stoichiometry! is the calculation of relative quantities of reactants and products in chemical...

Stoichiometry! is the calculation of relative quantities of reactants and products in chemical reactions.

SOURCE: Hands on Chemistry Activities…Herr

Stoichiometry is found in every branch of science!

Starting with Water!

2H2 + O2 2H20

Have No Fear Of Ice Cold Beer

LEFT RIGHTH O H O2 2 2 1

2 2 4 2

4 2 4 2

You have to balance equations by changing Coefficients!

H O

H H

O

H

+O H

O

H

H

H

Reading Chemical Equations!

2H2 + O2 2H20 What does this mean?

Two molecules of H2 gas combine with One molecule of O2 gas to produce Two molecules of Water.

Reading Chemical Equations!

2H2 + O2 2H20 What does this mean?

Two molecules of H2 gas combine with One molecule of O2 gas to produce Two molecules of Water.

2H2 + O2 2H20Two MOLES of H2 gas combine with One MOLE of O2 gas to produce Two MOLES of Water.

Reading Chemical Equations!

2H2 + O2 2H20 What does this mean?

Two molecules of H2 gas combine with One molecule of O2 gas to produce Two molecules of Water.

2H2 + O2 2H20Two MOLES of H2 gas combine with One MOLE of O2 gas to produce Two MOLES of Water.

Four GRAMS of H2 gas combine with Thirty-Two GRAMS of O2 gas to produce Thirty-Six GRAMS of Water.

Reading Chemical Equations!

CH4 + 2O2 CO2+ 2H20 What does this mean?

Reading Chemical Equations!

CH4 + 2O2 CO2+ 2H20 What does this mean?

One molecule of CH4 (methane) combines with Two molecules of O2 gas to produce One molecule of CO2 (carbon dioxide gas) and Two molecules of Water.

SOURCE Wiki: Stoichiometry

Reading Chemical Equations!

CH4 + 2O2 CO2+ 2H20 What does this mean?

One molecule of CH4 (methane) combines with Two molecules of O2 gas to produce One molecule of CO2 (carbon dioxide gas) and Two molecules of Water.

SOURCE Wiki: Stoichiometry

One MOLE of CH4 (methane) combines with Two MOLES of O2 gas to produce One MOLE of CO2 (carbon dioxide gas) and Two MOLES of Water.

Reading Chemical Equations!

CH4 + 2O2 CO2+ 2H20 What does this mean?

One molecule of CH4 (methane) combines with Two molecules of O2 gas to produce One molecule of CO2 (carbon dioxide gas) and Two molecules of Water.

SOURCE Wiki: Stoichiometry

One MOLE of CH4 (methane) combines with Two MOLES of O2 gas to produce One MOLE of CO2 (carbon dioxide gas) and Two MOLES of Water.

Sixteen GRAMS of CH4 combines with Sixty-Four GRAMS of O2 gas to produce Fourty-Four GRAMS of CO2 and Thirty-Six GRAMS of Water.

x =

Unit you want to change Unit you are changing into

Conversion

Factor

How many feet is 744 inches?

x =

Unit you want to change Unit you are changing into

Conversion

Factor

How many moles is 400g of Sodium?

UNIT CANCELLING WORKSHEET

• MOLE RATIO WORKSHEET

WATER: Stoichiometry

2H2 + O2 2H20If you have 80 molecules of H2, how much O2 do you need to react with all of it and how much water would you get?

WATER: Stoichiometry

2H2 + O2 2H20If you have 80 molecules of H2, how much O2 do you need to react with all of it and how much water would you get?

€

80 molecules H2 x 1 molecule 02

2 molecules H2

= 40 moleculesO2

WATER: Stoichiometry

2H2 + O2 2H20If you have 80 molecules of H2, how much O2 do you need to react with all of it and how much water would you get?

€

80 molecules H2 x 1 molecule 02

2 molecules H2

= 40 moleculesO2

€

80 molecules H2 x 2 molecules H2O

2 molecules H2

= 80 molecules H2O

WATER: Stoichiometry

2H2 + O2 2H20If you have 80 molecules of H2, how much O2 do you need to react with all of it and how much water would you get?

