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1 September, 21:17

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%?

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  1. 1 September, 22:06
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    Answer: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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