Methane is burned with air in a continuous steady-state combustion reactor to yield a mixture of carbon monoxide, carbon dioxide and water.

The feed to the reactor contains: $\pu{7.8 mole \% }\ce{CH4}$, $\pu{19.4 mole \% } \ce{O2}$, $\pu{72.8 mole \% } \ce{N2}$. The conversion of methane is $90\%$ and the gas leaving the reactor contains a ratio of $\pu{8 mol}\,\ce{CO2}/\pu{1 mol}\,\ce{CO}$.

My attempt:

  • $\ce{2 CH4 + 3 O2 -> 2 CO + 4 H2O}$

  • $\ce{CH4 + 2 O2 —> CO2 + 2 H2O}$

Both of these reactions have $90\%$ conversion.

From the ratio, can I say that approximately $88.88\%$ of the methane in the feed produces $\ce{CO2}$ while $11.11\%$ produces $\ce{CO}$?


1 Answer 1


Complicated little rascal! What we know: 1) The N2 is inert. 2) The oxygen is in excess. 3) 90% of the 7.8 mole % CH4 reacts; this is 7.02 mole % CH4. 4) This gives 8 moles CO2 per mole of CO, or 88.89% CO2 plus 11.11% CO.

The resultant output contains CO2: 7.02 mole % of the original CH4 x 88.89% = 6.24 mole % CO2.

The resultant output contains CO: 7.02 mole % of the original CH4 x 11.11% = 0.78 mole % CO.

The resultant output also contains 10% unreacted CH4. The 90% conversion refers to the starting material, not to the products. (Yield % would refer to one product.)

So the stoichiometry is 80% of the methane goes to CO2, 10% goes to CO, and 10% is unreacted.


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