Reactions Unit
Chemistry
Chemistry
1
Learning Objectives: Reactions
Essential knowledge and skills:





Classify types of chemical reactions as synthesis, decomposition, single replacement, double replacement, neutralization,
and/or combustion.
Transform word equations into chemical equations and balance chemical equations.
Recognize that there is a natural tendency for systems to move in a direction of randomness (entropy).
Recognize equations for redox reactions and neutralization reactions.
Determine the number of valence electrons and possible oxidation numbers from an element’s electron configuration.
Essential understandings:




Conservation of matter is represented in balanced chemical equations. A coefficient is a quantity that precedes a reactant or
product formula in a chemical equation and indicates the relative number of particles involved in the reaction.
Major types of chemical reactions are
synthesis (A+B → AB)
decomposition (BC → B+C)
single replacement (A+BC→ B+AC)
double replacement (AC+BD → AD+BC)
neutralization (HX+MOH → H2O + MX)
combustion (CxHy + O2 → CO2 + H2O).
Transition metals can have multiple oxidation states.
Reactivity is the tendency of an element to enter into a chemical reaction.
Exothermic and Endothermic Reactions
Almost all chemical and physical reactions involve energy (usually in the form of heat) being released or added. An
exothermic change is a reaction that releases energy. An endothermic change is one in which the energy must be
added for the reaction to occur. For exothermic reactions, energy can be thought of as a product in the reaction. For
endothermic changes, energy can be thought of as a reactant in the reaction.
If a chemical reaction occurs at constant pressure, as all of our chemical reactions do we can consider a property called
enthalpy. Enthalpy (H) is the energy (heat) content of a system at constant pressure. You cannot measure the actual
energy or enthalpy of a substance, but you can measure the change in enthalpy for a reaction. This change is symbolized
by ∆Hrxn.
For exothermic reactions, enthalpy values are always negative, that is the energy of the products is lower than that
of the reactants. This is because energy is released as new bonds are formed in the products and this amount of energy is
greater than the energy required to break the old bonds in the reactants.
∆Hrxn = Hproducts - Hreactants (small # - BIG#) = - negative #
For example: 4Fe + 3O2  2Fe2O3 ∆Hrxn = -1625 kJ
For endothermic reactions, enthalpy values are always positive, that is that energy of the products is greater than
that of the reactions. This is because the energy released as new bonds are formed in the products is less than the energy
required to break the bonds in the reactants. This energy must be supplied in order for the reaction to occur. The added
energy does not disappear, of course due to the Law of Conservation of Energy. Instead, it becomes stored in the chemical
bonds of the products.
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∆Hrxn = Hproducts - Hreactants (BIG# - small #) = positive #
For example: C + H2O  CO + H2
∆Hrxn = + 113 kJ
Complete the following chart:
Type of Reaction
Sign of ∆Hrxn
Exothermic
Endothermic
Reactivity Series
Entropy (∆S)
What is Entropy
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Which has more energy: reactants or
products?
Entropy is the degree of randomness or disorder in a substance. The symbol for change in entropy is ΔS.
Solids are very ordered and have low entropy. Liquids and aqueous ions have more entropy because they move
about more freely, and gases have an even larger amount of entropy. According to the Second Law of
Thermodynamics, nature is always proceeding to a state of higher entropy (more disorder). +ΔS means more
disorder. –ΔS means more order or less disorder.

Tendency in nature to move toward a state of higher disorder or randomness (Your bedroom gets messy
after a couple of days of neglect!)
 The Second Law of Thermodynamics states that there is an inherent direction in which processes occur.
This direction is towards a state of higher entropy (more disorder.)
For example, an egg falls to the floor and cracks, but it never falls back up and puts itself together.
Several factors can be assessed to determine if a chemical or physical reaction is likely to occur because of
increased entropy.
a) Phase Changes:
Solid (great order, low entropy) Liquid (more randomness, higher entropy)  Gas (max randomness,
highest entropy)
b) Physical Changes:
When a substance is divided in to parts entropy increases


A large crystal is broken in to smaller pieces
A solid is dissolved and dissociates
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c) Chemical Changes:
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

When there are more products than reactants in a chemical reaction, entropy increases.
When the products of a reaction are simpler than the reactants entropy is increased.
Compounds are much less random than their constituent elements
d) Temperature Changes:

