A hydrocarbon is burnt with excess air. The Orsat analysis of the flue gas shows 10.81% CO2, 3.78% O2 and 85.40 N2. Calculate the atomic ratio of C:H in the hydrocarbon and the % excess air.
Calculations:
100 moles of dry flue gas (i.e., excepting H2O)
The Orsat analysis shows the compositions of the flue gases by not taking into account of H2O.
From the composition of air(mole %),
Here, nitrogen is the tie component.
79 mole of N2 ≡ 21 mole of O2
Therefore, O2 that is entering the burner = 85.4 x 21/79 = 22.7
C + O2
→ CO21 mole of CO2 ≡ 1 mole of O2 ≡ 1 atom of C ( i.e.,1 mole of O2 reacts with 1 atom of C to produce 1 mole of CO2)
Therefore, O2 used up for reacting with carbon = 10.81 mole and,
Carbon in the hydrocarbon =
10.81 atomsO2 reacted with Hydrogen in the hydrocarbon = 22.7 - (10.81 + 3.78) = 8.11
4H + O2
→ 2H2O1 mole of O2 reacts with 4 atoms of hydrogen.
Therefore, hydrogen in the hydrocarbon = 8.11 x 4 =
32.44 atoms.C:H ratio in the hydrocarbon
= 1 : 32.44/10.81 = 1 : 3Theoretical air demand = air needed for complete conversion of carbon to carbon dioxide and hydrogen to water vapor
% excess air
= 100 x (actual air used - theoretical air demand) / theoretical air demand= 100 x (22.7 - (10.81 + 8.11))/( 10.81 + 8.11) =
20%Last Modified on: 04-Feb-2022
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