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# Gas calculation help please? Watch

1. 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
2. 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.
3. (Original post by joshed)
Lots of answers
It's better to help someone work out the answers, than give them the answers.
4. (Original post by gingerbreadman85)
It's better to help someone work out the answers, than give them the answers.
Which is why I gave them a step by step method to doing it, it is their choice if they take my answer or use my workings to understand it.

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