chapter 5 chemical quantities and reactions 5.1 the...

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Chapter 5 Chemical Quantities and Reactions

5.1The Mole

Copyright © 2009 by Pearson Education, Inc.

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Collection Terms

A collection term statesa specific number of items.

• 1 dozen donuts = 12 donuts

• 1 ream of paper = 500 sheets

• 1 case = 24 cans


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A mole is a collection that contains

• the same number of particles as there are carbon atoms in 12.0 g of carbon.

• 6.02 x 1023 atoms of an element (Avogadro’s number).

1 mole element Number of Atoms1 mole C = 6.02 x 1023 C atoms1 mole Na = 6.02 x 1023 Na atoms1 mole Au = 6.02 x 1023 Au atoms

A Mole of Atoms

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A mole • of a covalent compound has Avogadro’s number of

molecules.1 mole CO2 = 6.02 x 1023 CO2 molecules1 mole H2O = 6.02 x 1023 H2O molecules

• of an ionic compound contains Avogadro’s number of formula units.

1 mole NaCl = 6.02 x 1023 NaCl formula units1 mole K2SO4 = 6.02 x 1023 K2SO4 formula

units

A Mole of a Compound

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Samples of 1 Mole Quantities

1 mole of C atoms = 6.02 x 1023 C atoms

1 mole of Al atoms = 6.02 x 1023 Al atoms

1 mole of S atoms = 6.02 x 1023 S atoms

1 mole of H2O molecules = 6.02 x 1023 H2O molecules

1 mole of CCl4 molecules = 6.02 x 1023 CCl4 molecules

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Avogadro’s number, 6.02 x 1023, can be written as anequality and two conversion factors.

Equality:1 mole = 6.02 x 1023 particles

Conversion Factors:6.02 x 1023 particles and 1 mole

1 mole 6.02 x 1023 particles

Avogadro’s Number

Using Avogadro’s Number

Avogadro’s number is used to o convert moles to particleso convert particles to moles

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Converting Moles to Particles

Avogadro’s number is used to convertmoles of a substance to particles.

How many Cu atoms are in 0.50 mole of Cu?

0.50 mole Cu x 6.02 x 1023 Cu atoms1 mole Cu

= 3.0 x 1023 Cu atomsCopyright © 2009 by Pearson Education, Inc.

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Converting Particles to Moles

Avogadro’s number is used to convertparticles of a substance to moles.

How many moles of CO2 are in 2.50 x 1024 molecules CO2?

2.50 x 1024 molecules CO2 x 1 mole CO2

6.02 x 1023 molecules CO2

= 4.15 moles of CO2

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1. The number of atoms in 2.0 mole of Al atoms isA. 2.0 Al atoms.B. 3.0 x 1023 Al atoms.C. 1.2 x 1024 Al atoms.

2. The number of moles of S in 1.8 x 1024 atoms of S is A. 1.0 mole of S atoms.B. 3.0 moles of S atoms.C. 1.1 x 1048 moles of S atoms.

Learning Check

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Solution

C. 1.2 x 1024 Al atoms2.0 mole Al x 6.02 x 1023 Al atoms =

1 mole Al

B. 3.0 moles of S atoms1.8 x 1024 S atoms x 1 mole S =

6.02 x 1023 S atoms

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Subscripts and Moles

The subscripts in a formula show• the relationship of atoms in the formula.• the moles of each element in 1 mole of compound.

GlucoseC6H12O6

In 1 molecule: 6 atoms C 12 atoms H 6 atoms OIn 1 mole: 6 moles C 12 moles H 6 moles O

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Subscripts State Atoms and Moles

9 moles of C 8 moles of H 4 moles of O

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Factors from Subscripts

The subscripts are used to write conversion factors formoles of each element in 1 mole of a compound. Foraspirin, C9H8O4, the possible conversion factors are:

9 moles C 8 moles H 4 moles O 1 mole C9H8O4 1 mole C9H8O4 1 mole C9H8O4

and

1 mole C9H8O4 1 mole C9H8O4 1 mole C9H8O49 moles C 8 moles H 4 moles O

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Learning Check

A. How many moles of O are in 0.150 mole of aspirin, C9H8O4?

B. How many atoms of O are in 0.150 mole of aspirin, C9H8O4?

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Solution

A. Moles of O in 0.150 mole of aspirin, C9H8O4

0.150 mole C9H8O4 x 4 mole O = 0.600 mole of O1 mole C9H8O4

subscript factor

B. Atoms of O in 0.150 mole of aspirin, C9H8O4

0.150 mole C9H8O4 x 4 mole O x 6.02 x 1023 O atoms1 mole C9H8O4 1 mole O

Avogadro’s number

= 3.61 x 1023 atoms of O

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5.2Molar Mass


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Molar Mass

The molar mass

• is the mass of 1 mole of an element or compound.

