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15mL of a gaseous hydrocarbon required 4...

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"

A

`C_2H_6`

B

`C_2H_4`

C

`C_3H_6`

D

`C_2H_2`

Text Solution

Verified by Experts

The correct Answer is:
B
`C_xH_4+[x+y/4]O_2rarrxCO_2+y/2H_2O`
`{:(15,45,0,0),(0,0,15x,(15y)/2):}`
Given 15 x = 30 `:. X = 2 " "` Also `15[x+y/4] = 45 :. Y = 4`
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