1. Deals with amounts of reactants used & products formed.
11.1 STOICHIOMETRY
Deals with amounts of reactants used & products formed.

2. What is Stoichiometry?
The study of the relationship between amounts of reactants used and products formed in a chemical reaction
Based on the Law of Conservation of Mass and Matter
Matter is neither created nor destroyed
Mass of reactants equals the mass of the products

3. MOLE-MASS RELATIONSHIPS
Balanced equation:
4Fe (s) O2(g)  2Fe2O3(s)
Interpret: 4 atoms molecules formula units
(Particles)
Mole ratio: 4 moles moles moles
(Coefficients)
Note: Particles of ionic compounds are called
“formula units”
Calculate the mass of each reactant and product by multiplying the number of moles by the molar mass

4. SHOW MASS IS CONSERVED 4Fe (s) + 3O2(g)  2Fe2O3(s)
Mass reactants:
4 mol Fe (55.8g Fe) = g Fe
(1 mole Fe)
3 mol O2 (32.00 g O2) = g O2
(1 mol O2) ____________
319.2 g Total

5. MASS OF PRODUCTS: 4Fe (s) + 3O2(g)  2Fe2O3(s)
2 mol Fe2O3 (159.6 g Fe2O3) = g (1 mol Fe2O3)
Equals the mass of the reactants
(319.2g)
Law of Conservation of Matter

6. PROBLEM
Interpret in terms of particles, moles, and mass. Show that mass is conserved:
(Hint: look at coefficients for particles & moles)
4 Al O2  2 Al2O3
Particles:
Moles:
Mass:
Conserved?

7. SOLUTION: 4Al + 3O2  2 Al2O3 Law of Conservation of Matter shown.
Particles: molecules molecule 2 molecule
Moles: mole 3 mole 2 moles
Mass: (27.0 g) (32.0 g) = 2 (102.0 g)
Conserved? g = g YES!
Law of Conservation of Matter shown.

8. There are six mole ratios for the following:
MOLE RATIO – is a ratio between the number of moles of any two substances in a balanced chemical equation
There are six mole ratios for the following:
Ex Al(s) O2(g)  2 Al2O3(s)
Note: moles moles moles
4 mol Al and 3mol O2
3 mol O mol Al

9. 4 mol Al and 2 mol Al2O3 2 mol Al2O3 4 mol Al 3 mol O2 and 2 mol Al2O3
MOLE RATIO – is a ratio between the number of moles of any two substances in a balanced chemical equation
4 mol Al and mol Al2O3
2 mol Al2O mol Al
3 mol O and 2 mol Al2O3
2 mol Al2O mol O2
All stoichiometry calculations begin with a balanced equation and mole ratios!!

10. PROBLEM - FIND MOLE RATIOS FOR:
2 NH3  N H2
2 mol mol mol
Ans mol NH3 or mol N2
1 mol N mol NH3
2 mol NH3 or mol H2
3 mol H mol NH3

11. Ans NH3  N H2
1 mole N2 or 3 mole H2
3 mole H mole N2

12. 12.2 STOICHIOMETRIC CALCULATIONS
There are 3 Basic Stoichiometry Calculations

13. 1. Mole to Mole Conversions
A piece of magnesium burns in the presence
of oxygen forming magnesium oxide (MgO).
How many moles of oxygen are needed to produce 12 moles of magnesium oxide?
Step 1: Write a balanced equation
2 Mg (s) + O2 (g)  2 MgO (s)

14. Write mole ratios
Choose the correct mole ratio needed for this problem

15. Mole to Mole Conversion cont’d
Multiply the known number of moles of MgO by the mole ratio
6 mols of oxygen is needed to produce 12 mols of magnesium oxide

16. 2. Mole to Mass Conversions
The following reaction occurs in plants undergoing photosynthesis
CO2(g) + H2O(l) C6H12O6(s)+ O2(g)
How many grams of glucose (C6H12O6) are produced when 24.0 mols CO2 reacts in excess water?

17. Mole to Mass Conversion cont’d
Write a balanced equation
6 CO2(g) + 6 H2O(l)  C6H12O6(s) + 6 O2(g)
Use mole ratios to determine the number of moles of glucose produced by the given amount of carbon dioxide

18. Mole to Mass Conversion cont’d
Multiply by the molar mass
721 g glucose is produced from 24.0 moles carbon dioxide

19. 3. Mass to Mass Problems The only new step is first step: Convert grams of given substance to moles!
Massgiven  Molesgiv Molesdesired Massdes
÷ molar massgiven X mole ratio X molar massdesired
3 step problem!

20. Label: 25.0 g ? g NH4NO3  N2O + 2H2O Mole Ratio: 1 mol 1 mol 2 mol
Ammonium nitrate (NH4NO3) produces N2O gas and H2O when it decomposes. Determine the mass of water produced from the decomposition of 25.0 g of NH4NO3.
1) Find moles NH4NO3 (molar mass): g/mol
Use the inverse of the molar mass to convert grams of NH4NO3 to moles of NH4NO3

21. Label: 25.0 g ? g NH4NO3  N2O + 2H2O Mole ratio: 1 mol 1 mol 2 mol
Determine the mole ratio of mol H2O to mol NH4NO3 from the chemical equation. The desired substance is the numerator.
Multiply mol NH4NO3 by the mole ratio.
Calculate the mass of H2O using the molar mass.

