I. I.Stoichiometric Calculations Topic 9 Stoichiometry Topic 9 Stoichiometry.

I. Stoichiometric

Calculations

Topic 9 Topic 9 StoichiometryStoichiometry

StoichiometryStoichiometryStoichiometryStoichiometry

The study of quantitative relationships between amounts of reactants used and amount of products formed in a reaction.

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

Mole Ratio: Ratio between the number of moles of any two substances in a balanced chemical equation.

2 Mg + O2 2 Mg + O2 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.

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)

Steps to solving Steps to solving StoichiometryStoichiometrySteps to solving Steps to solving StoichiometryStoichiometry

Convert known quantity to moles.Convert moles of known to moles

of unknown using mole ratios.Convert unknown moles to

wanted quantity.

D. Stoichiometry D. Stoichiometry ProblemsProblemsD. Stoichiometry D. Stoichiometry ProblemsProblems

How many moles of KClO3 must decompose in order to produce 9 moles of oxygen gas?

9 mol O2 2 mol KClO3

3 mol O2

= 6 mol KClO3

2KClO3 2KCl + 3O2 ? mol 9 mol

How many grams of KClO3 are req’d to

produce 9.00 L of O2 at STP?

9.00 LO2

1 molO2

22.4 L O2

= 32.8 g KClO3

2 molKClO3

3 molO2

122.55g KClO3

1 molKClO3

? g 9.00 L


2KClO3 2KCl + 3O2


How many grams of silver will be formed from 12.0 g copper?

12.0g Cu

1 molCu

63.55g Cu

= 40.7 g Ag

Cu + 2AgNO3 2Ag + Cu(NO3)2

2 molAg

1 molCu

107.87g Ag

1 molAg

12.0 g ? g

63.55g Cu

1 molCu


How many grams of Cu are required to react with 1.5 L of 0.10M AgNO3?

1.5L

.10 molAgNO3

1 L= 4.8 g

Cu

Cu + 2AgNO3 2Ag + Cu(NO3)2

1 molCu

2 molAgNO3

? g 1.5L0.10M

E. Percent YieldE. Percent YieldE. Percent YieldE. Percent Yield

100yield ltheoretica

yield actualyield %

calculated on paper

measured in lab


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


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:


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

II. Stoichiometry in the Real World

Topic 9 Topic 9 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


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


1. Write a balanced equation.

2. For each reactant, calculate the

amount of product formed.

3. Smaller answer indicates:

• limiting reactant

• amount of product


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


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


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


Zn: 27.1 L H2 HCl: 25 L H2

Limiting reactant: HCl

Excess reactant: Zn

Product Formed: 25 L H2

left over zinc

I. I.Stoichiometric Calculations Topic 9 Stoichiometry Topic 9 Stoichiometry.

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