Chemical Reactions and the Mole….

Chemical Reactions and the Mole….

•1 Fe 1 Fe (s)(s) + 1 S + 1 S (s)(s) 1 FeS 1 FeS (s)(s)

•This chemical reaction means one atom This chemical reaction means one atom of iron reacts with one atom of sulfur to of iron reacts with one atom of sulfur to form one molecule of iron (II) sulfideform one molecule of iron (II) sulfide•Can we do this chemical reaction?Can we do this chemical reaction?•No - atoms are too small to see - No - atoms are too small to see - •We must perform the reaction many times We must perform the reaction many times larger to see it!larger to see it!•We always perform the reaction 6.02 x We always perform the reaction 6.02 x 101023 23 times larger, or 1 mole times largertimes larger, or 1 mole times larger

1 Fe 1 Fe (s)(s) + 1 S + 1 S (s)(s) 1 FeS 1 FeS (s)(s)

• If we doIf we do this reaction 6.02 x 10this reaction 6.02 x 102323 times bigger, it would be times bigger, it would be

written:written:

6.02 x 106.02 x 102323 Fe Fe (s)(s) + 6.02 x 10 + 6.02 x 102323 S S (s)(s) 6.02 x 10 6.02 x 102323 FeS FeS (s)(s)

• Since 6.02 x 10Since 6.02 x 102323 is 1 mole, we could also say: is 1 mole, we could also say:

1 mole Fe 1 mole Fe (s)(s) + 1 mole S + 1 mole S (s)(s) 1 mole FeS 1 mole FeS (s)(s)

• So we could view the 1 coefficient as either 1 atom, or 1 So we could view the 1 coefficient as either 1 atom, or 1 mole, depending on whether we were looking at the mole, depending on whether we were looking at the reaction on the atomic scale, or the large, human, molar reaction on the atomic scale, or the large, human, molar scale!scale!

1 Fe (s) + 1 S (s) 1 FeS (s)

• Here’s the problem -

• If you have 10 grams of Fe, how many grams of S do you need?

• Most would say 10 grams!

• It is a 1:1 ratio!

• But the units in the chemical reaction aren’t grams!

• The 1 coefficient does not stand for grams - it stands for moles!

• If you are going to Mexico, your dollars have to be converted into pesos before you go!

• When you look at the ratio in a chemical reaction, you must be in moles!

1 Fe (s) + 1 S (s) 1 FeS (s)

• Here’s the problem -

• If you have 10 grams of Fe, how many grams of S do you need?

• Most would say 10 grams!

• It is a 1:1 ratio!

• But the units in the chemical reaction aren’t grams!

• The 1 coefficient does not stand for grams - it stands for moles!

• If you are going to Mexico, your dollars have to be converted into pesos before you go!

• When you look at the ratio in a chemical reaction, you must be in moles!

• Let’s look at a sample problem…• Number one in your homework packet under Easy Stoichiometry Problems…

Carbon dioxide can be commercially prepared by heating chalk, or calcium carbonate. How many moles of carbon dioxide can be produced when 3.05 moles of calcium carbonate are heated?

• How do we approach this problem?•There are four steps we follow in any problem we do with chemical reactions and amounts…

• Let’s look at a sample problem…• Number one in your homework packet under Easy Stoichiometry Problems…

Carbon dioxide can be commercially prepared by heating chalk, or calcium carbonate. How many moles of carbon dioxide can be produced when 3.05 moles of calcium carbonate are heated?

• How do we approach this problem?•There are four steps we follow in any problem we do with chemical reactions and amounts…

1. Write a balanced chemical reaction

2. Make sure you are in the unit of moles (mole map)

3. Set up a ratio from the chemical reaction, putting what you want to solve for on top, and what you want to cancel on bottom

4. Convert out of the unit of moles, if necessary (mole map)

1. Write a balanced chemical reaction

2. Make sure you are in the unit of moles (mole map)

3. Set up a ratio from the chemical reaction, putting what you want to solve for on top, and what you want to cancel on bottom

4. Convert out of the unit of moles, if necessary (mole map)

Volume of A (liquid)

Volume of B (liquid)

Volume of A (gas)

Volume of B (gas)

Grams A Grams BMoles A Moles BUse

Molar Mass

Use

Molar Mass

STP

Use

22.

