1. Chapter 7 Stoichiometry: Calculations with Chemical Equations

2. Stoichiometric Calculations
The coefficients in the balanced equation give the ratio of moles of reactants and products

3. Using coefficients to deermine mole ratio
4 Fe O Fe2O3
How many moles of Fe2O3 are produced when 6.0 moles O2 react?
6.0 mol O2 x mol Fe2O3 = 4.0 mol Fe2O3
mol O2

4. Learning Check 2 4 Fe + 3 O2 2 Fe2O3
How many moles of Fe are needed to react with 12.0 mol of O2?
1) mol Fe
2) mol Fe
3) mol Fe

5. Solution S2 4 Fe + 3 O2 2 Fe2O3 12.0 mol O2 x mol Fe = 16.0 mol Fe

6. Converting mass and moles
4 Fe O Fe2O3
How many grams of O2 are needed to produce mol of Fe2O3?
1) g O2
2) g O2
3) g O2

7. Mass to Mass Stoichiometry Problems
Balance the equation and identify G and R
Convert Given to moles
Use coefficients to find moles of Required
Convert moles of R to mass in grams

8. Calculation
The reaction between H2 and O2 produces 13.1 g of water. How many grams of O2 reacted?
Write the equation
H2 (g) O2 (g) H2O (g)
Balance the equation
2 H2 (g) O2 (g) 2 H2O (g)

9. Organize data mol bridge 2 H2 (g) + O2 (g) 2 H2O (g) ? g 13.1 g
Plan g H2O mol H2O mol O O2
Setup
13.1 g H2O x 1 mol H2O x 1 mol O2 x g O g H2O 2 mol H2O 1 mol O2
= g O2

10. Learning Check S 4
How many O2 molecules will react with 505 grams of Na to form Na2O?
4 Na O Na2O
Complete the set up:
505 g Na x 1 mol Na x ________ x _______
23.0 g Na

11. Solution S 4
4 Na O Na2O
505 g Na x 1 mol Na x 1 mol O2 x 6.02 x 1023
23.0 g Na mol Na 1 mol O2
= x 1024 molelcules

12. Stoichiometric Calculations
From the mass of Substance A you can use the ratio of the coefficients of A and B to calculate the mass of Substance B formed (if it’s a product) or used (if it’s a reactant)

13. Stoichiometric Calculations
Example: 10 grams of glucose (C6H12O6) react in a combustion reaction. How many grams of each product are produced?
C6H12O6(s) + 6 O2(g)  6 CO2(g) + 6 H2O(l)
10.g ? ?
Starting with 10. g of C6H12O6…
we calculate the moles of C6H12O6…
use the coefficients to find the moles of H2O & CO2
and then turn the moles to grams

14. Stoichiometric calculations
C6H12O O2  6CO H2O
10.g ? ?
MW: 180g/mol g/mol g/mol
#mol: 10.g(1mol/180g)
0.055 mol (.055) (.055mol)
6(.055mol)44g/mol (.055mol)18g/mol
#grams: 15g g

15. Limiting Reactants

16. How Many Cookies Can I Make?
You can make cookies until you run out of one of the ingredients
Once you run out of sugar, you will stop making cookies

17. How Many Cookies Can I Make?
In this example the sugar would be the limiting reactant, because it will limit the amount of cookies you can make

18. Limiting Reactants
The limiting reactant is the reactant present in the smallest stoichiometric amount
2H O > H2O
#moles
Left:

19. Limiting Reactants
In the example below, the O2 would be the excess reagent

20. Limiting Reactants
If the amounts of two reactants are given, the reactant used up first determines the amount of product formed.

21. Analogy
Suppose you are preparing cheese sandwiches. Each sandwich requires 2 pieces of bread and 1 slice of cheese. If you have 4 slices of cheese and 10 pieces of bread, how many cheese sandwiches can you make?

22. Cheese Sandwich Products=
Sandwich 2 =

23. Learning Check S 6 How many sandwiches can you make?
____ slices of bread
+ ____ slices of cheese
= ____ sandwiches
What is left over? ________________
What is the limiting reactant?

