1. Limiting Reactant/Reagent Problems
Consider the balanced equation for the production of ammonia:
N2 + 3 H2  2NH3
If you have 2 moles of nitrogen and 3 moles of hydrogen, which reactant will be used up?
How much ammonia will you be able to make?
How much extra (and of which chemical) will you have?

2. Limiting Reactant/Reagent Problems
Spend 2 minutes discussing your thought process for the previous question with your neighbor.
How did you decide which reactant would be used up?
Write some simple math to explain how you figured out how much of the excess you would have left over. (label what your numbers mean)

3. Limiting Reactant/Reagent Problems p.7 in your notes
A stoichiometry problem where quantities of reactants are not available in the EXACT molar ratio.
This means one reactant is “in excess” and the other is the “limiting reactant.”
You will be given the mass/volume/moles of more than one reactant.
You find the actual theoretical amount of product(s) produced by determining which reactant LIMITS the reaction.

4. Steps for Identifying Limiting Reactants:
Write the balanced chemical equation.
Solve two (or more) stoichiometry problems – one for each reactant given.
Compare the results of step #2 above. The reactant that would produce the smaller amount of product is the limiting reactant; the other reactant is in excess.

5. OR Write the balanced chemical equation.
Find the number of moles of each reactant given. (convert from grams, volume, or particles if necessary)
Compare the ratio of moles from step #2 above to the ratio of moles in the balanced equation from step #1.
Use the limiting reactant to do one full stoichiometry problem to determine the theoretical yield from this reaction.

6. Example Problem:
What is the limiting reactant when 1.5 moles of hydrogen react with 0.5 moles of oxygen to produce water?
2H2 + O2  2H2O
1.5 mol H2 x 1 mol O2 = 0.75 mol O2
2 mol H2
This means, to completely react with 1.5 mol of hydrogen, you would need 0.75 mol of oxygen.

7. You don’t have that much oxygen!
Which means oxygen is the limiting reagent.
You will use up all of your oxygen and have hydrogen left over.
How much water will you be able to make?
0.5 mol O2 x 2 mol H2O = 1.0 mol H2O
1 mol O2
You could convert this to grams, particles, etc if necessary.

8. Example Problem
When 2.60 g of hydrogen react with g of oxygen, what mass of water will be produced? What is the limiting reactant?
First, find the moles of each reactant.
2.60 g H2 x 1 mol H2 = 1.29 mol H2
2.016 g H2
10.20 g O2 x 1 mol O2 = mol O2
32.00 g O2

9. Next, relate them to one another using the mole ratio
1.29 mol H2 x 1 mol O2 = mol O2
2 mol H2
This is how much oxygen you would need to exactly react with the hydrogen.
Do you have that much oxygen?
NO!
Oxygen is again the limiting reactant.

10. Now, figure out how many grams of water you can make.
Begin with oxygen, because we know we will use all of that up.
10.20 g O2 x 1 mol O2 x 2 mol H2O x g H2O = g H2O
32.00 g O mol O mol H2O