Chapter #3
description
Transcript of Chapter #3
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Chapter #3Chemical Composition
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Chemical Reactions• A chemical reactions is an abbreviated way to
show a physical orchemical change• A chemical change alters the physical and
chemical properties of a substance• Factors that indicate a chemical change
– Change in color– Temperature change– Change in odor– Change in taste (we do not taste chemicals)
• Reactions always contain an arrow that separates the reactants from the products
Reactants Products
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Types of Chemical Reactions
• Combination reaction (synthesis)– Elements for reactants– Examples:
H2 + O2 H2O
N2 + H2 NH3
Al + O2 Al2O3
The Law of Conservation of matter, states matter cannot be created nor destroyed, that means equations must be balanced.
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Types of Chemical Reactions
Balance the first equation
H2 + O2 H2O
Note two oxygen atoms on the reactant side and only one on the product side, therefore place a two in front of water
Combination reaction Continued
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Types of Chemical Reactions
Balance the first equation
H2 + O2 2H2O
Note two oxygen atoms on the reactant side and only one on the product side, therefore place a two in front of water
The two now doubles everything in water, thus 4 hydrogen and 2 oxygen. Now place a 2 in front of hydrogen.
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Types of Chemical Reactions
Balance the first equation
2H2 + O2 2H2O
Note two oxygen atoms on the reactant side and only one on the product side, therefore place a two in front of water
The two now doubles everything in water, thus 4 hydrogen and 2 oxygen. Now place a 2 in front of hydrogen.
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Types of Chemical Reactions
Now balance the second equation
N2 + H2 NH3
Note two nigrogen atoms on the reactant side and only one on the product side.
Place a 2 in front of ammonia
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Types of Chemical Reactions
Now balance the second equation
N2 + H2 2NH3
Note two nitrogen atoms on the reactant side and only one on the product side.
Place a 2 in front of ammonia. This makes 2 nitrogen atoms and 6 hydrogen atoms. Now place a 3 in front of hydrogen to balance hydrogen atoms.
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Types of Chemical Reactions
Now balance the second equation
N2 + 3 H2 2NH3
Note two nitrogen atoms on the reactant side and only one on the product side.
Place a 2 in front of ammonia. This makes 2 nitrogen atoms and 6 hydrogen atoms. Now place a 3 in front of hydrogen to balance hydrogen atoms.
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Types of Chemical Reactions
• Decomposition Reaction– Compounds form simpler compounds or
elements.– Examples
H2O H2 + O2
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Types of Chemical Reactions
• Decomposition Reaction– Compounds form simpler compounds or
elements.– Examples
2H2O H2 + O2
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Types of Chemical Reactions• Decomposition Reaction
– Compounds form simpler compounds or elements.
– Examples
2H2O 2H2 + O2
• Notice decomposition reactions are the opposite of combination reactions
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Types of Chemical Reactions
Single Replacement reactions have an element and a compound for reactants.
Example:
Zn + HCl
How do we predict the products? Trade places with the metal or nonmetal with the metal or nonmetal in the compound
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Types of Chemical Reactions
Single Replacement reactions have an element and a compound for reactants.
Example:
Zn + HCl
How do we predict the products? Trade places with the metal or nonmetal with the metal or nonmetal in the compound
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Types of Chemical Reactions
Single Replacement reactions have an element and a compound for reactants.
Example:
Zn + HCl ZnCl + H
Now make the products stable. Slide with Clyde
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Types of Chemical Reactions
Single Replacement reactions have an element and a compound for reactants.
Example:
Zn + HCl ZnCl2 + H2
Now make the products stable. Slide with Clyde
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Types of Chemical Reactions
Single Replacement reactions have an element and a compound for reactants.
Example:
Zn + HCl ZnCl2 + H2
Now make the products stable.
Now Balance
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Types of Chemical Reactions
Single Replacement reactions have an element and a compound for reactants.
Example:
Zn + 2HCl ZnCl2 + H2
Now make the products stable.
Now Balance
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Types of Chemical Reactions
Single Replacement reactions have an element and a compound for reactants.
Another Example:
Cl2 + MgBr2
How do we predict the products? Trade places with the metal or nonmetal with the metal or nonmetal in the compound. In this case we are trading nonmetals
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Types of Chemical Reactions
Single Replacement reactions have an element and a compound for reactants.
Another Example:
Cl2 + MgBr2 Br + MgCl
How do we predict the products? Trade places with the metal or nonmetal with the metal or nonmetal in the compound. In this case we are trading nonmetals
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Types of Chemical Reactions
Single Replacement reactions have an element and a compound for reactants.
Another Example:
Cl2 + MgBr2 Br2 + MgCl2
How do we predict the products? Trade places with the metal or nonmetal with the metal or nonmetal in the compound. In this case we are trading nonmetals
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Types of Chemical Reactions
Double Replacement reactions contain compounds as reactants.
HCl + Ca(OH)2 CaCl + HOH
Check formulas, and slide with Clyde when necessary
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Types of Chemical Reactions
Double Replacement reactions contain compounds as reactants.
HCl + Ca(OH)2 CaCl2 + HOH
Check formulas, and slide with Clyde when necessary
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Types of Chemical Reactions
Double Replacement reactions contain compounds as reactants.
2HCl + Ca(OH)2 CaCl2 + 2HOH
Check formulas, and slide with Clyde when necessary
Now Balance!
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Types of Chemical Reactions
Combustion Reactions occur when an element or compound combine with oxygen to produce oxides of each element.
H2 + O2
CH4 + O2
What is the oxide of hydrogen?
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Types of Chemical Reactions
Combustion Reactions occur when an element or compound combine with oxygen to produce oxides of each element.
H2 + O2
CH4 + O2
What is the oxide of hydrogen? Water
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Types of Chemical Reactions
Combustion Reactions occur when an element or compound combine with oxygen to produce oxides of each element.
H2 + O2 H2O
CH4 + O2
What is the oxide of hydrogen? Water
And the oxide of carbon?
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Types of Chemical Reactions
Combustion Reactions occur when an element or compound combine with oxygen to produce oxides of each element.
H2 + O2 H2O
CH4 + O2 CO2 + H2O
What is the oxide of hydrogen? Water
And the oxide of carbon? Carbon dioxide
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Types of Chemical Reactions
Combustion Reactions occur when an element or compound combine with oxygen to produce oxides of each element.
2H2 + O2 2H2O
CH4 + O2 CO2 + H2O
Now balance
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Types of Chemical Reactions
Combustion Reactions occur when an element or compound combine with oxygen to produce oxides of each element.
2H2 + O2 2H2O
CH4 + O2 CO2 + 2H2O
Now balance
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Types of Chemical Reactions
Combustion Reactions occur when an element or compound combine with oxygen to produce oxides of each element.
2H2 + O2 2H2O
CH4 + 2O2 CO2 + 2H2O
Now balance
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Ionic Solution Formation
KCN (S) K+ (aq) + CN- (aq) Ionic equation
Note: Not all ionic solutes are soluble in water.
How can we tell if an ionic solute is soluble in water?
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Ionic Equations
KCN (S) K+ (aq) + CN- (aq) Ionic equation
Note: Not all ionic solutes are soluble in water.
How can we tell if an ionic solute is soluble in water?
When some ionic solids are dissolved in water equations (physical or chemical) can be used to explain the process as shown below:
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Ionic Equations
KCN (S) K+ (aq) + CN- (aq) Ionic equation
Note: Not all ionic solutes are soluble in water.
How can we tell if an ionic solute is soluble in water?The solubility rules gives ionic solubility.
When some ionic solids are dissolved in water equations (physical or chemical) can be used to explain the process as shown below:
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There are some more specific rules that allows us to better estimate the solubility of ionic compounds.
You will be given these if you need them.
Solubility Rules
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Ionic EquationsUsing the solubility rules write the formula equation, the ionic equation and the net ionic equation when aqueous silver nitrate is combined with aqueous sodium chloride.
AgNO3 (aq) + NaCl (aq) AgCl (s) + NaNO3 (aq) Formula Equation
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Ionic EquationsUsing the solubility rules write the formula equation, the ionic equation and the net ionic equation when aqueous silver nitrate is combined with aqueous sodium chloride.
AgNO3 (aq) + NaCl (aq) AgCl (s) + NaNO3 (aq) Formula EquationAg+ (aq) + NO3
- (aq) + Na+ (aq) + Cl- (aq) AgCl (s) + Na+ (aq) + NO3- (aq)
Ionic Equation
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Ionic EquationsUsing the solubility rules write the formula equation, the ionic equation and the net ionic equation when aqueous silver nitrate is combined with aqueous sodium chloride.
