1. Stoichiometry

2. Stoichiometry Consider the chemical equation: 4NH3 + 5O2 → 6H2O + 4NO
“stochio” = Greek for element “metry” = measurement
Stoichiometry is about measuring the amounts of elements and compounds involved in a reaction.
Consider the chemical equation:
4NH3 + 5O2 → 6H2O + 4NO
There are several numbers involved.
What do they all mean?

3. Stoichiometry NH3 O2 H2O NO 1 nitrogen and 3 hydrogen atoms
Recall that Chemical formulas represent numbers of atoms
4NH3 + 5O2 → 6H2O + 4NO
NH3
1 nitrogen and 3 hydrogen atoms O2
2 oxygen atoms
H2O
2 hydrogen atoms and 1 oxygen atom NO
1 nitrogen atom and 1 oxygen atom

4. Recall that Chemical formulas have molar masses:
Stoichiometry
Recall that Chemical formulas have molar masses:
4NH3 + 5O2 → 6H2O + 4NO
NH3
17 g/mol O2
32 g/mol
H2O
18 g/mol NO
30 g/mol

5. Recall that Chemical formulas are balanced with coefficients
Stoichiometry
4NH3 + 5O2 → 6H2O + 4NO
Recall that Chemical formulas are balanced with coefficients
4 X NH3
= 4 nitrogen + 12 hydrogen
5 X O2
= 10 oxygen
6 X H2O
= 12 hydrogen + 6 oxygen
4 X NO
= 4 nitrogen + 4 oxygen

6. Stoichiometry 4NH3 + 5O2 → 6H2O + 4NO
With Stoichiometry we find out that
4 : 5 : 6 : 4
do more than just multiply atoms.
4 : 5 : 6 : 4
Are what we call a mole ratio.

7. 4 : 5 : 6 : 4 Stoichiometry 4NH3 + 5O2 → 6H2O + 4NO OR
Can mean either:
4 molecules of NH3 react with 5 molecules of O2
to produce 6 molecules of H2O and 4 molecules of NO OR
4 moles of NH3 react with 5 moles of O2
to produce 6 moles of H2O and 4 moles of NO

8. Question 1
4NH3 + 5O2 → 6H2O + 4NO
How many moles of H2O are produced if moles of O2 are used?
6 mol H2O
5 mol O2 =
2.40 mol H2O
2.00 mol O2
*Notice that a correctly balanced equation is essential to get the right answer

9. Question 2
4 NH O2 → 6 H2O NO
How many moles of NO are produced in the reaction if 15 mol of H2O are also produced?
4 mol NO
6 mol H2O
15 mol H2O =
10. mol NO

10. Stoichiometry Path
We always use the same type of information to make the jumps between steps:
grams (x) ↔ moles (x) ↔ moles (y) ↔ grams (y)
Molar mass of y
Molar mass of x
Mole ratio from balanced equation

11. We will be converting: Moles to Moles Grams to Grams
Stoichiometry Path
We will be converting:
Moles to Moles
Grams to Grams

12. Practice 1) How many moles of H2O can be made using 1.6 mol NH3?
Try these on your own NH3 + 5 O2 → 6 H2O + 4 NO
1) How many moles of H2O can be made using 1.6 mol NH3?

13. Answers 2.4 mol H2O 4 NH3 + 5 O2 → 6 H2O + 4 NO 1) 6 mol H2O 4 mol NH3x
2.4 mol H2O =
1.6 mol NH3
How many moles of H2O can be made using 1.6 mol NH3?

14. Practice 2) How many moles of NO can be made using 1.6 mol NH3?
Try these on your own NH3 + 5 O2 → 6 H2O + 4 NO
2) How many moles of NO can be made using 1.6 mol NH3?

15. Answers 1.6 mol NO 4 NH3 + 5 O2 → 6 H2O + 4 NO 2) 4 mol NO 1.6 mol NH3x
1.6 mol NH3
1.6 mol NO =
How many moles of NO can be made using 1.6 mol NH3?