When temperature increases, molecules move faster, increasing disorder and entropy.
Ex) Which direction is this reaction more likely to go in, and why?
2KClO3 (s)2KCl (s) + 3O2 (g) + 875 kJ
This reaction would go to the right because:
1) The Products are simpler  increased entropy
2) There are more Products  increased entropy
3) Reaction leads to a Gas  increased entropy
4) Exothermic Reaction  decreased enthalpy
Entropy Worksheet 1
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Determine whether the following reactions show an increase or decrease in entropy and write + or – to indicate
this.
1. 2KClO3(s) → 2KCl(s) + 2O2(g)
__________
2. H2O(l) → H2O(s)
__________
3. N2(g) + 3H2(g) → 2NH3(g)
__________
4. NaCl(s) → Na+(aq) + Cl-(aq)
__________
5. KCl(s) → KCl(l)
__________
6. CO2(s) → CO2(g)
__________
7. H+(aq) + C2H3O2-(aq) → HC2H3O2(l)
__________
8. C(s) + O2(g) → CO2(g)
__________
9. H2(g) + Cl2(g) → 2HCl(g)
__________
10. Ag+ + Cl-(aq) → AgCl(s)
__________
11. 2N2O5(g) →4NO2(g) + O2(g)
__________
12. 2Al(s) + 3I2(s) → 2AlI3(s)
__________
13. H+(aq) + OH-(aq) → H2O(l)
__________
14. 2NO(g) →N2(g) + O2(g)
__________
15. H2O(g) → H2O(l)
__________
Entropy Worksheet 2
1.
a.
b.
c.
d.
e.
f.
Predict whether the entropy change will be positive or negative for the following:
H2O (g)  H2O (l)
S_____
C6H12O6(s)  2C2H5OH(l) + 2CO2(g)
S_____
2NH3(g) + CO2(g)  H2O(l) + NH2CONH2(aq) S____
NaCl(s)  NaCl(aq)
S____
Cu(s) (100oC)  Cu(s) (25oC)
S____
2NH3(g) N2(g) + 3H2(g)
S____
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2. Which of the following reactions will have an increase in entropy? Choose all that apply.
a. SO3(g) → 2SO2(g) + O2(g)
b. H2O(l) → H2O(s)
c. Br2(l) → Br2(g)
d. H2O2(l) → H2O(l) + ½ O2(g)
3. Which of the following compounds has the lowest entropy at 25 oC?
a. CH3OH(l)
b. CO(g)
c. MgCO3(s)
d. H2O(l)
e. H2O(g)
Balancing Chemical Equations
How to balance chemical equations
Tips and Tricks!
A chemical equation describes what happens in a chemical reaction. The equation identifies the reactants
(starting materials) and products (resulting substance), the formulas of the participants, the phases of the
participants (solid, liquid, gas), and the amount of each substance. Balancing a chemical equation refers to
establishing the mathematical relationship between the quantity of reactants and products. The quantities are
expressed as grams or moles.
It takes practice to be able to write balanced equations. There are essentially three steps to the process:
1. Write the unbalanced equation.
 Chemical formulas of reactants are listed on the lefthand side of the equation.
 Products are listed on the righthand side of the equation.
 Reactants and products are separated by putting an arrow between them to show the direction of the
reaction. Reactions at equilibrium will have arrows facing both directions.
2. Balance the equation.
 Apply the Law of Conservation of Mass to get the same number of atoms of every element on each
side of the equation. Tip: Start by balancing an element that appears in only one reactant and product.
 Once one element is balanced, proceed to balance another, and another, until all elements are
balanced.
 Balance chemical formulas by placing coefficients in front of them. Do not add subscripts, because
this will change the formulas.
3. Indicate the states of matter of the reactants and products.
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Use (g) for gaseous substances.
Use (s) for solids.
Use (l) for liquids.
Use (aq) for species in solution in water.
Write the state of matter immediately following the formula of the substance it describes.
Worked Example Problem
Tin oxide is heated with hydrogen gas to form tin metal and water vapor. Write the balanced equation that
describes this reaction.
1. Write the unbalanced equation.
SnO2 + H2 → Sn + H2O
2. Balance the equation.
Look at the equation and see which elements are not balanced. In this case, there are two oxygen atoms on
the lefthand side of the equation and only one on the righthand side. Correct this by putting a coefficient of
2 in front of water:
SnO2 + H2 → Sn + 2 H2O
This puts the hydrogen atoms out of balance. Now there are two hydrogen atoms on the left and four
hydrogen atoms on the right. To get four hydrogen atoms on the right, add a coefficient of 2 for the
hydrogen gas. Remember, coefficients are multipliers, so if we write 2 H2O it denotes 2x2=4 hydrogen
atoms and 2x1=2 oxygen atoms.
SnO2 + 2 H2 → Sn + 2 H2O
The equation is now balanced. Be sure to double-check your math! Each side of the equation has 1 atom of
Sn, 2 atoms of O, and 4 atoms of H.
3. Indicate the physical states of the reactants and products.
To do this, you need to be familiar with the properties of various compounds or you need to be told what
the phases are for the chemicals in the reaction. Oxides are solids, hydrogen forms a diatomic gas, tin is a
solid, and the term 'water vapor' indicates that water is in the gas phase:
SnO2(s) + 2 H2(g) → Sn(s) + 2 H2O(g)
Balancing Chemical Equations
One of the most useful devices for communicating information related to chemical changes is the chemical
equation. The equation contains both qualitative and quantitative information related to the nature and quantity
of the substances involved in the chemical reaction. It may also include the energy change involved.
Atoms are fundamental building blocks of all matter. For the purpose of equation balancing we say that
they can be neither created nor destroyed. Thus the number of atoms at the beginning of a reaction (reactants left side of the equation) must equal the number of atoms at the end of the reaction (products - right side of the
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equation). Note that the number of atoms on each side of an equation must balance, therefore, the mass (number
of grams) must balance, but not the number of molecules nor the volume of gases.
The subscripts in a correct formula tell the number of atoms in one molecule. The coefficients (numbers
in front of a formula) in a correctly balanced equation tell the number of molecules involved in a reaction.
There is a particular order that you can follow in balancing. It is the MINOH method which is very
simple to use by inspection.
Where:
M - Metals.
I - Ions.
N - Non-metals.
O - Oxygen
H - Hydrogen.
Balance metals such as Fe or Na first.
Looks for polyatomic ions (such as PO43- or SO42- that cross from reactant to
product unchanged.
Balance them as a group.
Look for Cl or S, these are common ones.
Remember, oxygen by itself is O2
Remember, hydrogen by itself is H2
Often, balancing H and O will involve water on one side or the other. In some cases when balancing.
you might want to write water as HOH, instead of H2O. Also, look carefully for elements which occur in
only one place on each side of the arrow. These should be balanced before examining elements that are
spread over several compounds. Often, either H or O will be spread out over several compounds. This is the
one to leave to the last. Remember, you cannot change a subscript to balance the equation, nor can you add
in new compounds.
Finally, when an equation is difficult to balance, use the grid method covered in class. Look for
repeating numbers in your grid
and trying using
K
2
1
these numbers as coefficients to
balance the
equations. Remember, 2-3 and
3-3 combinations
CrO4
1
3
are very common due to the
polyatomic anions
Fe
1
2
having a 2- or 3- oxidation
number and
NO
3
1
3
polyvalent cations having a +2
or +3 oxidation
number.
K2CrO4 + Fe(NO3)3  Fe2(CrO4)3 + KNO3
Reactants
Products
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Reaction Classification
Chemical Reactions
Types of chemical reactions
There are 5 general types of reactions and two that are special cases of the following
1. Decomposition
2. Synthesis
3. Double displacement
4. Single displacement
5. Combustion
Special cases: Redox and Neutralization
Decomposition

