Question

When 5 litres of a gas mixture of methane and propane is perfectly combusted at $0^{\circ }C$ and 1 atmosphere, 16 litres of oxygen at the same temperature and pressure is consumed. The amount of heat released from this combustion is

$kJ\left(\Delta H_{comb.}\right(C{H}_4)=890kJmo{l}^{-1},\Delta H_{comb.}(C_3{H}_8)=2220kJmo{l}^{-1})$ is

• A. 38
• B. 317
• C. 477
• D. 32

Answer 1

Correct option: (C)

Step 1: Calculating the volume of methane and propane in the mixture

The balanced equation of combustion can be written as:

$C{H}_4\left(g\right)+2O_2\left(g\right)\to C{O}_2\left(g\right)+2H_{2}O\left(l\right)$

$C_3{H}_8\left(g\right)+5O_2\left(g\right)\to 3C{O}_2\left(g\right)+4H_{2}O\left(l\right)$

Let us assume the volume of methane$\left(C{H}_4\right)$ in $5L$ mixture $=xL$

Now, the volume of propane$\left(C_3{H}_8\right)$ in mixture $=5-xL$

From stoichiometry, 1 mole of methane consumes 2 moles of oxygen and 1 mole of propane consumes 5 moles of oxygen.

Hence, the total number of moles of oxygen consumed $=2x+5(5-x)=16$

So, $x=3$

Hence, the volume of methane $=3L$

The volume of propane $=2L$

Step 2: Calculating amount of heat evolved by combustion of mixture

$n_{C{H}_4}=\frac{3L}{22.4L}$

$n_{C_3{H}_8}=\frac{2L}{22.4L}$

$\text{Heat liberated}=n_{C{H}_4}\times \Delta H_{C{H}_4}+n_{C_3{H}_8}\times \Delta H_{C_3{H}_8}$

$\text{Heat liberated}=\frac{3\times 890+2\times 2220}{22.4}=317kJ$

Hence, the correct option is (C).