Answered

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A mixture of ethyne gas (C2H2) and methane gas (CH4) occupied a certain volume at a total pressure of 16.8 kPa. When the sample burned, the products were CO2 gas and H2O vapor. The CO2 was collected and its pressure found to be 25.2 kPa in the same volume and at the same temperature as the original mixture. What percentage of the original mixture was methane

Sagot :

ajeigbeibraheem

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%

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