of combustion of acetylene it is clear that one mole of acetylene will react with $2.5$ moles of oxygen to form $2$ moles of carbon dioxide. And we know that one mole of compound has $22.4L$. So $2.5$ moles of oxygen will have $2.5 \times 22.4 = 56L$ of oxygen.
Now, if $56L$ of oxygen reacts with $22.4L$ of acetylene.
Then $20L$ of oxygen will react with $\dfrac{{22.4}}{{56}} \times 20 = 8L$ of acetylene.
Now we have to calculate the percentage of acetylene in the mixture. Percentage is calculated as the ratio of amount of that particular substance to total amount of substance multiplied by $100$.
So the percentage of acetylene in the mixture will be $\dfrac{8}{{20}} \times 100 = 40\% $.

So option B is correct.

Note: Balance the equations carefully if you make any mistake in balancing the equation then your answer will be wrong. And also apply mole concept carefully as one molecule or atom is related to another molecule or atom by what factor.