Determining amounts of product and excess reagent remaining post reaction
Note the following relations from the equation you wrote: 
1 mol $\ce{FeS2}$ = (11/4) mol $\ce{O2}$;
1 mol $\ce{FeS2}$ = (1/2) mol $\ce{Fe2O3}$; 
1 mol $\ce{FeS2}$ = 2 mol $\ce{SO2}$.

You already determined that $\ce{FeS2}$ is the limiting reactant because we have 600/120 mol of iron sulfide and 800/32 mol of $\ce{O2}$;

This implies that 5 mol $\ce{FeS2}$ will require =5x (11/4) mol $\ce{O2}$.

Remaining mol oxygen = 25-13.75=11.25 

Now use the relations established above: 5 mol iron sulfide would produce (5/2) or 2.5 mol iron oxide, and 5 mol iron sulfide would produce 5x2 = 10 mol sulfur dioxide.