AgNO3 (aq) + NaCl (aq) AgCl (s) + NaNO3 (aq) Formula EquationAg+ (aq) + NO3
- (aq) + Na+ (aq) + Cl- (aq) AgCl (s) + Na+ (aq) + NO3- (aq)
Ionic Equation
Spectator ions are ions that are identical on the reactants and products side of the equation.
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Ionic EquationsUsing the solubility rules write the formula equation, the ionic equation and the net ionic equation when aqueous silver nitrate is combined with aqueous sodium chloride.
AgNO3 (aq) + NaCl (aq) AgCl (s) + NaNO3 (aq) Formula EquationAg+ (aq) + NO3
- (aq) + Na+ (aq) + Cl- (aq) AgCl (s) + Na+ (aq) + NO3- (aq)
Ionic EquationSpectator ions are ions that are identical on the reactants and products side of the equation. Place a around the spectator ions.
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Ionic EquationsUsing the solubility rules write the formula equation, the ionic equation and the net ionic equation when aqueous silver nitrate is combined with aqueous sodium chloride.
AgNO3 (aq) + NaCl (aq) AgCl (s) + NaNO3 (aq) Formula EquationAg+ (aq) + NO3
- (aq) + Na+ (aq) + Cl- (aq) AgCl (s) + Na+ (aq) + NO3- (aq)
Ionic EquationSpectator ions are ions that are identical on the reactants and products side of the equation. Place a around the spectator ions.
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Ionic EquationsUsing the solubility rules write the formula equation, the ionic equation and the net ionic equation when aqueous silver nitrate is combined with aqueous sodium chloride.
AgNO3 (aq) + NaCl (aq) AgCl (s) + NaNO3 (aq) Formula EquationAg+ (aq) + NO3
- (aq) + Na+ (aq) + Cl- (aq) AgCl (s) + Na+ (aq) + NO3- (aq)
Ionic EquationSpectator ions are ions that are identical on the reactants and products side of the equation. Eliminating the spectator ions produces the netionic equation.
Ag+ (aq) + Cl- (aq) AgCl (s) Net ionic equation
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Yet Another Ionic EquationWrite the formula, ionic and net ionic equationswhen aqueous sodium chloride combines withaqueous calcium bromide.
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Yet Another Ionic EquationWrite the formula, ionic and net ionic equationswhen aqueous sodium chloride combines withaqueous calcium bromide.
NaCl (aq) + CaBr2 (aq) CaCl2 (aq) + NaBr (aq)
Now balance
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Yet Another Ionic EquationWrite the formula, ionic and net ionic equationswhen aqueous sodium chloride combines withaqueous calcium bromide.
2NaCl (aq) + CaBr2 (aq) CaCl2 (aq) + 2NaBr (aq)
Now balance
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Yet Another Ionic EquationWrite the formula, ionic and net ionic equationswhen aqueous sodium chloride combines withaqueous calcium bromide.
NaCl(aq) + CaBr2 CaCl2(aq) + NaBr(aq)22
2Na+(aq) + 2Cl-(aq) + Ca2+(aq) + 2Br-(aq) Ca2+(aq) + 2Cl-(aq) + 2Na+(aq) +2 Br-(aq)Ionic equation
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Yet Another Ionic EquationWrite the formula, ionic and net ionic equationswhen aqueous sodium chloride combines withaqueous calcium bromide.
NaCl(aq) + CaBr2 CaCl2(aq) + NaBr(aq)Now balance
22
2Na+(aq) + 2Cl-(aq) + Ca2+(aq) + 2Br-(aq) Ca2+(aq) + 2Cl-(aq) + 2Na+(aq) +2 Br-(aq)Ionic equation
No net ionic equation No Reaction (NR)
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Types of Chemical Reactions
REDOX reactions where the oxidation number changes from reactants to products.
Oxidation is when the oxidation number increases, by losing of electrons.
Reduction is when the oxidation number decreases by gaining electrons.
Consider the following equation:
H2 + O2 H2O
What are the oxidation numbers of hydrogen and oxygen?
0 0
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Types of Chemical Reactions
REDOX reactions where the oxidation number changes from reactants to products.
Oxidation is when the oxidation number increases, by losing of electrons.
Reduction is when the oxidation number decreases by gaining electrons.
Consider the following equation:
H2 + O2 H2O
What are the oxidation numbers of hydrogen and oxygen?
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REDOX REACTIONS
H2 + O2 H2O
How about hydrogen and oxygen in water?
00 2(1+) 2- = 0
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REDOX REACTIONS
H2 + O2 H2O
How about hydrogen and oxygen in water?Oxidation is caused by the oxygen molecule,
so it is referred to as the oxidizing agent (OA)
Reduction is caused by the hydrogen molecule, so it is referred to as the reducing agent (RA)
00 2(1+) 2- = 0
reducedoxidized
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REDOX REACTIONSNote:
• All of the previously discussed reactions are REDOX except the double replacement reactions.
• The number of electrons lost is equal to the number of electrons gained in a reaction. Why?
• Most elements have variable oxidation numbers, except for hydrogen, oxygen, and the memorized polyatomic ions.
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REDOX REACTIONSOxidation numbers for a compound must
add up to equal zero, while the oxidation numbers for a polyatomic ion must up to equal the charge of that ion.
Consider the following chlorine compounds
HClO4, HClO3, HClO2, HClO, Cl2, HCl
What is the oxidation number of chlorine in each of these compounds, assuming
H 1+ and oxygen is 2-
1+ 4(2-)=0
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REDOX REACTIONSOxidation numbers for a compound must
add up to equal zero, while the oxidation numbers for a polyatomic ion must up to equal the charge of that ion.
Consider the following chlorine compounds
HClO4, HClO3, HClO2, HClO, Cl2, HCl
What is the oxidation number of chlorine in each of these compounds, assuming H is 1+ and oxygen is 2-
1+ 4(2-)=07+
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REDOX REACTIONSOxidation numbers for a compound must
add up to equal zero, while the oxidation numbers for a polyatomic ion must up to equal the charge of that ion.
Consider the following chlorine compounds
HClO4, HClO3, HClO2, HClO, Cl2, HCl
What is the oxidation number of chlorine in each of these compounds, assuming H is 1+ and oxygen is 2-
1+ 4(2-)=07+ 5+
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REDOX REACTIONSOxidation numbers for a compound must
add up to equal zero, while the oxidation numbers for a polyatomic ion must up to equal the charge of that ion.
Consider the following chlorine compounds
HClO4, HClO3, HClO2, HClO, Cl2, HCl
What is the oxidation number of chlorine in each of these compounds, assuming H is 1+ and oxygen is 2-
1+ 4(2-)=07+ 5+ 3+
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REDOX REACTIONSOxidation numbers for a compound must
add up to equal zero, while the oxidation numbers for a polyatomic ion must up to equal the charge of that ion.
Consider the following chlorine compounds
HClO4, HClO3, HClO2, HClO, Cl2, HCl
What is the oxidation number of chlorine in each of these compounds, assuming H is 1+ and oxygen is 2-
1+ 4(2-)=07+ 5+ 3+ 1+
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REDOX REACTIONSOxidation numbers for a compound must
add up to equal zero, while the oxidation numbers for a polyatomic ion must up to equal the charge of that ion.
Consider the following chlorine compounds
HClO4, HClO3, HClO2, HClO, Cl2, HCl
What is the oxidation number of chlorine in each of these compounds, assuming H is 1+ and oxygen is 2-
1+ 4(2-)=07+ 5+ 3+ 1+ 0
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REDOX REACTIONSOxidation numbers for a compound must
add up to equal zero, while the oxidation numbers for a polyatomic ion must up to equal the charge of that ion.
Consider the following chlorine compounds
HClO4, HClO3, HClO2, HClO, Cl2, HCl
What is the oxidation number of chlorine in each of these compounds, assuming H is 1+ and oxygen is 2-
1+ 4(2-)=07+ 5+ 3+ 1+ 0 1-
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REDOX REACTIONS
How about sulfur in SO3 2-
3(2-)=2-
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REDOX REACTIONS
How about sulfur in SO3 2-
How about carbon in C6H12O6
3(2-)=2-4+
12(1+) +6(2-)=0
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REDOX REACTIONS
How about sulfur in SO3 2-
How about carbon in C6H12O6
3(2-)=2-4+
12(1+) +6(2-)=00 +
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Types of Chemical Reactions
REDOX reactions where the oxidation number changes from reactants to products.
Oxidation is when the oxidation number increases, by losing of electrons.
