1. Chemistry
Limiting Reactions

2. Limiting Reactions
To drive a car, you need fuel and oxygen. When the fuel runs out, you can’t go any further even though there is still plenty of oxygen! (What is the limiting reactant?)

3. The Limiting Reactant Forms the Least Amount of Product
The limiting reactant is the substance that controls the amount of the product that can be formed from the reaction.
The excess reactant is the substance that is not used up completely in a reaction.

4. What Substance is the Limiting Reactant in this Reaction?

5. What is the Limiting Reactant in this Reaction?

6. Limiting Reactant Problem
If you have two GIVENs, it MUST be a limiting reactant problem.
Solve the stoichiometry problem twice—once for each GIVEN.
The correct answer is the SMALLEST answer—Why? Because the reaction will stop once one of the reactants is gone.

7. Example Problem
How many grams of Cu can be formed when g of Fe reacts with g of CuCl2?
2 Fe CuCl2 → 3 Cu + 2FeCl3
Solve the Stoichiometry problem twice, once for Fe and once for CuCl2.

8. Example Problem
How many grams of Cu can be formed when g of Fe reacts with g of CuCl2?
2 Fe + 3 CuCl2 → 3 Cu + 2FeCl3
167.4g Fe 1 mol Fe 3 mol Cu g Cu g Cu
55.8 g Fe 2 mol Fe 1 mol Cu
399.2g CuCl2 1 mol 3 mol Cu g Cu g Cu
mol mol Cu
CuCl2 CuCl2
This is the smallest! It is the answer!

9. Excess Reactant
To find how much excess you have, you must take the limiting reactant (Solve for that first!), then solve for the excess reactant.

10. Example Problem for Excess Reactant
In the previous problem, you determined that CuCl2 was the limiting reactant.
How did you determine this???
Because the amount of Cu formed by Fe and CuCl2 was limited by the CuCl2.
It was only g Cu from CuCl2 (compared to g from the Fe)
The Limiting Reactant is CuCl2; The Excess Reactant is Fe.

11. Example Problem for Excess Reactant:
How many grams of excess Fe is left over from the previous problem?
NOTE: Since CuCl2 was the limiting reactant, the amount of CuCl2 becomes the GIVEN:
399.2 g CuCl2 1 mol 2 mol Fe g Fe g Fe
134.5 g 3 mol mol Fe needed
CuCl2 CuCl2
We AREN’T Done Yet!!!!

12. Example Problem: Excess Reactant
The stoichiometry problem shows that you needed g of Fe. Since you started with g of Fe, all you need to do is subtract the two to find out how much excess you have.
167.4 g Fe (Given in original problem, step 1)
g Fe (Needed)
=57.0 g Fe (excess)

13. Percent Yield: Finding % Yield: Actual Yield X 100= % yield
Theoretical Yield

14. Actual Yield
Actual Yield Must be found experimentally. You must do the experiment yourself or you will be given the information. You will see words like: produced and formed in a problem that is giving you the actual yield.

15. Theoretical Yield
Theoretical Yield is done using stoichiometry. You would take the given and find out how much of the product should be formed.
In our problem, the yield of g Cu is the Theoretical Yield:
399.2g CuCl2 1 mol 3 mol Cu g Cu g Cu
mol mol Cu ↑
CuCl2 CuCl2
This is the Theoretical Yield because it is the smallest