Verified Solution • 2025–26 Session moderate

QJEEM16Chec29: JEE Main 2016 Chemistry - Class 11 JEEM16Chec29 Solution: At 300 K and 1 atm, 15 mL of a gaseous hydrocarbon requires 375 …

At 300 K and 1 atm, 15 mL of a gaseous hydrocarbon requires 375 mL air containing 20% O2 by volume for complete combustion. After combustion the gases occupy 345 mL. Assuming that the water formed is in liquid form and the volumes were measured at the same temperature and pressure, the formula of the hydrocarbon is:

Multiple Choice Options

  • A C3H8
  • B C4H8
  • C C4H10
  • D C3H6

Step-by-Step Verified Solution

CxHy(g)+(4x+y4)O2(g){C_x}{H_{y\left( g \right)}} + \left( {{{4x + y} \over 4}} \right){O_{2\left( g \right)}} →xCO2(g)+y2H2O(l)\to xC{O_{2\left( g \right)}} + {y \over 2}{H_2}O\left( l \right)

Volume of O2{O_2} used

=375×20100=75ml= 375 \times {{20} \over {100}} = 75ml

∴\therefore From the reaction of combination

1  ml CxHy  1\,\,ml\,{C_x}{H_y}\,\, requires =4x+y4ml O2= {{4x + y} \over 4}ml\,{O_2}

15(4x+y4)=7515\left( {{{4x + y} \over 4}} \right) = 75

So, 4x+y=204x+y=20

x=3x=3

y=8y=8

∴\therefore C3H8{C_3}{H_8}

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