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If by "air" you mean oxygen, then the mole ratio can by found by balancing a reaction equation. CH4 + 202 -> CO2 + 2H2O The ratio is 1:2.

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Q: What is the mole ratio of air to methane gas in the cumbustion of methane?
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What could cause incomplete cumbustion of a fuels?

Wrong air/fuel ratio Poor atomisation of fuel Poor spark Incorrect timing of spark


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What is the mass of air required for the complete combustion of 1g of methane?

CH4 + 2 O2 --> CO2 + 2 H2Ox / 64 = 1 / 16x = 4 g oxygenAssuming oxygen is 20% oxygen...4 = .2 xx = 20 g air that is one possible answer.A better answer--- air is 20% oxygen by volume and 80% nitrogen by volume.1g methane / 16 = 0.0625 mole0.0625 / 1mole meth = x / 2mole oxyx = 0.125 mole oxygen =20% of air--- 0.625 mole 'air'0.2 x 32 + 0.8 x 28 = 28.8 g/mole of air0.625 x 28.8 = 18 grams of airSince air is 20% oxygen by volume, the 2nd answer is best.


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