• is the atomic or molecular mass expressed in grams.


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Molar Mass from Periodic Table

Molar massis the atomic mass expressed in grams.

1 mole of Ag 1 mole of C 1 mole of S= 107.9 g = 12.01 g = 32.07 gCopyright © 2009 by Pearson Education, Inc.

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Give the molar mass for each (to the tenths decimal place).

A. 1 mole of K atoms = ________

B. 1 mole of Sn atoms = ________

Learning Check

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Give the molar mass for each (to the tenths decimal place).

A. 1 mole of K atoms = 39.1 g

B. 1 mole of Sn atoms = 118.7 g

Solution

Guide to Calculation Molar Mass of a Compound

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Molar Mass of a Compound

The molar mass of a compound is the sum of the molar masses of the elements in the formula.

Example: Calculate the molar mass of CaCl2.

Element Number of Moles

Atomic Mass Total Mass

Ca 1 40.1 g/mole 40.1 gCl 2 35.5 g/mole 71.0 gCaCl2 111.1 g

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Molar Mass of K3PO4

Calculate the molar mass of K3PO4.

Element Number of Moles

Atomic Mass Total Mass in K3PO4

K 3 39.1 g/mole 117.3 gP 1 31.0 g/mole 31.0 gO 4 16.0 g/mole 64.0 gK3PO4 212.3 g

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Some 1-mole Quantities

32.1 g 55.9 g 58.5 g 294.2 g 342.2 g

One-Mole Quantities


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What is the molar mass of each of the following?

A. K2O

B. Al(OH)3

Learning Check

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A. K2O 94.2 g/mole

2 moles of K (39.1 g/mole) + 1 mole of O (16.0 g/mole)78.2 g + 16.0 g = 94.2 g

B. Al(OH)3 78.0 g/mole

1 mole of Al (27.0 g/mole) + 3 moles of O (16.0 g/mole)+ 3 moles of H (1.01 g/mole) 27.0 g + 48.0 g + 3.03 g = 78.0 g

Solution

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Prozac, C17H18F3NO, is an antidepressant that inhibits the uptake of serotonin by the brain. What is the molar mass of Prozac?

1) 40.1 g/mole2) 262 g/mole 3) 309 g/mole

Learning Check

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Prozac, C17H18F3NO, is an antidepressant that inhibits the uptake of serotonin by the brain. What is the molar mass of Prozac?

3) 309 g/mole17 C (12.0) + 18 H (1.01) + 3 F (19.0) + 1 N (14.0) + 1 O (16.0) =

204 g C + 18.2 g H + 57.0 g F + 14.0 g N + 16.0 g O = 309 g/mole

Solution

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Molar mass conversion factors• are fractions (ratios) written from the molar mass.• relate grams and moles of an element or compound.• for methane, CH4, used in gas stoves and gas heaters

is1 mole of CH4 = 16.0 g (molar mass equality)

Conversion factors:16.0 g CH4 and 1 mole CH41 mole CH4 16.0 g CH4

Molar Mass Factors

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Acetic acid, C2H4O2, gives the sour taste to vinegar. Write two molar mass conversion factors for acetic acid.

Learning Check


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Calculate molar mass for acetic acid C2H4O2:24.0 g C + 4.04 g H + 32.0 g O = 60.0 g/mole C2H4O2

1 mole of acetic acid = 60.0 g acetic acid

Molar mass factors:1 mole acetic acid and 60.0 g acetic acid 60.0 g acetic acid 1 mole acetic acid

Solution

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Molar mass factors are used to convert between the grams of a substance and the number of moles.

Calculations Using Molar Mass

Grams Molar mass factor Moles

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Aluminum is often used to build lightweight bicycleframes. How many grams of Al are in 3.00 mole of Al?