Reduction is when the oxidation number decreases by gaining electrons.
Consider the following equation:
H2 + O2 H2O
What are the oxidation numbers of hydrogen and oxygen?
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Types of Chemical Reactions
REDOX reactions where the oxidation number changes from reactants to products.
Oxidation is when the oxidation number increases, by losing of electrons.
Reduction is when the oxidation number decreases by gaining electrons.
Consider the following equation:
H2 + O2 H2O
What are the oxidation numbers of hydrogen and oxygen?
0 0
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REDOX REACTIONS
H2 + O2 H2O
How about hydrogen and oxygen in water?
00 2(1-) 2- = 0
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REDOX REACTIONS
H2 + O2 H2O
How about hydrogen and oxygen in water?Oxidation is caused by the oxygen molecule,
so it is referred to as the oxidizing agent (OA)
Reduction is caused by the hydrogen molecule, so it is referred to as the reducing agent (RA)
00 2(1+) 2- = 0
reducedoxidized
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REDOX REACTIONSNote:
• All of the previously discussed reactions are REDOX except the double replacement reactions.
• The number of electrons lost is equal to the number of electrons gained in a reaction. Why?
• Most elements have variable oxidation numbers, except for hydrogen, oxygen, and the memorized polyatomic ions.
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Balancing Redox ReactionsI. Oxidation Number Method
a. Assign oxidation numbers to each elementb. Determine the elements oxidized and reducedc. Balance the atoms that are oxidized and reducedd. Balance the electrons lost or gained, to conform to the Law of
Conservation of Matter, by placing coefficients in front of the formulas containing the atoms oxidized and reduced to both sides of the equation.
e. The remaining atoms are balanced by inspectionf. Balance oxygen, or hydrogen by adding H2Og. Balance remaining hydrogen atoms by adding H+ h. Simplifyi. For basic reactions add the same number of OH- ions to both sides of the
equation as there are H+ ions.j. Combine H+ and OH- ions to make waterk. Simplify again if necessary.
HNO3 + Cu2O → Cu(NO3)2 + NO + H2O
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Balancing REDOX ReactionsI. Oxidation Number Method
a. Assign oxidation numbers to each elementb. Determine the elements oxidized and reducedc. Balance the atoms that are oxidized and reducedd. Balance the electrons lost or gained, to conform to the Law of
Conservation of Matter, by placing coefficients in front of the formulas containing the atoms oxidized and reduced to both sides of the equation.
e. The remaining atoms are balanced by inspectionf. Balance oxygen, or hydrogen by adding H2Og. Balance remaining hydrogen atoms by adding H+ h. Simplifyi. For basic reactions add the same number of OH- ions to both
sides of the equation as there are H+ ions.j. Combine H+ and OH- ions to make waterk. Simplify again if necessary.
HNO3 + Cu2O → Cu(NO3)2 + NO + H2O1+ 3(2-)=0?
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Balancing REDOX ReactionsI. Oxidation Number Method
a. Assign oxidation numbers to each elementb. Determine the elements oxidized and reducedc. Balance the atoms that are oxidized and reducedd. Balance the electrons lost or gained, to conform to the Law of
Conservation of Matter, by placing coefficients in front of the formulas containing the atoms oxidized and reduced to both sides of the equation.
e. The remaining atoms are balanced by inspectionf. Balance oxygen, or hydrogen by adding H2Og. Balance remaining hydrogen atoms by adding H+ h. Simplifyi. For basic reactions add the same number of OH- ions to both sides of
the equation as there are H+ ions.j. Combine H+ and OH- ions to make waterk. Simplify again if necessary.
HNO3 + Cu2O → Cu(NO3)2 + NO + H2O1+ 3(2-)=05+ 2-=0 2(?)+
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Balancing REDOX ReactionsI. Oxidation Number Method
a. Assign oxidation numbers to each elementb. Determine the elements oxidized and reducedc. Balance the atoms that are oxidized and reducedd. Balance the electrons lost or gained, to conform to the Law of
Conservation of Matter, by placing coefficients in front of the formulas containing the atoms oxidized and reduced to both sides of the equation.
e. The remaining atoms are balanced by inspectionf. Balance oxygen, or hydrogen by adding H2Og. Balance remaining hydrogen atoms by adding H+ h. Simplifyi. For basic reactions add the same number of OH- ions to both sides of
the equation as there are H+ ions.j. Combine H+ and OH- ions to make waterk. Simplify again if necessary.
HNO3 + Cu2O → Cu(NO3)2 + NO + H2O1+ 3(2-)=05+ 2-=0 2(1+)+
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Balancing Redox ReactionsI. Oxidation Number Method
a. Assign oxidation numbers to each elementb. Determine the elements oxidized and reducedc. Balance the atoms that are oxidized and reducedd. Balance the electrons lost or gained, to conform to the Law of
Conservation of Matter, by placing coefficients in front of the formulas containing the atoms oxidized and reduced to both sides of the equation.
e. The remaining atoms are balanced by inspectionf. Balance oxygen, or hydrogen by adding H2Og. Balance remaining hydrogen atoms by adding H+ h. Simplifyi. For basic reactions add the same number of OH- ions to both sides of
the equation as there are H+ ions.j. Combine H+ and OH- ions to make waterk. Simplify again if necessary.
HNO3 + Cu2O → Cu(NO3)2 + NO + H2O1+ 3(2-)=05+ 2-=0 2(1+)+ 2(1-)=0? +
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Balancing REDOX ReactionsI. Oxidation Number Method
a. Assign oxidation numbers to each elementb. Determine the elements oxidized and reducedc. Balance the atoms that are oxidized and reducedd. Balance the electrons lost or gained, to conform to the Law of
Conservation of Matter, by placing coefficients in front of the formulas containing the atoms oxidized and reduced to both sides of the equation.
e. The remaining atoms are balanced by inspectionf. Balance oxygen, or hydrogen by adding H2Og. Balance remaining hydrogen atoms by adding H+ h. Simplifyi. For basic reactions add the same number of OH- ions to both sides of
the equation as there are H+ ions.j. Combine H+ and OH- ions to make waterk. Simplify again if necessary.
HNO3 + Cu2O → Cu(NO3)2 + NO + H2O1+ 3(2-)=05+ 2-=0 2(1+)+ 2(1-)=02+ +
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Balancing REDOX ReactionsI. Oxidation Number Method
a. Assign oxidation numbers to each elementb. Determine the elements oxidized and reducedc. Balance the atoms that are oxidized and reducedd. Balance the electrons lost or gained, to conform to the Law of
Conservation of Matter, by placing coefficients in front of the formulas containing the atoms oxidized and reduced to both sides of the equation.
e. The remaining atoms are balanced by inspectionf. Balance oxygen, or hydrogen by adding H2Og. Balance remaining hydrogen atoms by adding H+ h. Simplifyi. For basic reactions add the same number of OH- ions to both sides of
the equation as there are H+ ions.j. Combine H+ and OH- ions to make waterk. Simplify again if necessary.
HNO3 + Cu2O → Cu(NO3)2 + NO + H2O1+ 3(2-)=05+ 2-=0 2(1+)+ 2(1-)=0 ? + 2- =0
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Balancing REDOX ReactionsI. Oxidation Number Method
a. Assign oxidation numbers to each elementb. Determine the elements oxidized and reducedc. Balance the atoms that are oxidized and reducedd. Balance the electrons lost or gained, to conform to the Law of
Conservation of Matter, by placing coefficients in front of the formulas containing the atoms oxidized and reduced to both sides of the equation.
e. The remaining atoms are balanced by inspectionf. Balance oxygen, or hydrogen by adding H2Og. Balance remaining hydrogen atoms by adding H+ h. Simplifyi. For basic reactions add the same number of OH- ions to both sides of
the equation as there are H+ ions.j. Combine H+ and OH- ions to make waterk. Simplify again if necessary.
HNO3 + Cu2O → Cu(NO3)2 + NO + H2O1+ 3(2-)=05+ 2-=0 2(1+)+ 2(1-)=0 2 + 2- =0
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Balancing REDOX ReactionsI. Oxidation Number Method
a. Assign oxidation numbers to each elementb. Determine the elements oxidized and reducedc. Balance the atoms that are oxidized and reducedd. Balance the electrons lost or gained, to conform to the Law of
Conservation of Matter, by placing coefficients in front of the formulas containing the atoms oxidized and reduced to both sides of the equation.
e. The remaining atoms are balanced by inspectionf. Balance oxygen, or hydrogen by adding H2Og. Balance remaining hydrogen atoms by adding H+ h. Simplifyi. For basic reactions add the same number of OH- ions to both sides of
the equation as there are H+ ions.j. Combine H+ and OH- ions to make waterk. Simplify again if necessary.
