Answer:
Equation:
[tex]\text{C}_{2} \text{H}_{6} + \dfrac{7}{2}\text{O}_{2} \rightarrow 2\text{CO}_{2} + 3\text{H}_{2}\text{O}[/tex]
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:
[tex] 0.233\text{N}_{air} = 7/2[/tex]
[tex]\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[/tex] and air is 0.233*32 + 0.767*28 = 28.9kg/kmol, the mass ratio is:
