Westonci.ca is the best place to get answers to your questions, provided by a community of experienced and knowledgeable experts. Explore our Q&A platform to find in-depth answers from a wide range of experts in different fields. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.
Sagot :
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%
Thanks for using our platform. We're always here to provide accurate and up-to-date answers to all your queries. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.