Question

Write down the equation for the stoichiometric combustion of ethane (C2H6) with oxygen, and determine the stoichiometric air fuel mass ratio. Atomic weights: H=1, C=12, O=16 (all kg/kmol), Air composition: 23.3% oxygen, 76.7% nitrogen (by mass).

Answer 1

Answer:

Equation:

$\text{C}_{2} \text{H}_{6} + \dfrac{7}{2}\text{O}_{2} \rightarrow 2\text{CO}_{2} + 3\text{H}_{2}\text{O}$

Supose we have 1 kmol of ethane, according to the equation we should have 7/2 kmol of oxygen, witch corresponds to a number of mols of air of:

$0.233\text{N}_{air} = 7/2$

$\text{N}_{air} = 15,0 \text{kmol} [\tex]As molar weight of ethane is [tex]\dfrac{M_{air}}{M_{ethane}} = \dfrac{N_{air}\cdot MW_{air}}{N_{ethane}\cdot MW_{etane}} = \dfrac{15kmol\cdot 28.9kg/kmol}{1kmol\cdot 30kg/kmol}=14.45$ and air is 0.233*32 + 0.767*28 = 28.9kg/kmol, the mass ratio is: