Ethane is burned with 50% excess air. The percentage conversion of ethane is 90%. Of the ethane burned, 25% reacts to form CO and the balance reacts to form carbon dioxide. Calculate the molar composition of the stack gas on a dry basis, and the mole ratio of water to dry stack gas.
The burning reaction (this determines the basis of the excess air) is:
C2H6+ 7/2 O2 -> 2 CO2 + 3 H2O (1)
Using 1 mol of ethane requires 7/2 mols of O2. But O2 (and implicitly air) is 50% in excess so one uses 7/2* (100+50)/100 =7/2 * 3/2 mols of O2. The corresponding amount of N2 (air is 20%O2, 80% N2) is 4 * 7/2 * 3/2 =21 mols of N2.
The reaction to CO is
C2H6+ 5/2 O2-> 2 CO + ...
The molar composition of a gas mixture is calculated.