Question: solve in a clear way please Ethane is burned with 50% excess air ([molesin - molesthooretcal)/molestheoretical)]. The percentage conversion (i.e. [molesin - molesout)/molesin)] of the
Ethane is burned with 50% excess air ([molesin - molesthooretcal)/molestheoretical)]. The percentage conversion (i.e. [molesin - molesout)/molesin)] of the ethane is 90%; of the ethane burned, 25% reacts to form CO and the balance reacts to form CO2. Carry out Degree-of-Freedom (DOF) analysis and calculate the molar composition of the stack gas ( n0 through n6 ) using extents of reaction method. C2H6+27O22CO2+3H2OC2H6+25O22CO+3H2O
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