Using R.I.C.E. Tables and stoichiometry with limiting reactants
II. Stoichiometry in the Real World * Limiting Reagents
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Transcript of II. Stoichiometry in the Real World * Limiting Reagents
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II. Stoichiometry in the Real World
* LimitingReagents
II. Stoichiometry in the Real World
* LimitingReagents
More Stoichiometry!More Stoichiometry!
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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
X
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Jam Sandwich EquationJam Sandwich EquationJam Sandwich EquationJam 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
X
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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
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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?
Example 1Example 1Example 1Example 1
2 Fe (s) + O2 (g) 2 FeO (s)
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2 Fe (s) + O2 (g) 2 FeO (s)
mass = 50 gmass = 50 g mass = 60 g mass = ?
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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 + 2HCl ZnCl2 + H2 79.1 g mass = ?0.90 L
2.5M
Example 2Example 2Example 2Example 2
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Zn + 2 HCl ZnCl2 +
H2 mass = 79.1 g mass = ?V = 0.90 L
of 2.5M
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Example 2 ResultsExample 2 ResultsExample 2 ResultsExample 2 Results
Limiting reactant: HCl
Excess reactant: Zn
Product Formed: x g ZnCl2
left over zinc
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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?
Student ExampleStudent ExampleStudent ExampleStudent Example
CaCO3 + 2 NaCl CaCl2 + Na2CO3
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CaCO3 + 2 NaCl CaCl2 + Na2CO3
mass = 5.0 g mass = 5.0 g mass = ?
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AssignmentAssignmentAssignmentAssignment
Stoichiometry: Limiting Reagent
Worksheet
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B. Percent YieldB. Percent YieldB. Percent YieldB. Percent Yield
100yield ltheoretica
yield actualyield %
calculated on paper
measured in lab
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Ivan
Buz
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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
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ExampleExampleExampleExample
K2CO3 + 2HCl 2KCl + H2O + CO2 mass = 45.8 g mass = ?
actual: 46.3 g
Theoretical Yield:
# moles =mm
sm
# moles =138.2 g/mol
45.8 g
# moles = 0.331 mol
# moles = 0.662 mol
mass = (# mol) (mm)
mass = (0.662 mol) (74.6 g/mol)
mass KCl = 49.4 g
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Example Continued Example Continued Example Continued Example Continued
Theoretical Yield = 49.4 g KCl
=46.3 g
49.4 g 100 %
= 93.7 %
% Yield =Actual Yield
Theoretical Yield 100 %
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AssignmentAssignmentAssignmentAssignment
Pg 295 # 2, 5, 7, 11, 12, 15, 18, 20, 25 - 30