CH. 9
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CH. 9. Stoichiometry. A. Proportional Relationships. Ratio of eggs to cookies. I have 5 eggs. How many cookies can I make?. 2 1/4 c. flour 1 tsp. baking soda 1 tsp. salt 1 c. butter 3/4 c. sugar. 3/4 c. brown sugar 1 tsp vanilla extract 2 eggs 2 c. chocolate chips

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Stoichiometry

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Stoichiometry

CH. 9

Stoichiometry


A proportional relationships

A. Proportional Relationships

Ratio of eggs to cookies

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

2 1/4 c. flour

1 tsp. baking soda

1 tsp. salt

1 c. butter

3/4 c. sugar

3/4 c. brown sugar

1 tsp vanilla extract

2 eggs

2 c. chocolate chips

Makes 5 dozen cookies.

5 eggs

5 doz.

2 eggs

= 12.5 dozen cookies


Proportional relationships

Proportional Relationships

  • Stoichiometry

    • mass relationships between substances in a chemical reaction

    • based on the mole ratio

  • Mole Ratio

    • indicated by coefficients in a balanced equation

2 Mg + O2 2 MgO


Stoichiometry steps

Stoichiometry Steps

1. Write a balanced equation.

2. Identify known & unknown.

3. Line up conversion factors.

  • Mole ratio - moles  moles

  • Molar mass -moles  grams

  • Molar volume -moles  liters gas

  • Mole ratio - moles  moles

Core step in all stoichiometry problems!!

4. Check answer.


Conversions

Conversions

MASS

IN

GRAMS

NUMBER

OF

PARTICLES

MOLES

Molar Mass

(g/mol)

6.02  1023

particles/mol


Stoichiometry problems moles only

Stoichiometry Problems: Moles Only

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

2KClO3 2KCl + 3O2

? mol

9 mol

2 mol KClO3

3 mol O2

x mol KClO3

9 mol O2

=

2 mol KClO3

3 mol O2

9 mol O2

= 6 mol KClO3


Stoichiometry problems mass

Stoichiometry Problems: Mass

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

2KClO3 2KCl + 3O2

? g

9 g

a) Molar mass

c) Molar mass

b) Mole ratio

Moles KClO3

Moles O2

Divide problem in steps


A calculate o 2 moles

a) Calculate O2 moles

2KClO3 2KCl + 3O2

? g

9 g

1mol

32.0g

9.00 g

= 0.281 mol O2


B calculate kclo 3 moles

b) Calculate KClO3 moles

2KClO3 2KCl + 3O2

? mol

0.281 mol

2 mol KClO3

3 mol O2

x mol KClO3

0.281 mol O2

=

0.281 mol O2

2 mol KClO3

3 mol O2

= 0.187 mol KClO3


C calculate kclo 3 mass

c) Calculate KClO3 mass

0.187mol

122.55 g

1 mol

= 22.9 g KClO3

2KClO3 2KCl + 3O2

22.9 g

9.00 g


Stoichiometry problems density

Stoichiometry Problems: density

  • How many grams of KClO3 must decompose in order to produce 6.3 mL of oxygen gas? (density of O2 1.429 g/mL)

2KClO3 2KCl + 3O2

? g

6.3 mL

a) Density

Mass O2

d) Molar mass

b) Molar mass

c) Mole ratio

Moles KClO3

Moles O2

Divide problem in steps


A calculate o 2 mass

a) Calculate O2 mass

2KClO3 2KCl + 3O2

? g

9 g

1.426 g

1 mL

6.3 mL

9 g O2


B calculate o 2 moles

b) Calculate O2 moles

2KClO3 2KCl + 3O2

? g

9 g

1mol

32.0g

9.00 g

= 0.281 mol O2


C calculate kclo 3 moles

c) Calculate KClO3 moles

2KClO3 2KCl + 3O2

? mol

0.281 mol

0.281 mol O2

2 mol KClO3

3 mol O2

= 0.187 mol KClO3


D calculate kclo 3 mass

d) Calculate KClO3 mass

0.187mol

122.55 g

1 mol

= 22.9 g KClO3

2KClO3 2KCl + 3O2

22.9 g

6.3 mL


Gas stoichiometry

Gas Stoichiometry

StandardTemperature&Pressure

0°C and 1 atm

  • Moles  Liters of a Gas:

    • STP - use 22.4 L/mol

    • Non-STP - use ideal gas law

1 mol of a gas=22.4 L

at STP


Molar volume at stp

Molar Volume at STP

LITERS

OF GAS

AT STP

Molar Volume

(22.4 L/mol)

MASS

IN

GRAMS

NUMBER

OF

PARTICLES

MOLES

Molar Mass

(g/mol)

6.02  1023

particles/mol


Stoichiometry problems volume

Stoichiometry Problems: Volume

  • How many grams of KClO3 are req’d to produce 9.00 L of O2 at STP?

2KClO3 2KCl + 3O2

? g

9 L

a) Molar volume

c) Molar mass

b) Mole ratio

Moles KClO3

Moles O2

Divide problem in steps


A calculate o 2 moles1

a): Calculate O2 moles

2KClO3 2KCl + 3O2

? g

9.00 L

1mol

22.4L

9.00 L

= 0.4 mol O2


B calculate kclo 3 moles1

b) Calculate KClO3 moles

2KClO3 2KCl + 3O2

? mol

0.4 mol

2 mol KClO3

3 mol O2

x mol KClO3

0.4 mol O2

=

0.4 mol O2

2 mol KClO3

3 mol O2

= 0.27 mol KClO3


C calculate kclo 3 mass1

c) Calculate KClO3 mass

0.27 mol

122.55 g

1 mol

= 33 g KClO3

2KClO3 2KCl + 3O2

33 g

9.00 L


Stoichiometry problems only gas reactions

Stoichiometry Problems: only gas reactions

  • 2 C2H2 (g) + 5 O2 (g)  4 CO2 (g) + 2 H2O (g)

  • If 79 L of carbon dioxide are produced in this reaction, how many liters of ethene (C2H2) were used?

2 L C2H2

4 L CO2

x L C2H2

78 L CO2

=


Stoichiometry problems particles

Stoichiometry Problems: particles

  • How many grams of KClO3 are req’d to produce 9.044 x 1023 molecules of O2?

2KClO3 2KCl + 3O2

? g

9.044 x 1023 molecules

a) Avogadro’s number

c) Molar mass

b) Mole ratio

Moles KClO3

Moles O2

Divide problem in steps


Step 1 calculate o 2 moles

Step 1: Calculate O2 moles

2KClO3 2KCl + 3O2

? g

9.044 x 1023 molecules

9.044 x 1023molecules

1mol

6.022 x 1023molecules

= 1.500 mol O2


Step 2 calculate kclo 3 moles

Step 2: Calculate KClO3 moles

2KClO3 2KCl + 3O2

? mol

1.500 mol

2 mol KClO3

3 mol O2

x mol KClO3

1.500 mol O2

=

1.500 mol O2

2 mol KClO3

3 mol O2

= 1.00 mol KClO3


Step 3 calculate kclo 3 mass

Step 3: Calculate KClO3 mass

1.00 mol

122.55 g

1 mol

= 122.55 g KClO3

2KClO3 2KCl + 3O2

122.33 g

9.044 x 1023 molecules


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