Answer:
[tex]x_{CO_2}^{out} =0.25[/tex]
Explanation:
Hello.
In this case, for the reactive scheme, it is very convenient to write each species' mole balance as shown below:
[tex]CH_4:f_{CH_4}^{out}=f_{CH_4}^{in}-\epsilon \\\\O_2:f_{O_2}^{out}=f_{O_2}^{in}-2\epsilon\\\\CO_2:f_{CO_2}^{out}=\epsilon\\\\H_2O:f_{H_2O}^{out}=2\epsilon[/tex]
Whereas [tex]\epsilon[/tex] accounts for the reaction extent. However, as all the methane is consumed, from the methane balance:
[tex]0=f_{CH_4}^{in}-\epsilon \\\\\epsilon=30gmol/s[/tex]
Thus, we can compute the rest of the outlet mole flows since not all the oxygen is consumed as it is in excess:
[tex]f_{O_2}^{out}=f_{O_2}^{in}-2\epsilon=75gmol/s-2*30gmol/s=15gmol/s\\\\f_{CO_2}^{out}=15gmol/s\\\\f_{H_2O}^{out}=2*15gmol/s=30gmol/s[/tex]
It means that the mole fraction of carbon dioxide in that output is:
[tex]x_{CO_2}^{out}=\frac{15}{15+15+30} =0.25[/tex]
Best regards.