Decomposition = one compound  two (or more pieces).
AB  A + B

Pieces can be elements or simpler compounds
i. Element examples:
1. HgO  Hg + O2
2. H2O  H2 + O2
3. MgCl2  Mg + Cl2
4. FeS  Fe + S
ii. Simpler compound examples
1. CaCO3  CaO + CO2
2. Na2CO3  Na2O + CO2
3. KClO3  KCl + O2
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
4. Ba(ClO3)2  BaCl2 + O2
iii. Acids and bases
1. (base) Ca(OH)2  CaO + H2O
2. (base) NaOH  Na2O + H2O
3. (acid) HNO3  N2O5 + H2O
4. (acid) H3PO4  P2O5 + H2O
Notice how, in every case so far, there is only one substance on the left-hand (reactant) side. This is
always the case in a decomposition reaction.
Single Replacement/Displacement
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Single displacement, one element replaces another element in a compound.
One reactant is always an element. It does not matter if the element is written first or second on the
reactant side. The other reactant will be a compound.
Two possibilities:
1. Cations switch.
AX + Y  YX + A


Element Y replaced A (in the compound AX) to form a new compound YX and the free element
A. Remember that A and Y are both cations (postively-charged ions) in this example.
Examples
1.
2.
3.
4.
Cu + AgNO3  Ag + Cu(NO3)2
Fe + Cu(NO3)2  Fe(NO3)2 + Cu
Ca + H2O  Ca(OH)2 + H2
Zn + HCl  ZnCl2 + H2
2. Anions switch places:
A + XY  XA + Y


Element A has replaced Y (in the compound XY) to form a new compound XA and the free
element Y. Remember that A and Y are both anions (negatively-charged ions) in this example.
Examples
1. Cl2 + NaBr  NaCl + Br2
2. Br2 + KI  KBr + I2
Double Replacement/Displacement



Double displacement, the cations and anions of two different compounds switch places.
Both reactants are compounds, each with a cation part and an anion part.
Diatomic elements do not count; they are included in the single displacement category.
AB + XY  AY + XB
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

A and X are the cations (postively-charged ions) in this example, with B and Y being the anions
(negatively-charged ions).
Examples:
1.
2.
3.
4.
KOH + H2SO4  K2SO4 + H2O
FeS + HCl  FeCl2 + H2S
NaCl + H2SO4  Na2SO4 + HCl
AgNO3 + NaCl  AgCl + NaNO3
Synthesis




Synthesis are, at this introductory level, almost always the reverse of a decomposition reaction.
Two pieces  one, more complex compound.
Complex means the product compound has more atoms than the reactant molecules. Usually!!
Pieces can be elements or simpler compounds.
A + B  AB


Examples: two elements are combining
1. Mg + O2  MgO
2. H2 + O2  H2O
3. K + Cl2  KCl
4. Fe + O2  Fe2O3
Examples: two compounds making a more complex compound (or a compound and an element joining
together):
1.
2.
3.
4.
CaO + CO2  CaCO3
Na2O + CO2  Na2CO3
KCl + O2  KClO3
BaCl2 + O2  Ba(ClO3)2
Combustion



Combustion, at its most general, can mean the reaction of oxygen gas (O2) with anything.
However, we define combustion as the reaction of oxygen with a compound containing carbon and
hydrogen.
A common synonym for combustion is burn.
CxHy + O2  CO2 + H2O




Examples:
1. CH4 + O2  CO2 + H2O
2. C2H6 + O2  CO2 + H2O
3. C6H12O6 + O2  CO2 + H2O
4. C2H5OH + O2  CO2 + H2O
Notice that some compounds contain carbon, hydrogen AND oxygen.
The products are all the same, in every reaction.
Variations include NO2 and SO2 Like this:
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1.
2.

C21H24N2O4 + O2  CO2 + H2O + NO2
C2H5SH + O2  CO2 + H2O + SO2
There are complexities with combustion as you get deeper into it.
o i.e. Not enough O2  CO instead of CO2
Neutralisation




