[JonathanEyoon]

Dear JonathanEyoon;

Sorry!,  ─   
I got 0.15625 Moles, because the MW of O₂ is 32.0 and not 16.0!
With an Excess of H₂ you can build 2 * 0.15625 Moles Water (from B.)).
How many Water will this now be?


Good Luck!
                    ARGOS⁺⁺

So 2 x 0.15625 is gonna be .3125 moles of water?

[ARGOS++]

Dear JonathanEyoon;

Exactly!,  ─    And how many Grams will that finally be?


Good Luck!
                    ARGOS⁺⁺

[JonathanEyoon]

also in simple terms, what does it mean that their is an excess of H2? Excess usually means it's left over.  Which should mean it should have no bearing at all to the problem right?

[JonathanEyoon]

Dear JonathanEyoon;

Exactly!,  ─    And how many Grams will that finally be?


Good Luck!
                    ARGOS⁺⁺

So the final answer to this problem is 5.625 grams?

[Arkcon]

also in simple terms, what does it mean that their is an excess of H2? Excess usually means it's left over.  Which should mean it should have no bearing at all to the problem right?

Yes, either they give you a stoichiometric amount, or something is in excess.

[ARGOS++]

Dear JonathanEyoon;

Bravo, You did it!

Yes, an Excess means that there is more then enough from that stuff, and so its amount will in no kind influence the Amount of the Product.
(Please remember the Recipe!)

Good Luck!
                    ARGOS⁺⁺

[JonathanEyoon]

I still don't understand how to do these types of problems    Can we try another one?

How about this one.

If 294 grams of FeS2 is allowed to react with 176 grams of O2 according to the following equation, how many grams of Fe2O3 are produced?


The equation: FeS2 + 02 -----> Fe203 + SO2

[ARGOS++]

Dear JonathanEyoon;

Will you try to do A.) and B.) for this Reaction?               

Good Luck!
                    ARGOS⁺⁺

[JonathanEyoon]

Dear JonathanEyoon;

Will you try to do A.) and B.) for this Reaction?               

Good Luck!
                    ARGOS⁺⁺

Ok so part A and B is to write the equation and balance it out right?


So the balanced equation should be

4FeS2 + 11O2 ------> 2Fe2O3 + 8SO2?  What should I do from here?

[Arkcon]

Great, you know the units of your balanced equation are given in moles, you've been given reactants in units of grams.  You have to convert.  Have you been given both reactants in stoichiometric amounts, or is one in excess?

[ARGOS++]

Dear JonathanEyoon;

(Hello again.)

Very well done!,    ─   
You know after A.) and B.), now C.) must follow:
So we need for C₁.)  all MW’s, and you write it just below each Educt and Product of your well found  “Reaction Equation”.
After that you write for C₂. )  all known masses of the Educts, but now translated into the corresponding Number of Moles under the MW’s.

Would you like to tell me the results of your line C₂. )?

Good Luck!
                    ARGOS⁺⁺

[JonathanEyoon]

Dear JonathanEyoon;

(Hello again.)

Very well done!,    ─   
You know after A.) and B.), now C.) must follow:
So we need for C₁.)  all MW’s, and you write it just below each Educt and Product of your well found  “Reaction Equation”.
After that you write for C₂. )  all known masses of the Educts, but now translated into the corresponding Number of Moles under the MW’s.

Would you like to tell me the results of your line C₂. )?

Good Luck!
                    ARGOS⁺⁺

Ok so  294 grams would be for FeS2 and 176 grams would be for the O2. So i change these two into moles? If that is so the

294 grams FeS2 = 294 / 120 = 2.45 Moles of FeS2

176 Grams O2 = 176 / 32 = 5.5 moles of O2

where do I go from here 

[ARGOS++]

Dear JonathanEyoon;

Excelent!,    ─   

Now you have to find the limiting Reagent and that we do in line D.).
For that you must “put” your calculated Moles from line C₂. ) on the same Denominator (to the same Basis) using the found indices from B.).
For Example:  Divide your 2.45 Moles by 4 (the Indices from B.) for FeS₂), and so on.

The lowest Number on line D.) tells you the limiting Reagent; with other words: All other Reagents of the Equation are in Excess and so you can calculate from the Moles of the limiting Reagent how many Moles of your asked Product you can build. Use again the Indices from line B.).
Now you have only to translate the Number of Moles back to Grams.

Do you think you can make the Desission and try to calculate ahead?
(I think I need now an eye full of sleep.)

Good Luck!
                    ARGOS⁺⁺

[JonathanEyoon]

Dear JonathanEyoon;

Excelent!,    ─   

Now you have to find the limiting Reagent and that we do in line D.).
For that you must “put” your calculated Moles from line C₂. ) on the same Denominator (to the same Basis) using the found indices from B.).
For Example:  Divide your 2.45 Moles by 4 (the Indices from B.) for FeS₂), and so on.

The lowest Number on line D.) tells you the limiting Reagent; with other words: All other Reagents of the Equation are in Excess and so you can calculate from the Moles of the limiting Reagent how many Moles of your asked Product you can build. Use again the Indices from line B.).
Now you have only to translate the Number of Moles back to Grams.

Do you think you can make the Desission and try to calculate ahead?
(I think I need now an eye full of sleep.)

Good Luck!
                    ARGOS⁺⁺

haha I'm not too sure how to proceed with this 

[JonathanEyoon]

Ok, i think I got the answer.  Is the answer 98 Grams?