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

## Full text

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

1 mole element Number of Atoms 1 mole C = 6.02 x 1023C atoms 1 mole Na = 6.02 x 1023Na atoms 1 mole Au = 6.02 x 1023Au 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 1023H

2O molecules

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

1 mole NaCl = 6.02 x 1023NaCl formula units 1 mole K2SO4 = 6.02 x 1023K2SO4formula units

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### Samples of 1 Mole Quantities

1 mole of C atoms = 6.02 x 1023C atoms

1 mole of Al atoms = 6.02 x 1023Al atoms

1 mole of S atoms = 6.02 x 1023S atoms

1 mole of H2O molecules = 6.02 x 1023H

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 1023particles

Conversion Factors:

6.02 x 1023particles and 1 mole 1 mole 6.02 x 1023particles

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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 1023Cu atoms

1 mole Cu

= 3.0 x 1023Cu 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 1024molecules CO 2? 2.50 x 1024molecules CO 2x 1 mole CO2 6.02 x 1023molecules 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 1024Al 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 1048moles of S atoms.

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### Solution

C. 1.2 x 1024Al atoms 2.0 mole Al x 6.02 x 1023Al atoms = 1 mole Al B. 3.0 moles of S atoms 1.8 x 1024S atoms x 1 mole S = 6.02 x 1023S 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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### 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

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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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### PO

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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.

### 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

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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 1023H 2O molecules 39

### Learning Check

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?

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### 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 1023molecules 114.2 g 1 mole = 1 x 109molecules 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 1023molecules 114.2 g 1 mole = 1 x 109molecules of C 6H10S are present 42

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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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## 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.

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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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### 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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### 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)

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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)

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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 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

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)

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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

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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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### 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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### 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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### 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)

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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

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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)

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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).

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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) + 1eFe2+(aq) __C. Cl2(g) + 2e2Cl-(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) + 1eFe2+(aq) 2 C. Cl2(g) + 2e2Cl-(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) + 1eoxidation

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

### Reactions

102

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

uv light

Ag+ + ClAg + 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+ + ClAg + Cl

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

B. Which reactant is reduced? Ag+ + 1eAg

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### Learning Check

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

A. Mg(s) + 2H+(aq) Mg2+(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+ + 2eH 2 B. Al is oxidized. Al Al3+ + 3e− Br is reduced. Br + e− Br − 106

## Chemical Equations

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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

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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

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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 H2 and 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

114

(20)

115

2

2

2

3

### 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

117

2

2

118

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

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

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 CO2 4 moles CO2 1 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

molar mole-mole molar mass C2H6 factor mass 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

molar mole-mole molar mass 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 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

## 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.

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

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

(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

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## References

Related subjects :