Neutralisation reactions are reactions involving and acid and a base.
When an acid and a base mix in the right proportion, they neutralize each other
The product is a salt and water – a neutral solution when measured with the pH scale
o Examples:
 NaOH + HCl  NaCl + H2O
 H2SO4 + 2 KOH  K2SO4 + 2 H2O
Neutralization is a special case of double replacement reactions
Redox
Simple definitions of oxidation and reduction are based on the loss/gain of oxygen or the loss/gain of hydrogen.
Oxidation is the gain of oxygen or the loss of hydrogen; reduction is the loss of oxygen or the gain of
hydrogen. These definitions can only be used when a chemical reaction involves hydrogen and oxygen, and
therefore their usefulness is limited.
A more basic and more useful definition of oxidation and reduction is based on the
loss/gain of electrons.
OXIDATION IS LOSS OF ELECTRONS
REDUCTION IS GAIN OF ELECTRONS
In reactions involving simple ions, it is usually easy to tell whether electrons are
lost or gained, but it is less easy to tell when complex ions or covalent molecules
are involved. Oxidation number is a useful concept for helping to decide in these
more awkward cases.
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Balancing Chemical Equations Worksheet
Balance the equations below:
1)
____ N2 + ____ H2  ____ NH3
2)
____ KClO3  ____ KCl + ____ O2
3)
____ NaCl + ____ F2  ____ NaF + ____ Cl2
4)
____ H2 + ____ O2  ____ H2O
5)
____ Pb(OH)2 + ____ HCl  ____ H2O + ____ PbCl2
6)
____ AlBr3 + ____ K2SO4  ____ KBr + ____ Al2(SO4)3
7)
____ CH4 + ____ O2  ____ CO2 + ____ H2O
8)
____ C3H8 + ____ O2  ____ CO2 + ____ H2O
9)
____ C8H18 + ____ O2  ____ CO2 + ____ H2O
10)
____ FeCl3 + ____ NaOH  ____ Fe(OH)3 + ____NaCl
11)
____ P + ____O2  ____P2O5
12)
____ Na + ____ H2O  ____ NaOH + ____H2
13)
____ Ag2O  ____ Ag + ____O2
14)
____ S8 + ____O2  ____ SO3
15)
____ CO2 + ____ H2O  ____ C6H12O6 + ____O2
16)
____ K + ____ MgBr2  ____ KBr + ____ Mg
17)
____ HCl + ____ CaCO3  ____ CaCl2 + ____H2O + ____ CO2
18)
____ HNO3 + ____ NaHCO3  ____ NaNO3 + ____ H2O + ____ CO2
19)
____ H2O + ____ O2  ____ H2O2
20)
____ NaBr + ____ CaF2  ____ NaF + ____ CaBr2
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21)
____ H2SO4 + ____ NaNO2  ____ HNO2 + ____ Na2SO4
Identify the type of reaction
For the following reactions, indicate whether the following are examples of synthesis, decomposition,
combustion, single displacement, double displacement, or acid-base reactions:
1)
Na3PO4 + 3 KOH  3 NaOH + K3PO4 _________________________
2)
MgCl2 + Li2CO3  MgCO3 + 2 LiCl _________________________
3)
C6H12 + 9 O2  6 CO2 + 6 H2O _________________________
4)
Pb + FeSO4  PbSO4 + Fe _________________________
5)
CaCO3  CaO + CO2 _________________________
6)
P4 + 3O2  2 P2O3 _________________________
7)
2 RbNO3 + BeF2  Be(NO3)2 + 2 RbF ________________________
8)
2 AgNO3 + Cu  Cu(NO3)2 + 2 Ag ________________________
9)
C3H6O + 4 O2  3 CO2 + 3 H2O _________________________
10)
2 C5H5 + Fe  Fe(C5H5)2 _________________________
11)
SeCl6 + O2  SeO2 + 3Cl2 _________________________
12)
2 MgI2 + Mn(SO3)2  2 MgSO3 + MnI4 _________________________
13)
O3  O + O2 _________________________
14)
2 NO2  2 O2 + N2_________________________
Section 2: Practicing equation balancing
1)
__ C6H6 + __ O2  __ H2O + __ CO2
2)
__ NaI + __ Pb(SO4)2  __ PbI4 + __ Na2SO4
3)
__ NH3 + __ O2 __ NO + __ H2O
4)
__ Fe(OH)3  __ Fe2O3 + __ H2O
5)
__ HNO3 + __ Mg(OH)2  __H2O + __ Mg(NO3)2
6)
__ H3PO4 + __ NaBr  __ HBr + __ Na3PO4
7)
__ C + __ H2  __ C3H8
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8)
__ CaO + __ MnI4  __ MnO2 + __ CaI2
9)
__ Fe2O3 + __ H2O  __ Fe(OH)3
10)
__ C2H2 + __ H2  __ C2H6
11)
__ VF5 + __ HI  __ V2I10 + __ HF
12)
__ OsO4 + __ PtCl4  __ PtO2 + __ OsCl8
13)
__ CF4 + __ Br2  __ CBr4 + __ F2
14)
__ Hg2I2 + __ O2  __ Hg2O + __ I2
15)
__ Y(NO3)2 + __ GaPO4  __ YPO4 + __ Ga(NO3)2
Section 3: Predicting the products of chemical reactions
Predict the products of the following reactions:
1)
__ Ag + __CuSO4 
Type:___________________________
2)
__ NaI + __ CaCl2 
Type:___________________________
3)
__ O2 + __ H2 
Type:___________________________
4) __ HNO3 + __ Mn(OH)2 
Type:___________________________
5) __ AgNO2 + __ BaSO4 
Type:___________________________
6) __ HCN + __ CuSO4 
Type:___________________________
7) __ H2O + __ AgI 
Type:___________________________
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8) __ HNO3 + __Fe(OH)3 
