Answer:
[tex]y_{O2} =4.3[/tex]%
Explanation:
The ethanol combustion reaction is:
[tex]C_{2}H_{5} OH+3O_{2}[/tex]→[tex]2CO_{2}+3H_{2}O[/tex]
If we had the amount (x moles) of ethanol, we would calculate the oxygen moles required:
[tex]x*1.10(excess)*\frac{3 O_{2}moles }{etOHmole}[/tex]
Dividing the previous equation by x:
[tex]1.10(excess)*\frac{3 O_{2}moles}{etOHmole}=3.30\frac{O_{2}moles}{etOHmole}[/tex]
We would need 3.30 oxygen moles per ethanol mole.
Then we apply the composition relation between O2 and N2 in the feed air:
[tex]3.30(O_{2} moles)*\frac{0.79(N_{2} moles)}{0.21(O_{2} moles)}=121.414 (N_{2} moles )[/tex]
Then calculate the oxygen moles number leaving the reactor, considering that 0.85 ethanol moles react and the stoichiometry of the reaction:
[tex]3.30(O_{2} moles)-0.85(etOHmoles)*\frac{3(O_{2} moles)}{1(etOHmoles)} =0.75O_{2} moles[/tex]
Calculate the number of moles of CO2 and water considering the same:
[tex]0.85(etOHmoles)*\frac{3(H_{2}Omoles)}{1(etOHmoles)}=2.55(H_{2}Omoles)[/tex]
[tex]0.85(etOHmoles)*\frac{2(CO_{2}moles)}{1(etOHmoles)}=1.7(CO_{2}moles)[/tex]
The total number of moles at the reactor output would be:
[tex]N=1.7(CO2)+12.414(N2)+2.55(H2O)+0.75(O2)\\ N=17.414(Dry-air-moles)[/tex]
So, the oxygen mole fraction would be:
[tex]y_{O_{2}}=\frac{0.75}{17.414}=0.0430=4.3[/tex]%
