Raining in TOKYO
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So i've had a major brain fart and have no clue how to go about this question.

https://ibb.co/J3Vxtkr

Part a) I think that I need to find the stoichiometric coefficient for O2 and then divide that by 21 and multiply by 100 to find the theoretical air.

So I went about making an equation for the complete combustion of the gas and ended up with 53 as the coefficient and therefore 252kmol but this is way off.
Last edited by Raining in TOKYO; 1 month ago
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GabiAbi84
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Do you have the answer to part i to check?
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Raining in TOKYO
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(Original post by GabiAbi84)
Do you have the answer to part i to check?
a)11.048 kmol
b) 3.2 mol% O2, 10.8 mol% CO2, 13.3 mol% H2O, 72.7 mol% N2
c) total inputs and outputs should be 411.31 kg.
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GabiAbi84
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Ooft well I got 14.3 for part 1 so I’m no use...
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golgiapparatus31
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For (a)
1 kmol of mixture contains
0.4 kmol of CH4
0.24 kmol of C2H4
0.16 kmol of C3H8
and 0.2 kmol of CO2

Write equations for each combustion
CH4 + 2O2 -> CO2 + 2H2O
C2H4 + 3O2 -> 2CO2 + 2H2O
C3H8 + 5O2 -> 3CO2 + 4H2O

So need 0.4 x 2 + 0.24 x 3 + 0.16 x 5 = 2.32 kmol of O2.
Percentage of O2 in air is 21%, so 2.32/0.21 = 11.048 kmol of air needed

For (b) calculate number of moles of CO2, H2O formed and number of moles of oxygen unused. Also calculate the number of moles of N2, then it is a percentage calculation.

For (c) it is dependent on answer to (b).
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GabiAbi84
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(Original post by golgiapparatus31)
For (a)
1 kmol of).
I did this the exact same way but somewhere along the way I must have put the wrong numbers in and messed it up.
Phew so glad to know I was doing it right after all. Thanks
Last edited by GabiAbi84; 1 month ago
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