Chapter 12 Stoichiometry 1 Section 12.1 The Arithmetic of - - PowerPoint PPT Presentation

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Chapter 12 Stoichiometry

Section 12.1

The Arithmetic

• f Equations

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Cookies and Chemistry…Huh!?!?

Lastly, instead of eggs, butter, sugar, etc. we use chemical compounds as ingredients Just like chocolate chip cookies have recipes, chemists have recipes as well Instead of calling them recipes, we call them reaction equations Furthermore, instead of using cups and teaspoons, we use moles

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Chemistry Recipes

Looking at a reaction tells us how much of something you need to react with something else to get a product (like the cookie recipe) Be sure you have a balanced reaction before you start!

Example: 2 Na + Cl2 → 2 NaCl

This reaction tells us that by mixing 2 moles of sodium with 1 mole of chlorine we will get 2 moles of sodium chloride What if we wanted 4 moles of NaCl? 10 moles? 50 moles?

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2H2 + O2 → 2H2O

◼ Two molecules of hydrogen and one molecule

• f oxygen form two molecules of water.

◼ 2 Al2O3 → Al + 3O2

2 formula units Al2O3 form 4 atoms Al and 3 molecules O2 Now try this: 2Na + 2H2O → 2NaOH + H2

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◼ Write the balanced reaction for hydrogen gas

reacting with oxygen gas.

2 H2 + O2 → 2 H2O

➢ How many moles of reactants are needed? ➢ What if we wanted 4 moles of water? ➢ What if we had 3 moles of oxygen, how much hydrogen would we need to react and how much water would we get? ➢ What if we had 50 moles of hydrogen, how much

• xygen would we need and how much water

produced?

Section 12.2

Chemical Calculations

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Mole Ratios

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These mole ratios can be used to calculate the moles of one chemical from the given amount of a different chemical Example: How many moles of chlorine are needed to react with 5 moles of sodium (without any sodium left over)?

Na + Cl2 → NaCl

2 mol 1 mol 5 mol x

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Cl mol 5 . 2 2 1 5 =  = x

2 2

Step 1: Write the chemical equation. Step 2: Balance the equation. Step 3: Step 4: Step 5: (Write the mole ratio) (Write the given)

Mole - Mole Conversions

◼ How many moles of sodium chloride will

be produced if you react 2.6 moles of chlorine gas with an excess (more than you need) of sodium metal?

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2 Na + Cl2 2 NaCl

1 mol 2 mol 2.6 mol x

NaCl mol 2 . 5 1 2 6 . 2 =  = x

Example:

Mole - Mole Conversions

◼ How many moles of O2 are

produced when 3.34 moles of Al2O3 decompose?

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Example:

Al2O3 → Al + O2

2 mol 3 mol

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O mol 01 . 5 2 3 34 . 3 =  = x

3.34 mol x

3  2

Mole - Mass Conversions

Most of the time in chemistry, the amounts are given in grams instead of moles We still go through moles and use the mole ratio, but now we also use molar mass to get to grams

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◼ Example: How many grams of chlorine

are required to react completely with 5.00 moles of sodium to produce sodium chloride?

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Na + Cl2 → NaCl 2 2

2 mol 1 mol 5 mol x

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Cl mol 5 . 2 2 1 5 =  = x

Change moles to grams: No. of moles X molar mass 2.5 X 71 = 178 g Cl2

◼ Calculate the mass in grams of Iodine

required to react completely with 0.50 moles of aluminum.

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Example:

I2 + Al → AlI3 2 3

3 mol 2 mol x 0.5 mol

2

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I mol 75 . 2 5 . 3 =  = x

Change moles to grams: No. of moles X molar mass 0.75 X 254 = 191 g I2

Mass – Mole Conversions

We can also start with mass and convert to moles of product or another reactant We use molar mass and the mole ratio to get to moles of the compound of interest

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◼ Calculate the number of moles of ethane (C2H6)

needed to produce 10.0 g of water

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Example:

2 C2H6 + 7 O2 → 4 CO2 + 6 H2O

2 mol 6 mol x 10 g

But before making cross multiplication, change 10 grams to moles 10 / 18.02 = 0.555 mol

6 2H

C 185 . 6 2 555 . mol x =  =

Mass – Mass Conversion

Example:

◼

Calculate how many grams of ammonia are produced when you react 2.00g of nitrogen with excess hydrogen.

