How much oxygen is needed to burn 7 liters of propane?

Let us write the equation for the reaction of propane combustion:
C3H8 + 5O2 = 3CO2 + 4H2O.
As you can see from the reaction, the amount of substance:
ν (C3H8) = ν (O2) / 5.
For gases, the amount of substance under normal conditions is:
ν (v-va) = V (v-va) / Vm, where Vm is the molar volume of gas, for all it is equal to 22.4 l / mol, under normal conditions.
Substituting the amount of substance into the equality, we get:
V (C3H8) / Vm = V (O2) / (5 * Vm).
Let us express the volume of oxygen from this expression:
V (O2) = V (C3H8) * 5 * Vm / Vm = V (C3H8) * 5.
Substitute the numerical values:
V (O2) = V (C3H8) * 5 = 7 * 5 = 35 liters.
Answer: It takes 35 liters of oxygen to burn propane.

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