€

80 molecules H2 x 1 molecule 02

2 molecules H2

= 40 moleculesO2

€

80 molecules H2 x 2 molecules H2O

2 molecules H2

= 80 molecules H2O

€

40 molecules O2 x 2 molecules H2O

1 molecules O2

= 80 molecules H2O

WATER: Stoichiometry

2H2 + O2 2H20If you have 80 MOLES of H2, how much O2 do you need to react with all of it and how much water would you get?

€

80 MOLES H2 x 1 MOLE 02

2 MOLE H2

= 40 MOLES O2

€

80 MOLES H2 x 2 MOLES H2O

2 MOLES H2

= 80 MOLES H2O

€

40 MOLES O2 x 2 MOLES H2O

1 MOLES O2

= 80 MOLES H2O

RUST: Iron (?) Oxide

2Fe + 3O2 2Fe203

If you have 444g of Iron, how many moles of O2 will you need to react with all of it and how much rust will you end up with?

First Convert Grams into Moles!!!

RUST: Iron (?) Oxide

2Fe + 3O2 2Fe203

If you have 444g of Iron, how many moles of O2 will you need to react with all of it and how much rust will you end up with?

26

FeIRON55.85

First Convert Grams into Moles!!!

RUST: Iron (?) Oxide

2Fe + 3O2 2Fe203

If you have 444g of Iron, how many moles of O2 will you need to react with all of it and how much rust will you end up with?

26

FeIRON55.85

First Convert Grams into Moles!!!Molar Mass = 55.85g/mol

RUST: Iron (?) Oxide

4Fe + 3O2 2Fe203

If you have 444g of Iron, how many moles of O2 will you need to react with all of it and how much rust will you end up with?

€

444g Fe x 1 mole Fe

55. 85g Fe= 7.9 moles Fe

26

FeIRON55.85

First Convert Grams into Moles!!!Molar Mass = 55.85g/mol

RUST: Iron (?) Oxide

4Fe + 3O2 2Fe203

If you have 444g of Iron, how many moles of O2 will you need to react with all of it and how much rust will you end up with?

€

444g Fe x 1 mole Fe

55. 85g Fe= 7.9 moles Fe

26

FeIRON55.85

Molar Mass = 55.85g/mol

RUST: Iron (?) Oxide

4Fe + 3O2 2Fe203

If you have 444g of Iron, how many moles of O2 will you need to react with all of it and how much rust will you end up with?

€

444g Fe x 1 mole Fe

55. 85g Fe= 7.9 moles Fe

26

FeIRON55.85

Molar Mass = 55.85g/mol

€

7.9mol Fe x 3 mol O2

4mol Fe= 5.9mol O2

RUST: Iron (?) Oxide

4Fe + 3O2 2Fe203

If you have 444g of Iron, how many moles of O2 will you need to react with all of it and how much rust will you end up with?

€

444g Fe x 1 mole Fe

55. 85g Fe= 7.9 moles Fe

26

FeIRON55.85

Molar Mass = 55.85g/mol

€

7.9mol Fe x 3 mol O2

4mol Fe= 5.9mol O2

RUST: Iron (?) Oxide

4Fe + 3O2 2Fe203

If you have 444g of Iron, how many moles of O2 will you need to react with all of it and how much rust will you end up with?

€

444g Fe x 1 mole Fe

55. 85g Fe= 7.9 moles Fe

26

FeIRON55.85

€

7.9mol Fe x 3 mol O2

4mol Fe= 5.9mol O2

€

7.9mol Fe x 2 mol Fe2O3

4mol Fe= 3.95 moles Fe2O3

RUST: Iron (?) Oxide

4Fe + 3O2 2Fe203

If you have 444g of Iron, how many moles of O2 will you need to react with all of it and how much rust will you end up with?

€

444g Fe x 1 mole Fe

55. 85g Fe= 7.9 moles Fe

26

FeIRON55.85

€

7.9mol Fe x 3 mol O2

4mol Fe= 5.9mol O2

€

7.9mol Fe x 2 mol Fe2O3

4mol Fe= 3.95 moles Fe2O3

Photochemical Smog!

Photochemical Smog!

Assignment:

1. Hydrogen Sulfide which smells like rotten eggs is found in volcanic ash. The balanced equation for the burning of hydrogen sulfide is 2H2S(g) + 3O2(g) 2SO2(g) + 2H2O(g). Express this equation using the masses of all the reactants and products.