HNO3 + Cu2O → Cu(NO3)2 + NO + H2O1+ 3(2-)=05+ 2-=0 2(1+)+ 2(1-)=0 2 + 2- =02+
oxidized
reduced
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Balancing REDOX ReactionsI. Oxidation Number Method
a. Assign oxidation numbers to each elementb. Determine the elements oxidized and reducedc. Balance the atoms that are oxidized and reducedd. Balance the electrons lost or gained, to conform to the Law of
Conservation of Matter, by placing coefficients in front of the formulas containing the atoms oxidized and reduced to both sides of the equation.
e. The remaining atoms are balanced by inspectionf. Balance oxygen, or hydrogen by adding H2Og. Balance remaining hydrogen atoms by adding H+ h. Simplifyi. For basic reactions add the same number of OH- ions to both sides of
the equation as there are H+ ions.j. Combine H+ and OH- ions to make waterk. Simplify again if necessary.
HNO3 + Cu2O → 2 Cu(NO3)2 + NO + H2O1+ 3(2-)=05+ 2-=0 2(1+)+ 2(1-)=0 2 + 2- =02+
oxidizedreduced
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Balancing REDOX ReactionsI. Oxidation Number Method
a. Assign oxidation numbers to each elementb. Determine the elements oxidized and reducedc. Balance the atoms that are oxidized and reducedd. Balance the electrons lost or gained, to conform to the Law of
Conservation of Matter, by placing coefficients in front of the formulas containing the atoms oxidized and reduced to both sides of the equation.
e. The remaining atoms are balanced by inspectionf. Balance oxygen, or hydrogen by adding H2Og. Balance remaining hydrogen atoms by adding H+ h. Simplifyi. For basic reactions add the same number of OH- ions to both sides of
the equation as there are H+ ions.j. Combine H+ and OH- ions to make waterk. Simplify again if necessary.
HNO3 + 3 Cu2O → 3 (2) Cu(NO3)2 + NO + H2O1+ 3(2-)=05+ 2-=0 2(1+)+ 2(1-)=0 2 + 2- =02+
Oxidized3( -2e)Reduced 2(+3)e
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Balancing REDOX ReactionsI. Oxidation Number Method
a. Assign oxidation numbers to each elementb. Determine the elements oxidized and reducedc. Balance the atoms that are oxidized and reducedd. Balance the electrons lost or gained, to conform to the Law of
Conservation of Matter, by placing coefficients in front of the formulas containing the atoms oxidized and reduced to both sides of the equation.
e. The remaining atoms are balanced by inspectionf. Balance oxygen, or hydrogen by adding H2Og. Balance remaining hydrogen atoms by adding H+ h. Simplifyi. For basic reactions add the same number of OH- ions to both sides of
the equation as there are H+ ions.j. Combine H+ and OH- ions to make waterk. Simplify again if necessary.
2HNO3 + 3 Cu2O → 3 (2) Cu(NO3)2 + 2NO + H2O1+ 3(2-)=05+ 2-=0 2(1+)+ 2(1-)=0 2 + 2- =02+
Oxidized3( -2e)Reduced 2(+3)e
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Balancing REDOX ReactionsI. Oxidation Number Method
a. Assign oxidation numbers to each elementb. Determine the elements oxidized and reducedc. Balance the atoms that are oxidized and reducedd. Balance the electrons lost or gained, to conform to the Law of
Conservation of Matter, by placing coefficients in front of the formulas containing the atoms oxidized and reduced to both sides of the equation.
e. The remaining atoms are balanced by inspectionf. Balance oxygen, or hydrogen by adding H2Og. Balance remaining hydrogen atoms by adding H+ h. Simplifyi. For basic reactions add the same number of OH- ions to both sides of
the equation as there are H+ ions.j. Combine H+ and OH- ions to make waterk. Simplify again if necessary.
2HNO3 + 3 Cu2O → 3 (2) Cu(NO3)2 + 2NO + H2O
1+ 3(2-)=05+ 2-=0 2(1+)+ 2(1-)=0 2 + 2- =02+
Oxidized3( -2e)Reduced 2(+3)e
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Balancing REDOX ReactionsI. Oxidation Number Method
a. Assign oxidation numbers to each elementb. Determine the elements oxidized and reducedc. Balance the atoms that are oxidized and reducedd. Balance the electrons lost or gained, to conform to the Law of
Conservation of Matter, by placing coefficients in front of the formulas containing the atoms oxidized and reduced to both sides of the equation.
e. The remaining atoms are balanced by inspectionf. Balance oxygen, or hydrogen by adding H2Og. Balance remaining hydrogen atoms by adding H+ h. Simplifyi. For basic reactions add the same number of OH- ions to both sides of
the equation as there are H+ ions.j. Combine H+ and OH- ions to make waterk. Simplify again if necessary.
14HNO3 + 3 Cu2O → 3 (2) Cu(NO3)2 + 2NO + 7 H2O
1+ 3(2-)=05+ 2-=0 2(1+)+ 2(1-)=0 2 + 2- =02+
Oxidized3( -2e)Reduced 2(+3)e
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Balancing REDOX ReactionsI. Oxidation Number Method
a. Assign oxidation numbers to each elementb. Determine the elements oxidized and reducedc. Balance the atoms that are oxidized and reducedd. Balance the electrons lost or gained, to conform to the Law of
Conservation of Matter, by placing coefficients in front of the formulas containing the atoms oxidized and reduced to both sides of the equation.
e. The remaining atoms are balanced by inspectionf. Balance oxygen, or hydrogen by adding H2Og. Balance remaining hydrogen atoms by adding H+ h. Simplifyi. For basic reactions add the same number of OH- ions to both sides of
the equation as there are H+ ions.j. Combine H+ and OH- ions to make waterk. Simplify again if necessary.
14 HNO3 + 3 Cu2O → 6 Cu(NO3)2 + 2 NO + 7 H2O
1+ 3(2-)=05+ 2-=0 2(1+)+ 2(1-)=0 2 + 2- =02+
Oxidized3( -2e)Reduced 2(+3)e
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OX # BALANCING EXAMPLE
MnO4 - + Cl- → Mn2+ + Cl2
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OX # BALANCING EXAMPLE
MnO4 - + Cl- → Mn2+ + Cl2
7+
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OX # BALANCING EXAMPLE
MnO4 - + Cl- → Mn2+ + Cl2
7+ 1-
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OX # BALANCING EXAMPLE
MnO4 - + Cl- → Mn2+ + Cl2
7+ 1- 0
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OX # BALANCING EXAMPLE
MnO4 - + Cl- → Mn2+ + Cl2
7+ 1- 0
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OX # BALANCING EXAMPLE
MnO4 - + Cl- → Mn2+ + Cl2
7+ 1- 0
reducedoxidized
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OX # BALANCING EXAMPLE
MnO4 - + Cl- → Mn2+ + Cl2
7+ 1- 0
reducedoxidized
Step C, balance atoms oxidized or reduced
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OX # BALANCING EXAMPLE
MnO4 - + Cl- → Mn2+ + Cl2
7+ 1- 0
reducedoxidized
Step C, balance atoms oxidized or reduced
2
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OX # BALANCING EXAMPLE
MnO4 - + Cl- → Mn2+ + Cl2
7+ 1- 0
reducedoxidized
Step d, balance electrons lost or gained. common denominator between 5 and
2 is 10. Therefore multiply Mn on both sides of the equation by two and Cl on both sides by 5.
2
+ 5 e-- 2 e-
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OX # BALANCING EXAMPLE
MnO4 - + Cl- → Mn2+ + Cl2
7+ 1- 0
reducedoxidized
Step d, balance electrons lost or gained. The common denominator between 5 and
2 is 10. Therefore multiply Mn on both sides of the equation by 2 and Cl on both sides by 5.
5(2)
+ 5 e-- 2 e-
2 2 5
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OX # BALANCING EXAMPLE
MnO4 - + Cl- → Mn2+ + Cl2
7+ 1- 0
reducedoxidized
Step e, balance remaining elements by inspection. There are 8 oxygen atoms on the left. Oxygen is balanced by adding water to the appropriate side. In this case since there are 8 oxygen atoms on the reactant side which requires adding 8 water molecules to the product side of the equation.