Type:___________________________
9) __ LiBr + __ CoSO3 
Type:___________________________
10) __ LiNO3 + __Ag 
Type:___________________________
11) __ N2 + __ O2 
Type:___________________________
12) __ H2CO3 
Type:___________________________
Classifying Reactions
Balance the following equations. Then classify the reactions as synthesis (S), decomposition (D), single
replacement (SR), double replacement (DR), or combustion (C). Write the corresponding letter(s) in the blank
on the left.
_____1.
____ Ca(OH)2 + ____ H2SO4  ____ CaSO4 + ____ H2O
_____2.
____ KClO3  ____ KCl + ____ O2
_____3.
____ HCl + ____ NaOH  ____ NaCl + ____ H2O
_____4.
____ Mg + ____ HCl  ____ MgCl2 + ____ H2
_____5.
____ N2 + ____ O2  ____ N2O5
_____6.
____ Al + ____ O2  ____ Al2O3
_____7.
____ Al + ____ NiBr2  ____ AlBr3 + ____ Ni
_____8.
____ NaCl  ____ Na + ____ Cl2
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_____9.
____ CaCl2 + ____ F2  ____ CaF2 + ____ Cl2
_____10.
____ (NH4)2SO4 + ____ Ba(NO3)2  ____ BaSO4 + ____ NH4NO3
_____11.
____ H2(g) + ____ O2(g)  ____ H2O (g)
_____12.
____ H2O(l)  ____ H2(g) + ____ O2(g)
_____13.
____ Zn (s) + ____ H2SO4(aq)  ____ ZnSO4(aq) + ____ H2 (g)
_____14.
____ CO + ____ O2  ____ CO2
_____15.
____ HgO  ____ Hg + ____ O2
_____16.
____ KBr + ____ Cl2  ____ KCl + ____ Br2
_____17.
____ CaO + ____ H2O  ____ Ca(OH)2
_____18.
____ AgNO3 (aq) + ____ NaCl(aq)  ____ AgCl (s) + ____ NaNO3 (aq)
_____19.
____ C4H8 (g) + ____ O2 (g)  ____ CO2 (g) + ____ H2O (g)
_____20.
____ H2O2 (l)  ____ H2O (g) + ____ O2 (g)
Write and balance equations for the following reactions AND classify them as synthesis (S), decomposition (D),
single replacement (SR), double replacement (DR), or combustion (C). Write the corresponding letter(s) in the
appropriate blank.
_____21. aluminum nitrate (aq) + sodium hydroxide (aq)  aluminum hydroxide (s) + sodium
nitrate (aq)
_____22. sulfur trioxide (g)  sulfur dioxide (g) + oxygen (g)
_____23. iron (s) + silver acetate (aq)  iron (II) acetate (aq) + ____ silver (s)
_____24. magnesium (s) + oxygen (g)  magnesium oxide (s)
18
_____25. ethanol (C2H5OH) (l) + oxygen (g)  carbon dioxide (g) + water (g)
Solubility Rules
19
Solubility Worksheet
Chemical Formula
Name
Solubility
1. NH4C2H3O2
2. Ba(OH)2
3.
Iron (II) Carbonate
4. NaOH
5. RbNO3
6.
Cesium Sulfate
7. MgSO4
8. ZnCl2
9.
Zinc Hydroxide
10. Zn3(PO4)2
11. AgBr
12. KNO3
13. Al2S3
14.
Silver Acetate
15. Sr2CrO4
16.
Aluminium Phosphate
17. BaSO4
18. Ca(OH)2
19. BaCO3
20. MgCrO4
21.
Lead (II) Chloride
22. NH4CN
23.
Silver Iodide
24. Hg2SO4
20
25.
Lithium Chloride
Net ionic equations
How to write net ionic equations
EXAMPLE: KCl(aq) + Pb(NO3)2(aq) 
1.
a. Take only one of the first cation(s) and match it with one of the second anion(s). (Write the
cation first)
b. Take only one of the second cation(s) and match it with one of the first anion(s). (Write the
cation first)
KCl(aq) + Pb(NO3)2(aq)  KNO3 +PbCl
2. Correct the formulas of the products based on the charges of the ions.
KCl(aq) + Pb(NO3)2(aq)  KNO3 +PbCl2 ◄
3. Balance the equation
2 KCl(aq) + Pb(NO3)2(aq)  2 KNO3 +PbCl2
4. Consult the solubility rules and assign the correct state symbol annotations. This should agree with
any observations concerning the formation of a precipitate which gets the symbol (s). If water is formed,
water is a molecule; it does not ionize to any significant extent. It is annotated (l).
2 KCl(aq) + Pb(NO3)2(aq)  2 KNO3(aq) ◄ +PbCl2(s) ◄
21
5. Write the Total Ionic Equation (T.I.E.). All compounds that are annotated (aq) break up into individual
cations and anions in that order.
2 K+(aq) + 2Cl-(aq) + Pb2+(aq) + 2NO3- (aq)  2 K+(aq) +2 NO3-(aq) +PbCl2(s)
6. Eliminate spectator ions. Spectator ions are in the same form on each side of the equation arrow.
2 K+(aq) + 2Cl-(aq) + Pb2+(aq) + 2NO3- (aq)  2 K+(aq) +2 NO3-(aq) +PbCl2(s)
7. Write the Net Ionic Equation (N.I.E.). The convention is to write the cation first followed by the
anion on the “reactants” side.
Pb2+(aq) + 2Cl-(aq)  +PbCl2(s)
Note that chemical equations are written using the lowest common coefficients.
If your NIE ended up as
2 H+(aq) + 2 OH-(aq)  2 H2O(l)
It would become
H+(aq) + OH-(aq)  H2O(l)
Net Ionic Equation Worksheet
Write the complete and net ionic equations for the following reactions.
If only the reactants are given, predict the products and balance the equation first. You also must include the
states of matter.
1. Pb(NO3)2 (aq) + 2 KCl (aq)