2. If you were given 7.5 x 1024 atoms of aluminum, how many moles of oxygen would you need so that all of the aluminum forms aluminum oxide? How many moles of aluminum oxide would be produced? ( Hint: atoms to moles via avogadro’s number and write a balanced equation for the formation of aluminum oxide).

3. How many molecules of oxygen are produced when 29.2 grams of H2O is decomposed by electrolysis into diatomic hydrogen and oxygen gas?

4. How many moles of ammonia are produced when 0.6mol nitrogen reacts with hydrogen? N2 + 3H2 2NH3

5. Calculate the number of grams of NH3 produced when 5.4grams of hydrogen reacts with an excess of nitrogen.

6. Phosphorous and hydrogen can be combined to form phosphine (PH3) according to the following equation: P4(s) + 6H2(g) 4PH3(g). How many liters of phosphine are formed when 0.42L of hydrogen reacts with an excess of phosphorous?

A Chemical Formula is Like a Recipe: BROWNIES

2cW + 4cS + 4cF + 4E + 8ozB + 8pcC 10 Brownies

A Chemical Formula is Like a Recipe: BROWNIES

We know that 4 eggs combined with everything else yields 10brownies but what if you only had 3 eggs?

2cW + 4cS + 4cF + 4E + 8ozB + 8pcC 10 Brownies

A Chemical Formula is Like a Recipe: BROWNIES

We know that 4 eggs combined with everything else yields 10brownies but what if you only had 3 eggs?

You have ¾ of the needed eggs so cut the entire recipe by ¾ and end up with ¾ the brownies.

2cW + 4cS + 4cF + 4E + 8ozB + 8pcC 10 Brownies

A Chemical Formula is Like a Recipe: BROWNIES

2cW + 4cS + 4cF + 4E + 8ozB + 8pcC 10 Brownies

€

3 eggs x10 brownies

4 eggs= 7.5 brownies

€

3 eggs x2 cups water

4 eggs=1.5 cups water

How many Brownies can you make? What limits your from making more?

A Chemical Formula is Like a Recipe: BROWNIES

Sometimes in chemistry we don’t have all the necessary ingredients

for complete reactions!

2cW + 4cS + 4cF + 4E + 8ozB + 8pcC 10 Brownies

Water

2H2 + O2 2H20Suppose you have 50moles of Oxygen and 40 moles of hydrogen. How much water can you make?

Water

2H2 + O2 2H20Suppose you have 50moles of Oxygen and 40 moles of hydrogen. How much water can you make?

By looking at the balanced equation we know that you need 2 moles of hydrogen for every mole of oxygen but we only have 40 moles of hydrogen.

Water

2H2 + O2 2H20Suppose you have 50moles of Oxygen and 40 moles of hydrogen. How much water can you make?

By looking at the balanced equation we know that you need 2 moles of hydrogen for every mole of oxygen but we only have 40 moles of hydrogen.

€

40 moles H2 x1 mole O2

2 moles H2

= 20 moles O2

Water

2H2 + O2 2H20Suppose you have 50moles of Oxygen and 40 moles of hydrogen. How much water can you make?

By looking at the balanced equation we know that you need 2 moles of hydrogen for every mole of oxygen but we only have 40 moles of hydrogen.

€

40 moles H2 x1 mole O2

2 moles H2

= 20 moles O2

Only 20 moles of the Oxygen will react and there will be an excess of 30 moles leftover.

Water

2H2 + O2 2H20Suppose you have 50moles of Oxygen and 40 moles of hydrogen. How much water can you make?

By looking at the balanced equation we know that you need 2 moles of hydrogen for every mole of oxygen but we only have 40 moles of hydrogen.

€

40 moles H2 x1 mole O2

2 moles H2

= 20 moles O2

Only 20 moles of the Oxygen will react and there will be an excess of 30 moles leftover.

Hydrogen: limiting reagentOxygen: excess reagent

Limiting and Excess

• limiting reagent is the reagent that determines the amount of product that can be formed by a reaction

• The reagent that is not used up is called the excess reagent.

Copper Reacts with sulfur to form copper (I) sulfide according to the following balanced equation.

2Cu(s) + S(s) Cu2S(s)

What is the limiting reagent when 80g Cu reacts with 25g S?

Copper Reacts with sulfur to form copper (I) sulfide according to the following balanced equation.

