1. II. Stoichiometry in the Real World * Limiting Reagents
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II. Stoichiometry in the Real World * Limiting Reagents

2. X 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. X Jam Sandwich Equation 2 loaves bread + 1 jar of jam  20 sandwiches
Mole Ratio 2 : 1 : 20
What would happen if we had 1 loaf of bread and 1 jar of jam?
Bread = limiting reactant
Jam = excess reactant (1/2 jar left)
Produce 10 sandwiches

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

5. Example 1 2 Fe (s) + O2 (g)  2 FeO (s)
Iron burns in air to form a solid black oxide (FeO). If you had 50.0 g of iron and 60.0 g of oxygen, what is the limiting and excess reactant? How much iron oxide could be produced?
2 Fe (s) + O2 (g)  2 FeO (s)

6. 2 Fe (s) + O2 (g)  2 FeO (s) mass = 50 g mass = 50 g mass = 60 g

7. Example 2 Zn + 2HCl  ZnCl2 + H2 79.1 g mass = ? 0.90 L 2.5M
79.1 g of zinc react with 0.90 L of 2.5M HCl. Identify the limiting and excess reactants. How many grams of zinc chloride would be produced?
Zn HCl  ZnCl H2
79.1 g
0.90 L
2.5M
mass = ?

8. Zn HCl  ZnCl H2
mass = 79.1 g
V = 0.90 L of 2.5M
mass = ?

9. Example 2 Results Limiting reactant: HCl Excess reactant: Zn
Product Formed: x g ZnCl2
left over zinc

10. Student Example CaCO3 + 2 NaCl  CaCl2 + Na2CO3
Sodium carbonate is needed in the manufacturing of glass, but very little occurs naturally. It can be made from the double replacement reaction between calcium carbonate and sodium chloride. If you had 5.00 g of each what is the limiting and excess reactant? How much sodium carbonate would be formed?
CaCO NaCl  CaCl2 + Na2CO3

11. CaCO NaCl  CaCl2 + Na2CO3
mass = 5.0 g
mass = 5.0 g
mass = ?

12. Assignment
Stoichiometry: Limiting Reagent
Worksheet

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

15. Ivan
Buz

16. When 45. 8 g of K2CO3 react with excess HCl, 46. 3 g of KCl are formed
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

17. Example K2CO3 + 2HCl  2KCl + H2O + CO2 Theoretical Yield: sm
mass = 45.8 g
mass = ?
actual: 46.3 g
# moles = mm sm
# moles = mol
mass = (# mol) (mm)
# moles =
138.2 g/mol
45.8 g
mass = (0.662 mol) (74.6 g/mol)
# moles = mol
mass KCl = 49.4 g
Theoretical Yield:

18. Example Continued Theoretical Yield = 49.4 g KCl Actual Yield
 100 %
% Yield = =
46.3 g
49.4 g
 100 %
= %

19. Assignment
Pg 295 # 2, 5, 7, 11, 12, 15, , 20,