16. Practice 3) How many grams of NO can be made from 47 g of NH3?
Try these on your own NH3 + 5 O2 → 6 H2O + 4 NO
3) How many grams of NO can be made from 47 g of NH3?

17. Answers 83 g NO 3. 4 NH3 + 5 O2 → 6 H2O + 4 NO 1 mol NH3 4 mol NO
How many grams of NO can be made from 47 g of NH3?
1 mol NH3
17.04 g NH3 x 3.
4 mol NO
4 mol NH3 x
30.01 g NO
1 mol NO x
47 g NH3
83 g NO =

18. Limiting Reactant

19. Limiting Reactant
The limiting reagent in a chemical reaction is the substance that is totally consumed when the chemical reaction is complete.

20. Limiting Reactant
The limiting reagent is the reactant that gets used up first during the reaction and also determines how much product can be made. We can find the limiting reagent using the stoichiometric ratios from the balanced chemical reaction.
Once we know the limiting reagent, we can calculate how much product will be produced, known as the theoretical yield.

22. Limiting Reactant Example
What is the limiting reagent if we start with 2.8 grams of Al and 4.25 grams of Cl2?
2Al + 3Cl2 → 2AlCl3

23. Limiting Reactant Example
What is the limiting reagent if we start with 2.8 grams of Al and 4.25 grams of Cl2?
2Al + 3Cl2 → 2AlCl3
Step 1: convert grams to moles
2.80 g x 1 mol = mol Al
27 g
4.25 g x 1 mol = mol Cl
70 g

24. Limiting Reactant Example
What is the limiting reagent if we start with 2.8 grams of Al and 4.25 grams of Cl2?
2Al + 3Cl2 → 2AlCl3
Step 1: mol Al & mol Cl
Step 2: Find the stoichiometric ratio of each
2 AlCl3
AlCl moles produced
2 AlCl3
0.104 mol Al
0.104 AlCl moles produced
0.059 mol Cl X = X =
2 Al
3 Cl
Cl2 is limiting since it is less!

25. Limiting Reactant Example
What is the limiting reagent if we start with 4 grams of CH4 and 80 grams of O2?
CH O2 → CO2 + 2 H2O
Step 1: convert grams to moles
Step 2: Find the stoichiometric ratio

26. Limiting Reactant Example
What is the limiting reagent if we start with 4 grams of CH4 and 80 grams of O2?
CH O2 → CO2 + 2 H2O
Step 1: CH4 = 0.25 moles O2 = 1.25 moles
Step 2:
0.25 CH4 X
1 mole CH4
2 mole H2O =
0.5 H2O moles produced
1.25 mol O2 X 2 mole H2O =
2 mol O2
1.25
H2O mol produced
CH4 = Limiting Reactant!

27. On the page before, try the same problem but use CO2 as your product.
How much H2O and CO2 would be produced?

28. Theoretical Yield

29. Theoretical Yield
Theoretical yield is the amount of product that will be created based on the limiting factor.

30. Theoretical Yield Example
What is the theoretical yield AlCl3 that the reaction can produce with 4.25 grams of Cl2, our limiting reagent?
2Al + 3Cl2 → 2AlCl3
Step 1 : find out how many moles of AlCl3 are created
0.059 mol Cl x 2 mole AlCl3 = moles AlCl3
3 mole Cl2

31. Theoretical Yield Example
What is the theoretical yield AlCl3 that the reaction can produce with 4.25 grams of Cl2, our limiting reagent?
2Al + 3Cl2 → 2AlCl3
Step 1 : find out how many moles of AlCl3 are created
Answer: moles AlCl3
Step 2: convert moles to grams of AlCl3
5.32 grams of AlCl3
moles AlCl3 x 133 grams (molar mass of AlCl3) =
1 mole