Molar mass equality: 1 mole of Al = 27.0 g of Al

Setup with molar mass as a factor:

3.00 mole Al x 27.0 g Al = 81.0 g of Al1 mole Al

molar mass factor for Al

Moles to Grams

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Learning Check

Allyl sulfide, C6H10S, is a compound that has the odor of garlic. How many moles of C6H10S are in 225 g?


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Calculate the molar mass of C6H10S. (6 x 12.0) + (10 x 1.01) + (1 x 32.1) = 114.2 g/mole

Set up the calculation using a mole factor. 225 g C6H10S x 1 mole C6H10S

114.2 g C6H10Smolar mass factor (inverted)

= 1.97 moles of C6H10S

Solution

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Grams, Moles, and Particles

A molar mass factor and Avogadro’s number convert…

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Learning Check

How many H2O molecules are in 24.0 g of H2O?

1) 4.52 x 1023 H2O molecules



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Solution

How many H2O molecules are in 24.0 g of H2O?

3) 8.02 x 1023

24.0 g H2O x 1 mole H2O x 6.02 x 1023 H2O molecules 18.0 g H2O 1 mole H2O

= 8.03 x 1023 H2O molecules

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Learning Check

If the odor of C6H10S can be detected from 2 x 10-13 g in 1 liter of air, how many molecules of C6H10S are present?

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Solution

If the odor of C6H10S can be detected from 2 x 10-13 g in 1 liter of air, how many molecules of C6H10S are present?

2 x 10-13 g x 1 mole x 6.02 x 1023 molecules 114.2 g 1 mole

= 1 x 109 molecules of C6H10S are present

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5.3 Chemical Changes


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Physical Change

In a physical change,

• the state, shape, or size of the material changes.

• the identity and composition of the substance do not change.


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Chemical Change

In a chemical change,

• reacting substances form new substances with different compositions and properties.

• a chemical reaction takes place.


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Some Examples of Chemical and Physical Changes


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Classify each of the following as a1) physical change or 2) chemical change.

A. ____Burning a candle.B. ____Ice melting on the street.C. ____Toasting a marshmallow.D. ____Cutting a pizza.E. ____Polishing a silver bowl.

Learning Check

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Classify each of the following as a1) physical change or 2) chemical change.

A. 2 Burning a candle.B. 1 Ice melting on the street.C. 2 Toasting a marshmallow.D. 1 Cutting a pizza.E. 2 Polishing a silver bowl.

Solution

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Chemical Reaction

In a chemical reaction• a chemical change

produces one or more new substances.

• there is a change in the composition of one or more substances.


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Chemical Reaction

In a chemical reaction• old bonds are broken and new

bonds are formed.

• atoms in the reactants are rearranged to form one or more different substances.

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Chemical Reaction

In a chemical reaction

• the reactants are Fe and O2.

• the new product Fe2O3 is called rust.


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5.4 Chemical Equations


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Chemical Equations

A chemical equation gives• the formulas of the reactants on the left of the arrow.• the formulas of the products on the right of the arrow.

Reactants Product

C(s)

O2 (g)CO2 (g)


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Symbols Used in Equations

Symbols in chemical equations show

• the states of the reactants.

• the states of the products.

• the reaction conditions.

TABLE

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Chemical Equations are Balanced

In a balancedchemical reaction• no atoms are

lost or gained.

• the number of reacting atoms is equal to the number of product atoms.


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A Balanced Chemical Equation

In a balanced chemical equation,• the number of each type of atom on the reactant side is

equal to the number of each type of atom on the product side.

• numbers called coefficients are used in front of one or more formulas to balance the number of atoms.

Al + S Al2S3 Not Balanced2Al + 3S Al2S3 Balanced using coefficients

2 Al = 2 Al Equal number of Al atoms3 S = 3 S Equal number of S atoms

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A Study Tip: Using Coefficients

When balancing a chemical equation, • use one or more coefficients to balance atoms.• never change the subscripts of any formula.

N2 + O2 NO Not Balanced

N2 + O2 N2O2 Incorrect formula

N2 + O2 2NO Correctly balanced using coefficients

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Learning Check

State the number of atoms of each element on thereactant side and the product side for each of thefollowing balanced equations.

