Question

A cylinder contains either ethylene or propylene. $12{\text{ }}ml$ of gas required $54{\text{ }}ml$ of oxygen for complete combustion. The gas is:
A. Ethylene
B. Propylene
C. $1:1$ mixture of two gases
D. $1:2$ mixture

Answer 1

This question is based on basic stoichiometry. In order to find out the answer to this question, we need to write down the balanced chemical equations of combustion of both ethylene (${C_2}{H_4}$) and propylene (${C_3}{H_6}$) and then interpret the given data.

Complete step by step solution:
Balanced chemical equation for the combustion of ethylene in the presence of oxygen gas is given by the following chemical equation:
${C_2}{H_4} + 3{O_2} \to 2C{O_2} + 2{H_2}O$
From the above mentioned balanced chemical equation, it is clear that $1{\text{ }}mol$ of Ethylene gas requires $3{\text{ }}mol$ of Oxygen gas to undergo complete combustion.
Hence, under NTP, $22400{\text{ }}ml$ of Ethylene would require $3 \times 22400{\text{ }}ml$ of Oxygen gas to burn completely. So, by using a unitary method, we can calculate that $12{\text{ }}ml$ of Ethylene would require $36{\text{ }}ml$ of Oxygen gas.
Now, the balanced chemical equation for the combustion of Propylene in the presence of Oxygen gas is given by the following chemical equation:
${C_3}{H_6} + \dfrac{9}{2}{O_2} \to 3C{O_2} + 3{H_2}O$
From the above mentioned balanced equation, it is clear that $1{\text{ }}mol$ of Propylene gas requires $\dfrac{9}{2}{\text{ }}mol$ of Oxygen gas for complete combustion.
Hence, under NTP, $22400{\text{ }}ml$ of Propylene would require $\dfrac{9}{2} \times 22400{\text{ }}ml$ of Oxygen gas to burn completely. Hence, by using a unitary method, we can calculate that $12{\text{ }}ml$ of Propylene would require $54{\text{ }}ml$ of Oxygen gas.

**So, option (B) is correct.

Note: **
At NTP, $1{\text{ }}mol$ of a gas occupies $22.4{\text{ }}L$ volume while at STP, $1{\text{ }}mol$ of a gas occupies $22.7{\text{ }}L$ volume. So students must not get confused between these two values.