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Stoichiometry. Assignments: 11.1 362/38cd, 39cd , 43acde, 40cd 11.2 363/46-48; 363/49,50ab,51 11.3 364/59-62 11.4 365/64 and 66. Chapter 11 Natural Approach to Chemistry. Learning Objectives.

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stoichiometry

Stoichiometry

Assignments:

11.1 362/38cd, 39cd, 43acde, 40cd

11.2 363/46-48; 363/49,50ab,51

11.3 364/59-62

11.4 365/64 and 66

Chapter 11

Natural Approach to

Chemistry

learning objectives
Learning Objectives
  • Apply the mole concept and the law of conservation of mass to calculate quantities of chemicals participating in reactions.
  • Important terms: stoichiometry, percent yield, actual yield, theoretical yield, limiting reactant
slide4

Chemical equations tell stories…

2CO(g) + O2(g) → 2CO2(g)

… and stories can be put into different categories

Nonfiction

Science fiction

Adventure

Romance

History

Psychology

Children’s literature

Synthesis / Decomposition

Single / Double replacement

Precipitate reaction

Polymerization reaction

slide5

Chemical equations tell stories…

But what exactly do they tell us?

2CO(g) + O2(g) 2CO2(g)

They tell us what compounds we start with:

Carbon monoxide (CO) gas

Oxygen (O2) gas

what compounds are formed:

Carbon dioxide (CO2) gas

slide6

2CO(g) + O2(g) 2CO2(g)

Chemical equations tell stories…

What else do they tell us?

2 CO molecules

1 O2 molecules

2 CO2 molecules

They tell us how much of each compound is involved

stoichiometry: the study of the amounts of substances involved in a chemical reaction.

slide7

2CO(g) + O2(g) 2CO2(g)

2 CO molecules

2 dozen CO molecules

2 moles CO molecules

2 x (6.023 x 1023) CO molecules

2 CO2 molecules

2 dozen CO2 molecules

2 moles CO2 molecules

2 x (6.023 x 1023) CO2 molecules

1 O2 molecules

1 dozen O2 molecules

1 mole O2 molecules

(1 x) 6.023 x 1023 O2 molecules

slide8

2CO(g) + O2(g) 2CO2(g)

Is that okay?

Number of moles is notconserved

+

2 moles

CO molecules

1 mole

O2 molecules

2 moles

CO2 molecules

Yes, as long as the chemical equation is balanced! The coefficients are important!!!

slide9

2CO(g) + O2(g) 2CO2(g)

These are important!

Coefficients

2 moles

CO molecules

1 mole

O2 molecules

2 moles

CO2 molecules

This chemical equation isbalanced

The coefficients are correct

slide10

Coefficients are important

1 bag

cake mix

+ 3 eggs + ¼ cup oil + 1 cup water 1 batch cupcakes

Write as a ratio:

slide11

Coefficients are important

1 bag

cake mix

+ 3 eggs + ¼ cup oil + 1 cup water 1 batch cupcakes

Write as a ratio:

slide12

Fermentation of sugar (glucose) into alcohol:

C6H12O6(aq) 2C2H5OH(aq) + 2CO2(g)

1 mole

glucose

2 moles

ethanol

2 moles

carbon dioxide

Write as a ratio:

These are stoichiometric equivalents

slide13

Fermentation of sugar (glucose) into alcohol:

C6H12O6(aq) 2C2H5OH(aq) + 2CO2(g)

1 mole

glucose

2 moles

ethanol

2 moles

carbon dioxide

Write as a ratio:

mole ratio: a ratio comparison between substances in a balanced equation. It is obtained from the coefficients in the balanced equation.

slide14

11.1 Analyzing a Chemical Reaction

Mole ratios

Fermentation of sugar (glucose) into alcohol:

C6H12O6(aq) 2C2H5OH(aq) + 2CO2(g)

1 mole

glucose

2 moles

ethanol

2 moles

carbon dioxide

mole ratios for this chemical equation

slide15

Mole ratios

Consider the following equation:

CO(g) + 2H2(g) CH3OH(l)

carbon monoxide

hydrogen

methanol

If the reaction produces 5 moles of CH3OH, how many moles of H2 are consumed?