5(2)
+ 5 e-- 2 e-
2 2 5
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OX # BALANCING EXAMPLE MnO4
- + Cl- → Mn2+ + Cl27+ 1- 0
reducedoxidized
Step e, balance remaining elements by inspection. There are 8 oxygen atoms on the left. Oxygen is balanced by adding water to the appropriate side. In this case since there are 8 oxygen atoms on the reactant side which requires adding 8 water molecules to the product side of the equation. Now the hydrogen atoms need to be balanced by adding 16 H+ to the reactant side.
10
+ 5 e-- 2 e-
2 2 5 + 8H2O
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OX # BALANCING EXAMPLE MnO4
-+ Cl- → Mn2++ Cl27+ 1- 0
reducedoxidized
Step e, balance remaining elements by inspection. There are 8 oxygen atoms on the left. Oxygen is balanced by adding water to the appropriate side. In this case since there are 8 oxygen atoms on the reactant side which requires adding 8 water molecules to the product side of the equation. Now the hydrogen atoms need to be balanced by adding 16 H+ to the reactant side.
10
+ 5 e-- 2 e-
2
2 5 + 8H2O16 H++
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Balancing REDOX Equationsby
The Half Reaction Method
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Half Reaction Steps1. Write separate equations (Half-reactions) for oxidized and reduced
substances.2. For each half-reaction balance all elements, except hydrogen and
oxygena. Balance oxygen using H2O
b. Balance hydrogen using H+
c. Balance charge in each half-reaction by adding electrons (reduction), or removing electrons (oxidation) to the appropriate half reaction.
3. Multiply each half-reaction by an integer so that the number of electrons lost equal the number of electrons gained
a. Add half-reactions, and simplifyb. For basic reactions add the same number of OH- ions to both
sides of the equation as there are H+ ions.c. Combine H+ and OH- ions to make waterd. Simplify again if necessary.
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Half Reaction Example
MnO4- + Fe2+ → Mn2+ + Fe3+
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Half Reaction Example
MnO4- + Fe2+ → Mn2+ + Fe3+
Step 1, Write half reactions
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Half Reaction Example
MnO4- + Fe2+ → Mn2+ + Fe3+
Step 1, Write half reactions
MnO4- → Mn2+
Fe2+ → Fe3+
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Half Reaction Example
MnO4- + Fe2+ → Mn2+ + Fe3+
Step 2a, Balance Oxygen by adding water.
MnO4- → Mn2+
Fe2+ → Fe3+
+ 4 H2O
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Half Reaction Example
MnO4- + Fe2+ → Mn2+ + Fe3+
Step 2b, Balance hydrogen by adding H+.
MnO4- → Mn2+
Fe2+ → Fe3+
+ 4 H2O8 H+ +
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Half Reaction Example
MnO4- + Fe2+ → Mn2+ + Fe3+
Step 2c, Balance charge by adding/removing e’s
MnO4- → Mn2+
Fe2+ → Fe3+
+ 4 H2O8 H+ +
In the top half equation the reactants have 7+ and the products 2+, adding 5 e’s to the reactant side gives 2+ on both sides.In the bottom half equation the reactants have 2+ and the products have 2+, removing 1 e from the reactant side gives 2+ on both sides.
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Half Reaction Example
MnO4- + Fe2+ → Mn2+ + Fe3+
Step 2c, Balance charge by adding/removing e’s
MnO4- → Mn2+
Fe2+ → Fe3+
+ 4 H2O8 H+ +
In the top half equation the reactants have 7+ and the products 2+, adding 5 e’s to the reactant side gives 2+ on both sides.In the bottom half equation the reactants have 2+ and the products have 2+, removing 1 e from the reactant side gives 2+ on both sides.
+ 5e-
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Half Reaction Example
MnO4- + Fe2+ → Mn2+ + Fe3+
Step 2c, Balance charge by adding/removing e’s
MnO4- → Mn2+
Fe2+ → Fe3+
+ 4 H2O8 H+ +
In the top half equation the reactants have 7+ and the products 2+, adding 5 e’s to the reactant side gives 2+ on both sides.In the bottom half equation the reactants have 2+ and the products have 2+, removing 1 e from the reactant side gives 2+ on both sides.
+ 5e-
- e-
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Half Reaction Example
MnO4- + Fe2+ → Mn2+ + Fe3+
Step 2c, Balance charge by adding/removing e’s
MnO4- → Mn2+
Fe2+ → Fe3+
+ 4 H2O8 H+ +
In the top half equation the reactants have 7+ and the products 2+, adding 5 e’s to the reactant side gives 2+ on both sides.In the bottom half equation the reactants have 2+ and the products have 2+, removing 1 e from the reactant side gives 2+ on both sides.
+ 5e-
- e-
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Half Reaction Example
MnO4- + Fe2+ → Mn2+ + Fe3+
Step 3, The common denominator between 5 and 1 is 5. Multiply the bottom half equation by 5
MnO4- → Mn2+
Fe2+ → Fe3+
+ 4 H2O8 H+ + + 5e-
- e-
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Half Reaction Example
MnO4- + Fe2+ → Mn2+ + Fe3+
Step 4, Add the two half equations together
MnO4- → Mn2+
Fe2+ → Fe3+
+ 4 H2O8 H+ + + 5e-
- e-5( )
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Half Reaction Example
MnO4- + Fe2+ → Mn2+ + Fe3+
Step 4, Add the two half equations together
MnO4- → Mn2+
Fe2+ → Fe3+
+ 4 H2O8 H+ + + 5e-
- e-5( )
8 H+ + MnO4- 5 Fe3++ 5 Fe2+ + Mn2+ + 4 H2O→
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Other REDOX ExamplesHNO2 + Cr2O7
2- → Cr2+ + NO3- (acidic)
CN- + MnO4- → CNO- + MnO2 (basic)
Al(s) + OH- (aq) → Al(OH)4- (aq) + H2 (g) (acidic or basic)
Cl2 (g) → Cl- (aq) + ClO- (aq) (basic)
Ag (s) + CN- + O2 → Ag(CN)2 - (aq) (basic)
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Real Life Examples of REDOX•REDOX reactions can be used to generate electricity.
•REDOX reactions can be used to protect metals from oxidation.
•REDOX reactions can be used to plate metals on to other metals or surfaces.
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The Chemical Package
• The baker uses a package called the dozen. All dozen packages contain 12 objects.
• The stationary store uses a package called a ream, which contains 500 sheets of paper.
• So what is the chemistry package?
About Packages
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The Chemical Package
• The baker uses a package called the dozen. All dozen packages contain 12 objects.
• The stationary store uses a package called a ream, which contains 500 sheets of paper.
• So what is the chemistry package? Well, it is called the mole (Latin for heap).
About Packages
Each of the above packages contain a number of objects that are convenient to work with, for that particular discipline.
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The atomic weights listed on the periodic chart are the weights of a mole of atoms. For example a mole of hydrogen atoms weighs 1.00797 g and a mole of carbon atoms weighs 12.01 g which are weighted averages of the natural abundance of isotopes for that element.
The MoleA mole contains 6.022X1023 particles, which is the number of carbon-12 atoms that will give a mass of 12.00 grams, which is a convenient number of atoms to work with in the chemistry laboratory.
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Moles of ObjectsSuppose we order a mole of marshmallows for a chemistry party. How much space here at Central would be required to store the marshmallows?
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Moles of Objects
Suppose we order a mole of marshmallows for a chemistry party. How much space here at Central would be required to store the marshmallows?Would cover the entire 50 states 60 miles deep
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Moles of Objects
Suppose we order a mole of marshmallows for a chemistry party. How much space here at Central would be required to store the marshmallows?Would cover the entire 50 states 60 miles deep
How about a mole of computer paper instead of a ream of computer paper, how far would that stretch?
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Moles of Objects
Suppose we order a mole of marshmallows for a chemistry party. How much space here at Central would be required to store the marshmallows?Would cover the entire 50 states 60 miles deep
How about a mole of computer paper instead of a ream of computer paper, how far would that stretch? Way past the planet Pluto!
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To calculate the molar mass of a compound we sum together the atomic weights of the atoms that make up the formula of the compound. This is called the formula weight (MW, M).
Formula weights are the sum of atomic weights of atoms making up the formula.
The following outlines how to find the formula weight of water
symbol weight numberHO
1.0116.0
21X
X== 2.02
16.018.0 g/mole
Formula Weight Calculation
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Percent Composition
Find the formula weight and the percent composition of
glucose (C6H12O6)
symbol weight number
HO
C
16.01.01
12.0
6
12
6
x
x
x
=
=
=
72.0
12.1296.0
180.1 g/mole
%C =
%H =
%O =
72.0
12.12
96.0
180.1
180.1
180.1
X =
X =
X =
40.0 %C
6.73 %H
53.3 %O
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A mole of glucose (C6H12O6) contains 6.022 X 1023 molecules of glucose. And 6 X 6.022 X 1023 atoms of C. Since a mole is 6.022 X 1023 particles then a mole of glucose must contain 6 moles of C atoms.
