1 chapter 5 chemical reactions chemical changes chemical equations balancing a chemical equation
TRANSCRIPT
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Chapter 5 Chemical Reactions
Chemical Changes
Chemical Equations
Balancing a Chemical Equation
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The physical properties of a substance are the characteristics we can observe or measure without changing the substance.
Physical Properties
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In a physical change, The identity and composition of the substance
do not change. The state can change or the material can be
torn into smaller pieces.
In a chemical change, New substances form with different
compositions and properties. A chemical reaction takes place.
Physical and Chemical Change
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Physical and Chemical Change
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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.
During a reaction, old bonds are broken and new bonds are formed.
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In a chemical reaction, atoms in the reactants are rearranged to form one or more different substances.
In this reaction, Fe and O2 react to form rust (Fe2O3).
4Fe + 3O2 2Fe2O3
Chemical Reaction
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A chemical equation Shows the chemical formulas of the reactants
to the left of an arrow and the products on the right. Reactants Products MgO + C CO + Mg
Can be read in words. “Magnesium oxide reacts with carbon to form carbon monoxide and magnesium.”
Writing a Chemical Equation
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Symbols Used in Equations
Symbols used in equations show the states of the reactants and products and the reaction conditions.
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4 NH3 + 5 O2 4 NO + 6 H2O
Four molecules of NH3 react with five molecules of O2 to produce four molecules of NO and six molecules of H2O.
or
Four moles of NH3 react with 5 moles of O2 to produce four moles of NO and six moles of H2O.
Quantities in A Chemical Reaction
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In any ordinary chemical reaction, matter is not created nor destroyed.
+ + H2 + Cl2 2 HCl Total atoms = Total atoms
2 H, 2 Cl 2H, 2 Cl
Total Mass = Total Mass2(1.0) + 2(35.5) 2(36.5)73.0 g = 73.0 g
Law of Conservation of Mass
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Balancing a Chemical Equation
A chemical equation is balanced when there are the same numbers of each type of atom on both sides of the equation.
Al + S Al2S3 Not Balanced
2Al + 3S Al2S3 Balanced
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To balance an equation, place coefficients in front of the appropriate formulas.
4 NH3 + 5 O2 4 NO + 6 H2O Check the balance by counting the atoms of
each element in the reactants and the products.
4 N (4 x 1N) = 4 N (4 x 1N)
12 H (4 x 3H) = 12 H (6 x 2H)
10 O (5 x 2O) = 10 O (4O + 6O)
Using Coefficients to Balance
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Balance one element at a time. Use only coefficients to balance.
Fe3O4 + H2 Fe + H2O
Fe: Fe3O4 + H2 3Fe + H2O
O: Fe3O4 + H2 3Fe + 4H2O
H: Fe3O4 + 4H2 3Fe + 4H2O
Steps in Balancing an Equation
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Check the balance of atoms in the following: Fe3O4 + 4 H2 3 Fe + 4 H2O
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 + 4 H2 3 Fe + 4 H2O
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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Balancing with Polyatomic Ions Polyatomic ions can be balanced as a unit
when they appear on both sides.Pb(NO3)2 + NaCl NaNO3 + PbCl2
Balance NO3- as a unit
Pb(NO3)2 + NaCl 2NaNO3 + PbCl2
2 NO3- = 2 NO3
-
Balance Na (or Cl)Pb(NO3)2 + 2NaCl 2NaNO3 + PbCl2
2Na+ = 2Na+
2Cl- = 2Cl-
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Balance each equation. The coefficients in the answers are read from left to right.
