Chemical Forums

Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: methic on October 17, 2010, 01:02:49 PM

Title: Mole fraction of a mixture. I'm stuck.
Post by: methic on October 17, 2010, 01:02:49 PM

I have no idea how to start this problem. Well, I started it; but I don't know if I started it correctly.

Quote

A gaseous fuel mixture stored at 744 mmHg and 298 K contains only methane, CH₄, and propane, C₃H₈. When 12.8 L of this mixture is burned, it produces 776 kJ of heat.

What is the mole fraction of methane in the mixture."

So, this is what I've done so far. I found the total number of moles burned through PV=nRT

(0.98atm)(12.8L)=n(0.08206)(298K) = 0.512 moles of mixture.

I'm not sure if I need these...
Molar mass of CH₄ is 16.04 g/mol
Molar mass of C₃H₈ is 44.10 g/mol

I'm basically stuck there. I have no idea how to set up the problem.  ???

Title: Re: Mole fraction of a mixture. I'm stuck.
Post by: Borek on October 17, 2010, 03:19:37 PM

You will need standard enthalpies of burning for both gases.

Title: Re: Mole fraction of a mixture. I'm stuck.
Post by: methic on October 17, 2010, 04:21:06 PM

Hi Borek. Thanks for replying.

So, I got them. Methane at -890.3 and Propane at -2219.1. Is it just a matter of solving for x in an equation like this:

-890.3x + -2219.1y = -776 kJ

And that will give me the moles of x, at which point I use the total moles I previously found to get the mole fraction?

y being (0.512-x),

so -890.3x + (-2219.1)(.512-x) = -776kJ

x = .27 mol / .512total moles = 53% CH₄

Is that correct?

//edit: Incorrect. Got it wrong. :( 'supposed to be 42.7%. Don't know where that came from

Title: Re: Mole fraction of a mixture. I'm stuck.
Post by: Borek on October 17, 2010, 06:04:24 PM

Your approach is correct - as far as I can tell there is nothing more to the question. Could be there is some problem with the enthalpies you used.