A. P4(s) + 6Br2(l) 4PBr3(g)

B. 2Al(s) + Fe2O3(s) 2Fe(s) + Al2O3(s)

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Solution

A. P4(s) + 6Br2(l) 4PBr3(g)4 P atoms = 4 P atoms12 Br atoms = 12 Br atoms

B. 2Al(s) + Fe2O3(s) 2Fe(s) + Al2O3(s)2 Al atoms = 2 Al atoms2 Fe atoms = 2 Fe atoms3 O atoms = 3 O atoms

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Learning Check

Determine if each equation is balanced or not.

A. Na(s) + N2(g) Na3N(s)

B. C2H4(g) + H2O(l) C2H6O(l)

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Solution

Determine if each equation is balanced or not.

A. Na(s) + N2(g) Na3N(s)No. 2 N atoms on reactant side; only 1 N atom on the product side. 1 Na atom on reactant side; 3 Na atoms on the product side.

B. C2H4(g) + H2O(l) C2H6O(l)Yes. 2 C atoms = 2 C atoms

6 H atoms = 6 H atoms1 O atom = 1 O atom

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Guide to Balancing a Chemical Equation

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To balance the following equation, Fe3O4(s) + H2(g) Fe(s) + H2O(l)

• work on one element at a time.• use only coefficients in front of formulas.• do not change any subscripts.Fe: Fe3O4(s) + H2(g) 3Fe(s) + H2O(l)

O: Fe3O4(s) + H2(g) 3Fe(s) + 4H2O(l)

H: Fe3O4(s) + 4H2(g) 3Fe(s) + 4H2O(l)

Steps in Balancing an Equation

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1. Write the equation with the correct formulas.NH3(g) + O2(g) NO(g) + H2O(g)

2. Determine if the equation is balanced.No, H atoms are not balanced.

3. Balance with coefficients in front of formulas. 2NH3(g) + O2(g) 2NO(g) + 3H2O(g)

Double the coefficients to give even number of O atoms.4NH3(g) + 7O2(g) 4NO(g) + 6H2O(g)

Balancing Chemical Equations

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4. Check that atoms of each element are equal in reactants and products.

4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)

4 N (4 x 1 N) = 4 N (4 x 1 N)12 H (4 x 3 H) = 12 H (6 x 2 H)10 O (5 x 2 O) = 10 O (4 O + 6 O)

Balancing Chemical Equations

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Equation for a Chemical Reaction


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Checking a Balanced Equation

Reactants Products1 C atom = 1 C atom4 H atoms = 4 H atoms4 O atoms = 4 O atoms


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Check the balance of atoms in the following:Fe3O4(s) + 4H2(g) 3Fe(s) + 4H2O(l)

A. Number of H atoms in products.1) 2 2) 4 3) 8

B. Number of O atoms in reactants.1) 2 2) 4 3) 8

C. Number of Fe atoms in reactants.1) 1 2) 3 3) 4

Learning Check

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Fe3O4(s) + 4H2(g) 3Fe(s) + 4H2O(l)

A. Number of H atoms in products.3) 8 (4H2O)

B. Number of O atoms in reactants.2) 4 (Fe3O4)

C. Number of Fe atoms in reactants.2) 3 (Fe3O4)

Solution

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Balance each equation and list the coefficients in the balanced equation going from reactants to products:

A. __Mg(s) + __N2(g) __Mg3N2(s)

1) 1, 3, 2 2) 3, 1, 2 3) 3, 1, 1

B. __Al(s) + __Cl2(g) __AlCl3(s)

1) 3, 3, 2 2) 1, 3, 1 3) 2, 3, 2

Learning Check

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A. 3) 3, 1, 13Mg(s) + 1N2(g) 1Mg3N2(s)

B. 3) 2, 3, 2 2Al(s) + 3Cl2(g) 2AlCl3(s)

Solution

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Equations with Polyatomic Ions

2Na3PO4(aq) + 3MgCl2 (aq) Mg3 ( PO4(s) + 6NaCl(aq)


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Balancing with Polyatomic Ions

Na3PO4(aq) + MgCl2(aq) Mg3(PO4)2(s) + NaCl(aq)

Balance PO43- as a unit

2Na3PO4(aq) + MgCl2(aq) Mg3(PO4)2(s) + NaCl(aq) 2 PO4

3- = 2 PO43-

Check Na + balance6 Na+ = 6 Na+

Balance Mg and Cl2Na3PO4(aq) + 3MgCl2(aq) Mg3(PO4)2(s) + 6NaCl(aq)

3 Mg2+ = 3 Mg2+

6 Cl- = 6 Cl-

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Balance and list the coefficients from reactants to products.