Asked: moles of H2

5 moles CH3OH x 2 moles H2= 10 moles H2

1 mole CH3OH

slide16

A mixture of aluminum metal and chlorine gas reacts to form aluminum chloride (AlCl3): 2Al(s) + 3Cl2(g) → 2AlCl3(s). How many moles of aluminum chloride will form when 5 moles of chlorine gas react with excess aluminum metal?

Asked:moles AlCl3

Given:moles Cl2

5 mole Cl2 x 2 mole AlCl3 = 3.3 mole AlCl3

3 mole Cl2

slide17

Potassium + Hydrogen Phosphate 

Finish the reaction in symbols and balance…

If 14.72 moles of hydrogen phosphate are consumed in the above reaction, how many moles of hydrogen are produced?

slide19

mass  moles….

There is no scale that measures in moles!

How do you convert from moles to grams?

By using the molar mass (g/mole)

The mass of 1 mole of Al is not the same as the mass of 1 mole of Cl2.

How do you convert from grams of Al to grams of Cl2?

By using the molar mass (g/mole) and mole ratios

slide21

If 45.0 g of calcium carbonate (CaCO3) decomposes in the reaction CaCO3(s) → CaO(s) + CO2(g), how many grams of CO2 are produced?

Asked:grams of CO2 Given:grams of CaCO3

Relationships:

mole ratios

molar mass of CaCO3 = 40.078 + 12.011 + (15.999 x 3) = 100.0 g/mole

molar mass of CO2 = 12.011 + (15.999 x 2) = 44.01 g/mole

Strategy:

slide22

If 45.0 g of calcium carbonate (CaCO3) decomposes in the reaction CaCO3(s) → CaO(s) + CO2(g), how many grams of CO2 are produced?

B

B

  • 0.45 mole CaC03 x 1 mole CO2 x 44.01 g CO2 = 19.8 g CO2
  • 1 mole CaC031 mole C02
slide23

CHAPTER 11

Stoichiometry

11.2 Percent Yield and Concentration

slide24

In theory, all 100 kernels should have popped.

Did you do something wrong?

+

100 kernels

82 popped

18 unpopped

slide25

In theory, all 100 kernels should have popped.

Did you do something wrong?

No

In real life (and in the lab) things are often not perfect

+

100 kernels

82 popped

18 unpopped

slide26

Percent yield

What you get to eat!

+

100 kernels

82 popped

18 unpopped

slide27

actual yield: the amount obtained in the lab in an actual experiment.

theoretical yield: the expected amount produced if everything reacted completely.

slide28

Percent yield in the lab

Decomposition of baking soda:

2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g)

Heating

slide29

Percent yield in the lab

Decomposition of baking soda:

2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g)

4.87 g

10.00 g

measured experimentally

Can you think of reasons why the final mass of Na2CO3 may not be accurate? (What could be sources of error?)

  • There is usually some human error, like not measuring exact amounts
  • carefully
  • Maybe the heating time was not long enough; not all the Na2HCO3
  • reacted
  • - Maybe Na2CO3 was not completely dry; some H2O(l) was measured too
  • - CO2 is a gas and does not get measured
slide30

Percent yield in the lab

Decomposition of baking soda:

2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g)

10.00 g

4.87 g

measured experimentally

Let’s calculate the percent yield

obtained in experiment

calculated

slide31

Percent yield in the lab

Decomposition of baking soda:

2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g)

10.00 g

4.87 g

measured experimentally

Let’s calculate the percent yield

calculated

slide32

Percent yield in the lab

Decomposition of baking soda:

2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g)

10.00 g

4.87 g

measured experimentally

Let’s calculate the percent yield

calculated

slide33

10.00 g

2NaHCO3(s) →Na2CO3(s) + H2O(l) + CO2(g)

This is a gram-to-gram conversion:

10.00g NaHCO3 1 mole NaHCO3 1 mole Na2CO3 1mole Na2CO3 =

84.01 g NaHCO3 2 mole NaHCO3 105.99 g Na2CO3

Answer: Mass of D = 6.306 g Na2CO3

slide34

2NaHCO3(s) →Na2CO3(s) + H2O(l) + CO2(g)

10.00 g

4.87 g

10.00 g

6.306 g

0.1190 moles

0.05950 moles

For 10.00 g of starting material (NaHCO3), the theoretical yield for Na2CO3 is 6.306 g.

The actual yield (measured) is 4.87 g.

slide35

2NaHCO3(s) →Na2CO3(s) + H2O(l) + CO2(g)

10.00 g

4.87 g

For 10.00 g of starting material (NaHCO3), the theoretical yield for Na2CO3 is 6.306 g.