How many moles of hydrogen atoms are contained in a mole of glucose?
In 5 moles of H2SO4 how many moles of oxygen atoms is there?
Mole Concepts
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A mole of glucose (C6H12O6) contains 6.022 X 1023 molecules of glucose. And 6 X 6.022 X 1023 atoms of C. Since a mole is 6.022 X 1023 particles then a mole of glucose must contain 6 moles of C atoms.
How many moles of hydrogen atoms are contained in a mole of glucose? 12 Moles of hydrogen.
How many moles of oxygen and hydrogen are in one mole of H2O contains:
Mole Concepts
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A mole of glucose (C6H12O6) contains 6.022 X 1023 molecules of glucose. And 6 X 6.022 X 1023 atoms of C. Since a mole is 6.022 X 1023 particles then a mole of glucose must contain 6 moles of C atoms.
How many moles of hydrogen atoms are contained in a mole of glucose? 12 Moles of hydrogen.
How many moles of oxygen and hydrogen are in one mole of H2O contains:One mole of oxygen atoms Two moles of hydrogen atoms
Mole Concepts
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A mole of glucose (C6H12O6) contains 6.022 X 1023 molecules of glucose. And 6 X 6.022 X 1023 atoms of C. Since a mole is 6.022 X 1023 particles then a mole of glucose must contain 6 moles of C atoms.
How many moles of hydrogen atoms are contained in a mole of glucose? 12 Moles of hydrogen.
How many moles of oxygen and hydrogen are in one mole of H2O contains:One mole of oxygen atoms Two moles of hydrogen atoms
In 5 moles of H2SO4 how many moles of oxygen atoms is there?
Mole Concepts
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A mole of glucose (C6H12O6) contains 6.022 X 1023 molecules of glucose. And 6 X 6.022 X 1023 atoms of C. Since a mole is 6.022 X 1023 particles then a mole of glucose must contain 6 moles of C atoms.
How many moles of hydrogen atoms are contained in a mole of glucose? 12 Moles of hydrogen.
How many moles of oxygen and hydrogen are in one mole of H2O contains:One mole of oxygen atoms Two moles of hydrogen atoms
In 5 moles of H2SO4 how many moles of oxygen atoms is there?
20 moles of O atoms.
Mole Concepts
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In 50.0g of H2SO4 how many moles of sulfuric acid are there?
Mole Conversions
50.0g of H2SO4
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In 50.0g of H2SO4 how many moles of sulfuric acid are there?
Mole Conversions
50.0g of H2SO4 =98.0g of H2SO4
mole H2SO4
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In 50.0g of H2SO4 how many moles of sulfuric acid are there?
Mole Conversions
50.0g of H2SO4 = 0.510 mole H2SO498.0g of H2SO4
mole H2SO4
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In 50.0g of H2SO4 how many moles of oxygen atoms are there?
Mole Conversions
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In 50.0g of H2SO4 how many moles of oxygen atoms are there?
Mole Conversions
50.0g of H2SO4 =
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In 50.0g of H2SO4 how many moles of oxygen atoms are there?
Mole Conversions
50.0g of H2SO4 =98.0g of H2SO4
mole H2SO4
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In 50.0g of H2SO4 how many moles of oxygen atoms are there?
Mole Conversions
50.0g of H2SO4 =mole H2SO4
4mole O98.0g of H2SO4
mole H2SO4
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In 50.0g of H2SO4 how many moles of oxygen atoms are there?
Mole Conversions
50.0g of H2SO4 =mole H2SO4
4mole O 2.04 mole O98.0g of H2SO4
mole H2SO4
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In 5 moles of H2SO4 how many atoms of oxygen are present?
Mole Conversions
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In 5 moles of H2SO4 how many atoms of oxygen are present?
Mole Conversions
5 moles H2SO4 =
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In 5 moles of H2SO4 how many atoms of oxygen are present?
Mole Conversions
5 moles H2SO4
mole H2SO4
4 mole O
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In 5 moles of H2SO4 how many atoms of oxygen are present?
Mole Conversions
5 moles H2SO4
mole H2SO4
4 mole O 6.02 x 1023 atoms O mole O
=
1.20 x 1025 atoms
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Empirical FormulasEmpirical formula is the smallest whole number ratio between atoms and can be calculated from the percent composition.
Molecular formulas happen to be the exact number of atoms making up a molecule, and may or may no be the simplest whole number ratio. Molecular formulas are whole number multiples of the empirical formula.
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Empirical Formula Steps1. Assume 100 g of compound.2. Convert percent to a mass number.3. Convert the mass to moles.4. Divide each mole number by the smallest mole
number.5. Rounding:
a. If the decimal is ≤ 0.1, then drop the decimalsb. If the decimal is ≥0.9, then round up.c. All other decimal need to be multiplied by a whole
number until roundable.
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Empirical Formula ExampleA compound is composed of 75.0% C and 15.0% H. Find its empirical formula.
Step #1 Assume 100 g of compound
75.0 g C
25.0 g H
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Empirical Formula ExampleA compound is composed of 75.0% C and 15.0% H. Find its empirical formula.
Step #2 Convert grams to moles.
75.0 g C
25.0 g H12.01 g
Mole C
1.008 g H
Mole H
= 6.225 mole C
= 24.802 mole
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Empirical Formula ExampleA compound is composed of 75.0% C and 15.0% H. Find its empirical formula.
Step #3 Divide each mole number by the smallest.
75.0 g C
25.0 g H12.01 g
Mole C
1.008 g H
Mole H
= 6.225 mole C
= 24.802 mole
6.225 6.2256.22524.802
= 1.00 = 3.98
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Empirical Formula ExampleA compound is composed of 75.0% C and 15.0% H. Find its empirical formula.
Step #4 Rounding; Decimal ≤ 0.1, drop decimals
75.0 g C
25.0 g H12.01 g
Mole C
1.008 g H
Mole H
= 6.225 mole C
= 24.802 mole
6.225 6.2256.22524.802
= 1.00 = 3.98
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Empirical Formula ExampleA compound is composed of 75.0% C and 15.0% H. Find its empirical formula.
Step #4 Rounding; Decimal ≤ 0.1, drop decimals
75.0 g C
25.0 g H12.01 g
Mole C
1.008 g H
Mole H
= 6.225 mole C
= 24.802 mole
6.225 6.2256.22524.802
= 1 C = 3.98
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Empirical Formula ExampleA compound is composed of 75.0% C and 15.0% H. Find its empirical formula.
Step #4 Rounding; Decimal ≥ 0.9, round up
75.0 g C
25.0 g H12.01 g
Mole C
1.008 g H
Mole H
= 6.225 mole C
= 24.802 mole
6.225 6.2256.22524.802
= 1 C = 3.98
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Empirical Formula ExampleA compound is composed of 75.0% C and 15.0% H. Find its empirical formula.
Step #4 Rounding; Decimal ≥ 0.9, round up
75.0 g C
25.0 g H12.01 g
Mole C
1.008 g H
Mole H
= 6.225 mole C
= 24.802 mole
6.225 6.2256.22524.802
= 1 C = 3.98
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Empirical Formula ExampleA compound is composed of 75.0% C and 15.0% H. Find its empirical formula.
Step #4 Rounding; Decimal ≥ 0.9, round up
75.0 g C
25.0 g H12.01 g
Mole C
1.008 g H
Mole H
= 6.225 mole C
= 24.802 mole
6.225 6.2256.22524.802
= 1 C = 4 H
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Empirical Formula ExampleA compound is composed of 75.0% C and 15.0% H. Find its empirical formula.
Step #4 Rounding; Decimal ≥ 0.9, round up
75.0 g C
25.0 g H12.01 g
Mole C
1.008 g H
Mole H
= 6.225 mole C
= 24.802 mole
6.225 6.2256.22524.802
= 1 C = 4 H
Empirical Formula = CH4
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Molecular Formulas
Empirical formula, is the smallest ratio between atoms in a molecular or formula unit.