A. __Fe2O3(s) + __C(s) __Fe(s) + __CO2(g)1) 2, 3, 2, 3 2) 2, 3, 4, 3 3) 1, 1, 2, 3

B. __Al(s) + __FeO(s) __Fe(s) + __Al2O3(s)1) 2, 3, 3, 1 2) 2, 1, 1, 1 3) 3, 3, 3, 1

C. __Al(s) + __H2SO4(aq) __Al2(SO4)3(aq) + __H2(g)1) 3, 2, 1, 2 2) 2, 3, 1, 3 3) 2, 3, 2, 3

Learning Check

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A. 2) 2, 3, 4, 3 2Fe2O3(s) + 3C(s) 4Fe(s) + 3CO2(g)

B. 1) 2, 3, 3, 1 2Al(s) + 3FeO(s) 3Fe(s) + 1Al2O3(s)

C. 2) 2, 3, 1, 3 2Al(s) + 3H2SO4(aq) 1Al2(SO4)3(aq) + 3H2(g)

Solution

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5.5Types of Reactions

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Type of Reactions

Chemical reactions can be classified as

• combination reactions.• decomposition reactions.• single replacement reactions.• double replacement reactions.

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Combination

In a combination reaction,• two or more elements form one product.• or simple compounds combine to form one product.

+

2Mg(s) + O2(g) 2MgO(s)

2Na(s) + Cl2(g) 2NaCl(s)

SO3(g) + H2O(l) H2SO4(aq)

A B A B

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Formation of MgO


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Decomposition

In a decomposition reaction, • one substance splits into two or more simpler

substances.

2HgO(s) 2Hg(l) + O2(g)2KClO3(s) 2KCl(s) + 3 O2(g)

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Decomposition of HgO


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Learning Check

Classify the following reactions as 1) combination or 2) decomposition.

___A. H2(g) + Br2(g) 2HBr(l)___B. Al2(CO3)3(s) Al2O3(s) + 3CO2(g)___C. 4Al(s) + 3C(s) Al4C3(s)

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Solution

Classify the following reactions as 1) combination or 2) decomposition.

1 A. H2(g) + Br2(g) 2HBr(l)2 B. Al2(CO3)3(s) Al2O3(s) + 3CO2(g)1 C. 4Al(s) + 3C(s) Al4C3(s)

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Single Replacement

In a single replacement reaction, • one element takes the place of a different element in

another reacting compound.

Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g) Fe(s) + CuSO4(aq) FeSO4(aq) + Cu(s)

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Zn and HCl is a Single Replacement Reaction


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Double Replacement

In a double replacement, • two elements in the reactants exchange places.

AgNO3(aq) + NaCl(aq) AgCl(s) + NaNO3(aq) ZnS(s) + 2HCl(aq) ZnCl2(aq) + H2S(g)

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Example of a Double Replacement


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Learning Check

Classify the following reactions as 1) single replacement or 2) double replacement.

A. 2Al(s) + 3H2SO4(aq) Al2(SO4)3(s) + 3H2(g)

B. Na2SO4(aq) + 2AgNO3(aq) Ag2SO4(s) + 2NaNO3(aq)

C. 3C(s) + Fe2O3(s) 2Fe(s) + 3CO(g)

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Solution

Classify the following reactions as 1) single replacement or 2) double replacement.

1 A. 2Al(s) + 3H2SO4(aq) Al2(SO4)3(s) + 3H2(g)

2 B. Na2SO4(aq) + 2AgNO3(aq) Ag2SO4(s) + 2NaNO3(aq)

1 C. 3C(s) + Fe2O3(s) 2Fe(s) + 3CO(g)

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Learning Check

Identify each reaction as 1) combination, 2) decomposition,3) single replacement, or 4) double replacement.