22. Label: 5.6g ? g PROBLEM: N2 + 3H2  2 NH3
If 5.6 g nitrogen reacts completely with hydrogen, what mass of ammonia is formed?
1) Find moles of nitrogen (given) – molar mass:
5.6 g N2_____ mass N2 = 2 x = g/mol
g N fill in units
5.6 g N2_1mol N2
g N2 fill in the rest

23. Label: 5.6 g ? g N2 + 3H2  2 NH3 Mole Ratio: 1mol 3mol 2mol
2) Find moles ammonia (desired) - mole ratio:
5.6 g N2 1mol N2 _2 mol NH3___
g N mol N2
3) Find grams ammonia (desired) –
molar mass: NH3 = = 17 g/mole
5.6 g N2 1mol N2 _2 mol NH g NH3
g N mol N mol NH3
given ÷ MM given mole ratio x MM desired
= 6.8 g NH3
Mass H2 = 6.8 g – 5.6 g = 1.2 g LOCOM

24. 12.3 LIMITING REACTANTS

25. What is a Limiting Reactant?
A limiting reactant is the reactant that limits (stops) a reaction and determines the amount of product
An excess reactant is any reactant that is left over after the reaction stops (all reactants except the limiting reactant)

26. Ingredients: You don’t always have the exact amounts.
IN ORDER FOR CHEMICAL REACTION TO OCCUR, YOU MUST HAVE A COMPLETE SET OF REACTANTS (REAGENTS):
You don’t always have the exact amounts.
Ex. Let’s make some McBurgers!!!
Ingredients:
2 buns; 1 burger patty; 1cheese slice; 1 tomato slice; 1 lettuce leaf; 3 pickles

27. How many McBurgers can you make?
RECIPE CALLS
FOR:
2 buns
1 burger patty
1 cheese slice
1 tomato slice
1 lettuce leaf
3 pickles
YOU HAVE AVAILABLE:
6 buns
3 burger patties
5 cheese slices
6 tomato slices
5 lettuce leafs
6 pickles
How many McBurgers can you make?

28. Based on the individual ingredients you have available:
You have enough:
Buns for (3)
Burger Patties for (3)
Cheese Slices for (5)
Tomato Slices for (6)
Lettuce Leafs for (5)
Pickles for (2)

29. (Remember each McBurger requires 3 pickles)
The Limiting Reactant (LR) for the McBurger making process in the pickles
This is the ingredient we ran out of first and were unable to continue with the McBurger making process
We were only able to make 2 McBurger from the 6 pickles we had available to use
(Remember each McBurger requires 3 pickles)

30. Excess reactants (XR) are the ingredients not used in the McBurger making process:
4 buns
1 burger patty
3 cheese slices
4 tomato slices
3 lettuce leafs
Note: You made 2 McBurgers

31. PRACTICE PROBLEM: N2 + 3H2  2 NH3
Using the above equation you are given 3.0 mols N2 and 5.0 mols H2
Determine the limiting reactant
Determine the excess reactant
Determine the amount of NH3 produced

32. l
Label: mol mol ?mol
N H  2 NH3 M.R mol mol mol
Draw 2 bridges. Smaller product is your answer. (Always use L.R. to find answer!)
3.0 mol N2|_2 mol NH3 = 6.0 mol NH3
| 1 mol N (N2 is XS reactant)
5.0 mol H2|_2 mol NH3 = 3.3 mol NH3
| 3 mol H Ans./H2 is L.R.

33. l
Label: mol mol ?mol
N H  2 NH3 M.R mol mol mol
Alternative way: Compare 2 mole ratios!
Note: 1mole N2 reacts with 3 mol H2
3.0 mol N2 reacts with 9.0 mol H2
Given – 5.0 mol H2 (Don’t have enough H2!)
Thus H2 is LR. Use H2 to solve for NH3.
5.0 mol H2|_2 mol NH3 = 3.3 mol NH3
| 3 mol H2

34. ANSWER: 3. 3 mol ammonia formed. Hydrogen (H2) is L. R
ANSWER: 3.3 mol ammonia formed Hydrogen (H2) is L.R Nitrogen (N2) is the XS.

35. IF A PAPER BURNS IN A ROOM:+
What is the limiting reactant?
What is in excess?
What would happen if the paper burned in a closed jar?
LR?
XS? +

36. 4.0mol 7.0mol PROBLEM: 2 Al + 3 Cl2  2 AlCl3
Given 4.0 mol Al and 7.0 mol Cl2, what is the maximum amount of aluminum chloride formed?
Step 1: 2 bridges:
4.0 mol Al|______________
| mol Al
7.0 mol Cl2 |________
| mol Cl2

37. Label: 4.0mol 7.0 mol ? Mol 2 Al + 3 Cl2  2 AlCl3
4.0 mol Al|__2 mol AlCl = 4.0 mol AlCl3
| mol Al
7.0 mol Cl2 |_2 mol AlCl = 4.7 mol AlCl3
| 3 mol Cl2
*Al is L.R Ans. 4.0 mol AlCl3
*Cl2 is XS

38. THE END