A. __Mg + __N2 __Mg3N2
1) 1, 3, 2 2) 3, 1, 2 3) 3, 1, 1
B.__Al + __Cl2 __AlCl3
1) 3, 3, 2 2) 1, 3, 1 3) 2, 3, 2
Learning Check
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A. 3) 3, 1, 1 3 Mg + 1 N2 1 Mg3N2
B. 3) 2, 3, 2
2 Al + 3 Cl2 2 AlCl3
Solution
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A. __Fe2O3 + __C __Fe + __CO2
1) 2, 3, 2,3 2) 2, 3, 4, 3 3) 1, 1, 2, 3
B. __Al + __FeO __Fe + __Al2O3
1) 2, 3, 3, 1 2) 2, 1, 1, 1 3) 3, 3, 3, 1
C. __Al + __H2SO4 __Al2(SO4)3 + __H2
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
2 Fe2O3 + 3 C 4 Fe + 3 CO2
B. 1) 2, 3, 3, 1
2 Al + 3 FeO 3 Fe + 1 Al2O3
C. 2) 2, 3, 1, 3
2 Al + 3 H2SO4 1 Al2(SO4)3 + 3 H2
Solution
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Types of Reactions
Chapter 5 Chemical Reactions
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Chemical reactions are classified into general types:
Combination Decomposition Single Replacement Double Replacement Combustion
Types of Reactions
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In a combination reaction, two or more elements or simple compounds combine to form one product.A + B AB
ExamplesH2 + Cl2 2HCl
2S + 3O2 2SO3
4Fe + 3O2 2Fe2O3
Combination Reactions
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In a combination reaction, magnesium and oxygen react to form magnesium oxide.2Mg + O2 2MgO
Combination Reactions
Mg
O2
MgO
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In a decomposition reaction, one substance is broken down (split) into two or more simpler substances.
AB A + B
2HgO 2Hg + O2
2KClO3 2KCl + 3 O2
Decomposition Reactions
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Classify the following reactions as
1) combination or 2) decomposition:
___A. H2 + Br2 2HBr
___B. Al2(CO3)3 Al2O3 + 3CO2
___C. 4 Al + 3C Al4C3
Learning Check
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Classify the following reactions as
1) combination or 2) decomposition:
1 A. H2 + Br2 2HBr
2 B. Al2(CO3)3 Al2O3 + 3CO2
1 C. 4 Al + 3C Al4C3
Solution
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In a single replacement, one element takes the place of an element in a reacting compound.A + BC AC + BZn(s) + HCl(aq) ZnCl2(aq) + H2(g)
Single Replacement
Zn HCl
H2
ZnCl2
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In a double replacement, the positive ions in the reacting compounds switch places.
AB + CD AD + CB
AgNO3 + NaCl AgCl + NaNO3
ZnS + 2HCl ZnCl2 +H2S
Double Replacement
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Example of a Double Replacement When solutions of sodium sulfate and barium
chloride are mixed, solid BaSO4 is produced.
BaCl2 + Na2SO4 BaSO4 + 2NaCl
BaSO4
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Classify each of the following reactions as a
1) single replacement or 2) double replacement
__A. 2Al + 3H2SO4 Al2(SO4)3 + 3H2
__B. Na2SO4 + 2AgNO3 Ag2SO4 + 2NaNO3
__C. 3C + Fe2O3 2Fe + 3CO
Learning Check
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Classify each of the following reactions as a
1) single replacement or 2) double replacement
1 A. 2Al + 3H2SO4 Al2(SO4)3 + 3H2
2 B. Na2SO4 + 2AgNO3 Ag2SO4 + 2NaNO3
1 C. 3C + Fe2O3 2Fe + 3CO
Solution
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In a combustion reaction, a reactant often containing carbon reacts with oxygen O2.
C + O2 CO2
CH4 + 2O2 CO2 + 2H2O
C3H8 + 5O2 3CO2 + 4H2O Many combustion reactions utilize fuels that
are burned in oxygen to produce CO2, H2O, and energy.