A. 3Ba(s) + N2(g) Ba3N2(s)

B. 2Ag(s) + H2S(aq) Ag2S(s) + H2(g)

C. SiO2(s) + 4HF(aq) SiF4(s) + 2H2O(l)

D. PbCl2(aq) + K2SO4(aq) 2KCl(aq) + PbSO4(s)

E. K2CO3(s) K2O(aq) + CO2(g)

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Solution

1 A. 3Ba(s) + N2(g) Ba3N2(s)

3 B. 2Ag(s) + H2S(aq) Ag2S(s) + H2(g)

4 C. SiO2(s) + 4HF(aq) SiF4(s) + 2H2O(l)

4 D. PbCl2(aq) + K2SO4(aq) 2KCl(aq) +PbSO4(s)

2 E. K2CO3(s) K2O(aq) + CO2(g)

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Learning Check

Each of the following reactions occur in the formation of smog or acid rain. Identify the type of reaction andbalance each.

A. NO(g) + O2(g) NO2(s)

B. N2(g) + O2(g) NO(g)

C. SO2(g) + O2(g) SO3(g)

D. NO2(g) NO(g) + O(g)

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Solution

Each of the following reactions occur in the formation of smog or acid rain. Identify the type of reaction andbalance each.

Combination

A. 2NO(g) + O2(g) 2NO2(s)

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

C. 2SO2(g) + O2(g) 2SO3(g)

Decomposition

D. NO2(g) NO(g) + O(g)

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5.6Oxidation-Reduction Reactions


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Oxidation and Reduction

An oxidation-reduction reaction

• provides us with energy from food.• provides electrical energy in

batteries.• occurs when iron rusts.

4Fe(s) + 3O2(g) 2Fe2O3(s)Copyright © 2009 by Pearson Education, Inc.

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An oxidation-reduction reaction involves

• a transfer of electrons from one reactant to another.

• oxidation as a loss of electrons (LEO).

• reduction as a gain of electrons (GER).

Electron Loss and Gain


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Oxidation and Reduction


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Zn and Cu2+

Zn(s) Zn2+(aq) + 2e- oxidationsilver metal

Cu2+(aq) + 2e- Cu(s) reduction blue solution orange


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Electron Transfer from Zn to Cu2+

Oxidation: loss of electrons

Reduction: gain of electrons


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Identify each of the following as 1) oxidation or 2) reduction.

__A. Sn(s) Sn4+(aq) + 4e−

__B. Fe3+(aq) + 1e− Fe2+(aq)__C. Cl2(g) + 2e− 2Cl-(aq)

Learning Check

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Identify each of the following as 1) oxidation or 2) reduction.

1 A. Sn(s) Sn4+(aq) + 4e−

2 B Fe3+(aq) + 1e− Fe2+(aq)2 C. Cl2(g) + 2e− 2Cl-(aq)

Solution

101

Write the separate oxidation and reduction reactions for the following equation.

2Cs(s) + F2(g) 2CsF(s)

A cesium atom loses an electron to form cesium ion.Cs(s) Cs+(s) + 1e− oxidation

Fluorine atoms gain electrons to form fluoride ions.F2(s) + 2e- 2F−(s) reduction

Writing Oxidation and Reduction Reactions

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In light-sensitive sunglasses, UV light initiatesan oxidation-reduction reaction.

uv lightAg+ + Cl− Ag + Cl

A. Which reactant is oxidized?

B. Which reactant is reduced?

Learning Check

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Solution

In light-sensitive sunglasses, UV light initiatesan oxidation-reduction reaction.

uv lightAg+ + Cl− Ag + Cl

A. Which reactant is oxidized? Cl− Cl + 1e−

B. Which reactant is reduced? Ag+ + 1e− Ag

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Learning Check

Identify the substances that are oxidized and reduced ineach of the following reactions.

A. Mg(s) + 2H+(aq) Mg2+(aq) + H2(g)

B. 2Al(s) + 3Br2(g) 2AlBr3(s)

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Solution

A. Mg is oxidized. Mg(s) Mg2+(aq) + 2e−H+ is reduced. 2H+ + 2e− H2

B. Al is oxidized. Al Al3+ + 3e−

Br is reduced. Br + e− Br −

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5.7 Mole Relationships in Chemical Equations


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Law of Conservation of Mass

The law of conservation of massindicates that in an

ordinary chemical reaction, • matter cannot be created or destroyed.• no change in total mass occurs in a

reaction.• mass of products is equal to mass of

reactants.

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Consider the following equation:4 Fe(s) + 3 O2(g) 2Fe2O3(s)

This equation can be read in “moles” by placing the

word “moles of” between each coefficient and formula.

