Calculate the volume of oxygen gas needed for complete combustion of 5L of methane gas. CH₄ + 2O₂ → CO₂ + 2H₂O
Answers
Answer:
The reaction is CH
4
( +2O2 →CO2 +2H2 O.)
At room temperature, water will condense to form liquid water.
1 mole of methane reacts with 2 moles of oxygen to form 1 mole of oxygen.
10 ml (0.445 moles) of methane will react with 20 ml (0.89 moles) of oxygen to form 10 ml (0.445 moles) of carbon dioxide.
Hence, the resulting mixture will contain 10 ml of unreacted methane and 10 ml of carbon dioxide.
The total volume of the mixture will be 20 ml.
Given volume of methane =0.25dm
3
=0.25L
Combustion of methane (CH
4
)-
(methane)
CH
4
+2O
2
⟶CO
2
+2H
2
O
As we know that volume of 1 mole of gas is 22.4 L.
Amount of oxygen required for the combustion of 22.4 L of methane =(2×22.4)L=44.8L
∴ Amount of oxygen required for the combustion of 0.25 L of methane =
22.4
44.8
×0.25=0.5L
Hence, 0.5 L of oxygen required for the complete combustion of 0.25dm
3
of methane.