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Octane is burnt with 40% excess O2, what is the percentage of CO2 in products?(a) 18.18%(b) 36.36%(c) 54.54%(d) 72.72% |
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Answer» Correct choice is (b) 36.36% The best I can explain: The reaction is C8H18 + 12.5O2 -> 8CO2 + 9H2O. Basis: 10 moles of C8H18, => moles of O2 reacted = 125, => moles of O2 entered the process = 175, => moles of CO2, H2O and O2 in products are 80, 90 and 50, => percentage of CO2 = 80/220*100 = 36.36%. |
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