Question

For the combustion of methane presented in Example 5.4, the chemical reaction is CH4 +2O2 →CO2 +2H2O Suppose that methane flows into a burner at 30 gmol/s, while oxygen flows into the same burner at 75 gmol/s. If all the meth- ane is burned and a single output stream leaves the burner, what is the mole fraction of CO2 in that output stream? Hint 1: Does the fact that all the methane is burned mean that all the oxygen is burned also? Hint 2: Find the molar flow rate of each component gas in the outlet gas ("flue gas").

Answer 1

Answer:

$x_{CO_2}^{out} =0.25$

Explanation:

Hello.

In this case, for the reactive scheme, it is very convenient to write each species' mole balance as shown below:

$CH_4:f_{CH_4}^{out}=f_{CH_4}^{in}-\epsilon \\\\O_2:f_{O_2}^{out}=f_{O_2}^{in}-2\epsilon\\\\CO_2:f_{CO_2}^{out}=\epsilon\\\\H_2O:f_{H_2O}^{out}=2\epsilon$

Whereas $\epsilon$ accounts for the reaction extent. However, as all the methane is consumed, from the methane balance:

$0=f_{CH_4}^{in}-\epsilon \\\\\epsilon=30gmol/s$

Thus, we can compute the rest of the outlet mole flows since not all the oxygen is consumed as it is in excess:

$f_{O_2}^{out}=f_{O_2}^{in}-2\epsilon=75gmol/s-2*30gmol/s=15gmol/s\\\\f_{CO_2}^{out}=15gmol/s\\\\f_{H_2O}^{out}=2*15gmol/s=30gmol/s$

It means that the mole fraction of carbon dioxide in that output is:

$x_{CO_2}^{out}=\frac{15}{15+15+30} =0.25$

Best regards.