PbCl2 (s) + 2 KNO3 (aq)
Complete Ionic:
22
Net Ionic:
2. 2 HCl (aq) + Ba(OH)2 (aq)  BaCl2 (aq) + 2 H2O (l)
Complete Ionic:
Net Ionic:
3. _____ K3PO4 (aq) + _____ Al(NO3)3 (aq) 
Complete Ionic:
Net Ionic:
4. _____ Cr(NO3)3 (aq) + _____ Ba (s) 
Complete Ionic:
Net Ionic:
23
5. beryllium iodide + strontium sulfate -- >
Molecular Equation:
Complete Ionic:
Net Ionic:
6. zinc + water 
Molecular Equation:
Complete Ionic:
Net Ionic:
24
7. barium hydroxide + sulfuric acid 
Molecular Equation:
Complete Ionic:
Net Ionic:
One way of accounting for electrons in equations is to use OXIDATION NUMBERS.
Oxidation number
The oxidation number of an atom shows the number of electrons which it has lost or gained as a
result of forming a compound
e.g. Fe2+ needs to gain two electrons for it to become neutral iron atom therefore its oxidation
number is +2.
Rules for Assigning Oxidation Numbers
This is a prioritized list. If two rules contradict each other, follow the rule that appears higher on the list.
1. The atoms in pure elements are assigned an oxidation number of zero.
2. Monatomic ions are assigned an oxidation number equal to their charge.
3. For atoms in covalent molecules and polyatomic ions:
a. The sum of all the oxidation numbers of the atoms in a covalent molecule must equal zero. The sum of all
the oxidation numbers of the atoms in a polyatomic ion must equal the charge on the ion.
b. Fluorine is assigned an oxidation number of –1.
25
c. Oxygen is assigned an oxidation number of –2 (an exception to this is when oxygen occurs as the peroxide
ion, O2-2, where it is assigned an oxidation number of –1).
d. Hydrogen is assigned an oxidation number of +1 apart from metal hydrides which have a value of -1
Examples
1. The oxidation number of S in H2SO4
H2
2 x +1
+2
+2
S
?
?
+6
s = +6
O4
4 x -2
-8
-8
=0
=0
=0
2. The oxidation number of S in S2O82S2
?
?
+14
S = +7
O4
8 x -2
-16
-16
= -2
= -2
= -2
3. The oxidation number of Cl in NaClO3.
Na
+1
+1
+1
Cl
?
?
+5
Cl = +5
O3
3 x -2
-6
-6
=0
=0
=0
4. The oxidation number of Mn in MnO4Mn
?
?
+7
Mn = +7
O4
4 x -2
-8
-8
= -1
= -1
= -1
How to assign oxidation numbers/states
Another tutorial on oxidation
numbers/states
26
Oxidation Numbers Worksheet
Rules for Assigning Oxidation Numbers
1. The oxidation number of any uncombined element is 0.
2. The oxidation number of a monatomic ion equals the charge on the ion.
3. The more-electronegative element in a binary compound is assigned the number equal to the charge it
would have if it were an ion.
4. The oxidation number of fluorine in a compound is always -1.
5. Oxygen has an oxidation number of -2 unless it is combined with F (when it is +2), or it is in a peroxide
(such as H2O2 or Na2O2), when it is -1.
6. The oxidation state of hydrogen in most of its compounds is +1 unless it is combined with a metal, in
which case it is -1.
7. In compounds, the elements of groups 1 and 2 as well as aluminum have oxidation numbers of +1, +2,
and +3 respectively.
8. The sum of the oxidation numbers of all atoms in a neutral compound is 0.
9. The sum of the oxidation numbers of all atoms in a polyatomic ion equals the charge of the ion.
Directions: Use the Rules for Assigning Oxidation Numbers to determine the oxidation number assigned to
each element in each of the given chemical formulas.
Formula
Element and
Oxidation Number
1.
Cl2
Cl
16.
Na2O2
Na
O
2.
Cl-
Cl
17.
SiO2
Si
O
3.
Na
Na
18.
CaCl2
Ca
Cl
4.
Na+
Na
19.
PO43-
P
O
5.
O2
O
20.
MnO2
Mn
O
6.
N2
N
21.
FeO
Fe
O
7.
Al+3
Al
22.
Fe2O3
Fe
O
8.
H2O
H
O
23.
H2O2
H
O
9.
NO3-
N
O
24.
CaO
Ca
O
10.
NO2
N
O
25.
H2S
H
S
11.
Cr2O72-
Cr
O
26.
H2SO4
H
S
O
12.
KCl
K
Cl
27.
NH4Cl
N
H
Cl
13.
NH3
N
H
28.
K3PO4
K
P
O
14.
15.
CaH2
SO42-
Ca
S
H
O
29.
30.
HNO3
KNO2
H
K
N
N
O
O
Formula
27
Element and Oxidation Number
Answer Key
1. Cl:0
7.
2.
Cl:-1
8.
3.
Na:0
9.
4.
Na:+1
10.
5.
O:0
11.
6.
N:0
12.
Al:+3
13.
H:+1
O:-2
N:+5
O:-2
N:+4
O:-2
Cr:+6
O:-2
K:+1
Cl:-1
14.
15.
16.
17.
18.
N:-3
H:+1
Ca:+2
H:-1
S:+6
O:-2
Na:+1
O:-1
Si:+4
O:-2
Ca:+2
Cl:-1
19.
20.
21.
22.
23.
24.
P:+5
O:-2
Mn:+4
O:-2
Fe:+2
O:-2
Fe:+3
O:-2
H:+1
O:-1
Ca:+2
O:-2
25.
H:+1
26.
28.
H:+1 S:+6
O:-2
N:-3 H:+1
Cl:-1
K:+1 P:+5 O:-2
29.
H:+1
N:+5
30.
K:+1
O:-2
N:+3
27.
S:-2
O:-2
Redox
The term REDOX stands for REDUCTION-OXIDATION.
Oxidation can be defined as gain of oxygen or loss of hydrogen.
Reduction can be defined as loss of oxygen or gain of hydrogen.
The most important definition is given in terms of electrons.
OXIDATION is LOSS of ELECTRONS
REDUCTION is GAIN of ELECTRONS
Using oxidation numbers it is possible to decide whether redox has occurred.
Increase in oxidation number is oxidation.
Decrease in oxidation number is reduction.
Redox Reactions
When magnesium is placed into a solution of copper (II) sulfate, a reaction occurs which in simple terms is
called a “displacement reaction”.
Chemical equation:
Ionic equation:
Mg + CuSO4  MgSO4 + Cu
Mg(s) + Cu2+(aq)  Mg2+(aq) + Cu(s)
The copper in this reaction is taking electrons from the magnesium.
The copper gains electrons (ox. no. has decreased)
The magnesium loses electrons (increase in ox. no.)
28
- it is REDUCED
- it is OXIDISED
So this is a REDOX reaction.
Whenever one substance gains an electron another substance must lose an electron, so reduction and oxidation
always go together.
Oxidising and reducing reagents
How to predict oxdising and reducing
agents
An oxidising agent causes another material to become oxidised. In the above example of adding magnesium to
copper sulfate, the magnesium is oxidised.
Since the copper ions in the copper sulfate cause this oxidation, they are the
oxidising agent.
In the same way the Mg causes the reduction of copper ions so it is the
reducing agent.
Mg(s)
+
reducing agent
Cu2+(aq)
Mg2+(aq)
+
Cu(s)
oxidising agent
In this example the oxidising agent (copper ions) is reduced and the reducing agent (magnesium) is oxidised.
This always happens with redox reactions: in a redox reaction the oxidising agent is reduced and the reducing
agent is oxidised.
electrons
REDUCING AGENT
+
MATERIAL
The reducing agent loses electrons
and so is oxidised.
Oxidation number and redox reactions
When a redox reaction occurs an electron transfer takes place and so the oxidation numbers of the substances
involved changes.
Consider the following reaction:
2HOBr + 2H+ + 2I-  Br2 + I2 + 2H2O
Reactants
Species
H in HOBr
O in HOBr
Br in HOBr
H+
I-
Ox. No.
+1
-2
+1
+1
-1
Products
Species
Br in Br2
I in I2
H in H2O
O in H2O
29
Ox. No.
0
0
+1
-2
The table shows us that the oxidation number of Br goes from +1 to 0, so it is reduced.
The iodine goes from -1 to 0, so this is oxidised.
Another example
3NaOCl
Reactants
Species
Na in NaOCl
O in NaOCl
Cl in NaOCl

