00:02
Question, it is asked to calculate the percent excess air use in the combustion and the composition of the winston.
00:11
So here, there is 100 moles of exit gas.
00:22
The reaction is ch4 plus 2 moles of oxygen leaving.
00:29
Co2 plus 2 moles of h2.
00:34
So, here 1 more of ch4 gives 1 ball of carbon dioxide and it is.
00:40
Uses 2 moles of oxygen.
00:45
So carbon dioxide present in the gases 8 .7 moves.
00:57
So since 1 mole of ch4 gives 1 mole of carbon dioxide, methane used is equal to 8 .7 moles.
01:15
And oxygen used is equal to the ratio is 2 is to 1.
01:20
So here it is 2 into 8 .7 which is equal to 17 .4 moles.
01:28
The other reaction is 2 points of ch4 plus 3 moles of oxygen giving 2 most of carbon monoxide plus 4 moles of h2.
01:43
Here carbon monoxide present is equal to 1 more.
01:56
Ch4 used is 1 more...