Molecular formula, is the exact number of atoms in a molecule; a whole number multiple of an empirical formula
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Possible Molecular Formulas
Assume an empirical formula of C3H5O
Empirical formula Integer Molecular Formula
C3H5O 1
2
3
4
5
C3H5O
C3H5O
C3H5O
C3H5O
C3H5O
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Possible Molecular Formulas
Assume an empirical formula of C3H5O
Empirical formula Integer Molecular Formula
C3H5O 1
2
3
4
5
C3H5O
C3H5O
C3H5O
C3H5O
C3H5O
C6H10O
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Possible Molecular Formulas
Assume an empirical formula of C3H5O
Empirical formula Integer Molecular Formula
C3H5O 1
2
3
4
5
C3H5O
C3H5O
C3H5O
C3H5O
C3H5O
C6H10O2
C9H15O3
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Possible Molecular Formulas
Assume an empirical formula of C3H5O
Empirical formula Integer Molecular Formula
C3H5O 1
2
3
4
5
C3H5O
C3H5O
C3H5O
C3H5O
C3H5O
C6H10O2
C9H15O3
C12H20O4
C15H25O5
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Sample ProblemCalculate the molecular formula of a molecule composed of 83.7%C and 16.3% H, with a molar mass of 86.0 g/mole
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Sample ProblemCalculate the molecular formula of a molecule composed of 83.7%C and 16.3% H, with a molar mass of 86.0 g/mole
Step #1 Assume 100g of compound
83.6 g C
16.3 g H
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Sample ProblemCalculate the molecular formula of a molecule composed of 83.7%C and 16.3% H, with a molar mass of 86.0 g/mole
Step #2 Convert grams to moles
83.6 g C
16.3 g H12.01 g Cmole
mole1.008 g H
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Sample ProblemCalculate the molecular formula of a molecule composed of 83.7%C and 16.3% H, with a molar mass of 86.0 g/mole
Step #2 Convert grams to moles
83.6 g C
16.3 g H12.01 g Cmole
mole1.008 g H
= 6.961 mole
= 16.17 mole
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Sample ProblemCalculate the molecular formula of a molecule composed of 83.7%C and 16.3% H, with a molar mass of 86.0 g/mole
Step #3 Divide each mole number by the smallest.
83.6 g C
16.3 g H12.01 g Cmole
mole1.008 g H
= 6.961 mole
= 16.17 mole
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Sample ProblemCalculate the molecular formula of a molecule composed of 83.7%C and 16.3% H, with a molar mass of 86.0 g/mole
Step #3 Divide each mole number by the smallest.
83.6 g C
16.3 g H12.01 g Cmole
mole1.008 g H
= 6.961 mole
= 16.17 mole
6.961 6.961
= 1.00
= 2.32
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Sample ProblemCalculate the molecular formula of a molecule composed of 83.7%C and 16.3% H, with a molar mass of 86.0 g/mole
Step #4 Round if---Not Roundable
83.6 g C
16.3 g H12.01 g Cmole
mole1.008 g H
= 6.961 mole
= 16.17 mole
6.961 6.961
= 1.00
= 2.32
Step #4, Multiply by an integer until roundable
1.00 X 3 = 3
2.32 X 3 = 7Empirical formula C3H7
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Molecular Formula IntegerDivide empirical weight into molecular weight
3x12 + 7x1 =43
43 862
Now multiply the empirical formula by 2
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Molecular Formula IntegerDivide empirical weight into molecular weight
3x12 + 7x1 =43
43 862
Now multiply the empirical formula by 2
Molecular Formula is C6 H14
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Stoichiometry
Sotichiometery is the process of converting quantities of reactants or products to other participants of a chemical or physical change using the coefficients of a balanced equation.
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STOICHIOMETRY
Stoichiometry is the use of balanced chemical equations in the conversion process.
Example
Calculate the mass of water formed from 6.33 g of hydrogen. A balanced equation is required.
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STOICHIOMETRY
Stoichiometry is the use of balanced chemical equations in the conversion process.
Example
Calculate the mass of water formed from 6.33 g of hydrogen. A balanced equation is required.
2 H2 + O2 2 H2O
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STOICHIOMETRY
Stoichiometry is the use of balanced chemical equations in the conversion process.
Examples
Calculate the mass of water formed from 6.33 g of hydrogen. A balanced equation is required.
2 H2 + O2 2H2O6.33 g H2
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STOICHIOMETRY
Stoichiometry is the use of balanced chemical equations in the conversion process.
Example
Calculate the mass of water formed from 6.33 g of hydrogen. A balanced equation is required.
2 H2 + O2 2 H2O6.33 g H2
2.016 g H2O
Mole H2
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STOICHIOMETRY
Stoichiometry is the use of balanced chemical equations in the conversion process.
Example
Calculate the mass of water formed from 6.33 g of hydrogen. A balanced equation is required.
2 H2 + O2 2 H2O6.33 g H2
2.016 g H2O
Mole H2 2 Mole H2O
2 Mole H2
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STOICHIOMETRY
Stoichiometry is the use of balanced chemical equations in the conversion process.
Example
Calculate the mass of water formed from 6.33 g of hydrogen. A balanced equation is required.
2 H2 + O2 2 H2O6.33 g H2
2.016 g H2O
Mole H2 2 Mole H2O
2 Mole H2 Mole H2O
18.02 g H2O
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STOICHIOMETRY
Stoichiometry is the use of balanced chemical equations in the conversion process.
Examples
Calculate the mass of water formed from 6.33 g of hydrogen. A balanced equation is required.
2 H2 + O2 2 H2O6.33 g H2
2.016 g H2O
Mole H2 2 Mole H2O
2 Mole H2 Mole H2O
18.02 g H2O= 56.6 g H2O
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Excess and Limiting
Reactants are substances that can be changed into something else. For example, nails and boards are reactants for carpenters, while thread and fabric are reactants for the seamstress. And for a chemist hydrogen and oxygen are reactants for making water.
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Building HousesSuppose, we want to build some houses, so we order 2 truck loads of boards and 2 truck loads of nails. If two truck loads of boards make one house and two truck loads of nails make 10 houses, then how many houses can we make?
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Building HousesOk, we want to build some houses, so we order 2 truck loads of boards and 2 truck loads of nails. If two truck loads of boards make one house and two truck loads of nails make 10 houses, then how many houses can we make?
Yes, only one house!
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Building HousesOk, we want to build some houses, so we order 2 truck loads of boards and 2 truck loads of nails. If two truck loads of boards make one house and two truck loads of nails make 10 houses, then how many houses can we make?
What reactant is in excess? And how many more houses could we use if we had enough boards?
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Building HousesOk, we want to build some houses, so we order 2 truck loads of boards and 2 truck loads of nails. If two truck loads of boards make one house and two truck loads of nails make 10 houses, then how many houses can we make?
What reactant is in excess? And how many more houses could we use if we have enough boards?
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Building HousesOk, we want to build some houses, so we order 2 truck loads of boards and 2 truck loads of nails. If two truck loads of boards make one house and two truck loads of nails make 10 houses, then how many houses can we make?
What reactant is in excess? And how many more houses could we use if we have enough boards?
Yes, nails are in excess!
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Building HousesOk, we want to build some houses, so we order 2 truck loads of boards and 2 truck loads of nails. If two truck loads of boards make one house and two truck loads of nails make 10 houses, then how many houses can we make?What reactant is in excess? And how many more houses could we use if we have enough boards?Yes, nails are in excess! Nine more houses if we have an adequate amount of boards.
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Making WaterIf we react 10.0g of hydrogen with 10.0g of oxygen, which, if any, reactant will be in excess?
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Making WaterIf we react 10.0g of hydrogen with 10.0g of oxygen, which, if any, reactant will be in excess?
Our conversion process can easily determine the excess reactant. We can convert 10.0 g of oxygen to grams of hydrogen to determine if there is enough hydrogen to consume the oxygen.
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Making WaterIf we react 10.0g of hydrogen with 10.0g of oxygen, which, if any, reactant will be in excess?
Our conversion process can easily determine the excess reactant. We can convert 10.0 g of oxygen to grams of hydrogen to determine if there is enough hydrogen to consume the oxygen.
2 H2 + O2 2 H2O
10.0 g O2
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Making WaterIf we react 10.0g of hydrogen with 10.0g of oxygen, which, if any, reactant will be in excess?
Our conversion process can easily determine the excess reactant. We can convert 10.0 g of oxygen to grams of hydrogen to determine if there is enough hydrogen to consume the oxygen.
2 H2 + O2 2 H2O
10.0 g O2
32.0 g O2
mole O2
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Making WaterIf we react 10.0g of hydrogen with 10.0g of oxygen, which, if any, reactant will be in excess?
Our conversion process can easily determine the excess reactant. We can convert 10.0 g of oxygen to grams of hydrogen to determine if there is enough hydrogen to consume the oxygen.
