1. Limiting Reagents Excess Reagents Percent yield

2. Limiting Reagents
The limiting reagent is the reactant that runs out first
Limits the amount of product that can be formed
Take both reactants to the same product
The answer is always the smallest amount of product

3. 4 KO2(s) + 2H2O(l) g 4KOH + 3O2(g)
Example
Potassium superoxide, KO2 is used in rebreathing gas masks to generate oxygen.
4 KO2(s) + 2H2O(l) g 4KOH + 3O2(g)
If a reaction vessel contains 0.25 mol KO2 and 0.15 mol H2O what is the limiting reagent? How many moles of oxygen can be produced?
Take both starting amounts to moles of oxygen, the limiting reagent is the one that produced the smallest amount.
0.25 mol KO2 | 3 mol O2 = 0.19 mol O2
| 4 mol KO2
0.15 mol H2O | 3 mol O2 = 0.23 mol O2
| 2 mol H2O
This is the smallest amount of O2 therefore KO2 is the limiting reagent and this is the amount of O2 produced.

4. Excess Reagent
The reagent (reactant) that is left over once the reaction stops.
Extra reactant
To determine the amount of left over (excess) reagent
Take both reactants to same product.
Find the difference between the two amounts of products
Use that a the starting number and convert back to the excess reagent.

5. Example
Potassium superoxide, KO2 is used in rebreathing gas masks to generate oxygen. 4 KO2(s) + 2H2O(l) g 4KOH + 3O2(g) If a reaction vessel contains 0.25 mol KO2 and 0.15 mol H2O what is the excess reagent? How many moles were in excess?
Take both starting amounts to moles of oxygen, find the difference between the two amounts, convert back to the excess reagent
0.25 mol KO2 | 3 mol O2 = 0.19 mol O2
| 4 mol KO2
0.15 mol H2O | 3 mol O2 = 0.23 mol O2
| 2 mol H2O
0.23 – .19 molO2 | 2 mol H2O = 0.03 mol H2O
| 3 mol O2
This is the smallest amount of O2 therefore KO2 is the limiting reagent and H2O is the excess reagent.
This is the amount of excess

6. Percent Yield Is found using the formula: Actual Theoretical *100
The actual yield is what is actually produced in the lab
The theoretical yield is what is supposed to be formed based on stoichiometry
*100

7. Example
Potassium superoxide, KO2 is used in rebreathing gas masks to generate oxygen. 4 KO2(s) + 2H2O(l) g 4KOH + 3O2(g) If a reaction vessel contains 0.25 mol KO2 and 0.15 mol H2O what is the limiting reagent? How many moles of oxygen can be produced? In the lab 0.15 mol of O2 were actually produced, what is the percent yield?
Take both starting amounts to moles of oxygen, the limiting reagent is the one that produced the smallest amount. Divide actual yield by the theoretical yield based on limiting reagent.
0.25 mol KO2 | 3 mol O2 = 0.19 mol O2
| 4 mol KO2
0.15 mol H2O | 3 mol O2 = 0.23 mol O2
| 2 mol H2O
For % yield, 0.15 / 0.19 *100 = 79%
This is not the greatest % yield, this of the percentage as your test score.
This is the smallest amount of O2 therefore KO2 is the limiting reagent and this is the amount of O2 produced.