4 moles of Fe + 3 moles of O2 2

Reading Equations in Moles

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A mole-mole factor is a ratio of the moles for any two

substances in an equation. 4Fe(s) + 3O2(g) 2Fe2O3(s)

Fe and O2 4 moles Fe and 3 moles O2

3 moles O2 4 moles Fe

Fe and Fe2O3 4 moles Fe and 2 moles Fe2O3

Writing Mole-Mole Factors

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Consider the following equation:3H2(g) + N2(g) 2NH3(g)

A. A mole-mole factor for H2 and N2 is1) 3 moles N2 2) 1 mole N2 3)

1 mole N21 mole H2 3 moles H2

2 moles H2

B. A mole-mole factor for NH3 and H2 is1) 1 mole H2 2) 2 moles NH3 3)

3 moles N2

Learning Check

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3H2(g) + N2(g) 2NH3(g)

A. A mole-mole factor for H2 and N2 is

2) 1 mole N23 moles H2

B. A mole-mole factor for NH3 and H2 is

2) 2 moles NH3

Solution

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How many moles of Fe2O3 can form from 6.0 moles of O2?

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

Relationship: 3 mole O2 = 2 mole Fe2O3

Use a mole-mole factor to determine the moles of Fe2O3.

Calculations with Mole Factors

114

Guide to Using Mole Factors


115

How many moles of Fe are needed for the reaction of 12.0 moles of O2?4Fe(s) + 3O2(g) 2Fe2O3(s)

1) 3.00 moles of Fe 2) 9.00 moles of Fe3) 16.0 moles of Fe

Learning Check

116

In a problem, identify the compounds given and needed.

How many moles of Fe are needed for the reaction of

12.0 moles of O2?4Fe(s) + 3O2(g) 2Fe2O3(s)

The possible mole factors for the solution are 4 moles Fe and 3 moles O23 moles O2 4 moles Fe

Study Tip: Mole Factors

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3) 16.0 moles of Fe

12.0 moles O2 x 4 moles Fe = 16.0 moles of Fe

3 moles O2

Solution

118


5.8Mass Calculations for Reactions


119

Moles to Grams

Suppose we want to determine the mass (g) of NH3that can form from 2.50 moles of N2.

N2(g) + 3H2(g) 2NH3(g)

The plan needed would be

moles N2 moles NH3 grams NH3

The factors needed would be: mole factor NH3/N2 and the molar mass NH3

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Moles to Grams

The setup for the solution would be:

2.50 mole N2 x 2 moles NH3 x 17.0 g NH31 mole N2 1 mole NH3

given mole-mole factor molar mass

= 85.0 g of NH3

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How many grams of O2 are needed to produce 0.400 mole of Fe2O3 in the following reaction?

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

1) 38.4 g of O2

2) 19.2 g of O23) 1.90 g of O2

Learning Check

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Solution

2) 19.2 g of O2

0.400 mole Fe2O3 x 3 mole O2 x 32.0 g O2= 19.2 g of O2

2 mole Fe2O3 1 mole O2mole factor molar mass

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The reaction between H2 and O2 produces 13.1 g of water.How many grams of O2 reacted?

2 H2(g) + O2(g) 2 H2O(g) ? g 13.1 g

The plan and factors would be

g H2O mole H2O mole O2 g of O2molar mole-mole molarmass H2O factor mass O2

Calculating the Mass of a Reactant

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Calculating the Mass of a Reactant

The setup would be:

13.1 g H2O x 1 mole H2O x 1 mole O2 x 32.0 g O218.0 g H2O 2 moles H2O 1 mole O2molar mole-mole molarmass H2O factor mass O2

= 11.6 g of O2

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Learning Check

Acetylene gas, C2H2, burns in the oxyacetylene torch for welding. How many grams of C2H2 are burned if the reaction produces 75.0 g of CO2?

2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(g)

1) 88.6 g of C2H22) 44.3 g of C2H23) 22.2 g of C2H2

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Solution

3) 22.2 g of C2H2

2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(g)

75.0 g CO2 x 1 mole CO2 x 2 moles C2H2 x 26.0 g C2H244.0 g CO2 4 moles CO2 1 mole C2H2

molar mole-mole molarmass CO2 factor mass C2H2

= 22.2 g of C2H2

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Calculating the Mass of Product

When 18.6 g of ethane gas, C2H6, burns in oxygen, howmany grams of CO2 are produced?