Combustion
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Balance the combustion equation:
___C5H12 + ___O2 ___CO2 + ___H2O
Learning Check
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Balance the combustion equation:
1 C5H12 + 8 O2 5 CO2 + 6 H2O
Solution
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Summary of Reaction Types
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Learning Check
Identify each reaction as 1) combination 2) decomposition 3) combustion4) single replacement 5) double replacementA. 3Ba + N2 Ba3N2
B. 2Ag + H2S Ag2S + H2
C. SiO2 + 4HF SiF4 + 2H2O
D. PbCl2 + K2SO4 2KCl + PbSO4
E. K2CO3 K2O + CO2
F. C2H4 + 3O2 2CO2 + 2H2O
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Solution
Identify each reaction as 1) combination 2) decomposition 3) combustion4) single replacement 5) double replacement 1 A. 3Ba + N2 Ba3N2
4 B. 2Ag + H2S Ag2S + H2
5 C. SiO2 + 4HF SiF4 + 2H2O
5 D. PbCl2 + K2SO4 2KCl + PbSO4
2 E. K2CO3 K2O + CO2
3 F. C2H4 + 3O2 2CO2 + 2H2O
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Chapter 5 Chemical Reactions
Oxidation-Reduction Reactions
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Oxidation and reduction Are an important type of
reaction. Provide us with energy
from food. Provide electrical energy
in batteries. Occur when iron rusts.
4Fe + 3O2 2Fe2O3
Oxidation and Reduction
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An oxidation-reduction reaction involves the transfer of electrons from one reactant to another.
In oxidation, electrons are lost. Zn Zn2+ + 2e- (loss of electrons) In reduction, electrons are gained.
Cu2+ + 2e- Cu (gain of electrons)
Electron Loss and Gain
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Half-Reactions for Oxidation-Reduction
In the oxidation-reduction reaction of zinc and copper(II) sulfate, the zinc is oxidized and the Cu2+ (from Cu2+ SO4
2-) is reduced.
Zn Zn2+ + 2e- oxidationCu2+ + 2e- Cu reduction
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Identify each of the following as an
1) oxidation or a 2) reduction:
__A. Sn Sn4+ + 4e-
__B. Fe3+ + 1e- Fe2+
__C. Cl2 + 2e- 2Cl-
Learning Check
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Identify each of the following as an
1) oxidation or a 2) reduction:
1 A. Sn Sn4+ + 4e-
2 B. Fe3+ + 1e- Fe2+
2 C. Cl2 + 2e- 2Cl-
Solution
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In a balanced oxidation-reduction equation, the loss of electrons is equal to the gain of electrons.Zn + Cu2+ Zn2+ + Cu
The loss and gain of two electrons is shown in the separate oxidation and reduction reactions.Zn Zn2+ + 2e- oxidationCu2+ + 2e- Cu reduction
Balanced Red-Ox Equations
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In light-sensitive sunglasses, UV light initiatesan oxidation-reduction reaction. uv light
Ag+ + Cl- Ag + ClA. Which reactant is oxidized?
1) Ag+ 2) Cl- 3) AgB. Which reactant is reduced?
1) Ag+ 2) Cl- 3) Cl
Learning Check
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In light-sensitive sunglasses, UV light initiatesan oxidation-reduction reaction.
uv light
Ag+ + Cl- Ag + ClA. Which reactant is oxidized
2) Cl- Cl- Cl + e-
B. Which reactant is reduced?1) Ag+ Ag+ + e- Ag
Solution
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Write the separate oxidation and reduction reactions for the following equation.
2Cs + F2 2CsF
Learning Check
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Write the separate oxidation and reduction reactions for the following equation.
2Cs + F2 2CsF
Cs Cs+ + 1e- oxidation
F + 1e- F- reduction
Solution
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An early definition of oxidation is the addition of oxygen O2 to a reactant.
A metal or nonmetal is oxidized while the O2 is reduced to O2-.
4K + O2 2K2O
C + O2 CO2
2SO2 + O2 2SO3
Oxidation with Oxygen
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In organic and biological reactions, oxidation involves the loss of hydrogen atoms and reduction involves a gain of hydrogen atoms.
oxidation = Loss of H
reduction = Gain of H
CH3OH H2CO + 2H (loss of H)
Methanol Formaldehyde
Gain and Loss of Hydrogen
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Summary
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Learning Check
Identify the substances that are oxidized andreduced in the following reactions.
