Question

Please explain. pls i need explanation,,,,!!! pls In a closed container, combustion of 50 mL methane is carried out with 150 ml of air containing 21% of oxygen. What will be the total volume of gaseous mixture after reaction ? CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

Answer 1

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1 vol of methane will react with 2 vol of oxygen to give one vol of CO₂ and 2 vol of water, so, 50 ml of methane will react with 100 ml of oxygen,

but the amount of oxygen present is 21%, = (21/100) x (150) = 31.5 ml of oxygen

so, half of the volume of oxygen will react with methane, = 31.5/2 = 15.75 ml of methane. So, unreacted methane will be: 50- 15.75 = 34.25 ml

The rest of the gas will remain same, = 150-31.5 = 118.5 ml of air

15.75 ml of CO₂ and 31.5 ml of water will be produced (it becomes liquid)

Now, total volume = 118.5 (left gas) + (unreacted methane) + 15.75 (carbon dioxide) + 31.5 (oxygen) = 200 ml of gas

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Answer 2

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200 ml of gas