c. johannesson ii. stoichiometry in the real world (p. 288-294) stoichiometry – ch. 9
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C. Johannesson
II. Stoichiometry in the Real World
(p. 288-294)
II. Stoichiometry in the Real World
(p. 288-294)
Stoichiometry – Ch. 9Stoichiometry – Ch. 9
C. Johannesson
A. Limiting ReactantsA. Limiting ReactantsA. Limiting ReactantsA. Limiting Reactants
Available IngredientsAvailable Ingredients• 4 slices of bread• 1 jar of peanut butter• 1/2 jar of jelly
Limiting ReactantLimiting Reactant• bread
Excess ReactantsExcess Reactants• peanut butter and jelly
C. Johannesson
A. Limiting ReactantsA. Limiting ReactantsA. Limiting ReactantsA. Limiting Reactants
Limiting ReactantLimiting Reactant• used up in a reaction• determines the amount of product
Excess ReactantExcess Reactant• added to ensure that the other
reactant is completely used up• cheaper & easier to recycle
C. Johannesson
A. Limiting ReactantsA. Limiting ReactantsA. Limiting ReactantsA. Limiting Reactants
1. Write a balanced equation.
2. For each reactant, calculate the
amount of product formed.
3. Smaller answer indicates:
• limiting reactant
• amount of product
C. Johannesson
A. Limiting ReactantsA. Limiting ReactantsA. Limiting ReactantsA. Limiting Reactants
79.1 g of zinc react with 81g HCl. Identify the limiting and excess reactants. How many grams of hydrogen are formed at STP?
Zn + 2HCl ZnCl2 + H2 79.1 g ? L0.90 L
2.5M
C. Johannesson
A. Limiting ReactantsA. Limiting ReactantsA. Limiting ReactantsA. Limiting Reactants
79.1g Zn
1 molZn
65 g Zn
= 2.4 g H2
1 molH2
1 molZn
2gH2
1 molH2
Zn + 2HCl ZnCl2 + H2 79.1 g ? g81.0g
C. Johannesson
A. Limiting ReactantsA. Limiting ReactantsA. Limiting ReactantsA. Limiting Reactants
2g H2
1 molH2
81.Og
1 molHCl
36g= 2.3 g
H2
1 molH2
2 molHCl
Zn + 2HCl ZnCl2 + H2 79.1 g ? g81.0 g
C. Johannesson
A. Limiting ReactantsA. Limiting ReactantsA. Limiting ReactantsA. Limiting Reactants
Zn: 2.4 g H2 HCl: 2.3 gH2
Limiting reactant: HCl
Excess reactant: Zn
Product Formed: 2.3 g H2
left over zinc
C. Johannesson
B. Percent YieldB. Percent YieldB. Percent YieldB. Percent Yield
100yield ltheoretica
yield actualyield %
calculated on paper
measured in lab
C. Johannesson
B. Percent YieldB. Percent YieldB. Percent YieldB. 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
C. Johannesson
B. Percent YieldB. Percent YieldB. Percent YieldB. Percent Yield
45.8 gK2CO3
1 molK2CO3
138.21 gK2CO3
= 49.4g KCl
2 molKCl
1 molK2CO3
74.55g KCl
1 molKCl
K2CO3 + 2HCl 2KCl + H2O + CO2 45.8 g ? g
actual: 46.3 g
Theoretical Yield:
C. Johannesson
B. Percent YieldB. Percent YieldB. Percent YieldB. Percent Yield
Theoretical Yield = 49.4 g KCl
% Yield =46.3 g
49.4 g 100 =93.7%
K2CO3 + 2HCl 2KCl + H2O + CO2 45.8 g 49.4 g
actual: 46.3 g