A. 4Fe + 3O2 2Fe2O3
B. 6Na + N2 2Na3N
C. 2K + I2 2KI
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Solution
3+ 2-
A. 4Fe + 3O2 2Fe2O3
Fe oxidized; O2 reduced1+ 3-
B. 6Na + N2 2Na3N
Na oxidized: N2 reduced 1+ 1-
C. 2K + I2 2KI
K oxidized: I2 reduced
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The Mole
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A collection term indicates a specific number of items.
For example, 1 dozen doughnuts contains 12 doughnuts.
1 ream of paper means 500 sheets.
1 case is 24 cans.
Collection Terms
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A mole contains 6.02 x 1023 particles, which is the number of carbon atoms in 12.01 g of carbon.
1 mole C = 6.02 x 1023 C atoms The number 6.02 x 1023 is known as Avogadro’s
number. One mole of any element contains Avogadro’s
number of atoms.1 mole Na = 6.02 x 1023 Na atoms1 mole Au = 6.02 x 1023 Au atoms
A Mole
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Avogadro’s number is also the number of molecules and formula units in one mole of a compound.
One mole of a covalent compound contains Avogadro’s number of molecules.1 mole CO2 = 6.02 x 1023 CO2 molecules
1 mole H2O = 6.02 x 1023 H2O molecules One mole of an ionic compound contains
Avogadro’s number of formula units.1 mole NaCl = 6.02 x 1023 NaCl formula units
A Mole of Molecules
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Samples of One Mole Quantities
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Avogadro’s number is written as conversion factors.
6.02 x 1023 particles and 1 mole 1 mole 6.02 x 1023 particles The number of molecules in 0.50 mole of CO2
molecules is calculated as
0.50 mole CO2 molecules x 6.02 x 1023 CO2 molecules
1 mole CO2 molecules
= 3.0 x 1023 CO2 molecules
Avogadro’s Number
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A. Calculate the number of atoms in 2.0 moles of Al. 1) 2.0 Al atoms
2) 3.0 x 1023 Al atoms 3) 1.2 x 1024 Al atoms
B. Calculate the number of moles of S in 1.8 x 1024 S. 1) 1.0 mole S atoms 2) 3.0 mole S atoms 3) 1.1 x 1048 mole S atoms
Learning Check
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A. Calculate the number of atoms in 2.0 moles of Al. 3) 1.2 x 1024 Al atoms
2.0 moles Al x 6.02 x 1023 Al atoms1 mole Al
B. Calculate the number of moles of S in 1.8 x 1024 S. 2) 3.0 mole S atoms
1.8 x 1024 S atoms x 1 mole S
6.02 x 1023 S atoms
Solution
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The mass of one mole is called molar mass.
The molar mass of an element is the atomic mass expressed in grams.
Molar Mass
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Give the molar mass to the nearest 0.1 g.
A. 1 mole of K atoms = ________
B. 1 mole of Sn atoms = ________
Learning Check
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Give the molar mass to the nearest 0.1 g.
A. 1 mole of K atoms =39.1 g
B. 1 mole of Sn atoms = 118.7 g
Solution
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Molar Mass of CaCl2
For a compound, the molar mass is the sum of the molar masses of the elements in the formula. We calculate the molar mass of CaCl2 to the nearest 0.1 g as follows.
Element Number of Moles
Atomic Mass Total Mass
Ca 1 40.1 g/mole 40.1 g
Cl2 2 35.5 g/mole 71.0 g
CaCl2 111.1 g
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Molar Mass of K3PO4
Determine the molar mass of K3PO4 to 0.1 g.
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
K3PO4 212.3 g
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One-Mole Quantities
32.1 g 55.9 g 58.5 g 294.2 g 342.3 g
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A. 1 mole of K2O = ______g
B. 1 mole of antacid Al(OH)3 = ______g
Learning Check
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A. 1 mole of K2O
2 moles K (39.1 g/mole) + 1 mole O (16.0 g/mole)
78.2 g + 16.0 g = 94.2 g
B. 1 mole of antacid Al(OH)3
1 mole Al (27.0 g/mole) + 3 moles O (16.0 g/mole)+ 3 moles H (1.0 g/mole)
27.0 g + 48.0 g + 3.0 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) 40.0 g/mole2) 262 g/mole 3) 309 g/mole
Learning Check
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Prozac, C17H18F3NO, is a widely used antidepressant that inhibits the uptake of serotonin by the brain. What is the molar mass of Prozac?
