If pure oxygen is fed in excess by 25%, what would the fractional conversion of methane be for the final concentration of CO2 in the outlet stream, in mole percent, to be 10%?

the fractional conversion of methane is 12.5%

Explanation:The reaction represent the combustion of methane to produce Co2 and steam.

CH4 +2O2_CO2 + 2H2O

From gay lussac law of proportionality

1mol of CH4 requires 2mol of Oxygen to produce 1 mol of CO2 and 2mol of H2O

So from the combining ratio,25% of O2 will fractional produce 25×1/2% of CH4.

While 12.5% of CO2 and 25% of steam is also produced .so in essence 2.5% of CO2 was lost in the reaction.

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