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Date Posted: 05:08:49 11/04/01 Sun
Author: Jac
Subject: y liddat??!

Mr Yeo,
tys (19)17 organic chem no 3c(ii)
my ans for c(ii):vol of air required to burn 2g of methane is 9dm cube but ans booklet say 28.5dm cube leh. y? i tot if methane need 3 dm cube, oxygen need 6dm cube, then total air needed is 9 dm cube? or izit the air is not referring to that only or wad? dun understand. can teach me? thanx

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[> Re: y liddat??! -- Mr. Yeo Woei Ter, 19:52:08 11/04/01 Sun

Hi Jac!

I don't really understand what you say.

But here's my solution:
c(i):

no. of moles of methane = 2/(12+4) = 0.125 mol

from equation,

no. of moles of O2/no. of moles of CH4 = 2/1

no. of moles of O2 = 2 * 0.125 = 0.25 mol

Thus, vol. of O2 = 0.25 * 24 dm3
= 6 dm3

c(ii):
(Air contains about 1/5 by volume of O2
Thus, volume of air required = 5 * 6 = 30 dm3

The answer provided in your TYS is correct also. They consider that the percentage of oxygen by volume in air is 21%.

(Rough paper:
21% --- 6 dm3
100% --- ?? dm 3)

Thus volume of air required = 100/21 * 6 = 28.5 dm3

Both answers should be acceptable.

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