The actual yield (measured) is 4.87 g.

slide36

Stoichiometry with solutions

Decomposition of baking soda: (We just looked at this.)

2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g)

10.00 g

Convert to moles

Reaction of solid zinc with hydrochloric acid:

Zn(s) + 2HCl(aq) → H2(g) + ZnCl2(aq)

50.0 mL of a 3.0 M solution

Reactions in solution

Convert to moles

slide37

A sample of zinc metal (Zn) reacts with 50.0 mL of a 3.0 M solution of hydrochloric acid (HCl) according to:

Zn(s) + 2HCl(aq) → H2(g) + ZnCl2(aq). How many grams of hydrogen gas (H2) will be produced? Assume zinc metal is present in excess.

Relationships: M = mole/L

Mole ratio: 2 moles HCl ~ 1 mole H2

Molar mass of H2 = 1.0079 x 2

= 2.02 g/mole

Asked: grams of H2produced

Given:50.0 mL of 3.0 M HCl

Reacting with excess zinc

Solve:

0.0500L HCl x 3.0mole HCl x 1 mol H2 x 2.02 g H2= 0.15 g H2

L HCl 2 mol HCl 1 mol H2

Answer:0.15 grams of H2 are produced

slide39

Reaction of solid zinc with hydrochloric acid:

Zn(s) + 2HCl(aq) → H2(g) + ZnCl2(aq)

50.0 mL of a 3.0 M solution

Convert molarity to moles

Sometimes the concentration is written in mass percent

Vinegar is 5% acetic acid by mass

slide40

Commercial vinegar is reported to be 5% acetic acid (C2H4O2) by mass. How many grams of acetic acid are in 120 mL of commercial vinegar? (Assume the density of vinegar is the same as pure water, 1.0 g/mL.)

Asked: grams of acetic acid in 120 mL of vinegar

Given: 120 mL of vinegar and 5% acetic acid by mass

Relationships: 120 mL = 120 g,

given a density of 1.0 g/mL

Solve:

Answer: 6.0 g of acetic acid.

slide41

Let’s Review:

Obtained from the experiment

Calculate using molar masses and mole ratios

slide44

Suppose you want to make 2 ham & cheese sandwiches

Can you still make 2 ham & cheese sandwiches if you have 4 slices of bread, 4 slices of ham, and 1 slice of cheese?

Limiting factor

No, you are limited by the cheese!

You can only get 1 ham & cheese sandwich.

slide45

Limiting reactant

Excess reactant

limiting reactant: the reactant that “runs out” first in a chemical reaction.

excess reactant: the reactant that is remaining after the reaction is complete.

slide47

Sample Problem: 365/58.

Iron can be produced from the following reaction:

Fe2O3 + 2Al  2Fe + Al2O3

a. If 100.0 g of Fe2O3 reacts with 30.0 g Al, which one will be used up first?

slide48

Sample Problem: 365/58.

Iron can be produced from the following reaction:

Fe2O3 + 2Al  2Fe + Al2O3

b. For this next step, use the smaller mole answer from a. to find the needed amount

slide49

How much Fe can be produced?

1.12 mole Al 2 mole Fe 55.85 g Fe = 62.10 g Fe

2 mole Al 1 mole Fe

How much of the excessive reactant is remaining?

Fe2O3 is the excessive reactant.

mole Al  mole Fe2O3  mass Fe2O3

Have – needed = excess

1.112 mole Al 1 mole Fe2O3 159.70 g Fe2O3 = 88.795 Fe2O3

2 mole Al 1 mole Fe2O3

100.0g – 88.80 g = 11.20 g remaining (excess)

slide53

Section 11.1

Analyzing a Chemical Reaction

Section 11.2

Percent Yield and Concentration

Section 11.3

Limiting Reactants

Section 11.4

Solving Stoichiometric Problems

  • Use what we’ve learned to answer these questions:
  • - What is the limiting reactant?
  • - What is the theoretical yield?
  • What is the percent yield?
  • - How much excess reactant is left?
  • - How much reactant is used if it’s in a solution?
slide54

Section 11.1

Analyzing a Chemical Reaction

Section 11.2

Percent Yield and Concentration

Section 11.3

Limiting Reactants

Section 11.4

Solving Stoichiometric Problems

  • Use what we’ve learned to answer these questions:
  • - What is the limiting reactant?
  • - What is the theoretical yield?
  • What is the percent yield?
  • - How much excess reactant is left?
  • - How much reactant is used if it’s in a solution?
slide55

What is the limiting reactant?

Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction:

When 48.0 g of Li reacts with 46.5 g of N2, which reactant is the limiting reactant? Lithium is the only group 1 metal that is capable of reacting directly with nitrogen gas.

6Li(s) + N2(g) → 2Li3N(s)

  • Solve: 1) Moles of each reactant?
  • 2) Apply the mole ratio
  • 3) Compare what we have
  • with what we need
slide56

What is the limiting reactant?

Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction:

When 48.0 g of Li reacts with 46.5 g of N2, which reactant is the limiting reactant? Lithium is the only group 1 metal that is capable of reacting directly with nitrogen gas.

6Li(s) + N2(g) → 2Li3N(s)

Asked: Limiting reactant

Given: 48.0 g of Li (have)

46.5 g of N2 (have)

  • molar mass of Li = 6.941 g/mole
  • molar mass of N2 = 28.01 g/mole
  • mole ratio: 6 moles Li ~ 1 mole N2
slide57

What is the limiting reactant?

Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction:

48.0 g of Li reacts with 46.5 g of N2

6Li(s) + N2(g) → 2Li3N(s)

slide58

What is the limiting reactant?

Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction:

6Li(s) + N2(g) → 2Li3N(s)

?

2) Apply the mole ratio

We have: 6.92 moles Li; 1.66 moles N2

How much N2 do we need to react with 6.92 moles Li?

  • Do we have enough N2?
  • Yes, we have more than enough N2. That means we will run out of Li before we run out of N2
slide59

Section 11.1

Analyzing a Chemical Reaction

Section 11.2

Percent Yield and Concentration

Section 11.3

Limiting Reactants

Section 11.4

Solving Stoichiometric Problems

  • Use what we’ve learned to answer these questions:
  • - What is the limiting reactant?
  • - What is the theoretical yield?
  • What is the percent yield?
  • - How much excess reactant is left?
  • - How much reactant is used if it’s in a solution?
slide60

What is the theoretical yield?

Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction:

How much lithium nitride (LiN3) can be produced from this reaction?

6Li(s) + N2(g) → 2Li3N(s)

slide61

What is the theoretical yield?

Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction:

How much lithium nitride (Li3N) can be produced from this reaction?

6Li(s) + N2(g) → 2Li3N(s)

Asked: Amount of Li3N produced

Given: Li is the limiting reactant

6.92 moles Li (have)

Relationships:

molar mass of Li3N = 34.83 g/mole

mole ratio: 6 moles Li ~ 2 moles Li3N

From

the last problem

slide62

What is the theoretical yield?

Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction:

How much lithium nitride (Li3N) can be produced from this reaction?

6Li(s) + N2(g) → 2Li3N(s)

Asked: Amount of Li3N produced

Given: Li is the limiting reactant

6.92 moles Li (have)

Relationships:

molar mass of Li3N = 34.83 g/mole

mole ratio: 6 moles Li ~ 2 moles Li3N

Solve: 1) Find moles of Li3N

2) Convert moles to grams

slide63

What is the theoretical yield?

Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction:

6Li(s) + N2(g) → 2Li3N(s)

Asked: Amount of Li3N produced

Given: Li is the limiting reactant

6.92 moles Li (have)

Relationships:

molar mass of Li3N = 34.83 g/mole

mole ratio: 6 moles Li ~ 2 moles Li3N

Solve:1) Find moles of Li3N

2) Convert moles to grams

slide64

What is the theoretical yield?

Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction:

6Li(s) + N2(g) → 2Li3N(s)

Asked: Amount of Li3N produced

Given: Li is the limiting reactant

6.92 moles Li (have)

Relationships:

molar mass of Li3N = 34.83 g/mole

mole ratio: 6 moles Li ~ 2 moles Li3N

Solve: 1) Find moles of Li3N

2) Convert moles to grams

slide65

What is the theoretical yield?

Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction:

6Li(s) + N2(g) → 2Li3N(s)

Asked: Amount of Li3N produced

Given: Li is the limiting reactant

6.92 moles Li (have)

Relationships:

molar mass of Li3N = 34.83 g/mole

mole ratio: 6 moles Li ~ 2 moles Li3N

Solve: 1) Find moles of Li3N

2) Convert moles to grams

Asked:80.46 g of Li3N are produced

slide66

Section 11.1

Analyzing a Chemical Reaction

Section 11.2

Percent Yield and Concentration

Section 11.3

Limiting Reactants

Section 11.4

Solving Stoichiometric Problems

  • Use what we’ve learned to answer these questions:
  • - What is the limiting reactant?
  • - What is the theoretical yield?
  • What is the percent yield?
  • - How much excess reactant is left?
  • - How much reactant is used if it’s in a solution?
slide67

What is the percent yield?

Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction:

Calculate the percent yield of an experiment that actually produced 62.5 g of Li3N.

6Li(s) + N2(g) → 2Li3N(s)

slide68

What is the percent yield?

Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction:

Calculate the percent yield of an experiment that actually produced 62.5 g of Li3N.

6Li(s) + N2(g) → 2Li3N(s)

Asked: Percent yield

Given: Theoretical yield: 80.46 g Li3N

Actual yield: 62.5 g Li3N

Relationships:

Solve: Use the percent yield formula

From the

last problem

slide69

What is the percent yield?

Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction:

6Li(s) + N2(g) → 2Li3N(s)

Asked: Percent yield

Given: Theoretical yield: 80.46 g Li3N

Actual yield: 62.5 g Li3N

Relationships:

Solve:Use the percent yield formula

slide70

What is the percent yield?

Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction:

6Li(s) + N2(g) → 2Li3N(s)

Answer: The percent yield in this particular experiment is 77.7%

Asked: Percent yield

Given: Theoretical yield: 80.46 g Li3N

Actual yield: 62.5 g Li3N

Relationships:

Solve: Use the percent yield formula

slide71

Section 11.1

Analyzing a Chemical Reaction

Section 11.2

Percent Yield and Concentration

Section 11.3

Limiting Reactants

Section 11.4

Solving Stoichiometric Problems

  • Use what we’ve learned to answer these questions:
  • - What is the limiting reactant?
  • - What is the theoretical yield?
  • What is the percent yield?
  • - How much excess reactant is left?
  • - How much reactant is used if it’s in a solution?
slide72

What is left?

Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction:

How much of the excess reactant remains after the limiting reactant is completely consumed?

6Li(s) + N2(g) → 2Li3N(s)

slide73

What is left?

Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction:

How much of the excess reactant remains after the limiting reactant is completely consumed?

6Li(s) + N2(g) → 2Li3N(s)

Asked: Amount of excess reactant left

Given: N2 is the excess reactant

1.15 moles N2 (need)

1.66 moles N2 (have)

Relationships:

Molar mass of N2 = 28.01 g/mole

Solve: 1) How many moles N2 remain?

2) Convert moles to grams

slide74

What is left?

Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction:

6Li(s) + N2(g) → 2Li3N(s)

Asked: Amount of excess reactant left

Given: N2 is the excess reactant

1.15 moles N2 (need)

1.66 moles N2 (have)

Relationships:

28.01 g/mole N2

Solve:1) How many moles N2 remain?

2) Convert moles to grams

slide75

What is left?

Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction:

6Li(s) + N2(g) → 2Li3N(s)

Asked: Amount of excess reactant left

Given: N2 is the excess reactant

1.15 moles N2 (need)

1.66 moles N2 (have)

Relationships:

28.01 g/mole N2

Solve: 1) How many moles N2 remain?

2) Convert moles to grams

slide76

What is left?

Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction:

6Li(s) + N2(g) → 2Li3N(s)

Asked: Amount of excess reactant left

Given: N2 is the excess reactant

1.15 moles N2 (need)

1.66 moles N2 (have)

Relationships:

28.01 g/mole N2

Solve: 1) How many moles N2 remain?

2) Convert moles to grams

Answer:14 g of N2 will remain at the end of the reaction.

slide77

365/64. Aluminum metal reacts with oxygen in the air to form a layer of aluminum oxide according to the equation:4Al + 3O2 --> 2Al203(s)

  • If 28.0 g of Al reacts with excess oxygen in the air, what mass of aluminum oxide is formed?
  • Molar Mass Al203 = 2Al+3(0) = 2(27)+3(16) = 102 g/mole

28.0gAl 1mole Al 2 mole Al203 102 g Al203 = 52.9 g Al203

27g Al 4 mole Al 1 mole Al203

slide78

365/64. Aluminum metal reacts with oxygen in the air to form a layer of aluminum oxide according to the equation:4Al + 3O2 --> 2Al203(s)

B How many moles of oxygen are consumed during this reaction?

