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

Options

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

Answer

B

Explanation

`C_(x)H_(y) + (x+(Y)/(4))O_(2)rarr xCO_(2)+(y)/(2)H_(2)O`
1 vol. `(x+(y)/(4))vol`
`16 cm^(3) 16 (x+(y)/(4))cm^(3)`
Now, we have `16(x+(y)/(4)) = 48`
Putting `x = 2`
`16(2+(y)/(4))=48`
or `y = 4`
Hence, the hydrocarbon is `C_(2)H_(4)`.