2C2H6(g) + 7O2(g) 4CO2(g) + 6H2O(g)18.6 g ? g

The plan and factors would be

g C2H6 mole C2H6 mole CO2 g of CO2molar mole-mole molarmass C2H6 factor mass CO2

128

Calculating the Mass of Product

2C2H6(g) + 7O2(g) 4CO2(g) + 6H2O(g)

The setup would be:18.6 g C2H6 x 1 mole C2H6 x 4 moles CO2 x 44.0 g CO2

30.1 g C2H6 2 moles C2H6 1 mole CO2molar mole-mole molarmass C2H6 factor mass CO2

= 54.4 g of CO2

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Study Tip: Check Units

Be sure to check that all units cancel to give the needed unit. Note each cancelled unit in the following setup:

needed unitg C2H6 x mole C2H6 x moles CO2 x g CO2

g C2H6 moles C2H6 mole CO2

molar mole-mole molarmass C2H6 factor mass CO2

130

Learning Check

How many grams of H2O are produced when 35.8 g of C3H8 react by the following equation?

C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)

1) 14.6 g of H2O2) 58.6 g of H2O3) 117 g of H2O

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Solution

2) 58.6 g of H2O

C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)

35.8 g C3H8x 1 mole C3H8 x 4 mole H2O x 18.0 g H2O44.0 g C3H8 1 mole C3H8 1 mole H2Omolar mole-mole molarmass C3H8 factor mass H2O

= 58.6 g of H2O

132

Chapter 5Chemical Quantities and Reactions

5.9 Energy in Chemical

Reactions


133

Collision Theory of Reactions

A chemical reaction occurs when• collisions between molecules have

sufficient energy to break the bonds in the reactants.

• bonds between atoms of the reactants (N2 and O2) are broken and new bonds (NO) can form.


134

Activation Energy

• The activation energy is the minimum energy needed for a reaction to take place.

• When a collision provides energy equal to or


135

C(s) + 2H2(g) CH4(g) + 18 kcal

In an exothermic reaction,

• heat is released.• the energy of the

products is less than the energy of the reactants.

• heat is a product.

Exothermic Reactions


136

Endothermic Reactions

In an endothermicreaction

• Heat is absorbed.• The energy of the

products is greater than the energy of the reactants.

• Heat is a reactant (added).

N2(g) + O2 (g) + 43.3 kcal 2NO(g)Copyright © 2009 by Pearson Education, Inc.

137

Learning Check

Identify each reaction as 1) exothermic or 2) endothermic.

A. N2 + 3H2 2NH3 + 22 kcal

B. CaCO3 + 133 kcal CaO + CO2

C. 2SO2 + O2 2SO3 + heat

138

Solution

Identify each reaction as 1) exothermic or 2) endothermic.

1 A. N2 + 3H2 2NH3 + 22 kcal

2 B. CaCO3 + 133 kcal CaO + CO2

1 C. 2SO2 + O2 2SO3 + heat

139

Rate of Reaction

Reaction rate

• is the speed at which reactant is used up.

• is the speed at which product forms.

• increases when temperature rises because reacting molecules move faster, providing more colliding molecules with energy of activation.

140

Reaction Rate and Catalysts

A catalyst

• increases the rate of a reaction.

• lowers the energy of activation.

• is not used up during the

ti


141

Summary


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Learning Check

State the effect of each on the rate of reaction as:

1) increases 2) decreases 3) no change

A. increasing the temperature.B. removing some of the reactants.C. adding a catalyst.D. placing the reaction flask in ice.E. increasing the concentration of one of

143

Solution

State the effect of each on the rate of reaction as:

1) increases 2) decreases 3) no change

1 A. increasing the temperature.2 B. removing some of the reactants.1 C. adding a catalyst.2 D. placing the reaction flask in ice.1 E. increasing the concentration of one of

the reactants

144

Learning Check

Indicate the effect of each factor listed on the rate of the

following reaction as:1) increases 2) decreases 3) none

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

A. raising the temperatureB. adding O2C. adding a catalystD. lowering the temperature

145

Solution

Indicate the effect of each factor listed on the rate of the

following reaction as1) increases 2) decreases 3) none

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

1 A. raising the temperature1 B. adding O21 C. adding a catalyst2 D. lowering the temperature

146

Summary of Factors That Increase Reaction Rate


chapter 5 chemical quantities and reactions 5.1 the...

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