The gas molar volume is 24dm3 at rtp. Air is 20% oxygen. What is the minimum value for complete combustion of 2.4dm3 of methane?
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The combustion formula is CH4 + 2O2 ---> CO2 + 2H2O. You need 2 moles of O2 to react with one mole of methane. There is 0.1 mole of methane (0.1 molar volume), so you need 0.2 moles of O2, or 0.2 molar volumes or 4.8 dm³; if this represents 20% of air, the volume is 5*4.8 = 24.0 dm³
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If we know how much oxygen is used,
we'll know how much air is used.
2.4 dm^3 methane = 0.1 mol methane.
We need a balanced combustion equation:
CH4 + 202 ----> CO2 + 2H2O.
0.1 mol CH4 reacts with 0.2 mol oxygen
so 0.2 * (100 / 20) =1.0 mol air
= 22.4 dm^3 air is the minimum value.
we'll know how much air is used.
2.4 dm^3 methane = 0.1 mol methane.
We need a balanced combustion equation:
CH4 + 202 ----> CO2 + 2H2O.
0.1 mol CH4 reacts with 0.2 mol oxygen
so 0.2 * (100 / 20) =1.0 mol air
= 22.4 dm^3 air is the minimum value.