Determining amounts of product and excess reagent remaining post reaction


> 
> $$\ce{2FeS2 + 11/2 O2 -> Fe2O3 + 4 SO2}$$
> 
> 
> 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.
> 
> 
> 


### 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.


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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...*


 1