JEE MAIN - Chemistry (2016 (Offline) - No. 5)
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:
C3H8
C4H8
C4H10
C3H6
Wyjaśnienie
$${C_x}{H_{y\left( g \right)}} + \left( {{{4x + y} \over 4}} \right){O_{2\left( g \right)}}$$
$$ \to xC{O_{2\left( g \right)}} + {y \over 2}{H_2}O\left( l \right)$$
Volume of $${O_2}$$ used
$$ = 375 \times {{20} \over {100}} = 75ml$$
$$\therefore$$ From the reaction of combination
$$1\,\,ml\,{C_x}{H_y}\,\,$$ requires $$ = {{4x + y} \over 4}ml\,{O_2}$$
$$15\left( {{{4x + y} \over 4}} \right) = 75$$
So, $$4x+y=20$$
$$x=3$$
$$y=8$$
$$\therefore$$ $${C_3}{H_8}$$
Volume of $${O_2}$$ used
$$ = 375 \times {{20} \over {100}} = 75ml$$
$$\therefore$$ From the reaction of combination
$$1\,\,ml\,{C_x}{H_y}\,\,$$ requires $$ = {{4x + y} \over 4}ml\,{O_2}$$
$$15\left( {{{4x + y} \over 4}} \right) = 75$$
So, $$4x+y=20$$
$$x=3$$
$$y=8$$
$$\therefore$$ $${C_3}{H_8}$$