28 g Al 1mole Al 3mole O2= 0.778 mole O2

27 g Al 4 mole Al

slide79

365/64. Aluminum metal reacts with oxygen in the air to form a layer of aluminum oxide according to the equation:4Al + 3O2 --> 2Al203(s)

C. If 250 g Al203 of is formed, how much Al reacted?

250 g Al203 1 mole Al203 4 mole Al 27.0 g Al

102 g Al203 2 mole Al203 1 mole Al

Answer: 132.4 g Al

slide80
365/66. Wine can spoil when ethanol is converted to acetic acid by oxidation:C2H5OH + O2 --> CH3COOH + H2O
  • Determine the limiting reactant when 5.00 g of ethanol (C2H5OH) and 2.0 g of oxygen are sealed in a bottle.

2C + 6H + O = 24 + 6 + 16 = 46.1 g/mole

5.00g eth. 1 mole eth. = 0.108 moles ethanol

46.1 g eth

2.0 g oxygen 1 mole oxy. = 0.0625 moles oxygen

32.0g oxy.

O2 is the limiting reactant.

We converted reactants to moles individually.

slide81
365/66. Wine can spoil when ethanol is converted to acetic acid by oxidation:C2H5OH + O2 --> CH3COOH + H2O

b. Calculate how much acetic acid will form in grams. Start with the oxygen because it is the limiting reactant.

0.0625 moles O2 1 mole CH3COOH 60 g CH3COOH

1 mole O2 1 mole CH3COOH

Answer: 3.75 g

Molar mass: 2C + 4H + 2(O) =24+4+32 = 60 g/mole

slide82
365/66. Wine can spoil when ethanol is converted to acetic acid by oxidation:C2H5OH + O2 --> CH3COOH + H2O

c. Calculate the amount of excess reactant remaining in grams.

Excess reactant is ethanol.

Available less used:

0.109 moles – 0.0625 moles = 0.0465 moles

0.0465 moles eth. 46 g eth. = 2.139 g eth.

1 mole eth.

Amount of excess ethanol is 2.139 g

slide83

Section 11.1

Analyzing a Chemical Reaction

Section 11.2

Percent Yield and Concentration

Section 11.3

Limiting Reactants

Section 11.4

Solving Stoichiometric Problems

  • Use what we’ve learned to answer these questions:
  • - What is the limiting reactant?
  • - What is the theoretical yield?
  • What is the percent yield?
  • - How much excess reactant is left?
  • - How much reactant is used if it’s in a solution?
slide84

Reactants in solution

The concentration of Cu2+ ions [found as CuSO4(aq)] in the water discharged from an industrial plant is found by adding an excess of sodium sulfide (Na2S) solution to 1.0 L of the contaminated water. What is the concentration of CuSO4 in the water if 0.021 g of CuS precipitate is formed? The reaction is:

CuSO4(aq) + Na2S(aq) → Na2SO4(aq) + CuS(s)

slide85

Reactants in solution

The concentration of Cu2+ ions [found as CuSO4(aq)] in the water discharged from an industrial plant is found by adding an excess of sodium sulfide (Na2S) solution to 1.0 L of the contaminated water. What is the concentration of CuSO4 in the water if 0.021 g of CuS precipitate is formed? The reaction is:

CuSO4(aq) + Na2S(aq) → Na2SO4(aq) + CuS(s)

Asked: Concentration of CuSO4(aq)

Given: 1.0 L of solution is tested

0.021 g CuS (formed)

Relationships:

Molar mass of CuS = 95.61 g/mole

Mole ratio: 1 mole CuSO4 ~ 1 mole CuS

slide86

Reactants in solution

The concentration of Cu2+ ions [found as CuSO4(aq)] in the water discharged from an industrial plant is found by adding an excess of sodium sulfide (Na2S) solution to 1.0 L of the contaminated water. What is the concentration of CuSO4 in the water if 0.021 g of CuS precipitate is formed? The reaction is:

CuSO4(aq) + Na2S(aq) → Na2SO4(aq) + CuS(s)

Asked: Concentration of CuSO4(aq)

Given: 1.0 L of solution is tested

0.021 g CuS (formed)

Relationships:

Molar mass of CuS = 95.61 g/mole

Mole ratio: 1 mole CuSO4 ~ 1 mole CuS

Solve: 1) How many moles of CuS?

