Question

Where the oxygen comes from the air (21% O2 and 79% N2). If oxygen is fed from air in excess of the stoichiometric amount required to combust all of the ethanol by 10%, and 85% of the ethanol reacts, calculate the mole fraction of oxygen leaving the reactor.

Answer 1

`y_(O2) =4.3`%

Step-by-step explanation:

The ethanol combustion reaction is:

`C_(2)H_(5) OH+3O_(2)`→
`2CO_(2)+3H_(2)O`

If we had the amount (x moles) of ethanol, we would calculate the oxygen moles required:

`x*1.10(excess)*(3 O_(2)moles )/(etOHmole)`

Dividing the previous equation by x:

`1.10(excess)*(3 O_(2)moles)/(etOHmole)=3.30(O_(2)moles)/(etOHmole)`

We would need 3.30 oxygen moles per ethanol mole.

Then we apply the composition relation between O2 and N2 in the feed air:

`3.30(O_(2) moles)*(0.79(N_(2) moles))/(0.21(O_(2) moles))=121.414 (N_(2) moles )`

Then calculate the oxygen moles number leaving the reactor, considering that 0.85 ethanol moles react and the stoichiometry of the reaction:

`3.30(O_(2) moles)-0.85(etOHmoles)*(3(O_(2) moles))/(1(etOHmoles)) =0.75O_(2) moles`

Calculate the number of moles of CO2 and water considering the same:

`0.85(etOHmoles)*(3(H_(2)Omoles))/(1(etOHmoles))=2.55(H_(2)Omoles)`

`0.85(etOHmoles)*(2(CO_(2)moles))/(1(etOHmoles))=1.7(CO_(2)moles)`

The total number of moles at the reactor output would be:

`N=1.7(CO2)+12.414(N2)+2.55(H2O)+0.75(O2)\\ N=17.414(Dry-air-moles)`

So, the oxygen mole fraction would be:

`y_{O_(2)}=(0.75)/(17.414)=0.0430=4.3`%