1. II. Stoichiometry in the Real World (p. 368 – 375)
Stoichiometry – Ch. 12
II. Stoichiometry in the Real World (p. 368 – 375)

2. A. Limiting Reactants Available Ingredients 4 slices of bread
1 jar of peanut butter
1/2 jar of jelly
Limiting Reactant
bread
Excess Reactants
peanut butter and jelly

3. A. Limiting Reactants Available Ingredients 24 graham cracker squares
1 bag of marshmallows
12 pieces of chocolate
Limiting Reactant
chocolate
Excess Reactants
Marshmallows and graham crackers

4. A. Limiting Reactants Limiting Reactant
one that is used up in a reaction
determines the amount of product that can be produced
Excess Reactant
added to ensure that the other reactant is completely used up
cheaper & easier to recycle

5. A. Limiting Reactants Write the balanced equation.
Identify and label known and unknown.
For each reactant, calculate the amount of product formed.
Smaller answer indicates:
limiting reactant
maximum amount of product actually possible

6. A. Limiting Reactants ? LR ? ER Zn + 2HCl  ZnCl2 + H2 79.1 g ? g
79.1 g of zinc react with 68.1 g HCl. Identify the limiting and excess reactants. How many grams of hydrogen can be formed?
? LR
? ER
Zn HCl  ZnCl H2
79.1 g
68.1 g
? g

7. A. Limiting Reactants Zn + 2HCl  ZnCl2 + H2 79.1 g 68.1 g ? g 79.1
g Zn
1 mol Zn
65.39
g Zn
1 mol H2 Zn
2.02 g H2
1 mol
= 2.44 g H2

8. A. Limiting Reactants Zn + 2HCl  ZnCl2 + H2 79.1 g 68.1 g ? g 68.1
g HCl
1 mol
HCl
g HCl
1 mol H2
2 mol
HCl
2.02 g H2
1 mol H2
= 1.89 g H2

9. A. Limiting Reactants Zn: 2.44 g H2 HCl: 1.89 g H2
Limiting reactant: HCl
Excess reactant: Zn
Product Formed: 1.89 g H2
left over zinc

10. A. Limiting Reactants #2 ? LR ? ER 2Mg + O2  2MgO 5.42 g 4.00 g ? g
5.42 g of magnesium ribbon react with 4.00 g of oxygen gas. Identify the limiting and excess reactants. How many grams of magnesium oxide are formed?
? LR
? ER
2Mg O2  2MgO
5.42 g
4.00 g
? g

11. A. Limiting Reactants #2 2Mg + O2  2MgO 5.42 g 4.00 g ? g 5.42 g Mg
1 mol Mg
24.31
g Mg
2 mol
MgO Mg
40.31 g
MgO
1 mol
= 8.99 g
MgO

12. A. Limiting Reactants #2 2Mg + O2  2MgO 5.42 g 4.00 g ? g 4.00 g O2
1 mol O2
g O2
2 mol
MgO
1 mol O2
40.31
g MgO
1 mol
MgO
= 10.1 g MgO

13. A. Limiting Reactants #2 Limiting reactant: Mg Excess reactant: O2
Mg: 8.99 g MgO O2: 10.1 g MgO
Limiting reactant: Mg
Excess reactant: O2
Product Formed: 8.99 g MgO
Excess oxygen

14. A. Limiting Reactants
What other information could you find in these problems?
How much of each reactant is used – in grams, liters, moles
How much of excess reactant is left over – in grams, liters, moles

15. A. Limiting Reactants – Shortcut
5.00 mol of magnesium ribbon react with 4.00 mol of oxygen gas. Identify the limiting and excess reactants. How many moles of excess reactant are left over?
2Mg O2  2MgO
5.00 mol
4.00 mol
? mol ER left over

16. A. Limiting Reactants – ShortcutLR ER
2Mg O2
divide
divide M 2 1 A U L
Mole Ratio
multiply
multiply
5.00
4.00
Subtract
Subtract
Available
bring down
bring down
8.00
2.50
Answer
Answer
Used
1.50
Left Over
1.50 mol of O2 left over

17. B. Percent Yield
Percent of theoretical yield that is actually produced in the lab
Theoretical yield = amount of product calculated on paper using stoichiometry
Actual yield = amount of product actually formed in the lab

18. B. Percent Yield
measured in lab
calculated on paper

19. B. Percent Yield
When 45.8 g of K2CO3 react with excess HCl, 46.3 g of KCl are formed. Calculate the theoretical and % yields of KCl.
K2CO3 + 2HCl  2KCl + H2O + CO2
45.8 g
? g
actual: 46.3 g

20. B. Percent Yield K2CO3 + 2HCl  2KCl + H2O + CO2 45.8 g ? g
actual: 46.3 g
Theoretical Yield:
45.8 g
K2CO3
1 mol
K2CO3 g
2 mol
KCl
1 mol
K2CO3
74.55
g KCl
1 mol
KCl
= 49.4
g KCl

21. B. Percent Yield K2CO3 + 2HCl  2KCl + H2O + CO2 45.8 g 49.4 g
actual: 46.3 g
Theoretical Yield = 49.4 g KCl
46.3 g
49.4 g
% Yield =
 100 =
93.7%