Orsat Analysis of Flue Gas

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 → CO₂

1 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 atoms

O₂ reacted with Hydrogen in the hydrocarbon = 22.7 - (10.81 + 3.78) = 8.11

4H + O₂ → 2H₂O

1 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 : 3

Theoretical 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%