Determine stoichiometric and actual air required as well as the combustion products generated?
150 KMOLES of CO with 30% excess air
- Roger the MoleLv 72 years agoFavorite Answer
2 CO + O2 → 2 CO2
There is only one combustion product: carbon dioxide.
(150 kmol CO) x (1 mol O2 / 2 mol CO) = 75 kmol O2
Supposing air to be 20% O2:
(75 kmol O2) / (0.20) = 375 kmoles air
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