15mL of a gaseous hydrocarbon required 45 mL of oxygen for complete combustion. 30 mL of `CO_(2)` is formed. The formula of hydrocarbon is"

Options

A. `C_(2)H_(6)`
B. `C_(2)H_(4)`
C. `C_(3)H_(6)`
D. `C_(2)H_(2)`

Answer

B

Explanation

`C_(x)H_(y)+[x+(y)/(4)]O_(2)rarrxCO_(2)+(y)/(2)H_(2)O`
`{:(15,45,0,0),(0,0,15x,(15y)/(2)):}`
Given, `15x=30 therefore x=2`
Also, `15[x+(y)/(4)]=45 therefore y=4`