1 chapter 5 chemical reaction. mole and avogadro's number just as a grocer sells rice by weight...

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Chapter 5 chemical reaction

Mole and Avogadro's number Mole and Avogadro's number

• Just as a grocer sells rice by weight rather than by counting grains; a chemist uses weight to count for atoms

• As a dozen of anything contains 12 a mole of anything contains 6.022x1023

A mole is a quantity that contains 6.02 x 1023 items.

Use of the mole?• 1mole = Avogadro's number• This graph will help you with most of chapter 5 calculations

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The Mole and Avogadro’s Number

It can be used as a conversion factor to relate thenumber of moles of a substance to the number ofatoms or molecules:

1 mol6.02 x 1023 atoms

or 6.02 x 1023 atoms1 mol

1 mol6.02 x 1023 molecules

or 6.02 x 1023 molecules1 mol

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Coefficients are used to form mole ratios, which canserve as conversion factors.

N2(g) + O2(g) 2 NO(g)

Mole ratios:

1 mol N2

1 mol O2

1 mol N2

2 mol NO

1 mol O2

2 mol NO

Mole Calculations in Chemical Equations

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Mass Calculations in Chemical Equations

HOW TO Convert Grams of Reactant to Grams of Product

Moles ofreactantMoles ofreactant

Grams ofproduct

Grams ofproduct

mole–moleconversion

factor

mole–moleconversion

factor

molar massconversion

factor

molar massconversion

factor

Moles ofproduct

Moles ofproduct

Grams ofreactant

Grams ofreactant

molar massconversion

factor

molar massconversion

factor [1]

[2]

[3]

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5.7 Percent Yield

•The theoretical yield is the amount of product expected from a given amount of reactant based on the coefficients in the balanced chemical equation.

•The actual yield is the amount of product isolated from a reaction.

•Usually, however, the amount of product formed is less than the maximum amount of product predicted.

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5.7 Percent Yield

Percent yield =actual yield (g)

theoretical yield (g)x 100%

Sample Problem 5.14

If the reaction of ethylene with water to form ethanol has a calculated theoretical yield of 23 g of ethanol, what is the percent yield if only 15 g of ethanol are actually formed?

=15 g23 g x 100% = 65%

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5.8 Limiting Reactants

•The limiting reactant is the reactant that is completely used up in a reaction.

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Limiting Reactant

A limiting reactant in a chemical reaction is the

substance that

• Is used up first.• Stops the reaction.• Limits the amount of product that can form.

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Determining the Limiting Reactant

Sample Problem 5.18

[1]: Determine how much of one reactant is needed to react with a second reactant.

2 H2(g) + O2(g) 2 H2O(l)

chosen to be“Original Quantity”

chosen to be“Unknown Quantity”

There are 4molecules of H2

in the picture.

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Determining the Limiting Reactant

Sample Problem 5.18

[2]: Write out the conversion factors that relate the numbers of moles (or molecules) of reactants

2 H2(g) + O2(g) 2 H2O(l)

2 molecules H2

1 molecule O2

1 molecule O2

2 molecules H2

Choose this conversion factor to cancel molecules H2

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Determining the Limiting Reactant

Sample Problem 5.18

[3]: Calculate the number of moles (molecules) of the second reactant needed for complete reaction.

2 H2(g) + O2(g) 2 H2O(l)

1 molecule O2

2 molecules H2

4 molecules H2 x = 2 molecules O2

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Determining the Limiting Reactant

Sample Problem 5.18

[4]: Analyze the two possible outcomes:

• If the amount present of the second reactantis less than what is needed, the secondreactant is the limiting reagent.

• If the amount present of the second reactant isgreater than what is needed, the secondreactant is in excess.

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Determining the Limiting Reactant

Sample Problem 5.18

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Determining the Limiting Reactant Using the Number of Grams

Sample Problem 5.20

Using the balanced equation, determine the limitingreactant when 10.0 g of N2 (MM = 28.02 g/mol) reactwith 10.0 g of O2 (MM = 32.00 g/mol).

N2(g) + O2(g) 2 NO(g)

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Determining the Limiting Reactant Using the Number of Grams

Sample Problem 5.20

[1] Convert the number of grams of each reactant into moles using the molar masses.