2NaCl + NaClO3
Products
Species
Na in NaCl
Na in NaClO3
Cl in NaCl
Cl in NaClO3
O in NaClO3
Oxid’n No
+1
-2
+1
Oxid’n No
+1
+1
-1
+5
-2
In this reaction the Cl in NaOCl is oxidised in one reaction to +5 and in another reaction is reduced to -1. Such
an occurrence is called disproportionation.
Disproportionation takes place a particular species undergoes simultaneous
oxidation and reduction.
Oxidation Reduction Worksheet 1
1. Determine the oxidation number of each atom in the following substances
a. NF3
N
F
b. K2CO3
K
C
c. NO3-
N_________
O__________
d. HIO4
H
I
O
O
2. For the following balanced redox reaction answer the following questions
2 Fe2+(aq) + H2O2(aq)  2Fe3+(aq) + 2 OH-(aq)
a. What is the oxidation state of oxygen in H2O2?
b. What is the element that is oxidized?
c. What is the element that is reduced?
d. What is the oxidizing agent?
30
e. What is the reducing agent?
3. For the following balanced redox reaction answer the following questions
4NaOH(aq) + Ca(OH)2(aq) + C(s) + 4ClO2(g)  4NaClO2(aq) + CaCO3(s) + 3H2O(l)
a. What is the oxidation state of Cl in ClO2(g)?
b. What is the oxidation state of C in C(s)?
______
c. What is the element that is oxidized?
d. What is the element that is reduced?
e. What is the oxidizing agent?
f. What is the reducing agent?
4. For the following balanced redox reaction answer the following questions
16 HCl(aq) + 5 SnCl2(aq) + 2 KMnO4(aq)  2 MnCl2(aq) 5 SnCl4(aq) + 8 H2O(l) + 2 KCl(aq)
a. What is the oxidation state of Mn in KMnO4(aq)?
b. What is the oxidation state of Cl in SnCl2(aq)?
c. What is the element that is oxidized?
d. What is the element that is reduced?
e. What is the oxidizing agent?
f. What is the reducing agent?
g. How many electrons are transferred in the reaction as it is balanced?
5. Determine which element is oxidized and which is reduced when lithium reacts with nitrogen to form lithium
nitride.
6 Li(s) + N2(g)
a. element oxidized:
31
2 Li3N(s)
b. element reduced:
6. Determine which atom is oxidized and which is reduced in the following reaction
Sr(s) + 2 H2O(l)
Sr2+(aq) + 2 OH-(aq) + H2(g)
a. element oxidized:
b. element reduced:
Oxidation/Reduction Worksheet 2
1)
4Fe (s)
+ 3 O2 (g)
→ 2 Fe2O3(s)
__________was oxidized __________was the oxidizing agent
__________was reduced
2)
Cu(s) + AgNO3(aq)
_________ was the reducing agent
→ Ag(s) + CuNO3(aq)
__________was reduced
_________ was the reducing agent
__________was oxidized __________was the oxidizing agent
3)
2Na(s) + Cl2(g)
→
2NaCl(s)
__________was reduced and was the __________________agent
__________was oxidized and was the __________________agent
4)
2HNO3(aq) + 6HI(aq)
→
2NO(g) + 3I2(s)
+ 4 HOH(l)
__________was oxidized _________ was the reducing agent
__________was reduced
__________was the oxidizing agent
32
Oxidation/reduction Worksheet 3
1. Assign oxidation numbers to chlorine in each of the following chemicals.
HCl(aq),
Cl2(g),
NaClO(s),
Cl–(aq),
ClO3–(aq),
KClO2(s),
ClO2(g),
HClO4(aq)
2.
HClO3(aq),
Assign oxidation numbers to manganese in each of the following chemicals.
MnO2(s),
KMnO4(s),
Mn2O7(s)
Mn2+(aq)
Mn(s),
MnO42–(aq),
For the following reaction equations, use oxidation numbers to identify the oxidation
(the atom oxidized) and the reduction (the atom reduced).
3. AsO33–(aq) + IO3–(aq) AsO43–(aq) + I–(aq)
4. CuO(s) + NH3(g) N2(g) + H2O(l) + Cu(s)
5. MnO4–(aq) + H2Se(g) + H+(aq) Se(s) + Mn2+(aq) + H2O(l)
33
MnCl2(s),
Polyatomic Ions
AsO43-
arsenate
Fe(CN)64-
ferrocyanide
NH4+
ammonium
BO33-
borate
HCO3-
NO2-
nitrite
B4O72-
tetraborate
hydrogen
carbonate
(bicarbonate)
NO3-
nitrate
BrO3-1
bromate
Hg22+
mercury (I)
O22-
peroxide
CHO2-
formate
H3O+
hydronium
OCN-
cyanate
C2H3O2-
acetate
H2PO4-
dihydrogen
