Answer
0 mol $S$, 1.5 mol $O_{2}$, 5 mol $SO_{3}$
Work Step by Step
Calculating how much product can be made based on the amount of each reactant, we have
$5\,mol\,S\times\frac{2\,mol\,SO_{3}}{2\,mol\,S}=5\,mol\,SO_{3}$
$9\,mol\,O_{2}\times\frac{2\,mol\,SO_{3}}{3\,mol\,O_{2}}=6\,mol\,SO_{3}$
Since $5\,mol\,S$ makes the least amount of product, it is the limiting reactant.
Therefore, moles of $S$ after the reaction= 0.
Now, 2 moles of $S$ react completely with 3 moles of $O_{2}$.
$\implies 3\times\frac{5}{2}=7.5$ moles of $O_{2}$ react with 5 moles of $S$.
So, after the reaction, moles of $O_{2}=9-7.5=1.5$
Moles of $SO_{3}$= theoretical yield (maximum amount of product that can be made in the reaction based on the amount of limiting reactant)= $5\,mol$
