Determining amounts of product and excess reagent remaining post reaction $$\ce{2FeS2 + 11/2 O2 -&gt; Fe2O3 + 4 SO2}$$
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.

My Attempt
$$\frac{n_\ce{FeS2}}{2} = \frac{600/120}{2} = \frac{5}{2} \lt \frac{n_\ce{O2}}{11/2} = \frac{800/32}{11/2} = \frac{50}{11}\implies \ce{FeS2}~\text{is limiting reagent}$$
By Mole-Mole Analysis, we have
$$\frac{n_\ce{FeS2}}{2} = \frac{n_\ce{Fe2O3}}{1} = \frac{n_\ce{SO2}}{4}\implies \begin{bmatrix}n_\ce{Fe2O3} \\ n_\ce{SO2}\end{bmatrix} = \begin{bmatrix}5/2 ~\text{mol} \\ 10~\text{mol}\end{bmatrix}$$
$$\text{Excess reagent remaining} = \frac{800}{32} - \dfrac{5}{2}\cdot\dfrac{11}{2} = \pu{11.25 mol}$$
However, my answer does not match with my other classmates. What am I doing wrong?
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.