OH-
hydroxide
C2O42-
oxalate
PO33-
phosphite
PO43-
phosphate
P2O74-
pyrophosphate
SCN-
thiocyanate
SeO42-
selenate
phosphate
HPO42-
hydrogen
C4H4O62-
tartrate
ClO-
hypochlorite
HSO3-
hydrogen sulfite
ClO2-
chlorite
HSO4-
hydrogen sulfate
ClO3-
chlorate
I3-
triiodide
ClO4-
perchlorate
IO3-
iodate
CN-
cyanide
MnO4-
permanganate
CO32-
carbonate
MnO42-
manganate
CrO42-
chromate
MoO42-
molybdate
Cr2O72-
dichromate
N3-
azide
Fe(CN)63-
ferricyanide
phosphate
34
SiF62- hexafluorosilicate
SiO32-
silicate
SO32-
sulfite
SO42-
sulfate
S2O32-
thiosulfate
Reactions Worksheet
Write balanced chemical equations from the following word equations
1. Potassium nitrate  Potassium nitrite + oxygen (gas)
2. Zinc + hydrochloric acid  Zinc chloride + hydrogen (gas)
3. Potassium Chlorate  Potassium Chloride + Oxygen (gas)
4. Ammonium Nitrate  Nitrogen (gas) + oxygen (gas) + Water
5. Calcium Oxide + hydrochloric acid  Calcium Chloride + Water
6. Ammonia + Oxygen  Nitrogen Monoxide + Water
7. Iron (III) Oxide + Carbon Monoxide  Iron + Carbon Dioxide
8. Calcium Oxide + diphosphorous pentoxide  Calcium Phosphate
9. Aluminium Hydroxide + acetic acid  Aluminium Acetate + Water
10. Aluminium Hydroxide + Cupric Chloride  Aluminium Chloride + Copper (II) hydroxide
11. Iron + Silver Acetate  Iron (III) Acetate + Silver
12. Bromine + Calcium Iodide  Calcium Bromide + Iodine
13. Sodium Hydroxide + Sulfuric Acid  Sodium Sulfate + Water
14. Lithium + Water  Lithium Hydroxide + Hydrogen
15. Magnesium + Oxygen  Magnesium Oxide
16. Mercury (II) Oxide  Mercury + Oxygen
17. Fluorine + Potassium Chloride  Potassium Fluoride + Chlorine
35
18. Oxygen + Iron  Ferrous oxide
19. Calcium Carbonate  Calcium Oxide + Carbon Dioxide
20. Aluminium Oxide  Aluminium + Oxygen
21. Magnesium Oxide + Carbon Dioxide  Magnesium Carbonate
22. Copper + Sulfuric Acid  Cupric Sulfate + Sulfur Dioxide + Water
23. Calcium Hydroxide + Phosphoric Acid  Calcium Phosphate + Water
24. Magnesium Nitrate + Sulfuric Acid  Magnesium Sulfate + Nitric Acid
25. Potassium Carbonate + Barium Chloride  Potassium Chloride + Barium Carbonate
26. Aluminium Chloride + Sulfuric Acid  Aluminium Sulfate + Hydrogen Chloride
27. Cadmium Phosphate + Ammonium Sulfate  Cadmium Sulfide + Ammonium Phosphate
28. Manganese (IV) Oxide + Hydrochloric Acid  Manganese (II) Chloride + Water + Chlorine
29. Magnesium hydroxide + ammonium phosphate magnesium phosphate + ammonium hydroxide
30. Ferric bromide + ammonium sulfide  ferric sulfide + ammonium bromide
31. Calcium oxide + diphosphorous pentoxide  calcium phosphate
32. Magnesium chloride + silver nitrate  magnesium nitrate + silver chloride
33. Sodium carbonate + sulfuric acid  sodium sulfate + carbon dioxide + water
34. Aluminium hydroxide + acetic acid  aluminium acetate + water
36
35. Plumbous nitrate + copper (II) sulfate  plumbous sulfate + copper (II) nitrate
36. Aluminium + cupric chloride  aluminium chloride + copper
Reaction Predictions
Label each equation according to its reaction type, then predict the products and balance the equations
1. Aqueous silver nitrate reacts with aqueous calcium chloride.
2. _____ Cl2 (g) + _____ NaI (aq) 
3. Solid zinc reacts with aqueous copper (II) nitrate.
4. Predict the reactant isotope that yields iridium-181 during alpha decay.
5. Solid magnesium is placed in a beaker of hydrochloric acid.
6. _____ P4 (s) + _____ O2 (g) 
7. _____ Ca(OH)2 (aq) + _____ HgCl2 (aq) 
8. Barium chlorate is heated.
9.
_____ CaBr2 (aq) + _____ KOH (aq) 
37
10. _____ NH4OH (aq) 
11. Benzene (C6H6) is ignited in the presence of oxygen gas.
12. Nitrogen gas reacts with oxygen gas.
13. _____ Al2(SO4)3 (aq) + _____ Ca(OH)2 (aq) 
209
14.
Bi
83
+
58
26
Fe
+
1
n
0
38