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N2 + H2 → NH3

1 mol 2 mol 2 g x

3 2

But before making cross multiplication, change 2 grams to moles 2 / 28 = 0.0714 mol

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NH mol 143 . 1 2 0714 . =  = x

Change moles to grams: No. of moles X molar mass 0.143 X 17.03 = 2.44 g NH3

◼ Calculate how many grams of oxygen are

required to make 10.0 g of aluminum oxide

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Example:

Al2O3 3  2 Al + O2→

3 mol 2 mol x 10 g

But before making cross multiplication, change 10 grams to moles 10 / 102 = 0.098 mol

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O mol 147 . 2 3 098 . =  = x

Change moles to grams: No. of moles X molar mass 0.147 X 32 = 4.7 g O2

• If 3.84 moles of C2H2 are burned, how many

moles of O2 are needed?

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(9.6 mol)

Example:

2 C2H2 + 5 O2 → 4 CO2 + 2 H2O

2 mol 5 mol 3.84 mol x

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O mol 6 . 9 2 5 84 . 3 =  = x

• How many moles of C2H2 are needed to produce

8.95 mole of H2O? (8.95 mol)

Example:

2 C2H2 + 5 O2 → 4 CO2 + 2 H2O

2 mol 2 mol x 8.95 mol

2 2H

C mol 95 . 8 2 2 95 . 8 =  = x

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• If 2.47 moles of C2H2 are burned, how many

moles of CO2 are formed? (4.94 mol)

Example:

2 C2H2 + 5 O2 → 4 CO2 + 2 H2O

2 mol 4 mol 2.47 mol x

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CO mol 94 . 4 2 4 47 . 2 =  = x

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Mass – Mass Problem:

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6.50 grams of aluminum reacts with an excess of

• xygen. How many grams of aluminum oxide are

formed?

4 Al + 3 O2 → 2Al2O3

4 mol 2 mol 6.5 g x

But before making cross multiplication, change 6.5 grams to moles 6.5 / 27 = 0.241 mol Al

3 2O

Al mol 121 . 4 2 241 . =  = x

Change moles to grams: No. of moles X molar mass 0.121 X 102 = 12.3 g Al2O3

Another example:

2 Fe + 3 CuSO4 → Fe2(SO4)3 + 3 Cu

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If 10.1 g of Fe are added to a solution of Copper (II) Sulfate, how many grams of solid copper would form? 2 mol 3 mol 10.1 g x

before making cross multiplication, change 10.1 grams to moles 10.1 / 55.8 = 0.181 mol Fe

Cu mol 272 . 2 3 181 . =  = x

Change moles to grams: No. of moles X molar mass 0.272 X 63.5 = 17.3 g Cu

Volume – Volume Calculations:

◼ How many liters of CH4 at STP are required to

completely react with 17.5 L of O2 ?

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CH4 + 2 O2 → CO2 + 2 H2O

1 mol 2 mol x 17.5 L

before making cross multiplication, change 17.5 L to moles 17.5 / 22.4 = 0.781 mol O2

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CH mol 391 . 2 1 781 . =  = x

Change moles to liters:

• No. of moles X 22.4

0.391 X 22.4 = 8.76 L CH4

Section 12.3

Limiting Reactant and Yield

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Limiting Reactant: Cookies

1 cup butter, 1/2 cup white sugar 1 cup packed brown sugar 1 teaspoon vanilla extract 2 eggs 2 1/2 cups all-purpose flour 1 teaspoon baking soda 1 teaspoon salt 2 cups semisweet chocolate chips Makes 3 dozen

If we had the specified amount of all ingredients listed, could we make 4 dozen cookies? What if we had 6 eggs and twice as much of everything else, could we make 9 dozen cookies? What if we only had one egg, could we make 3 dozen cookies?

Limiting Reactant

◼ Most of the time in chemistry we have more

• f one reactant than we need to completely

use up other reactant.

◼ That reactant is said to be in excess

(there is too much).

◼ The other reactant limits how much product

we get. Once it runs out, the reaction s. This is called the limiting reactant.