24. Solution S 6 How many sandwiches can you make? __10__ slices of bread
+ __4__ slices of cheese
= __4__ sandwiches
What is left over? _2 slices of bread
What is the limiting reactant? cheese

25. Limiting reagent, example:
Soda fizz comes from sodium bicarbonate and citric acid (H3C6H5O7) reacting to make carbon dioxide, sodium citrate (Na3C6H5O7) and water. If 1.0 g of sodium bicarbonate and 1.0g citric acid are reacted, which is limiting? How much carbon dioxide is produced?
3NaHCO3(aq) + H3C6H5O7(aq) > 3CO2(g) + 3H2O(l) + Na3C6H5O7(aq)
1.0g g
84g/mol g/mol g/mol
1.0g(1mol/84g) (1mol/192g)
0.012 mol mol
(if citrate limiting)
0.0052(3)= mol
So bicarbonate limiting:
0.012 mol (1/3)=.0040mol moles CO2
44g/mol(0.012mol)=0.53g CO2
=.0012mol left
mol(192 g/mol)=
0.023 g left.

26. Theoretical Yield
The theoretical yield is the amount of product that can be made
In other words it’s the amount of product possible from stoichiometry. The “perfect reaction.”
This is different from the actual yield, the amount one actually produces and measures

27. Percent Yield Actual Yield Theoretical Yield Percent Yield = x 100
A comparison of the amount actually obtained to the amount it was possible to make
Actual Yield
Theoretical Yield
Percent Yield = x 100

28. Example C6H6 + Br2 ------> C6H5Br + HBr
Benzene (C6H6) reacts with Bromine to produce bromobenzene (C6H6Br) and hydrobromic acid. If 30. g of benzene reacts with 65 g of bromine and produces 56.7 g of bromobenzene, what is the percent yield of the reaction?
C6H Br > C6H5Br HBr
30.g g g
78g/mol g/mol g/mol
30.g(1mol/78g) g(1mol/160g)
0.38 mol mol
(If Br2 limiting)
0.41 mol mol
(If C6H6 limiting)
0.38 mol mol mol(157g/1mol) = 60.g
56.7g/60.g(100)=94.5%=95%

29. Percent Yield
You prepared cookie dough to make 5 dozen cookies. The phone rings while a sheet of 12 cookies is baking. You talk too long and the cookies burn. You throw them out (or give them to your dog.) The rest of the cookies are okay.
How many cookies could you have made (theoretical yield)?
How many cookies did you actually make to eat? (Actual yield)

30. Vocabulary
Actual yield is the amount of product actually recovered from an experiment
Theoretical (possible) yield is the maximum
amount of product that could be produced from
the reactant.
Percent Yield is the actual yield compared to the maximum (theoretical yield) possible.

31. Percent Yield Calculation
What is the percent yield of cookies?
Percent Yield = Actual Yield (g) recovered X Possible Yield (g)
% cookie yield = cookies x 100 = 80% yield
60 cookies

32. Example, one more
React 1.5 g of NH3 with 2.75 g of O2. How much NO and H2O is produced? What is left?
4NH O > NO H2O
1.5g g ? ?
17g/mol g/mol g/mol g/mol
1.5g(1mol/17g)= g(1mol/32g)=
.088mol
(If NH3 limiting):
.088mol (5/4)=.11
O2 limiting:
.086(4/5)= mol mol(4/5)= (6/5)=
.069mol mol mol
.069mol(17g/mol) mol(30.g/mol) .10mol(18g/mol)
1.2g g g g

34. Gun powder reaction And heat. What is interesting about this reaction?
10KNO3(s) + 3S(s) + 8C(s) ---- 2K2CO3(s) + 3K2SO4(s) + 6CO2(g) + 5N2(g)
Salt peter sulfur charcoal
And heat.
What is interesting about this reaction?
What kind of reaction is it?
What do you think makes it so powerful?

35. Gun powder reaction And heat. What is interesting about this reaction?
Oxidizing
agent
Oxidizing
agent
Reducing
agent
10KNO3(s) + 3S(s) + 8C(s) ---- 2K2CO3(s) + 3K2SO4(s) + 6CO2(g) + 5N2(g)
Salt peter sulfur charcoal
And heat.
What is interesting about this reaction?
Lots of energy, no oxygen
What kind of reaction is it?
Oxidation reduction
What do you think makes it so powerful and explosive?
Makes a lot of gas!!!!

36. White phosphorous and Oxygen under water