Chapter 5 Chemical Quantities and Reactions. Collection Terms. 5.1 The Mole. A Mole of a Compound. A Mole of Atoms.

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Chapter 5 Chemical Quantities and Reactions

5.1 The Mole

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Collection Terms

A collection termstates a specific number of items.

• 1 dozen donuts = 12 donuts

• 1 ream of paper = 500 sheets

• 1 case = 24 cans

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A moleis 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 Atoms 1 mole C = 6.02 x 1023_{C atoms} 1 mole Na = 6.02 x 1023_{Na atoms} 1 mole Au = 6.02 x 1023_{Au atoms}

A Mole of Atoms

4 A mole

• of a covalent compound has Avogadro’s number of molecules.

1 mole CO2= 6.02 x 1023CO2molecules 1 mole H₂O = 6.02 x 1023_H

2O molecules

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

1 mole NaCl = 6.02 x 1023_{NaCl formula units} 1 mole K2SO4 = 6.02 x 1023K2SO4formula 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_H

2O molecules

1 mole of CCl4 molecules = 6.02 x 1023CCl4molecules

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Avogadro’s number, 6.02 x 1023_{, can be written as an}

equality 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

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Using Avogadro’s Number

Avogadro’s number is used to

o convert moles to particles

o convert particles to moles

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Converting Moles to Particles

Avogadro’s number is used to convert moles 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 atoms}

1 mole Cu

= 3.0 x 1023_{Cu atoms}

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Converting Particles to Moles

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

How many moles of CO2are in

2.50 x 1024_{molecules CO} 2? 2.50 x 1024_{molecules CO} 2x 1 mole CO2 6.02 x 1023_{molecules CO} 2 = 4.15 moles of CO2 10

1. The number of atoms in 2.0 mole of Al atoms is A. 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 atoms} 2.0 mole Al x 6.02 x 1023_{Al atoms} ₌ 1 mole Al B. 3.0 moles of S atoms 1.8 x 1024_{S atoms x 1 mole S} ₌ 6.02 x 1023_{S atoms} 12

Subscripts and Moles

The subscripts in a formulashow

• the relationship of atoms in the formula.

• the moles of each element in 1 mole of compound.

Glucose C₆H₁₂O₆

In 1 molecule: 6 atoms C 12 atoms H 6 atoms O In 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 for moles of each element in 1 mole of a compound. For aspirin, 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 C9H8O4

9 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 O

1 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 1023O atoms

1 mole C₉H₈O₄ 1 mole O Avogadro’s number = 3.61 x 1023 _{atoms of O}

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Chapter 5 Chemical Quantities and Reactions

5.2 Molar 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 mass is 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 g

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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 g Cl 2 35.5 g/mole 71.0 g

CaCl₂ 111.1 g

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Molar Mass of K

₃

PO

₄

Calculate the molar mass of K3PO4.

Element Number of Moles

Atomic Mass Total Mass in K3PO4

K 3 39.1 g/mole 117.3 g P 1 31.0 g/mole 31.0 g O 4 16.0 g/mole 64.0 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)378.0 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/mole 2) 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/mole 17 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

is

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

Conversion factors:

16.0 g CH₄ and 1 mole CH₄ 1 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 bicycle frames. 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 Al 1 mole Al

molar mass factor for Al

Moles to Grams

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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 C6H10S

molar mass factor (inverted) = 1.97 moles of C6H10S

Solution

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Grams, Moles, and Particles

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Learning Check

How many H2O molecules are in 24.0 g of H2O?

1) 4.52 x 1023 _H 2O molecules 2) 1.44 x 1025 _H 2O molecules 3) 8.03 x 1023 _H 2O molecules 38

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 1023H2O molecules 18.0 g H2O 1 mole H2O = 8.03 x 1023_H 2O molecules 39

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 10}23_molecules 114.2 g 1 mole = 1 x 109_{molecules of C} 6H10S are present 41

Solution

If the odor of C6H10S can be detected from 2 x 10-13g

in 1 liter of air, how many molecules of C₆H₁₀S are present? 2 x 10-13 _{g x 1 mole x 6.02 x 10}23_molecules 114.2 g 1 mole = 1 x 109_{molecules of C} 6H10S are present 42

Chapter 5 Chemical Quantities and Reactions

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 a 1) 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 a 1) 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

• 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

• the reactants are Fe and O2.