2) How many moles of CuSO4?

3) What is the concentration of CuSO4?

slide87

Reactants in solution

The concentration of Cu2+ ions [found as CuSO4(aq)] in the water discharged from an industrial plant is found by adding an excess of sodium sulfide (Na2S) solution to 1.0 L of the contaminated water. The reaction is:

CuSO4(aq) + Na2S(aq) → Na2SO4(aq) + CuS(s)

Asked: Concentration of CuSO4(aq)

Given: 1.0 L of solution is tested

0.021 g CuS (formed)

Relationships:

95.61 g/mole CuS

1 mole CuSO4 ~ 1 mole CuS

Solve:1) How many moles of CuS?

2) How many moles of CuSO4?

3) What is the concentration of CuSO4?

slide88

Reactants in solution

The concentration of Cu2+ ions [found as CuSO4(aq)] in the water discharged from an industrial plant is found by adding an excess of sodium sulfide (Na2S) solution to 1.0 L of the contaminated water. The reaction is:

CuSO4(aq) + Na2S(aq) → Na2SO4(aq) + CuS(s)

Asked: Concentration of CuSO4(aq)

Given: 1.0 L of solution is tested

0.021 g CuS (formed)

Relationships:

95.61 g/mole CuS

1 mole CuSO4 ~ 1 mole CuS

Solve: 1) How many moles of CuS?

2) How many moles of CuSO4?

3) What is the concentration of CuSO4?

Have:

slide89

Reactants in solution

The concentration of Cu2+ ions [found as CuSO4(aq)] in the water discharged from an industrial plant is found by adding an excess of sodium sulfide (Na2S) solution to 1.0 L of the contaminated water. The reaction is:

CuSO4(aq) + Na2S(aq) → Na2SO4(aq) + CuS(s)

Asked: Concentration of CuSO4(aq)

Given: 1.0 L of solution is tested

0.021 g CuS (formed)

Relationships:

95.61 g/mole CuS

1 mole CuSO4 ~ 1 mole CuS

Solve: 1) How many moles of CuS?

2) How many moles of CuSO4?

3) What is the concentration of CuSO4?

Have:

slide90

Reactants in solution

The concentration of Cu2+ ions [found as CuSO4(aq)] in the water discharged from an industrial plant is found by adding an excess of sodium sulfide (Na2S) solution to 1.0 L of the contaminated water. The reaction is:

CuSO4(aq) + Na2S(aq) → Na2SO4(aq) + CuS(s)

Asked: Concentration of CuSO4(aq)

Given: 1.0 L of solution is tested

0.021 g CuS (formed)

Relationships:

95.61 g/mole CuS

1 mole CuSO4 ~ 1 mole CuS

Solve: 1) How many moles of CuS?

2) How many moles of CuSO4?

3) What is the concentration of CuSO4?

Answer: The concentration of CuSO4 is 2.20 x 10-4 M.

slide91

Reactants in solution

The concentration of Cu2+ ions [found as CuSO4(aq)] in the water discharged from an industrial plant is found by adding an excess of sodium sulfide (Na2S) solution to 1.0 L of the contaminated water. The reaction is:

CuSO4(aq) + Na2S(aq) → Na2SO4(aq) + CuS(s)

Discussion:

If the legal limit is 5.0 x 10-4 M of Cu2+, is the industrial plant following the environmental guidelines?

Answer: The concentration of CuSO4 is 2.20 x 10-4 M.

slide92

Reactants in solution

The concentration of Cu2+ ions [found as CuSO4(aq)] in the water discharged from an industrial plant is found by adding an excess of sodium sulfide (Na2S) solution to 1.0 L of the contaminated water. The reaction is:

CuSO4(aq) + Na2S(aq) → Na2SO4(aq) + CuS(s)

Discussion:

If the legal limit is 5.0 x 10-4 M of Cu2+, is the industrial plant following the environmental guidelines?

Yes, because 2.20 x 10-4 M is less than the legal limit.

Answer: The concentration of CuSO4 is 2.20 x 10-4 M.

CuSO4(aq) →Cu2+(aq) + SO42–(aq)

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