Consider the reaction between reactants S and O2:
2 S(s) + 3 O2(g) → 2 SO3(g)
If a reaction vessel initially contains 5 mol S and 9 mol O2, how many moles of S, O2, and SO3 will be in the reaction vessel after the reactants have reacted as much as possible? (Assume 100% actual yield.)
Here, we are going to calculate the number of moles of S, O2 and SO3 present in the reaction vessel after the reactants have reacted as much as possible.
The reactant which is totally consumed during a chemical reaction and limits the product formation is called limiting reagent.
The given reaction is:
2S(s) + 3O2(g) → 2SO3(g)
Here, 2 moles of S requires 3 moles of O2 for complete reaction.
Therefore, 5 mol of S will require (3/2 x 5 = 7.5) mol of O2 for reaction.
But, given amount of O2 = 9 mol, which is more than the required amount for complete reaction.
Thus, S will be completely consumed during the reaction and hence behaves as a limiting reagent.
Amount of O2 remained after the reaction = Initial amount - amount consumed
= 9 mol - 7.5 mol
= 1.5 mol