1. Limiting Reactants
ABClark-Grubb

2. Limiting vs. Excess Excess Reactant Limiting Reactant
The reactant in a chemical reaction that limits the amount of product that can be formed.  The reaction will stop when all of the limiting reactant is consumed.
Excess Reactant
The reactant in a chemical reaction that remains when the reaction stops. The excess reactant remains because there is nothing left with which it can react.

3. Real Life Example… Limiting Reactant Excess Reactant
If there are only 8 car bodies, then only 8 cars can be made.  Likewise with chemistry, if there is only a certain amount of one reactant available for a reaction, the reaction must stop when that reactant is consumed whether or not the other reactant has been used up.

4. Example Problem #1 4 NH3+ 5 O2  4 NO + 6 H2O
A 2.00 mol sample of ammonia is mixed with 4.00 mol of oxygen. 
Which is the limiting reactant and which is in excess?

5. Step 1: Solve for moles of one product for ALL reactants
4 NH3+ 5 O2  4 NO + 6 H2O
Step 1: Solve for moles of one product for ALL reactants
NH3 is Limiting Reactant =
2.00 mol NO
2.00 mol NH3
4 mol NO
4 mol NH3
O2 is Excess Reactant =
3.20 mol NO
4.00 mol O2
4 mol NO
5 mol O2

6. Example Problem #2 4 NH3+ 5 O2  4 NO + 6 H2O
A 2.00 g sample of ammonia is mixed with 4.00 g of oxygen. 
What mass of nitrogen monoxide produced?

7. Step 1: Solve for moles of one product for ALL reactants
4 NH3+ 5 O2  4 NO(g) + 6 H2O
Step 1: Solve for moles of one product for ALL reactants
NH3 is Excess Reactant
0.117 mol NO =
2.00 g NH3
4 mol NO
1 mol NH3
17.0 g NH3
4 mol NH3
O2 is Limiting Reactant
0.100 mol NO =
4.00 g O2
1 mol O2
4 mol NO
30.0 g NO =
32.0 g O2
5 mol O2
5 mol O2
5 mol O2
1 mol NO
Once you determine the limiting reactant, use that to solve any other problems.
3.00 g NO produced

8. Example Problem #3 4 K + O2  2 K2O Step 1:
When 0.65 moles of oxygen reacts with
0.56 moles of K how many grams of potassium oxide will be produced?
4 K + O2  2 K2O
1.3 mol K2O =
Step 1:
0.65 mol O2
2 mol K2O
O2 is Excess Reactant
1 mol O2
0.28 mol K2O =
2 mol K2O
0.56 mol K
94.2 g K2O
Step 2:
4 mol K
1 mol K2O
K is Limiting Reactant
26 g K2O produced =