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Determining the Limiting Reactant Using the Number of Grams

Sample Problem 5.20

[2] Determine the limiting reactant by choosing N2

as the original quantity and converting to mol O2.

1 mol O2

1 mol N2

0.357 mol N2 x = 0.357 mol O2

mole–moleConversion factor

mole–moleConversion factor

The amount of O2 we started with (0.313 mol) isless than the amount we would need (0.357 mol) so O2 is the limiting reagent.

End of chapter 5 question 87

• Question 87

• The local anesthetic ethyl chloride (C2H5Cl, molar mass is 64.51g/mol) can be prepared by reaction of ethylene (C2H4 molar mass 28.05g/mol) according to the balanced equation

• A. if 8.00g of ethylene and 12.0g of HCl are used, how many moles of each reactant are used?

• What is the limiting reactant• How many grams of the product formed?• If a 10.6g of product are formed, what is the percent yield of

the reaction?

C2H4 + HCl C2H5Cl

Problem 5.87

• The local anesthetic ethyl chloride (C2H5Cl, molar mass 64.51g/mole) can be prepared by reaction of ethylene (C2H4, molar mass 28.05g/mole) with HCl (molar mass 36.46g/mole), according to the balanced equation,

• a. if 8.00g of ethylene and 12.0g of HCl are used, how many moles of each reacted are used?

C2H4 + HCl C2H5Cl

1 mol C2H4

28.05g C2H4

8.00g C2H4 x = 0.285 mol C2H4

1 mol HCl

36.46g HCl12.0g HCl x = 0.329 mol HCl

• b. What is the limiting reactant

• 0.285 mol C2H4 is completely used up in the reaction so it is the limiting reactant

• c. how many moles of product are formed

Problem 5.87

C2H4 + HCl C2H5Cl

1 mol C2H5Cl 1mol C2H4

0.285mol C2H4 x = 0.285 mol C2H5Cl

• d. How many grams of the product formed

• e. if 10.6g of product are formed, what is the percent yield of the reaction?

Problem 5.87

C2H4 + HCl C2H5Cl

64.51gC2H5Cl 1mol C2H5Cl

0.285mol C2H5Cl x = 18.4 g C2H5Cl

Percent yield =actual yield (g)

theoretical yield (g)x 100%

Percent yield =10.6 g C2H5Cl 18.4 g C2H5Cl

x 100%

Percent yield = 57.6%

Alka seltzer Calculations

• 1. NaHCO3(s) Na+(aq) + HCO3-(aq) 

• 2. HCO3-(aq) + H3O+(aq) 2H2O(l) + CO2 (g)

• For example if you determined the mass of CO2 lost is 0.512g

determine moles of CO2 (g) lost

Use mole ratio to calculate moles of NaHCO3(s)

1 mol CO2

44.01g CO2

0.50g CO2 x = 0.0163 mol CO2

1 mole NaHCO3(s)1 mole CO2

0.0163moles CO2 x = 0.0163 mol NaHCO3(s)

• Mass of NaHCO3(s)

• Calculated Mass% of NaHCO3 reacted in tablet (printed on label is 1.916g of NaHCO3

83.00g NaHCO3(s)1 mole NaHCO3(s)

0.0163moles NaHCO3 x = 1.35g NaHCO3(s)

1.35g NaHCO3(s)1.916g in tablet

x 100% = 70 % NaHCO3(s) reacted

Alka seltzer Calculations

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5.9 Oxidation and Reduction

•Oxidation is the loss of electrons from an atom.

•Reduction is the gain of electrons by an atom.

•Both processes occur together in a single reaction called an oxidation−reduction or redox reaction. Thus, a redox reaction always has two components, one that is oxidized and one that is reduced.

•A redox reaction involves the transfer of electrons from one element to another.

A. General Features

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Zn + Cu2+ Zn2+ + Cu

Zn loses 2 e–

Cu2+ gains 2 e−

•Zn loses 2 e− to form Zn2+, so Zn is oxidized.

•Cu2+ gains 2 e− to form Cu, so Cu2+ is reduced.