• the new product Fe2O3 is

called rust.

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Chapter 5 Chemical Quantities and Reactions

5.4 Chemical Equations

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Chemical Equations

A chemical equationgives

• 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) _CO

2(g)

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Symbols Used in Equations

Symbolsin 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 balanced chemical 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 coefficientsare used in front of one or more formulas to balance the number of atoms.

Al + S Al2S3 Not Balanced

2Al + 3S Al2S3 Balanced using coefficients

2 Al = 2 Al Equal number of Al atoms 3 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 57

Learning Check

State the number of atoms of each element on the reactant side and the product side for each of the following balanced equations.

A. P4(s) + 6Br2(l) 4PBr3(g) B. 2Al(s) + Fe2O3(s) 2Fe(s) + Al2O3(s) 58

Solution

A. P4(s) + 6Br2(l) 4PBr3(g) 4 P atoms = 4 P atoms 12 Br atoms = 12 Br atoms B. 2Al(s) + Fe2O3(s) 2Fe(s) + Al2O3(s) 2 Al atoms = 2 Al atoms 2 Fe atoms = 2 Fe atoms 3 O atoms = 3 O atoms 59

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 atoms 1 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 onlycoefficients 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 Products 1 C atom = 1 C atom 4 H atoms = 4 H atoms 4 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

70 A. 3) 3, 1, 1 3Mg(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 PO₄3- ₌ _{2 PO}

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3-Check Na + _balance

6 Na+ _{= 6 Na}+

Balance Mg and Cl

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

3 Mg2+ _{= 3 Mg}2+ 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

74 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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Chapter 5 Chemical Quantities and Reactions

5.5 Types 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 78

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) 82

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) 83

Single Replacement

In a single replacementreaction,

• 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)

90

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)

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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 and balance 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) 92

Solution

Each of the following reactions occur in the formation of smog or acid rain. Identify the type of reaction and balance 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) 93

Chapter 5 Chemical Quantities and Reactions

5.6 Oxidation-Reduction Reactions

94

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)

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An oxidation-reduction reaction involves

• a transfer of electrons from one reactant to another.

• oxidationas 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 Cu

2+

Zn(s) Zn2+_{(aq) + 2e-} _oxidation

silver metal

Cu2+_{(aq) + 2e-} _Cu(s) _reduction

blue solution orange

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Electron Transfer from Zn to Cu

2+

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

100

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

102

In light-sensitive sunglasses, UV light initiates an oxidation-reduction reaction.

uv light

Ag+ _{+ Cl}− _{Ag + Cl}

A. Which reactant is oxidized?

B. Which reactant is reduced?

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Solution

In light-sensitive sunglasses, UV light initiates an oxidation-reduction reaction.

uv light

Ag+ _{+ Cl}− _{Ag + Cl}

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

B. Which reactant is reduced? Ag+ _{+ 1e}− _Ag

104

Learning Check

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

A. Mg(s) + 2H+_{(aq) Mg}2+_{(aq) + H} 2(g) B. 2Al(s) + 3Br2(g) 2AlBr3(s) 105

Solution

A. Mg is oxidized. Mg(s) Mg2+_{(aq) + 2e}− H+_{is reduced.} _2H+ _{+ 2e}− _H 2 B. Al is oxidized. Al Al3+ _{+ 3e}− Br is reduced. Br + e− _Br− 106

Chapter 5 Chemical Quantities and Reactions

5.7 Mole Relationships in

Chemical Equations

107

Law of Conservation of Mass

The

law of conservation of mass

indicates 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.

108

Conservation of Mass

2 moles of Ag + 1 mole of S = 1 mole of Ag2S

2 (107.9 g) + 1(32.1 g) = 1 (247.9 g) 247.9 g reactants = 247.9 g product

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109

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 ofO2 2 moles

of Fe₂O₃

Reading Equations in Moles

110

A mole-mole factoris 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

2 moles Fe2O3 4 moles Fe

O2and Fe2O3 3 moles O2 and 2 moles Fe2O3

2 moles Fe2O3 3 moles O2

Writing Mole-Mole Factors

111

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

A. A mole-mole factor for H2 and N2is

1) 3 moles N2 2) 1 mole N2 3) 1 mole N2

1 mole H2 3 moles H2 2 moles H2

B. A mole-mole factor for NH3and H2is

1) 1 mole H2 2) 2 moles NH3 3) 3 moles N2

2 moles NH3 3 moles H2 2 moles NH3

Learning Check

112

3H₂(g) + N₂(g) 2NH₃(g) A. A mole-mole factor for H₂and N₂is

2) 1 mole N₂

3 moles H2

B. A mole-mole factor for NH₃and H₂is 2) 2 moles NH₃

3 moles H₂

Solution

113

How many moles of Fe₂O₃can 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.

