Question

A natural gas may be assumed to be a mixture of methane and ethane only. On complete combustion of $10ltr$ of gas at STP , the heat evolved was $474.6kJ$ . Assuming $\Delta {H_{comb}}C{H_4} = - 894kJmo{l^{ - 1}}$ and $\Delta {H_{comb}}{C_2}{H_6} = - 1500kJ$. Calculate the percentage composition of the mixture by volume.

Answer 1

Hint: Standard temperature and pressure (STP) are a useful set of benchmark conditions to compare other properties of gases. At STP, gases have a volume of 22.4 L per mole. The ideal gas law can be used to determine densities of gases.

Complete step by step answer:
Let $x$ litre of $C{H_4}$ is present in the natural gas.
So moles of $C{H_4}$ present is equal to $\dfrac{x}{{22.4}}$ [since combustion is taking place at STP].
Since total gas is $10ltr$ in amount.
Left amount for ${C_2}{H_6} = (10 - x)ltr$
Moles present in ${C_2}{H_6}$$ = \dfrac{{10 - x}}{{22.4}}$
Now we are given that the total heat evolved during combustion of natural gas is $474.6kJ$.
And we know that heat evolution will take place due to methane and ethane.
Heat evolved by per mole combustion of methane is:
$\Delta {H_{comb}}C{H_4} = - 894kJmo{l^{ - 1}}$
Heat evolved by $\dfrac{x}{{22.4}}$ moles combustion of methane is:
$\dfrac{x}{{22.4}} \times 894$

Heat evolved by per mole combustion of ethane is:
$\Delta {H_{comb}}{C_2}{H_6} = - 1500kJ$
Heat evolved by $\dfrac{{10 - x}}{{22.4}}$ moles combustion of ethane is:
$\dfrac{{10 - x}}{{22.4}} \times 1500$
But we are given that the total heat evolved during combustion of natural gas is $474.6kJ$.
So-
$\dfrac{x}{{22.4}} \times 894$$ + $ $\dfrac{{10 - x}}{{22.4}} \times 1500$$ = $ $474.6$
On solving we will get $x = 0.745$ .
So, $74.5\% $ is the answer.

Note: Standard enthalpy of combustion- It is the heat evolved when 1 mol of a substance burns completely in oxygen at standard conditions. It is also represented as \[\Delta {H_c}^ \circ \] . Here in this question it is denoted as $\Delta {H_{comb}}$