The oxidation reaction of propane with oxygen occurs in accordance with the following chemical reaction equation:
C3H8 + 5O2 = 3CO2 + 4H2O;
Combustion of 1 mol of propane requires 5 mol of oxygen.
Let’s find the required amount of propane.
M C3H8 = 12 x 3 + 8 = 44 grams / mol;
N C3H8 = 8.8 / 44 = 0.2 mol;
To oxidize this amount of propane, 0.2 x 5 = 1 mole of oxygen is required.
Let’s find its volume.
One mole of gas under normal conditions fills a volume of 22.4 liters.
The oxygen volume will be equal to:
V O2 = 1 x 22.4 = 22.4 liters;
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