ii. stoichiometry in the real world limiting reagents and % yield (p. 268-378)
DESCRIPTION
Stoichiometry – Ch. 12. II. Stoichiometry in the Real World Limiting Reagents and % yield (p. 268-378). 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. - PowerPoint PPT PresentationTRANSCRIPT
II. Stoichiometry in the Real World
Limiting Reagents and % yield
(p. 268-378)
Stoichiometry – Ch. Stoichiometry – Ch. 1212
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
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• reactant left over… • cheaper & easier to recycle, added to
ensure that the other reactant is completely used up
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
A. Limiting ReactantsA. Limiting ReactantsA. Limiting ReactantsA. Limiting Reactants
79.1 g of zinc react with 0.90 L of 2.5M HCl. Identify the limiting and excess reactants. How many liters of hydrogen are formed at STP?
Zn + 2HCl ZnCl2 + H2 79.1 g ? L82.0g
Summary of Limiting Summary of Limiting ReactantReactantSummary of Limiting Summary of Limiting ReactantReactant
Do stoichiometry twice to go from each reactant to same product
Now you know LR and actual amount produced
Do stoichio once more from LR to ER to determine amount used…then subtract
A. Limiting ReactantsA. Limiting ReactantsA. Limiting ReactantsA. Limiting Reactants
79.1g Zn
1 molZn
65.39g Zn
= 27.1 L H2
1 molH2
1 molZn
22.4 LH2
1 molH2
Zn + 2HCl ZnCl2 + H2 79.1 g ? L82.0g
A. Limiting ReactantsA. Limiting ReactantsA. Limiting ReactantsA. Limiting Reactants
22.4L H2
1 molH2
82.0g
1 molHCl
36.5gHCl
= 25 L H2
1 molH2
2 molHCl
Zn + 2HCl ZnCl2 + H2 79.1 g ? L82.0
A. Limiting ReactantsA. Limiting ReactantsA. Limiting ReactantsA. Limiting Reactants
Zn 27.1 L H2 HCl 25 L H2
Limiting reactant: HCl
Excess reactant: Zn
Product Formed: 25 L H2
left over zinc
Choose smallest #
How many grams of excess How many grams of excess reactant remain?reactant remain?How many grams of excess How many grams of excess reactant remain?reactant remain?
Use stoichiometry to convert from L.R. (HCl) to amount of E.R. (Zinc) USED..getting…
Now subtract amount ER (73.5g) used from starting ER (79.1g) to get amount remaining (79.1 -73.5) =
73.5g Zn used
5.6grams of Zinc remaining
B. Percent YieldB. Percent YieldB. Percent YieldB. Percent Yield
100yield ltheoretica
yield actualyield %
calculated on paper
measured in lab
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
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:
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