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CH4 + 2 O2 --> CO2 + 2 H2O

x / 64 = 1 / 16

x = 4 g oxygen

Assuming oxygen is 20% oxygen...

4 = .2 x

x = 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 mole

0.0625 / 1mole meth = x / 2mole oxy

x = 0.125 mole oxygen =20% of air

--- 0.625 mole 'air'

0.2 x 32 + 0.8 x 28 = 28.8 g/mole of air

0.625 x 28.8 = 18 grams of air

Since air is 20% oxygen by volume, the 2nd answer is best.

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