How much air is required to completely burn 5.6 m3 of methane (CH4)? The volume fraction of oxygen in the air is 21%. Gas volumes are measured under normal conditions.
The methane oxidation reaction is described by the following chemical reaction equation.
CH4 + 2O2 = CO2 + 2H2O;
According to the coefficients of this equation, the oxidation of 1 methane molecule requires 2 oxygen molecules. This synthesizes 1 molecule of carbon dioxide.
The molar volume of the gas is standard and fills a volume of 22.4 liters under normal conditions.
Since the molar volume of gas is standard, twice the amount of oxygen will require twice the volume of oxygen.
The required volume of oxygen will be 5.6 x 2 = 11.2 m3.
With the condition that the volume fraction of oxygen in the air is 21%, the required air volume will be 11.2 / 0.21 = 53.3 m3;
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