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How many liters of propane gas (C3H8) will undergo complete combustion with 37.0 L of oxygen gas? (20pts) C3H8(g) + 5O2(g) ⇨ 3CO2(g) + 4H2O(g)
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How many liters of propane gas (C3H8) will undergo complete combustion with 37.0 L of oxygen gas? (20pts) C3H8(g) + 5O2(g) ⇨ 3CO2(g) + 4H2O(g)

User Simurg
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Since we are working with gases, the liters can be taken as moles.

According to the given balanced reaction, 1 liter of propane reacts completely with 5 liters of oxygen, use this ratio to find the liters needed to react with 37 liters of oxygen:


37LO_2\cdot(1LC_3H_8)/(5LO_2)=7.4LC_3H_8

It means that 7.4 liters of propane will be needed.

User Fetus
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