I. Stoichiometric

Calculations

I. Stoichiometric

Calculations

StoichiometryStoichiometry

A. Proportional A. Proportional RelationshipsRelationshipsA. Proportional A. Proportional RelationshipsRelationships

I have 5 eggs. How many cookies can I make?

3/4 c. brown sugar1 tsp vanilla extract2 eggs2 c. chocolate chipsMakes 5 dozen cookies.

2 1/4 c. flour1 tsp. baking soda1 tsp. salt1 c. butter3/4 c. sugar

5 eggs 5 doz.

2 eggs= 12.5 dozen cookies

Ratio of eggs to cookies

A. Proportional A. Proportional RelationshipsRelationshipsA. Proportional A. Proportional RelationshipsRelationships

StoichiometryStoichiometry• mass relationships between

substances in a chemical reaction• based on the mole ratio

Mole RatioMole Ratio• indicated by coefficients in a

balanced equation

2 Mg + O2 Mg + O22 2 MgO 2 MgO

B. Stoichiometry StepsB. Stoichiometry StepsB. Stoichiometry StepsB. Stoichiometry Steps

1. Write a balanced equation.2. Identify known & unknown.3. Line up conversion factors.

• Mole ratio - moles moles• Molar mass - moles grams• Molarity - moles liters soln• Molar volume - moles liters gas

Core step in all stoichiometry problems!!

• Mole ratio - moles moles

4. Check answer.

1 mol of a gas=22.4 Lat STP

C. Molar Volume at STPC. Molar Volume at STPC. Molar Volume at STPC. Molar Volume at STP

Standard Temperature & Pressure0°C and 1 atm

C. Molar Volume at STPC. Molar Volume at STPC. Molar Volume at STPC. Molar Volume at STP

Molar Mass(g/mol)

6.02 1023

particles/mol

MASSIN

GRAMSMOLES

NUMBEROF

PARTICLES

LITERSOF

SOLUTION

Molar Volume (22.4 L/mol)

LITERSOF GASAT STP

Molarity (mol/L)

II. Stoichiometry in the Real World

II. Stoichiometry in the Real World

StoichiometryStoichiometry

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• added to ensure that the other

reactant is completely used up• cheaper & easier to recycle

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 ? L0.90 L

2.5M

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 ? L0.90 L

2.5M

A. Limiting ReactantsA. Limiting ReactantsA. Limiting ReactantsA. Limiting Reactants

22.4L H2

1 molH2

0.90L

2.5 molHCl

1 L= 25 L

H2

1 molH2

2 molHCl

Zn + 2HCl ZnCl2 + H2 79.1 g ? L0.90 L

2.5M

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

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