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Unit 08 – Moles and StoichiometryPowerPoint Presentation

Unit 08 – Moles and Stoichiometry

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Unit 08 – Moles and Stoichiometry

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Unit 08 – Moles and Stoichiometry

- Molar Conversions

VERY

A large amount!!!!

- A counting number (like a dozen)
- Avogadro’s number (6.02 1023 particles) (SI unit)
- 1 mol = molar mass

HOW LARGE IS IT???

- 1 mole of pennies would cover the Earth 1/4 mile deep!

- 1 mole of hockey pucks would equal the mass of the moon!

- 1 mole of basketballs would fill a bag the size of the earth!

- Molar Mass- the mass of a mole of any element or compound (in grams)
- Round to 2 decimal places

- Gram Formula mass – sum of the atomic masses of all the atoms in a formula of a compound
- Formula weight

- water
- sodium chloride

- H2O
- (1.01g x 2) + 16.00g = 18.02 g
- NaCl
- 22.99g + 35.45g = 58.44 g

- sodium hydrogen carbonate
- sucrose

- NaHCO3
- 22.99g + 1.01g + 12.01g + (16.00g x 3) = 84.01 g
- C12H22O11
- (12.01g x12) + (1.01g x 22) + (16.00g x11)= 342.34 g

1 mole = 6.02 × 10 23representative particles

(also called Avogadro’s Number)

- Atom- rep. particle for most elements
- Ions – if atom is charged
- Molecule- rep. particle for covalent compounds and diatomic molecules “BrINCl HOF”
- Formula unit- rep. particle for ionic compounds

What is a representative particle?

How the substance normally exists:

- The Volume of a gas varies with a change in temperature or pressure.
- Measured at standard temperature and pressure (STP)
- 0°C at 1 atmosphere (atm)

Molecule

Atoms (ions)

Formula unit

- How many moles of carbon are in 26 g of carbon?

26 g C

1 mol C

12.01 g C

= 2.2 mol C

- How many molecules are in 2.50 moles of C12H22O11?

6.02 1023

molecules

1 mol

2.50 mol

= 1.51 1024

molecules

C12H22O11

- Find the number of molecules of 12.00 L of O2 gas at STP.

6.02 x 1023 molecules

1 mol

1 mol

22.4 L

12.00 L

= 3.225 x 1023 molecules

II. Stoichiometric Calculations

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. Convert known to moles (IF NECESSARY) Line up conversion factors.

4. Use Mole ratio – from equation

5. Convert moles to unknown unit

(IF NECESSARY)

6. Calculate and write units.

C. Stoichiometry Problems

- Calculate the number of grams of NH3 produced by the reaction of 5.40 g of hydrogen with an excess of nitrogen.
N2 + 3H2→ 2NH3

III. Stoichiometry in the Real World

- Available Ingredients
- 4 slices of bread
- 1 jar of peanut butter
- 1/2 jar of jelly

- Limiting Reactant
- bread

- Excess Reactants
- peanut butter and jelly

- Limiting Reactant
- used up in a reaction
- determines the amount of product

- Excess 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

Using the following equation identify the limiting reagent.

- How many moles of ammonia (NH3) can be produced from the reaction of 28.2 L of nitrogen and 25.3 L of hydrogen?

N2 + 3H2 2NH3

28.2 L

? mol

25.3 L

N2 + 3H2 2NH3

28.2 L

? mol

25.3 L

28.2

L N2

1 mol

N2

22.4

L N2

2 mol

NH3

1 mol

N2

= 2.5 mol NH3

N2 + 3H2 2NH3

28.2 L

? mol

25.3 L

25.3

L H2

1 mol

H2

22.4 L H2

2 mol

NH3

3 mol

H2

= 0.753 mol NH3

N2: 2.5 mol NH3 H2: 0.753 mol NH3

Limiting reactant: H2

Excess reactant: N2

Product Formed: 0.753 mol NH3

Mg + 2HCl → MgCl2 + H2

How many grams of magnesium chloride are produced from the reaction of 2.08 mol of Mg and 2.08 mol of HCl?

measured in lab

calculated on paper

- 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

K2CO3 + 2HCl 2KCl + H2O + CO2

Theoretical Yield:

45.8 g

? g

actual: 46.3 g

45.8 g

K2CO3

1 mol

K2CO3

138.21 g

K2CO3

2 mol

KCl

1 mol

K2CO3

74.55

g KCl

1 mol

KCl

= 49.4

g KCl

46.3 g

49.4 g

K2CO3 + 2HCl 2KCl + H2O + CO2

Theoretical Yield = 49.4 g KCl

45.8 g

49.4 g

actual: 46.3 g

100 =

93.7%

% Yield =