Exam in january and I'm stessing out, finding gas calculations really difficult.
The Q is:
18g of pentane is completely burnt in a car engine to form c02 and water.
What volume of 02 is needed, assuming that one mole occupies 24dm^3?
What volume of air is needed (assume that air contains 20% 02 by volume)
What volume of carbon dioxide is produced?
Thank you so much!
P.S. Please write down step-by-step method and perhaps explaining why a certain step is used
Thanks again, much appreciated
Gas calculation help please? Watch
- Thread Starter
- 23-12-2010 21:37
- 23-12-2010 22:03
First you need to work out the moles of pentane used. Which is 18 / ((12 X 5) + (1 X 12) = 0.25 moles of pentane used.
Now work out a balance equation:
C5H12 + 5O2 >>>>>> 5 CO2 + 5 H2O
So that means for everyone one mole of pentane you need 5 moles of O2 thus:
0.25 * 5 = 1.25 moles of O2 needed.
Next you need to times 24dm^3 by the number of moles of O2 used (1.25) thus: 1.25 * 24 = 30dm^3 of O2 used.
As you know 20% is 30dm^3 you need to do 30dm^3 / 20 * 100 = 150dm^3 of air needed.
Finally the volume of CO2 produced is equal to the volume of O2 produce so 30dm^3
I hope that makes sense, if not please do quote me and ask.
- 23-12-2010 22:06
- 24-12-2010 11:38