Question

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)

Answer 1

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:

$37LO_2\cdot\frac{1LC_3H_8}{5LO_2}=7.4LC_3H_8$

It means that 7.4 liters of propane will be needed.