[Sev]

No, too low.

You have 2.45mol FeS₂ and 5.5mol O₂.  You have to find which reagent is limiting - consumed first.

From the rxn eqn 4 mole FeS₂ reacts with 11 mole O₂.  That means 2 moles of FeS₂ reacts with 5.5 moles of O₂ (not 2.45!).  So O₂ is the limiting reagent (it is completely consumed in the reaction).

Now, look at the coefficients from the rxn eqn.  How many moles of Fe₂O₃ are formed from 5.5 mole of O₂?

[JonathanEyoon]

so what should the answer be?  I'm asking because this way i can work backwards to towards something and try to understand it myself while doing it.  I'm still so lost on how to do this stuff with the explanations on here.  Can't understand some of the stuff you guys are saying.  So I'm gonna try to work towards the answer or backwards from it and try to figure out a simple way to do this stuff since i'm sure their is a way method to do all of this.  Thanks and hope to hear it soon

[Sev]

You end up with 1 mole of Fe₂O₃.  Do you know what a limiting reagent is?

[JonathanEyoon]

no idea what that is

[Sev]

The limiting reagent is completely consumed in the rxn.  In this case, it is O₂.
FeS₂ is in excess - there is still some left after all of O₂ has reacted.

See: http://www.chembuddy.com/?left=balancing-stoichiometry&right=limiting-reagents 

[Borek]

First of all - on the same site - read how to read balanced reaction equation and what it means:

http://www.chembuddy.com/?left=balancing-stoichiometry&right=stoichiometric-calculations

[ARGOS++]

Dear JonathanEyoon;

Let’s collect together in the Diagram what you have calculated till jet:

A.)    Reaction:       FeS₂   +          O₂    ----->    Fe₂O₃   +     SO₂. 
B.)    Balancing¹⁾:  4FeS₂   +      11O₂    ----->   2Fe₂O₃  +    8SO₂.   

C₁.)   MW’s:         120.0 g/m      32.0 g/m            158.7 g/m     64.0 g/m
C_m.)  Masses:       294.0  g        176.0 g
C₂.)   Moles:         2.45 m          5.50  m

Now I was asking for the Decision following exactly my Example above:
You should result in:   2.45 m /   4  =  0.6125 ;    and  -
                              5.50 m /  11  =  0.5000.
As you don’t know any other Masses or Moles of any Reactant, your line D.) must look like:
D.)   Multiplier:      0.6125          0.5000

Now it should be clear that you can only produce the Amount, what the smallest Multiplier telling you, all other is in Excess. All Excess can NOT react, because it has NO “Reaction Partner”!
So you can copy the smallest Multiplier to all your missing/(not known till yet) Multipliers of your “Unknowns”  on line D. ).  That must result in:   
D.)   Multiplier:      0.6125           0.5000                   0.50             0.50 

The final step is now to complete lines C₂.) and  C_m.) for all “Unknowns” from the bottom up, because you have figured out the Multipliers for all.
That gives you:   
C₂.)   Moles:         2.45 m           5.5  m           2 * 0.50 m    8 * 0.50m.

And to get the final masses you have only to back-translate the new values from line C₂.) into masses, to complete also line C_m.) as you did it for the water in the first Example. 

Yet you know how much of each Product you were able to produce.

¹⁾ For “How to Balance a Chemical Reaction” you may read on: "Balancing Chemical Equations”

Will you try to complete your Diagram this way and tell me your final results? 
(Finally we put the whole Diagram together.)

Good Luck!
                    ARGOS⁺⁺

[ARGOS++]

Dear JonathanEyoon;

Maybe you’re anyway coming once back, but it’s also for all who try to learn from this Topic!
(I also like to “hold” my contracts.)

Here is the resulting Diagram for your Stoichiometry Problem:

A.)    Reaction:       FeS₂   +          O₂    ----->    Fe₂O₃   +     SO₂. 
B.)    Balancing:     4FeS₂   +     11O₂    ----->   2Fe₂O₃  +    8SO₂.   

C₁.)   MW’s:         120.0 g/m      32.0 g/m            158.7 g/m     64.0 g/m
C_m.)  Masses:       294.0  g        176.0 g                  158.7 g          256.0 g
C₂.)   Moles:         2.45 m          5.50  m          2 * 0.50 m    8 * 0.50 m

D.)   Multiplier:      0.6125           0.5000                   0.50             0.50 

As you can see: All Mass results of you Stoichiometry Problem are located on the C_m.)  line.

Surprise: Also all Excesses can be calculated in the same easy manner: 
Add on the left an additional Column for each Excess and calculate from the bottom to the top.

For your particular Stoichiometry Problem it looks like:
A.)    Reaction:       Excess    (FeS₂)
B.)    Balancing:     

C₁.)   MW’s:           120.0 g/m     
C_m.)  Masses:           54.0  g        
C₂.)   Moles:     4 * 0.1125 m

D.)    Multiplier:       0.1125  (= 0.6125 – 0.5000)

Now the Diagram gives you the “whole” Overview about your Stoichiometry Problem and can easy be controlled/validated, because the line C_m.) must end in identity for the left and the right side, except all (multiplied) small rounding errors.
(Remember: Excesses you have to subtract on the left Side!)

I hope it may be of help.

Good Luck!
                    ARGOS⁺⁺

[Borek]

http://www.chembuddy.com/?left=balancing-stoichiometry&right=ratio-proportions

[ARGOS++]

Dear Mr.Borek;

Bravo!  “Chembuddy.com”,   ─   Quite a good Page with good Explanations.

But it doesn’t answers the initial Stoichiometry Problems, and it covers only a very small part of the Stoichiometry Problems at all.
It doesn’t handle “Limiting Reagent” of Stoichiometry Problems.
It doesn’t address/calculate “Excesses” of Stoichiometry Problems.
Also the control/validation of the calculations of the “whole” Stoichiometry Problems is missing.

So I still believe that the Diagram is the smallest, simplest, and most complete calculation of such and most other related Stoichiometry Problems.
And it gives you at once the “whole” Overview about your Stoichiometry Problem.

But the Page is a good start point for beginners of Stoichiometry Problems.

Good Luck!
                    ARGOS⁺⁺

[Borek]

But it doesn’t answers the initial Stoichiometry Problems, and it covers only a very small part of the Stoichiometry Problems at all.

http://www.chembuddy.com/?left=balancing-stoichiometry&right=toc

It doesn’t handle “Limiting Reagent” of Stoichiometry Problems.

http://www.chembuddy.com/?left=balancing-stoichiometry&right=limiting-reagents

[ARGOS++]

Dear Mr. Borek;

I beg your Pardon!, ─  I did not recognise the last Link on the left of your first presented Page. (Maybe I took the wrong glasses, Sorry!)

But the Diagram has still some advantages, - but your Pages have much more Explanations!
Sorry once again.

Good Luck!
                    ARGOS⁺⁺