Limiting Reactant

◼ To find the correct answer, we have to try all of the

• reactants. We have to calculate how much of a

product we can get from each of the reactants to determine which reactant is the limiting one.

◼ The lower amount of a product is the correct

answer.

◼ The reactant that makes the least amount of product

is the limiting reactant. Once you determine the limiting reactant, you should ALWAYS start with it!

◼ Be sure to pick a product! You can’t compare to see

which is greater and which is lower unless the product is the same!

Example

◼ 10.0g of aluminum reacts with 35.0 grams of chlorine gas to

produce aluminum chloride. Which reactant is limiting, which is in excess, and how much product is produced?

2 Al + 3 Cl2 2 AlCl3

◼ Start with Al: ◼ Now Cl2:

2 mol 2 mol 10 g x

But before making cross multiplication, change 10 grams to moles 10 / 27 = 0.370 mol

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C A mol 370 . 2 2 370 . l l x =  = 3 mol 2 mol 35 g x

But before making cross multiplication, change 35 grams to moles 35 / 71 = 0.493 mol

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C A mol 329 . 3 2 493 . l l x =  =

0.370 x 133.5 = 49.4 g 0.329 x 133.5 = 43.9 g

LR Example Continued

◼ We get 49.4 g of aluminum chloride from

the given amount of aluminum, but only 43.9 g of aluminum chloride from the given amount of chlorine. Therefore, chlorine is the limiting reactant. Once the 33.6 g of chlorine is used up, the reaction comes to a complete .

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◼ If 10.6 g of copper reacts with 3.83 g sulfur, how many grams of product (cuprous sulfide) will be formed?

Practice

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◼ 15.0 g of potassium reacts with 15.0 g of

• iodine. Calculate which reactant is limiting

and how much product is made.

Finding the Amount of Excess

◼ By calculating the amount of the excess

reactant needed to completely react with the limiting reactant, we can subtract that amount from the given amount to find the amount of excess.

◼ Can we find the amount of excess

potassium in the previous problem?

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Practice

15.0 g of potassium reacts with 15.0 g of iodine. If 19.6 g of potassium iodide is produced, find the limiting and excess reactants then how much of the excess amount is left over?

Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen.

4KO2(s) + 2H2O(l) → 4KOH(s) + 3O2(g)

• a. How many moles of O2 can be produced from 0.15 mol KO2

and 0.10 mol H2O?

• b. Determine the limiting and excess reactants.
• c. How much of the excess is left over?

Limiting Reactant: Recap

• 1. You can recognize a limiting reactant problem

because there is MORE THAN ONE GIVEN AMOUNT.

• 2. Convert ALL of the reactants to the SAME product

(pick any product you choose.)

• 3. The lowest answer is the correct answer.
• 4. The reactant that gave you the lowest answer is

the LIMITING REACTANT.

• 5. The other reactant(s) are in EXCESS.
• 6. To find the amount of excess, subtract the

amount used from the given amount.

• 7. If you have to find more than one product, be

sure to start with the limiting reactant. You don’t have to determine which is the LR over and over again!

Yield

◼ The amount of product made in a

chemical reaction.

◼ There are three types ◼ Actual yield- what you get in the lab when

the chemicals are mixed

◼ Theoretical yield- what the balanced

equation tells you should make.

◼ Percent yield = Actual x 100 %

Theoretical

Percent Yield

◼ To determine percentage yield, you need

two pieces of information:

1) Theoretical yield = 2) Actual yield = Comes from the stoichiometry Comes from experimental data Percent Yield = Actual Yield Theoretical Yield 100%

Example

◼ 6.78 g of copper is produced when 3.92 g

• f Al are reacted with excess copper (II)

sulfate.

◼ 2Al + 3 CuSO4 → Al2(SO4)3 + 3Cu ◼ What is the actual yield? ◼ What is the theoretical yield? ◼ What is the percent yield? ◼ If you had started with 9.73 g of Al, how

much copper would you expect?

Details

◼ Percent yield tells us how “efficient” a

reaction is.

◼ Percent yield can not be bigger than 100 %.

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12.75 moles of NaClO3 will produce how many grams of O2? Calculate the percent yield of oxygen if 20 grams is produced. NaClO3 NaCl + O2