|By Mo222 (Mo222) on Saturday, October 04, 2003 - 08:13 pm: Edit|
A 1.00-L sample of a gaseous mixture at 0 C and 1.00 atm evolves, upon complete combustion at constant pressure, 82.58 kJ of heat. If the gas is a mixture of ethane (C2H6) and propane (C3H8), what is the mole fraction of ethane in the mixture?
|By Mo222 (Mo222) on Saturday, October 04, 2003 - 10:07 pm: Edit|
|By Mo222 (Mo222) on Sunday, October 05, 2003 - 10:23 am: Edit|
THIS IS THE ONLY ONE OF THE PROBLEMS I POSTED THAT I HAVENT BEEN ABLE TO FIGURE OUT. HERE's MY WORK SO FAR. CAN SOMEONE PLEASE HELP ME FIGURE IT OUT? Jason, fairy, mazzo, bigman, chenfei, anyone?
So far I figured that you had to find out how many moles of gas you
(1 atm)(1.00 L)=n(.0821 atm*L / mol*K)(273 K)
0.0446 mol = n
And I also wrote the equation for the combsution of propane and ethane,
by writing & balancing the individual combustion reactions and adding
2C2H6 (g) + C3H8 (g) +12O2 (g) ----> 7CO2 (g) + 10H2O (l)
After that im stuck.
|By Mo222 (Mo222) on Sunday, October 05, 2003 - 11:44 am: Edit|
|By Mo222 (Mo222) on Sunday, October 05, 2003 - 01:18 pm: Edit|
|By Jszab (Jszab) on Sunday, October 05, 2003 - 02:16 pm: Edit|
Hmmm. Hard.I really must go now, so write me, tommorow, or thursday.
Now I can give You one hint:
X * Heat + !-X * heat = 82,58 kj.
You know the number of moles, so it would be easier.
Is it opened, or in utensil? If it is closed, you need to add water, to it which will be increasing pressure, by dcreasing volume, so You will add energy from pressure decresing , p dV.
Thus:X * Heat + !-X * heat = 82,58 kj - p dV ,
where V is V of water.
|By Mo222 (Mo222) on Sunday, October 05, 2003 - 04:34 pm: Edit|
Jzab thanks for helping.
It seems like everyone is ignoring this problem for some reason.
However I am having a really hard time following your work. WHat does !-X mean? What does "in utensil" mean? Why would you be adding water?
|By Jszab (Jszab) on Sunday, October 05, 2003 - 05:07 pm: Edit|
Umm. Sorry. ! = 1.
Water is producing during reaction. Yes?
If it is closed beaker, the water will occupy volume. So pressure will become higher [pv/t = const]. If it is so, we need to take some energy, to balance the pressure as the same level.
I meant - beaker.
|By Mo222 (Mo222) on Sunday, October 05, 2003 - 07:22 pm: Edit|
Jzab Im really confused -- I don't get your steps.
|By Mo222 (Mo222) on Sunday, October 05, 2003 - 09:31 pm: Edit|
Ok everybody by Jszab seems to be ignoring this post! I've been bumping it for 2 days! =*(
|By Mo222 (Mo222) on Monday, October 06, 2003 - 12:04 pm: Edit|
PLEASE HELP ME! If you can't get it at least post some work so I can get an idea...
|By Mo222 (Mo222) on Monday, October 06, 2003 - 02:09 pm: Edit|
|By Mo222 (Mo222) on Monday, October 06, 2003 - 07:21 pm: Edit|
OK I KNOW SOMEONE CAN ANSWER THIS! WHY ARE YOU IGNORING THIS POST??? HELP =*(
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