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

Equation:

`\text{C}_(2) \text{H}_(6) + (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]</p><p></p><p>As molar weight of ethane is [tex](M_(air))/(M_(ethane)) = (N_(air)\cdot MW_(air))/(N_(ethane)\cdot MW_(etane)) = (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: