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Stoichiometry

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CH. 9

Stoichiometry

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

- 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

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.

MASS

IN

GRAMS

NUMBER

OF

PARTICLES

MOLES

Molar Mass

(g/mol)

6.02 1023

particles/mol

- 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

- 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

2KClO3 2KCl + 3O2

? g

9 g

1mol

32.0g

9.00 g

= 0.281 mol O2

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

0.187mol

122.55 g

1 mol

= 22.9 g KClO3

2KClO3 2KCl + 3O2

22.9 g

9.00 g

- 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

2KClO3 2KCl + 3O2

? g

9 g

1.426 g

1 mL

6.3 mL

9 g O2

2KClO3 2KCl + 3O2

? g

9 g

1mol

32.0g

9.00 g

= 0.281 mol O2

2KClO3 2KCl + 3O2

? mol

0.281 mol

0.281 mol O2

2 mol KClO3

3 mol O2

= 0.187 mol KClO3

0.187mol

122.55 g

1 mol

= 22.9 g KClO3

2KClO3 2KCl + 3O2

22.9 g

6.3 mL

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

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

- 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

2KClO3 2KCl + 3O2

? g

9.00 L

1mol

22.4L

9.00 L

= 0.4 mol O2

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

0.27 mol

122.55 g

1 mol

= 33 g KClO3

2KClO3 2KCl + 3O2

33 g

9.00 L

- 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

=

- 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

2KClO3 2KCl + 3O2

? g

9.044 x 1023 molecules

9.044 x 1023molecules

1mol

6.022 x 1023molecules

= 1.500 mol O2

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

1.00 mol

122.55 g

1 mol

= 122.55 g KClO3

2KClO3 2KCl + 3O2

122.33 g

9.044 x 1023 molecules