3) 309 g/mole
17C (12.0) + 18H (1.0) + 3F (19.0) + 1N (14.0) + 1 O (16.0) =
204 + 18 + 57.0 + 14.0 + 16.0
Solution
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Methane CH4 known as natural gas is used in gas cook tops and gas heaters.
1 mole CH4 = 16.0 g
The molar mass of methane can be written as 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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Acetic acid C2H4O2 gives the sour taste to vinegar. Write two molar mass factors for acetic acid.
1 mole of acetic acid = 60.0 g acetic acid
1 mole acetic acid and 60.0 g acetic acid 60.0 g acetic acid 1 mole acetic acid
Solution
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Mole factors are used to convert between the grams of a substance and the number of moles.
Calculations with Molar Mass
Grams Mole factor Moles
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Aluminum is often used for the structure oflightweight bicycle frames. How many gramsof Al are in 3.00 moles of Al?
3.00 moles Al x 27.0 g Al = 81.0 g Al1 mole Al
mole factor for Al
Calculating Grams from Moles
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The artificial sweetener aspartame (Nutri-Sweet) C14H18N2O5 is used to sweeten diet foods, coffee and soft drinks. How many moles of aspartame are present in 225 g of aspartame?
Learning Check
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Calculate the molar mass of C14H18N2O5.
(14 x 12.0) + (18 x 1.0) + (2 x 14.0) + (5 x 16.0) = 294 g/mole
Set up the calculation using a mole factor. 225 g aspartame x 1 mole aspartame
294 g aspartame
mole factor(inverted)
= 0.765 mole aspartame
Solution
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Conservation of Mass In a chemical reaction, the mass of the reactants
is equal to the mass of the products.
2 moles Ag + 1 mole S = 1 mole Ag2S2 (107.9 g) + 1(32.1 g) = 1 (247.9 g)
247.9 g reactants = 247.9 g product
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We can read the equation in “moles” by placing the word “moles” between each coefficient and formula.4 Fe + 3 O2 2 Fe2O3 4 moles Fe + 3 moles O2 2 moles Fe2O3
Moles in Equations
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A mole-mole factor is a ratio of the coefficients for two substances. 4 Fe + 3 O2 2 Fe2O3
Fe and O2 4 mole Fe and 3 mole O2
3 mole O2 4 mole Fe
Fe and Fe2O3 4 mole Fe and 2 mole Fe2O3
2 mole Fe2O3 4 mole Fe
O2 and Fe2O3 3 mole O2 and 2 mole Fe2O3
2 mole Fe2O3 3 mole O2
Writing Mole-Mole Factors
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Consider the following equation:
3 H2 + N2 2 NH3
A. A mole factor for H2 and N2 is 1) 3 mole N2 2) 1 mole N2 3) 1 mole N2
1 mole H2 3 mole H2 2 mole H2
B. A mole factor for NH3 and H2 is 1) 1 mole H2 2) 2 mole NH3 3) 3 mole N2
2 mole NH3 3 mole H2 2 mole NH3
Learning Check
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3 H2 + N2 2 NH3
A. A mole factor for H2 and N2 is 2) 1 mole N2
3 mole H2
B. A mole factor for NH3 and H2 is 2) 2 mole NH3
3 mole H2
Solution
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Consider the following reaction:
4 Fe + 3 O2 2 Fe2O3
How many moles of Fe2O3 are produced when 6.0moles O2 react?
Use the appropriate mole factor to determine themoles Fe2O3.6.0 mole O2 x 2 mole Fe2O3 = 4.0 mole Fe2O3
3 mole O2
Calculations with Mole Factors
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Consider the following reaction:
4 Fe + 3 O2 2 Fe2O3
How many moles of Fe are needed to react
with 12.0 moles of O2?
