Ten millilitre of a gaseous hydrocarbon was burnt completely in 80 ml of `O_(2)` at STP. The volume of the remaining gas is 70 ml. The volume became 50 ml on treatment with NaOH. The formula of the hydrocarbon is:

Options

A. `C_2H_6`
B. `C_2H_4`
C. `C_3H_8`
D. `C_3H_6`

Answer

B

Explanation

`C_xH_y+(x+y/4)O_2 rarrxCO_2 +y/2H_2O`
`{:( 1ml,(x+y/4)ml,xml,"__"),(10 ml,10(x+y/4)ml,10xml,"__"):}`
Volume of `CO_2=(70–50)=20 ml`
10x=20, Therefore x=2
Volume of `CO_2` + Volume of `O_2` (left)=70 ml
Volume of `O_2` (left)= 70 - 20 = 50 ml
Volume of `O_2` (used) = 80-50= 30 ml
`:. 10 (x+y/4)=30`
Solve for y, putting x = 2, y = 4 Hence the formula is `C_2 H_4`