4 L

1 m

ole STP

Use 22.4 L

1 mole

Use M

olarityU

se M

olar

ity

moles

Liter mol

es

Lite

r

USE YOUR “CHEMICAL”RECIPE HERE

SET UP ARATIO FROMYOUR BALANCEDEQUATION

X moles you want moles you have

• Step 1: Write a balanced chemical Step 1: Write a balanced chemical equation:equation:

• 1 CaCO1 CaCO3 (s) 3 (s) 1 CaO 1 CaO (s) (s) + 1 CO+ 1 CO2 (g)2 (g)

• Step 2: Convert into the unit of moles:Step 2: Convert into the unit of moles:

• Done! We already have 3.05 moles Done! We already have 3.05 moles CaCOCaCO33

• Step 3: Set up a ratio:Step 3: Set up a ratio:• The reaction says that I can create 1 The reaction says that I can create 1 mole of COmole of CO22 for every 1 mole of CaCO for every 1 mole of CaCO33 I I

start with start with • It is just like a recipe!It is just like a recipe!

1 CaCO1 CaCO3 (s) 3 (s) 1 CaO 1 CaO (s) (s) + 1 CO+ 1 CO2 (g)2 (g)

• 3.05 moles CaCO3.05 moles CaCO3 3 x 1 mole COx 1 mole CO2 =2 =

1 mole CaCO1 mole CaCO33

3.05 moles CaCO3.05 moles CaCO33

• No step 4 is required, because the No step 4 is required, because the

problem asks for the answer in problem asks for the answer in moles!moles!

• Let’s look at problem #10….

• Group 1 metals are explosive when they come into contact with water. If 20.0 grams of potassium were to explode with excess water, how many moles of hydrogen would be produced?

• Let’s write a balanced chemical equation - step 1!

• 2 K (s) + 2 HOH (l) 2 KOH (aq) + 1 H2 (g)

• Writing reactions will be CRITICAL on the test!!!

• Let’s look at problem #10….

• Group 1 metals are explosive when they come into contact with water. If 20.0 grams of potassium were to explode with excess water, how many moles of hydrogen would be produced?

• Let’s write a balanced chemical equation - step 1!

• 2 K (s) + 2 HOH (l) 2 KOH (aq) + 1 H2 (g)

• Writing reactions will be CRITICAL on the test!!!

2 K (s) + 2 HOH (l) 2 KOH (aq) + 1 H2 (g)

• Step 2: Convert into moles:

• 20.0 g K x 1 mole = .51 moles K

39.1 grams

• I MUST be in the unit of the mole to be able to look at the ratios or amounts from the chemical reaction!

• Step 3: Set up a ratio from the reaction:

• .51 moles K x 1 mole H2 = .255 moles H2

2 mole K

• No step 4 is required because the problem asks for moles of hydrogen

2 K (s) + 2 HOH (l) 2 KOH (aq) + 1 H2 (g)

• Step 2: Convert into moles:

• 20.0 g K x 1 mole = .51 moles K

39.1 grams

• I MUST be in the unit of the mole to be able to look at the ratios or amounts from the chemical reaction!

• Step 3: Set up a ratio from the reaction:

• .51 moles K x 1 mole H2 = .255 moles H2

2 mole K

• No step 4 is required because the problem asks for moles of hydrogen

• Let’s look at problem #29….

• Cigarette lighters use butane, or C4H10, as their fuel. If 120 grams of butane burn with only 55 liters of oxygen gas at STP, how many liters of carbon dioxide are produced? Which reactant is your limiting reactant?

• Let’s write a balanced chemical equation - step 1!

• 2 C4H10 (g) + 13 O2 (g) 10 HOH (g) + 8 CO2 (g)

• There is a problem here….

•One of our starting chemicals is going to run out!

