1. Limiting Reactants

2. Limiting Reactants You have a lit candle You put a jar over the candle
What will happen?
It will go out
Why?
It runs out of oxygen

3. Limiting Reactants Limiting reactant –
A chemical reaction will stop when you run out of one of your products
Limiting reactant –
limits the extent of the reaction.
Determines the amount of product that is formed. It runs out first
Excess reactant – left over reactant

4. Limiting Reactant Problems
A limiting reactant problem can be easily identified because you have TWO givens
You are going to do 2 stoichiometry problems

5. Steps Write and balance the chemical equation
Start with given #1 and go to desired product
Start with given #2 and go to the same product
The one that formed the least amount of grams is the limiting reactant

6. Example
S8 + 4Cl2  4S2Cl2
g of S8 and g of Cl2 are combined in a flask. How much S2Cl2 will you get?
Start with given #1
g S8 x 1 mol S8 x 4 mol S2Cl2 x g S2Cl2 = g S2Cl2
256.5 g S mol S mol S2Cl2

7. Example Now do given #2 Now we have a decision to make.
S8 + 4Cl2  4S2Cl2
g of S8 and g of Cl2 are combined in a flask. How much S2Cl2 will you get?
Now do given #2
g Cl2 x 1 mol Cl2 x 4 mol S2Cl2 x g S2Cl2 = g S2Cl2
70.91 g Cl mol Cl mol S2Cl2
Now we have a decision to make.
Did we make g S2Cl2 or g S2Cl2
The answer is g.
After that amount, the reaction stopped

8. Other questions What was the limiting reactant? Cl2
What was the excess reactant? S8
How much excess did we have left over after the reaction was completed?

9. Other questions
Before we can determine how much excess was left over, we must figure out how much was used.
There are several ways to do this.
Using stoichiometry, start with how much product you formed OR
the total amount of the limiting reactant and go to the amount (g) of excess

10. Other questions This will tell you how much excess was used
Now you subtract how much you used from how much you started with to give you the amount left over

11. Other questions In the previous example
g Cl2 x 1 mol Cl2 x 1 mol S8 x g S8 = g S8
70.91 g Cl mol Cl mol S8
This tells you how much S8 was used. We want to know how much was left over
We started with g S8 and we used g S8.
Therefore, we must have g S8 left over

12. Another Example P4 + 5O2  P4O10
If we used 25.4 g P4 and 50.0 g O2 answer the following questions.
What is the limiting reactant?
What is the excess reactant?
How much P4O10 will you get?
How much excess did you use?
How much excess will you have left over?

13. Another Example What is the limiting reactant? P4
What is the excess reactant?O2
How much P4O10 will you get? g P4O10
How much excess did you use? g O2 used
How much excess will you have left over? g O2 left over

14. % Yield Remember % = (part / whole) x 100
When performing an experiment, things do not always go exactly perfect
Some product may get spilled, some may get sneezed on, or the reaction may not have gone to completion

15. % Yield
Theoretical yield –amount of product you should get if the experiment went perfectly. You get this number from stoichiometry
Actual Yield – this is what you actually got in the lab. You measure this on a balance
% yield – how close you were to the correct answer
% yield = (actual / theoretical) x 100

16. % Yield Example K2CrO4 + 2AgNO3  Ag2CrO4 + 2KNO3
What is the theoretical yield of Ag2CrO4 formed from g AgNO3
0.500 g AgNO3 x 1 mol AgNO3 x 1 mol Ag2CrO4 x g Ag2CrO4 = g
169.9 g AgNO mol AgNO mol Ag2CrO Ag2CrO4
What is the % yield if g is actually formed?
% Yield = (actual / theoretical) x 100
% Yield = (0.455 / 0.488) x 100 = 93.2%