1. The substance that limits the amount of product that can be made
Limiting Reagents
The substance that limits the amount of product that can be made
How many people can you invite to your dinner party? What is the limiting reagent?

2. Sample Problem
What is the limiting reactant when 7.0 g copper react with 25.0 g of silver nitrate?
2AgNO Cu Cu(NO3) Ag
Knowns:
Mass of Cu = 7.0 g
Mass of AgNO3 = 25.0 g
2 mol AgNO3 = 1 mol Cu
Unknowns:
Limiting Reagent

3. 1. Find the number of moles of each reactant you HAVE
g AgNO3 = ? mol AgNO3
g Cu = ? mol Cu
you HAVE
= mol AgNO3
1 mol AgNO3
25.0 g AgNO3
? mol AgNO3 =
170 g AgNO3
you HAVE
= 0.11 mol Cu
1 mol Cu
7.0 g Cu
? mol Cu =
63.5 g Cu

4. 2. Take moles of the 1st reactant and determine the number of moles of the second reactant you NEED. (Use mole ratio from balanced equation.)
mol AgNO3 = ? mol Cu
2AgNO Cu Cu(NO3) Ag
you NEED
= mol Cu
1 mol Cu
0.147 mol AgNO3
2 mol AgNO3

5. Do you have enough moles of Cu? YES!
Just to check, do you have enough moles of AgNO3?
NO!
you NEED
= 0.22 mol AgNO3
0.11 mol Cu
2 mol AgNO3
1 mol Cu
So, what is the limiting reagent?
AgNO3
(you NEED 0.22 mol and you only HAVE mol)

6. Sample Problem
What is the limiting reactant when 8.00g HCl reacts with 5.0g Zinc.
2HCl + Zn H2 + ZnCl2
Knowns:
Mass of HCl = 6.00 g
Mass of Zn = 5.0 g
2 mol HCl = 1 mol Zn
Unknowns:
Limiting Reagent

7. 1. Find the number of moles of each reactant you HAVE
g HCl = ? mol HCl
g Zn = ? mol Zn
you HAVE
= 0.22 mol HCl
1 mol HCl
8.00 g HCl
? mol HCl =
36 g HCl
you HAVE
= mol Zn
1 mol Zn
5.0 g Zn
? mol Zn =
65 g Zn

8. 2. Take moles of the 1st reactant and determine the number of moles of the second reactant you NEED. (Use mole ratio from balanced equation.)
mol HCl = ? mol Zn
2HCl + Zn H2 + ZnCl2
you NEED
= 0.11 mol Zn
1 mol Zn
0.22 mol HCl
2 mol HCl

9. Do you have enough moles of Zn?
So, Zinc is the limiting reagent
So…how much ZnCl2 will you form?

10. Percent Yield
When 84.8g of iron (III) oxide reacts with an excess of carbon monoxide, 54.3g of iron is produced.
Fe2O CO Fe + 3CO2
1. Find theoretical yield (change grams of reactant that you’re starting with to grams of the product that you want to find)
84.8 g Fe2O3
1 mol Fe2O3
2 mol Fe
56 g Fe
= 59.4 g Fe
160 g Fe2O3
1 mol Fe2O3
1 mol Fe

11. % Yield = (actual / theoretical) * 100%
54.3 g Fe
= 91.4% Fe recovered
% Yield =
* 100%
59.4 g Fe