• Let’s look at problem #29….

• Cigarette lighters use butane, or C4H10, as their fuel. If 120 grams of butane burn with only 55 liters of oxygen gas at STP, how many liters of carbon dioxide are produced? Which reactant is your limiting reactant?

• Let’s write a balanced chemical equation - step 1!

• 2 C4H10 (g) + 13 O2 (g) 10 HOH (g) + 8 CO2 (g)

• There is a problem here….

•One of our starting chemicals is going to run out!

• 2 C2 C44HH1010 (g) (g) + 13 O+ 13 O22 (g)(g) 10 HOH 10 HOH (g) (g) + 8 CO+ 8 CO2 (g)2 (g)

• To figure out which starting material runs out, called our To figure out which starting material runs out, called our limiting reactantlimiting reactant, we must first convert both into moles:, we must first convert both into moles:

• 120.0 g C120.0 g C44HH1010 x 1 mole = 2.07 moles C x 1 mole = 2.07 moles C44HH1010

58 grams58 grams

• 55 L O55 L O22 x 1 mole = 2.46 moles O x 1 mole = 2.46 moles O22

22.4 liters22.4 liters

•Which one runs out?

• Is it always the one we have less of?

• What if I was making sandwiches, and used two pieces of bread, and one piece of bologna, per sandwich?

•With 8 pieces of bologna, and 10 pieces of bread, what runs out….?

•Which one runs out?

• Is it always the one we have less of?

• What if I was making sandwiches, and used two pieces of bread, and one piece of bologna, per sandwich?

•With 8 pieces of bologna, and 10 pieces of bread, what runs out….?

• The bread!The bread!

• I need twice as much bread as I need twice as much bread as bologna to make a sandwich….bologna to make a sandwich….

• In this problem, what runs out?In this problem, what runs out?

• 2 C4H10 (g) + 13 O2 (g) 10 HOH (g) + 8 CO2 (g)

• 120.0 g C4H10 x 1 mole = 2.07 moles C4H10

58 g C4H10

• 55 L O2 x 1 mole O2 = 2.46 moles O2

22.4 L O2

• To see which one runs out, we divide each number by the number needed in the reaction…

• 2.07/2 for the C4H10, and 2.46/13 for the O2

• Which of these numbers is less?

• That is the one that runs out…

• The O2 is less, so it runs out!

• We finish the problem with the amount of O2, and ignore the C4H10

• All problems are done this way - finish the problem with the chemical that runs out, and ignore the one you have extra of!

• 2 C4H10 (g) + 13 O2 (g) 10 HOH (g) + 8 CO2 (g)

• 120.0 g C4H10 x 1 mole = 2.07 moles C4H10

58 g C4H10

• 55 L O2 x 1 mole O2 = 2.46 moles O2

22.4 L O2

• To see which one runs out, we divide each number by the number needed in the reaction…

• 2.07/2 for the C4H10, and 2.46/13 for the O2

• Which of these numbers is less?

• That is the one that runs out…

• The O2 is less, so it runs out!

• We finish the problem with the amount of O2, and ignore the C4H10

• All problems are done this way - finish the problem with the chemical that runs out, and ignore the one you have extra of!

• 2 C4H10 (g) + 13 O2 (g) 10 HOH (g) + 8 CO2 (g)

• Step 3: Set up a ratio from the reaction:

• 2.46 moles O2 x 8 moles CO2 = 1.51 moles CO2

13 moles O2

• Step 4: Convert out of the mole:

• 1.51 moles CO2 x 22.4 L = 33.85 L CO2

1 mole CO2

• Now try some problems on your own!

• 2 C4H10 (g) + 13 O2 (g) 10 HOH (g) + 8 CO2 (g)

• Step 3: Set up a ratio from the reaction:

• 2.46 moles O2 x 8 moles CO2 = 1.51 moles CO2

13 moles O2

• Step 4: Convert out of the mole:

• 1.51 moles CO2 x 22.4 L = 33.85 L CO2

1 mole CO2

• Now try some problems on your own!