The complete balanced equation for the given reaction is:
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
This means that
Amount of oxygen required for 1 mole of propane = 5 moles
So, volume of oxygen required by 500 mL propane = 5 × 500 = 2500 m L
Available oxygen = 300 mL
Thus , oxygen is the limiting reagent.
Now, 2500 mL oxygen will produce CO2= 3 × 500 = 1500 m L
So, 300 mL oxygen will produce CO2= 1500 m L 2500 m L × 300 m L = 180 mL
Similarly, 2500 mL O2 will produce H2O = 4 × 500 = 2000 m L
So, 300 mL O2 will produce H2O = 2000 m L 2500 m L × 300 m L = 240 m L
Total volume of products that should be formed = 180 + 240 mL = 420 mL
But, total volume of products produced = 50% of 420 mL
= 50 100 × 420 = 210 m L
Now, since O2 is the limiting reactant, thus it is present in lesser amount than required for propane. So, some amount of propane will also be left in the reaction mixture after reaction.
So, 2500 mL oxygen requires propane = 500 mL
Thus, 300 mL oxygen will react with propane = 500 m L 2500 m L × 300 m L = 60 m L
So volume of propane left unreacted = 500 - 60 mL = 440 mL
Oxygen is completely consumed.
So, total volume of all gases that will be there at the end of the reaction = 210 + 440 mL
= 650 mL
