Given the problem as stated, you did the problem correctly. However I would have done it a bit differently to make it easier to check. I like to do the problems in steps. I also dislike carrying a lot of fractions in intermediate calculations since I get confused easily. (This is basically M. Farooq's answer with some explanation.)
Given
$$\ce{2FeS2 + 11/2 O2 -> Fe2O3 + 4 SO2}$$
Mass of $\ce{FeS2}$ is $\pu{600 g}$ and mass of $\ce{O2}$ is $\pu{800 g}$. Find the amounts of $\ce{Fe2O3}$ and $\ce{SO2}$ and the remaining amount of the excess reagent.
Molar Quantities
For $\ce{Fe}$:
$n_\ce{FeS2} = \frac{600}{120.} = 5.0000 $
For $\ce{O2}$
$n_\ce{O2} = \frac{800}{32.0} = 25.000$
Here I calculate the decimal amounts rather than carrying fractions. Since it would seem that the problem has 3(?) significant figures, I carried two extra for the intermediate calculation.
Stoichiometric Quantities
For $\ce{Fe}$:
$\frac{n_\ce{FeS2}}{2} = \frac{5.0000}{2} = 2.5000 $
For $\ce{O2}$
$\frac{n_\ce{O2}}{11/2} = \frac{25.000}{11/2} = \frac{50.000}{11} =4.4545$
Thus $\ce{FeS2}$ is limiting reagent
Here I could have reached the conclusion in my head, but I do the math anyway so that I don't make a stupid mistake and also so that I can check the problem easily.
Rewrite the chemical equation
Since we now know that $\ce{FeS2}$ is the limiting reagent, rewrite the chemical equation in terms of $\ce{FeS2}$.
$\ce{FeS2 + 11/4 O2 -> 1/2Fe2O3 + 2 SO2}$
Now for each mole of $\ce{FeS2}$:
- You use 11/4 mole $\ce{O2}$ so $\ce{O2}$ remaining is:
$25.0000 - (5.0000*11/4) = 25.0000 - 13.75 = 11.25$
- You get 1/2 mole $\ce{Fe2O3}$ so:
$5.0000 / 2 = 2.50$
- You get 2 mole $\ce{SO2}$ so:
$5.0000 * 2 = 10.0$
Now for a final check I note that at the start there were 25.0 moles of $\ce{O2}$, but I have only 11.25 left. That is good since some was consumed. If I had ended up with more oxygen I obviously would have a mistake...