1. Using the Balanced Equation
Stoichiometry
Using the Balanced Equation

2. What does the balanced equation really mean?
2H2 + O2  2H2O
Particles:
Moles:
2 molecules H molecule O2  2 molecules H2O
2 moles H mole O2  2 moles H2O
Which is more useful?
moles

3. 2H2 + O2  2H2O
Like a recipe, the balanced equation tells you in what ratios the ingredients must be mixed.
Therefore, for every 2 moles of H2 you will need 1 mole of O2.
Mathematically, this can be written
2 moles H2 = 1 mole O2
What can you do with equalities?
Dimensional Analysis

4. 2H2 + O2  2H2O
How many conversion factors (called mole ratios) can be written from this 1 equation?
2 moles H2 = 1 mole O2
2 moles H2 = 2 mole H2O
1 mole O2 = 2 moles H2O

5. How many moles of O2 are needed to react with 6.4 mol H2?
2H2 + O2  2H2O
How many moles of O2 are needed to react with 6.4 mol H2? =
6.4 mol H2 1
mol O2
3.2
mol O2 2
mol H2

6. How many moles of H2O are produced from 6.4 mol H2?
2H2 + O2  2H2O
How many moles of H2O are produced from 6.4 mol H2? =
6.4 mol H2 2
mol H2O
6.4
mol H2O 2
mol H2

7. Multi-step Stoichiometry
EVERY stoichiometry problem involves moving from 1 substance to another!
How many molecules of H2 does it take to make 647 g of H2O?
2H2 + O2  2H2O
? molecules
647 g
The only way to move from 1 substance to another is the mole ratio!

8. Multi-step Stoichiometry
You have three tools for stoichiometry problems!
Tools of Stoichiometry
1. Mole Ratio balanced equation coefficients moles A  moles B
2. Molar Mass periodic table mass = 1 mol mass A (g)  1 mole A
Avogadro's #  1023 particles = 1 mol particles particles A  1 mole A
Volume L at STP mole A  22.4 L of A
Every stoichiometry problem uses the mole ratio!!!

9. Multi-step Stoichiometry
To answer the original question, you need to use all 3 tools.
How many molecules of H2 does it take to make 647 g of H2O?
2H2 + O2  2H2O
647 g
? molecules
This problem not only involves changing substances (mole ratio), but also mass (molar mass) and molecules (Avogadro’s number).

10. 2H2 + O2  2H2O
? molecules
647 g
First, you want to set up a dimensional analysis problem. Put your given and target in the equation.
647 g H2O
= ? molecules H2

11. Next, convert g of H2O to moles of H2O using molar mass.
2H2 + O2  2H2O
? molecules
647 g
Next, convert g of H2O to moles of H2O using molar mass.
647 g H2O
= ? molecules H2 1
mol H2O
g H2O

12. Next, EVERY problem must have a mole ratio step, so put that next!
2H2 + O2  2H2O
? molecules
647 g
Next, EVERY problem must have a mole ratio step, so put that next!
647 g H2O
= ? molecules H2 2
mol H2 1
mol H2O 2
mol H2O
g H2O
You don’t want H2O, so put moles H2O in the bottom of the middle step (remember you have to be in moles to move from 1 substance to another).
You want to change to H2, so put moles of H2 on top (mole ratio is moles to moles).
The numbers come from the balanced equation!

13. Answer = 2.161025 molecules H2 2H2 + O2  2H2O
647 g
Since, we need molecules and the middle step gets us to moles H2, we need to use Avogadro’s number.
647 g H2O
= ? molecules H2
mol H2O
mol H2 2
g H2O 1
6.021023
molecule H2 1
mol H2
Answer = 2.161025 molecules H2

14. Calculating Stoichiometric Problems
Balance the equation
Convert given to moles
Set up mole ratios.
Use mole ratios to calculate moles of desired chemical.
Convert moles to target.

15. Another Example
How many grams of O2 are needed to use up 13.6 moles of H2?
2H2 + O2  2H2O
13.6 mol
g ?
g O2
13.6 mol H2 1
mol O2
= g O2
218 2
mol H2 1
mol O2

16. 12.3 Limiting Reactants Cake Recipe 2 cups flour 1 cup butter
1tsp. salt
1/2 cup sugar
If I have 6 cups of flour, 4 cups of butter, 112 tsp. of salt, and 429 cups of sugar, how many cakes can I make?

17. 12.3 Limiting Reactants Cake Recipe You Have in the Cupboard
2 cups flour 6 cups flour
1 cup butter 4 cup butter
1tsp. Salt tsp. salt
1/2 cup sugar cups sugar
6 cups flour 1
cakes
= cakes 3 2
cups flour
4 cups butter 1
cakes
= cakes 4 1
cups butter

18. 12.3 Limiting Reactants Cake Recipe You Have in the Cupboard
2 cups flour 6 cups flour
1 cup butter 4 cup butter
1tsp. Salt tsp. salt
1/2 cup sugar cups sugar
112 tsp. salt 1
cakes
= cakes
112 1
tsp. salt 1
cakes
429 cups sugar
= cakes
858 .5
cups sugar

19. It also controls how much of the other ingredients we need!
12.3 Limiting Reactants
Flour controls or limits how many cakes we can make because after we make 3 cakes we run out of flour.
It also controls how much of the other ingredients we need!

20. 12.3 Limiting Reactants How much butter is left over?
First find how much was used and then subtract how much is left.
FLOUR controls how much we use.
6 cups flour 1
cup butter
= cups butter 3 2
cups flour
4 cups butter – 3 cups butter = 1 cup butter left

21. Chemical reactions work the same way!!!!
12.3 Limiting Reactants
Limiting reactant - the reactant that makes the least. It controls everything about the reaction because when it is used up the reaction stops.
Chemical reactions work the same way!!!!

22. 12.3 Limiting Reactants 2H2 + O2  2H2O 2.50 g 3.40 g
How much water can be made?
2.50 g H2 1
mol H2 2
mol H2O
g H2O
You must find which reactant controls (limiting reactant) the reaction.
= g H2O
22.3
g H2 2
mol H2 1
mol H2O
3.40 g O2 1
mol O2 2
mol H2O
g H2O
= g H2O
3.83 1
mol O2 1
mol H2O
g O2
O2 is therefore the limiting reactant and 3.83 g of water is produced!

23. 12.3 Limiting Reactants 2H2 + O2  2H2O How much H2 is left over?
2.50 g g
How much H2 is left over?
Find out how much H2 was used up by the 3.40 g O2
Then subtract that amount from 2.50 g to find out how much H2 is left

24. 12.3 Limiting Reactants 2H2 + O2  2H2O How much H2 is left over?
2.50 g g
How much H2 is left over?
3.40 g O2 1
mol O2 2
mol H2
g H2
= g H2
.428 1
mol O2 1
mol H2
g O2
2.50 g
.428 g -
2.07 g H2 left over

25. 12.4 Percent Yield

26. We made theoretical prediction of how much could be made.
Percent Yield
When we calculated that 3.83 g of H2O could be made in the previous sample problem, we didn’t actually perform the reaction and measure the resulting water.
We made theoretical prediction of how much could be made.
Actual Yield – do the reaction to find the yield
Theoretical Yield – do the math to predict the yield

27. Under ideal conditions, this is exactly what will happen.
12.4 Percent Yield
When doing stoichiometric calculations, you are calculating what is known as a theoretical yield - the maximum amount that could be produced during the reaction.
Under ideal conditions, this is exactly what will happen.
In reality, however, the actual yield will be less than what is predicted.