6.0 mole O₂ x 2 mole Fe2O3 = 4.0 moles of Fe2O3

3 mole O2

Calculations with Mole Factors

114

Guide to Using Mole Factors

(20)

115

How many moles of Fe are needed for

the reaction of 12.0 moles of O

₂

?

4Fe(s) + 3O

₂

(g) 2Fe

₂

O

₃

(s)

1) 3.00 moles of Fe

2) 9.00 moles of Fe

3) 16.0 moles of Fe

Learning Check

116

In a problem, identify the compounds given and needed.

How many moles of Feare 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 O2

3 moles O2 4 moles Fe

Study Tip: Mole Factors

117

3) 16.0 moles of Fe

12.0 moles O

₂

x 4

moles Fe

=

16.0 moles of Fe

3 moles O

₂

Solution

118

Chapter 5 Chemical Quantities and Reactions

5.8 Mass Calculations for Reactions

119

Moles to Grams

Suppose we want to determine the mass (g) of NH₃ that 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/N2and the molar mass NH3

120

Moles to Grams

The setup for the solution would be: 2.50 mole N2x 2 moles NH3 x 17.0 g NH3

1 mole N2 1 mole NH3

given mole-mole factor molar mass

(21)

121

How many grams of O₂are 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 O2 3) 1.90 g of O2

Learning Check

122

Solution

2) 19.2 g of O2 0.400 mole Fe2O3x 3 mole O2 x 32.0 g O2= 19.2 g of O2 2 mole Fe2O3 1 mole O2

mole factor molar mass

123

The reaction between H2and O2produces 13.1 g of water.

How many grams of O2reacted?

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 O2

molar mole-mole molar mass H2O factor mass O2

Calculating the Mass of a Reactant

124

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 O2

18.0 g H₂O 2 moles H₂O 1 mole O₂

molar mole-mole molar mass H2O factor mass O2

= 11.6 g of O2

125

Learning Check

Acetylene gas, C₂H₂, burns in the oxyacetylene torch for welding. How many grams of C₂H₂are burned if the reaction produces 75.0 g of CO2?

2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(g) 1) 88.6 g of C2H2 2) 44.3 g of C2H2 3) 22.2 g of C2H2 126

Solution

3) 22.2 g of C2H2 2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(g) 75.0 g CO2x 1 mole CO2 x 2 moles C2H2x 26.0 g C2H2 44.0 g CO₂4 moles CO₂1 mole C₂H₂

molar mole-mole molar mass CO2 factor mass C2H2

(22)

127

Calculating the Mass of Product

When 18.6 g of ethane gas, C2H6, burns in oxygen, how

many grams of CO2are 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 CO2

molar mole-mole molar mass 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 C2H6x 1 mole C2H6x 4 moles CO2x 44.0 g CO2

30.1 g C2H6 2 moles C2H6 1 mole CO2

= 54.4 g of CO2

129

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 unit g C2H6x mole C2H6 x moles CO2 x g CO2

g C₂H₆ moles C₂H₆ mole 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 H2O 2) 58.6 g of H2O 3) 117 g of H2O 131

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 H2O 44.0 g C3H8 1 mole C3H8 1 mole H2O

molar mole-mole molar mass C3H8 factor mass H2O

= 58.6 g of H₂O

132

Chapter 5 Chemical Quantities and Reactions

5.9 Energy in Chemical

Reactions

(23)

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 greater than the activation energy, product can form.

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 endothermic

reaction

• 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)

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

(24)

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

141

Summary

142

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 the reactants.

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 temperature

B. adding O₂ C. adding a catalyst D. lowering the temperature

(25)

145

Solution

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) 1 A. raising the temperature 1 B. adding O₂

1 C. adding a catalyst 2 D. lowering the temperature

146