II. Stoichiometry in the Real World

* LimitingReagents

II. Stoichiometry in the Real World

* LimitingReagents

More Stoichiometry!More Stoichiometry!

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

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

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

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)

2 Fe (s) + O2 (g) 2 FeO (s)

mass = 50 gmass = 50 g mass = 60 g mass = ?

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

Zn + 2 HCl ZnCl2 +

H2 mass = 79.1 g mass = ?V = 0.90 L

of 2.5M

Example 2 ResultsExample 2 ResultsExample 2 ResultsExample 2 Results

Limiting reactant: HCl

Excess reactant: Zn

Product Formed: x g ZnCl2

left over zinc

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

CaCO3 + 2 NaCl CaCl2 + Na2CO3

mass = 5.0 g mass = 5.0 g mass = ?

AssignmentAssignmentAssignmentAssignment

Stoichiometry: Limiting Reagent

Worksheet

B. Percent YieldB. Percent YieldB. Percent YieldB. Percent Yield

100yield ltheoretica

yield actualyield %

calculated on paper

measured in lab

Ivan

Buz

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

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

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 %

AssignmentAssignmentAssignmentAssignment

Pg 295 # 2, 5, 7, 11, 12, 15, 18, 20, 25 - 30