2 H2 + O2 2 H2O
10.0 g O2
32.0 g O2
mole O2
mole O2
2 mole H2
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Making WaterIf we react 10.0g of hydrogen with 10.0g of oxygen, which, if any, reactant will be in excess?
Our conversion process can easily determine the excess reactant. We can convert 10.0 g of oxygen to grams of hydrogen to determine if there is enough hydrogen to consume the oxygen.
2 H2 + O2 2 H2O
10.0 g O2
32.0 g O2
mole O2
mole O2
2 mole H2
mole H2
2.02 g H2
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Making WaterIf we react 10.0g of hydrogen with 10.0g of oxygen, which, if any, reactant will be in excess?
Our conversion process can easily determine the excess reactant. We can convert 10.0 g of oxygen to grams of hydrogen to determine if there is enough hydrogen to consume the oxygen.
2 H2 + O2 2 H2O
10.0 g O2
32.0 g O2
mole O2
mole O2
2 mole H2
mole H2
2.02 g H2 = 1.26 g H2
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Making WaterOnly 1.26 g of hydrogen are required to react with 10.0 g of oxygen. Since there are 10.0 g of hydrogen available, then hydrogen must be the excess reactant and oxygen is the limiting reactant. The remainder of hydrogen 10.0 -1.26 = 8.7 g is called the amount in excess. The amount of water produced is determined by using the limiting reactant and converting it into water.
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Making WaterOnly 1.26 g of hydrogen are required to react with 10.0 g of oxygen. Since there are 10.0 g of hydrogen available, then hydrogen must be the excess reactant and oxygen is the limiting reactant. The remainder of hydrogen 10.0 -1.26 = 8.7 g is called the amount in excess. The amount of water produced is determined by using the limiting reactant and converting it into water.
10.0 g O2mole O2
32.0 g O2
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Making WaterOnly 1.26 g of hydrogen are required to react with 10.0 g of oxygen. Since there are 10.0 g of hydrogen available, then hydrogen must be the excess reactant and oxygen is the limiting reactant. The remainder of hydrogen 10.0 -1.26 = 8.7 g is called the amount in excess. The amount of water produced is determined by using the limiting reactant and converting it into water.
10.0 g O2mole O2
32.0 g O2 mole O2
2 mole H2O
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Making WaterOnly 1.26 g of hydrogen are required to react with 10.0 g of oxygen. Since there are 10.0 g of hydrogen available, then hydrogen must be the excess reactant and oxygen is the limiting reactant. The remainder of hydrogen 10.0 -1.26 = 8.7 g is called the amount in excess. The amount of water produced is determined by using the limiting reactant and converting it into water.
10.0 g O2mole O2
32.0 g O2 mole O2
2 mole H2O 18.0 g H2Omole H2O
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Making WaterOnly 1.26 g of hydrogen are required to react with 10.0 g of oxygen. Since there are 10.0 g of hydrogen available, then hydrogen must be the excess reactant and oxygen is the limiting reactant. The remainder of hydrogen 10.0 -1.26 = 8.7 g is called the amount in excess. The amount of water produced is determined by using the limiting reactant and converting it into water.
10.0 g O2mole O2
32.0 g O2 mole O2
2 mole H2O 18.0 g H2Omole H2O
= 11.3 g H2O
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Percentage YieldThe percent yield is a comparison of the laboratory answer to the correct answer which is determined by the conversion process. Suppose a student combined 10.0 g of oxygen and 10.0 g of hydrogen in the lab and recovered 8.66 g of water. What would be the percent yield?
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Percentage YieldThe percent yield is a comparison of the laboratory answer to the correct answer which is determined by the conversion process. Suppose a student combined 10.0 g of oxygen and 10.0 g of hydrogen in the lab and recovered 8.66 g of water. What would be the percent yield?
percent yield = Yield (the lab amount)Theoretical Yield (by conversions)
X 100
percent yield = 8.6611.3
X 100 = 76.6%
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Combustion AnalysisEmpirical formulas of hydrocarbons can be determined by combustion analysis. The complete combustion of a hydrocarbon produces carbon dioxide and water. Measuring the mass of the carbon dioxide and water produced can give the mass of hydrogen and carbon present in the compound. Subtracting the mass of carbon and hydrogen from the weight of the starting hydrocarbon gives the mass of the oxygen. Ascarite™ a commercial name for sodium or potassium hydroxide absorbs between 0-1 ppm of the carbon dioxide.
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Combustion AnalysisAscarite the weight increase of Ascarite gives the mass of the carbon dioxide according to the following equation.
CO2 + 2 KOH K2CO3 + 2 H2O
Vitamin C is essential for the prevention of scurvy. Combustion of a 0.2000 g sample of this hydrocarbon, which may or may not contain oxygen, gave 0.2998 g of carbon dioxide and 0.0819 g of water. What is the empirical formula of vitamin C?
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Combustion AnalysisVitamin C is essential for the prevention of scurvy. Combustion of a 0.2000 g sample of this hydrocarbon, which may or may not contain oxygen, gave 0.2998 g of carbon dioxide and 0.0819 g of water. What is the empirical formula of vitamin C?
First covert the mass of carbon dioxide and water into grams of carbon and of carbon and hydrogen. Subtract these masses from the sample weight, if the difference is zero then vitamin C does not contain any oxgen.
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Combustion Analysis
0.2298 g CO2
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Combustion Analysis
0.2298 g CO2
44.010 g CO2
Mole CO2
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Combustion Analysis
0.2298 g CO2
44.010 g CO2
Mole CO2
Mole CO2
Mole C
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Combustion Analysis
0.2298 g CO2
44.010 g CO2
Mole CO2
Mole CO2
Mole C
Mole C
12.011 g C= 0.08182 g C
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Combustion Analysis
0.2298 g CO2
44.010 g CO2
Mole CO2
Mole CO2
Mole C
Mole C
12.011 g C= 0.08182 g C
0.0819 g H2O
18.02 g H2O
mole H2O
mole H2O
2 mole H
mole H
1.008 g H= 0.00916 g H
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Combustion Analysis
0.2298 g CO2
44.010 g CO2
Mole CO2
Mole CO2
Mole C
Mole C
12.011 g C= 0.08182 g C
0.0819 g H2O
18.02 g H2O
mole H2O
mole H2O
2 mole H
mole H
1.008 g H= 0.00916 g H
0.2000 - 0.08182 - 0.00916 = 0.1090 g O
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Combustion Analysis
0.2298 g CO2
44.010 g CO2
Mole CO2
Mole CO2
Mole C
Mole C
12.011 g C= 0.08182 g C
0.0819 g H2O
18.02 g H2O
mole H2O
mole H2O
2 mole H
mole H
1.008 g H= 0.00916 g H
0.2000 - 0.08182 - 0.00916 = 0.1090 g O
0.08182 g C
12.011 g C
Mole C= 0.006812 mole
0.00916 g H
1.008 g H
mole H= 0.00909 mole
0.1090 g O
16.00 g O
Mole O= 0.006813 mole
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Combustion Analysis
0.2298 g CO2
44.010 g CO2
Mole CO2
Mole CO2
Mole C
Mole C
12.011 g C= 0.08182 g C
0.0819 g H2O
18.02 g H2O
mole H2O
mole H2O
2 mole H
mole H
1.008 g H= 0.00916 g H
0.2000 - 0.08182 - 0.00916 = 0.1090 g O
0.08182 g C
12.011 g C
Mole C= 0.006812 mole
0.00916 g H
1.008 g H
mole H= 0.00909 mole
0.1090 g O
16.00 g O
Mole O= 0.006813 mole
0.006812 0.006812
0.006812 0.006813
0.006812 0.00909
= 1.000 X 3 = 3
= 1.000 X 3 = 3
= 1.333 X 3 = 4
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Combustion Analysis
0.2298 g CO2
44.010 g CO2
Mole CO2
Mole CO2
Mole C
Mole C
12.011 g C= 0.08182 g C
0.0819 g H2O
18.02 g H2O
mole H2O
mole H2O
2 mole H
mole H
1.008 g H= 0.00916 g H
0.2000 - 0.08182 - 0.00916 = 0.1090 g O
0.08182 g C
12.011 g C
Mole C= 0.006812 mole
0.00916 g H
1.008 g H
mole H= 0.00909 mole
0.1090 g O
16.00 g O
Mole O= 0.006813 mole
0.006812 0.006812
0.006812 0.006813
0.006812 0.00909
= 1.000 X 3 = 3
= 1.000 X 3 = 3
= 1.333 X 3 = 4
C3H4O3
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The End