Question

When 5 litres of a gas mixture of methane and propane is perfectly combusted at 0°C and 1 atmosphere, 16 litres of oxygen at the same temperature and pressure is consumed. The amount of heat released from this combustion in kJ $\left(\Delta H_{comb}. \left(CH_{4}\right)=890 kJ mol^{-1}, DH_{comb}\left(C_{3}H_{8}\right)=2220 kJ mol^{-1} \right)$ is :-

• A. 38
• B. 317
• C. 477
• D. 32

Answer 1

Answer: (B)

The balanced equations of combustion reactions are:
$CH _{4}( g )+2 O _{2}( g ) \rightarrow CO _{2}( g )+2 H _{2} O ( l ),$
$\Delta H _{ CH _{4}( g )}=890\, kJ / mol$
$C _{3} H _{8}( g )+5 O _{2}( g ) \rightarrow 3 CO _{2}( g )+4 H _{2} O ( l ),$
$\Delta H _{ C _{3} H _{8}( g )}=2220\, kJ / mol$
Let here $x\, L\, CH _{4}$ and $(5- x ) L\, C _{3} H _{8}$ in $5\, L$ gas mixture at $STP$
So total volume of oxygen consumed as per stoichiometry of the reaction involved will be
$\Rightarrow 2 x+5(5-x)=16$
$\Rightarrow 2 x+25-5 x=16$
$\Rightarrow 3 x=9$
$\Rightarrow x=3\, L$
So volume of $CH _{4}( g )=3\, L$
and volume of $C _{3} H _{8}( g )=2\, L$ at $STP$
So corresponding number of moles of gaese present in $5\, L$ gas mixture can be obtained by dividing the volumes of gases at STP with $22.4\, L$ as any gas of $1 mol$ occupied $22.4\, L$ at STP.
$n _{ CH _{4}( g )}=\frac{3}{22.4} mol$
$n _{ C _{3} H _{3}( g )}=\frac{2}{22.4} mol$
So the amount of heat released due to combustion of the $5 L$ gas mixture
$\Delta H _{\text {mixture }}=\frac{3}{22.4} \Delta H _{ CH _{4}( g )}+\frac{2}{22.4} \Delta H _{ C _{3} H _{8}( g )}$
$\Delta H _{\text {mixture }}=\frac{3 \times 890+2 \times 2220}{22.4} \approx 317\, kJ$