5.9 Oxidation and Reduction

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Oxidation half reaction: Zn Zn2+ + 2 e−

Each of these processes can be written as an individual half reaction:

Zn + Cu2+ Zn2+ + Cu

Zn loses 2 e–

Cu2+ gains 2 e−

loss of e−

Reduction half reaction: Cu2+ + 2e− Cu

gain of e−

5.9 Oxidation and Reduction

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Zn + Cu2+ Zn2+ + Cu

•Zn acts as a reducing agent because it causes Cu2+ to gain electrons and become reduced.

A compound that is oxidized while causing anothercompound to be reduced is called a reducing agent.

oxidized reduced

5.9 Oxidation and Reduction

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Zn + Cu2+ Zn2+ + Cu

A compound that is reduced while causing anothercompound to be oxidized is called an oxidizing agent.

•Cu2+ acts as an oxidizing agent because it causes Zn to lose electrons and become oxidized.

oxidized reduced

5.9 Oxidation and Reduction

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5.9 Oxidation and Reduction

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Iron Rusting

4 Fe(s) + 3 O2(g) 2 Fe2O3(s)

Fe3+ O2–neutral Fe neutral O

Fe loses e– and is oxidized.

O gains e– and is reduced.

Examples of Oxidation–Reduction Reactions

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Inside an Alkaline Battery

Zn + 2 MnO2 ZnO + Mn2O3

neutral Zn Mn4+ Zn2+ Mn3+

Zn loses e− and is oxidized.

Mn4+ gains e− and is reduced.

Examples of Oxidation–Reduction Reactions

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Zn + 2 MnO2 ZnO + Mn2O3

Examples of Oxidation–Reduction Reactions

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Oxidation results in the: Reduction results in the:

•Gain of oxygen atoms

•Loss of hydrogen atoms

•Loss of oxygen atoms

•Gain of hydrogen atoms

Examples of Oxidation–Reduction Reactions

Oxidation occurs when a molecule does any of the following:

redox chemistry

• Loses electrons

• Gains oxygen

If a molecule undergoes oxidation, it is the reducing agent.

Reduction occurs when a molecule does any of the following:

Gains electrons

Loses oxygen

If a molecule undergoes reduction, it is the oxidizing agent.

redox chemistry

Question 5.92

• Identify the species that is oxidized and the species that is reduced in each reaction. Write out two half reactions to show how many electrons are gained or lost by each species.

Mg + Fe2+ Mg2+ + Fe

Sn + Cu2+ Sn2+ + Cu

4Na + O2

2Na2O

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Oxidation half reaction: Mg Mg2+ + 2 e−

Each of these processes can be written as an individual half reaction:

Mg + Fe2+ Mg2+ + Fe

Mg loses 2 e–

Fe2+ gains 2 e−

loss of e−

Reduction half reaction: Fe2+ + 2e− Fe

gain of e−

Mg + Fe2+ Mg2+ + Fe

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Oxidation half reaction: Sn Sn2+ + 2 e−

Each of these processes can be written as an individual half reaction:

Sn + Cu2+ Sn2+ + Cu

Sn loses 2 e–

Cu2+ gains 2 e−

loss of e−

Reduction half reaction: Cu2+ + 2e− Cu

gain of e−

Sn + Cu2+ Sn2+ + Cu

Chapter 5 question 91

• Identify the species that is oxidized and the species that is reduced in each reaction. Write out two half reactions to show how many electrons are gained or lost by each species.

Fe + Cu2+ Fe 2+ + Cu

Cl2 + 2I- I2 + 2Cl-

2Na + Cl2 2NaCl

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Oxidation half reaction: 4Na 4Na1+ + 4 e−

Each of these processes can be written as an individual half reaction:

4Na + O2 2Na2O

gain of oxygen

Loss of oxygen

loss of e−

Reduction half reaction: O2 + 4e− 2O-2

gain of e−

4Na + O2

2Na2O

Question 5.92

• Identify the species that is oxidized and the species that is reduced in each reaction. Write out two half reactions to show how many electrons are gained or lost by each species.

Mg is oxidized becomes the reducing agent

Fe2+ is reduced becomes the oxidizing agent

Zn is oxidized becomes the reducing agent

Cu2+ is reduced becomes the oxidizing agent

Na is oxidized becomes the reducing agent

O is reduced becomes the oxidizing agent

Mg + Fe2+ Mg2+ + Fe

Sn + Cu2+ Sn2+ + Cu

4Na + O2

2Na2O