1) 3.00 moles Fe
2) 9.00 moles Fe
3) 16.0 moles Fe
Learning Check
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3) 16.0 moles Fe
Consider the following reaction:
4 Fe + 3 O2 2 Fe2O3
How many moles of Fe are needed to react with 12.0 moles of O2?
12.0 mole O2 x 4 mole Fe = 16.0 moles Fe 3 mole O2
Solution
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Mass Calculations
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How many grams of O2 are needed to produce
0.400 mole of Fe2O3?
4 Fe + 3 O2 2 Fe2O3
1) 38.4 g O2
2) 19.2 g O2
3) 1.90 g O2
Learning Check
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2) 19.2 g O2
0.400 mole Fe2O3 x 3 mole O2 x 32.0 g O2
2 mole Fe2O3 1 mole O2
mole factor molar mass
= 19.2 g O2
Solution
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The reaction between H2 and O2 produces 13.1 g of
water. How many grams of O2 reacted?
2H2 + O2 2H2O
? g 13.1 gPlan: g H2O mole H2O mole O2 g O2
13.1 g H2O x 1 mole H2O x 1 mole O2 x 32.0 g O2
18.0 g H2O 2 mole H2O 1 mole O2
= 11.6 g O2
Calculating the Mass of a Reactant
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Learning Check
Acetylene gas C2H2 burns in the oxyactylene torch for welding. How many grams of C2H2 are burned if the reaction produces 75.0 g of CO2?
2 C2H2 + 5 O2 4 CO2 + 2 H2O
1) 88.6 g C2H2
2) 44.3 g C2H2
3) 22.2 g C2H2
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3) 22.2 g C2H2
2 C2H2 + 5 O2 4 CO2 + 2 H2O
75.0 g CO2 x 1 mole CO2 x 2 moles C2H2 x 26.0 g C2H2
44.0 g CO2 4 moles CO2 1 mole C2H2
= 22.2 g C2H2
Solution
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You prepared cookie dough to make 5 dozen cookies. The phone rings and you answer. While you talk, a sheet of 12 cookies burns. You have to throw them out. The rest of the cookies are okay. The results of our baking can be described as follows:
Theoretical yield 60 cookies possible
Actual yield 48 cookies to eat
Percent yield 48 cookies x 100 = 80% yield 60 cookies
Percent Yield
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Percent Yield
The theoretical yield is the maximum amount of product calculated using the balanced equation.
The actual yield is the amount of product obtained when the reaction is run.
Percent yield is the ratio of actual yield compared to the theoretical yield.Percent Yield = Actual Yield (g) x 100
Theoretical Yield (g)
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Sample Exercise % Yield
Without proper ventilation and limited oxygen, the reaction of carbon and oxygen produces carbon monoxide.
2C + O2 2CO
What is the percent yield if 40.0 g of CO are produced from the reaction of 30.0 g O2?
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Sample Exercise % Yield (cont.)1. Calculate theoretical yield of CO. 30.0 g O2 x 1 mole O2 x 2 mole CO x 28.0 g CO
32.0 g O2 1 mole O2 1 mole CO= 52.5 g CO (theoretical)
2. Calculate the percent yield. 40.0 g CO (actual) x 100 = 76.2 % yield
52.5 g CO(theoretical)
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Learning Check
In the lab, N2 and 5.0 g of H2 are reactedand produce 16.0 g of NH3. What is thepercent yield for the reaction?
N2(g) + 3H2(g) 2NH3(g)
1) 31.3 %
2) 56.5 %
3) 80.0 %
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Solution
2) 56.5 %
N2(g) + 3H2(g) 2NH3(g)
5.0 g H2 x 1 mole H2 x 2 moles NH3 x 17.0 g NH3
2.0 g H2 3 moles H2 1 mole NH3
= 28.3 g NH3 (theoretical)
Percent yield = 16.0 g NH3 x 100 = 56.5 %
28.3 g NH3