2Cu(s) + S(s) Cu2S(s)

What is the limiting reagent when 80g Cu reacts with 25g S?

You need twice as much Copper as sulfur in terms of moles but our problem gives us values in grams.

First we need to convert 80g Cu moles and 25g S moles.

Copper Reacts with sulfur to form copper (I) sulfide according to the following balanced equation.

2Cu(s) + S(s) Cu2S(s)

What is the limiting reagent when 80g Cu reacts with 25g S?

€

80g Cu x1mol Cu

63.5gCu=1.26mol Cu

Copper Reacts with sulfur to form copper (I) sulfide according to the following balanced equation.

2Cu(s) + S(s) Cu2S(s)

What is the limiting reagent when 80g Cu reacts with 25g S?

€

80g Cu x1mol Cu

63.5gCu=1.26mol Cu

€

25g S x1mol S

32.1g S= 0.779mol S

Copper Reacts with sulfur to form copper (I) sulfide according to the following balanced equation.

2Cu(s) + S(s) Cu2S(s)

What is the limiting reagent when 80g Cu reacts with 25g S?

€

80g Cu x1mol Cu

63.5gCu=1.26mol Cu

€

25g S x1mol S

32.1g S= 0.779mol S

We Know that we need twice as much copper as sulfur from the balanced equation. We don’t have twice as much so Copper is the limiting reagent! There will be an excess of sulfur.

Copper Reacts with sulfur to form copper (I) sulfide according to the following balanced equation.

2Cu(s) + S(s) Cu2S(s)

What is the maximum amount of Cu2S that can be formed in grams when 80g of Cu reacts with 25g of S?

€

80g Cu x1mol Cu

63.5gCu=1.26mol Cu

€

25g S x1mol S

32.1g S= 0.779mol S

Copper Reacts with sulfur to form copper (I) sulfide according to the following balanced equation.

2Cu(s) + S(s) Cu2S(s)

What is the maximum amount of Cu2S that can be formed in grams when 80g of Cu reacts with 25g of S?

€

80g Cu x1mol Cu

63.5gCu=1.26mol Cu

€

25g S x1mol S

32.1g S= 0.779mol S

We know that Copper is the limiting reagent and there will be an excess of sulfur. So ALL of the copper will react!

Copper Reacts with sulfur to form copper (I) sulfide according to the following balanced equation.

2Cu(s) + S(s) Cu2S(s)

What is the maximum amount of Cu2S that can be formed in grams when 80g of Cu reacts with 25g of S?

€

80g Cu x1mol Cu

63.5gCu=1.26mol Cu

€

25g S x1mol S

32.1g S= 0.779mol S

We know that Copper is the limiting reagent and there will be an excess of sulfur. So ALL of the copper will react!

€

1.26mol Cu x 1mol Cu2S

2mol Cu= 0.63mol Cu2S

Copper Reacts with sulfur to form copper (I) sulfide according to the following balanced equation.

2Cu(s) + S(s) Cu2S(s)

What is the maximum amount of Cu2S that can be formed in grams when 80g of Cu reacts with 25g of S?

€

80g Cu x1mol Cu

63.5gCu=1.26mol Cu

€

25g S x1mol S

32.1g S= 0.779mol S

We know that Copper is the limiting reagent and there will be an excess of sulfur. So ALL of the copper will react!

€

1.26mol Cu x 1mol Cu2S

2mol Cu= 0.63mol Cu2S

Convert Moles Grams!

Copper Reacts with sulfur to form copper (I) sulfide according to the following balanced equation.

2Cu(s) + S(s) Cu2S(s)

What is the maximum amount of Cu2S that can be formed in grams when 80g of Cu reacts with 25g of S?

€

80g Cu x1mol Cu

63.5gCu=1.26mol Cu

€

25g S x1mol S

32.1g S= 0.779mol S

We know that Copper is the limiting reagent and there will be an excess of sulfur. So ALL of the copper will react!

€

1.26mol Cu x 1mol Cu2S

2mol Cu= 0.63mol Cu2S

Convert Moles Grams!

€

0.63mol Cu2S x 159.1g Cu2S

1mol Cu2S=100g Cu2S

Percent Yield

• The theoretical yield is the maximum amount of product that could be formed from given amounts of reactants.

• In contrast, the amount of product that actually forms when the reaction is carried out in the laboratory is called the actual yield.

Assignment: