Answer:
50%
Explanation:
Let assume that there are:
"a" moles of CH4 & "b" moles of C2H2
Then by applying the ideal gas equation:
PV = nRT
[tex]16.8 \times 10^3 \times V = (a+b)RT[/tex]
Make (a+b) the subject of the formula:
[tex](a+b) = \dfrac{16800 \ V }{RT} mol[/tex] --- (1)
Since 1 mole of CH₄ yields 1 mol of CO₂ & 1 mol of C₂H₂ yields 2 moles of CO₂
Then;
the total moles of CO₂ = (a +2b)
Now:
[tex]25.3 \times 10^3 \times V = (a+ 2b)RT[/tex]
[tex](a + 2b) = \dfrac{25200 \ V}{RT \ mol}[/tex] ---- (2)
∴
By solving the above equations
[tex]a = \dfrac{8400 \ V}{RT} \\ \\ b = \dfrac{8400 \ V}{RT}[/tex]
Hence, the estimate of the percentage of methane in the original mixture is:
[tex]= 100 \times \dfrac{a}{(a+b)}[/tex]
[tex]= 100 \times \dfrac{8400\dfrac{ \ V}{RT} }{8400\dfrac{ \ V}{RT} + 8400\dfrac{ \ V}{RT}}[/tex]
= 50%
