Question

Formaldehyde is produced by the oxidation of methane in a reactor. The following two parallel reactions occur: \[ CH_4 + O_2 \rightarrow HCHO + H_2O \] \[ CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O \] Methane and oxygen are fed to the reactor. The product gases leaving the reactor include methane, oxygen, formaldehyde, carbon dioxide and water vapor.
60 mol/s of methane enters the reactor. The molar flowrate (in mol/s) of CH₄, O₂ and CO₂ leaving the reactor are 26, 2, and 4, respectively. The molar flowrate of oxygen entering the reactor is \(\underline{\hspace{1cm}}\) mol/s.

Hint:

For parallel reactions, conserve material balances on each flow component.

Answer 1

Correct Answer: 40

Let the molar flowrates be:
- \( \dot{n}_{CH_4} = 60 \) mol/s entering the reactor
- \( \dot{n}_{CH_4, exit} = 26 \) mol/s leaving the reactor
- \( \dot{n}_{O_2, exit} = 2 \) mol/s (for both reactions)
- \( \dot{n}_{CO_2, exit} = 4 \) mol/s The total oxygen used in both reactions is calculated as:
- Reaction 1: \( \dot{n}_{CH_4} = 26 \) mol/s, thus oxygen consumed = 26 mol/s.
- Reaction 2: \( \dot{n}_{CH_4} = 34 \), oxygen consumed = 34. \[ \dot{n}_{O_2, entering} = \text{oxygen consumed} + exit = 60
\] Final result: 40 to 42 mol/s.