Question

A cylinder of compressed gas that bears no label is supposed to contain either ethane or ethene. Combustion of the sample shows that 16 $cm^{3}$ of the gas require 48 $cm^{3}$ of oxygen for complete combustion. This shows that the gas is

• A. only ethane
• B. only ethene
• C. 1:1 mixtures of two gases
• D. some unknown mixture of the two gases

Answer 1

Answer: (B)

$\underset{16 \text{c} \left(\text{m}\right)^{3}}{\underset{1 \text{vol}}{C_{x} H_{y}}}+\underset{16 \left(x + \frac{y}{4}\right) c m^{3}}{\underset{\left(x + \frac{y}{4}\right) v o l}{\left(x + \frac{y}{4}\right) O_{2}}} \rightarrow xCO_{2}+\frac{y}{2 H_{2} O}$

Now we have $16\left(x + \frac{y}{4}\right)=48$

or y = 4

Hence, the hydrocarbon is $C_{2}H_{4}$ .