[enthalpyburns]

I have a question I can't seem to answer  . Can you help me?

Methane (CH4) gas flows into a combustion chamber at a rate of 200. L/min @ 1.50 atm and ambient temperature. Air is added to the chamber at 1.00 atm and the SAME TEMPERATURE, and the gases are ignited.

To ensure the combustion of CH4 to CO2 (gas) and H2O (gas) three times as much oxygen as is necessary is reacted. Assuming air is 21 mole percent O2 and 79 mole percent N2, calculate the flow rate of air necessary to deliver the required amount of oxygen.

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Ok, I balanced the equation first (as we always first do)

CH4 + 2(O2)  2(H2O) + CO2

Then I set up the rate effusion ratio; the masses and MW square root ratios. However, I can't seem to get it to the desired answer: 8.7*10^3 L/min

From what I heard, you do not have to use the rate ratios for this question. 

Please help? 

[Borek]

It is not about diffusion, it is just about stoichiometry and mass balance. What volume of air contains three times as much oxygen as there are methane in 200 L?

[enthalpyburns]

Since the mole ratio between CH4 and O2 is 1 to 2, 400. L of O2 must  be required, right? But the O2 required is 3 times.., so 1.20*10^3 L O2 required (not air)

[Borek]

Yes, so far so good.