Determine stoichiometric and actual air required as well as the combustion products generated?

Update:

150 KMOLES of CO with 30% excess air

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  • 2 years ago
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    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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