How many liters of air containing 20% oxygen is needed to burn 1 mole of propane?

univerkov | December 10, 2020 | education

propane formula C3H8
C3H8 + 5O2 = 3CO2 + 4H2O
n (propane) = 1 mol; n (o2) = 1 * 5 = 5 mol
v (o2) = 5 * 22.4 = 112 liters
and since there is 20% oxygen in the air by volume, the air volume will be 5 times greater = 112 * 5 = 560 liras

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