[StillLearning]

The equation:

2NO + 5H₂  2NH₃ + 2H₂O

86.3 grams of NO
25.6 grams of H₂

I converted NO and H₂ to moles and got 2.876 and 12.8 respectively.  Can someone explain what the next steps are.  I'm failing to understand the logic in my book.

[Yggdrasil]

To figure out which reagent is limiting, divide the number of moles of each reagent by its stoichiometric coefficient in the balanced chemical reaction.  Whichever number is smaller after this is the limiting reagent.

[Samwich717]

Now how would you find the amount of excess reagent with this problem?

[StillLearning]

To figure out which reagent is limiting, divide the number of moles of each reagent by its stoichiometric coefficient in the balanced chemical reaction.  Whichever number is smaller after this is the limiting reagent.

Looking at Nitrogen Oxide first:  I have 2.876 moles.  Which one of the following is correct?

a)  2.876 mol NO X (2 mol NH₃ / 2 mol NO)   OR
b)  2.876 mol NO X (2 mol H₂O / 2 mol NO)

In this case the coefficients of water and ammonia are the same so the answer would be the same.  But if they were different, which one would I use?  I'm still a little confused.  

[Samwich717]

a)  2.876 mol NO X (2 mol NH₃ / 2 mol NO)   OR
b)  2.876 mol NO X (2 mol H₂O / 2 mol NO)

I think you are just looking for the precipitate so it would be the Ammonia coefficient in the numerator.

[Samwich717]

Oh and when you write this problem out I find it a time-saver to assign each molecule a different letter.  This makes it so that when you multiply by your converters, you can just write: molA, molB, gA, gB, etc.

[StillLearning]

a)  2.876 mol NO X (2 mol NH₃ / 2 mol NO)   OR
b)  2.876 mol NO X (2 mol H₂O / 2 mol NO)

I think you are just looking for the precipitate so it would be the Ammonia coefficient in the numerator.

Yes, i'm looking for the theoretital yield of Ammonia.  Ok, so in all theoretital yield problems, it will say which compound i'm looking for, and that's the one I take the coefficients from?  That sounds SO simple, yet the book does not not emphasize this at all.  Am I understanding correctly what you're saying?

[Samwich717]

You have to know which one is insoluble in water.  That would be your precipitate.

[Yggdrasil]

Looking at Nitrogen Oxide first:  I have 2.876 moles.  Which one of the following is correct?

a)  2.876 mol NO X (2 mol NH₃ / 2 mol NO)   OR
b)  2.876 mol NO X (2 mol H₂O / 2 mol NO)

In this case the coefficients of water and ammonia are the same so the answer would be the same.  But if they were different, which one would I use?  I'm still a little confused.  

You're comparing the wrong values.  What you want to compare is the amount of ammonia produced if you assume NO is the limiting reagent (a) and the amount of ammonia produced if you assume H₂ is the limiting reagent (b):

a) 2.876 mol NO * (2 mol NH₃ / 2 mol NO)
vs
b) 12.8 mol H₂ * (2 mol NH₃ / 5 mol H₂)

[StillLearning]

You have to know which one is insoluble in water.  That would be your precipitate.

The books says nothing about that.  The entire problem is written at the top of this thread.  I used ammonia and I got the right answer.  I have another question now though.

The book says that once I convert the reactants from grams to moles, then whichever is smaller is the limiting reagent.  It seems to me that you have to use the stoichiometric coefficients before you can determine the limiting reagent.  Am I correct?  

[Samwich717]

Yeah.  Once you convert it into moles of the precipitate then you have to multiply by the molecular weight of said precipitate.  The path I use to go from gAgB is      gAAmolBmolgB.....if that makes sense.

[StillLearning]

Yeah.  Once you convert it into moles of the precipitate then you have to multiply by the molecular weight of said precipitate.  The path I use to go from gA-->gB is      gA-->Amol-->Bmol-->gB.....if that makes sense.

Ok, I think i'm understanding this now.  Thanks to all of you who provided assistance!   

[Samwich717]

Glad to help 

[Yggdrasil]

You have to know which one is insoluble in water.  That would be your precipitate.

The books says nothing about that.  The entire problem is written at the top of this thread.  I used ammonia and I got the right answer.  I have another question now though.

You are correct.  This problem has nothing to do with precipitates.

The book says that once I convert the reactants from grams to moles, then whichever is smaller is the limiting reagent.  It seems to me that you have to use the stoichiometric coefficients before you can determine the limiting reagent.  Am I correct?  

Yes.