Reaction Types: Single Replacement
Important notes to remember: (1) NONE of the equations are balanced!! and (2) make
sure to write correct formulas. DO NOT just copy the subscripts from the reactants over
into the products.
During single replacement, one element replaces another element in a compound. There
are two different possibilities:
1. One cation replaces another. Written using generic symbols, it is:
AX + Y ---> YX + A
Element Y has 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.
Some examples are:
Cu + AgNO3 ---> Ag + Cu(NO3)2
Fe + Cu(NO3)2 ---> Fe(NO3)2 + Cu
Ca + H2O ---> Ca(OH)2 + H2
Zn + HCl ---> ZnCl2 + H2
Notice how, when hydrogen gets displaced, I write it as a diatomic. I do that because
elemental hydrogen is diatomic. Don't forget that!!
2. One anion replaces another. Written using generic symbols, it is:
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.
Cl2 + NaBr ---> NaCl + Br2
Br2 + KI ---> KBr + I2
In single replacement, 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.
Typically, you will be given the left-hand (reactant side) and asked to provide the
products to the reaction. You need to be able to recognize single replacement reactions
AND be able to break a formula apart into proper cations and anions as well as write
correct formulas
Here are several examples which are solved below:
1) ZnS + O2 --->
2) K + H2O --->
3) Fe + HCl --->
4) NaI + Br2 --->
Example #1
How to figure out the right (or product side):
(1) Decide if the reactant element (O2 in this case) is normally positive or negative. (It is
a negative 2. Just use O, not O2.)
(2) Identify the opposite charged (from step one) portion of the compound and its charge.
(Zn has a charge of +2.)
(3) Write a formula using information from step one & two. (ZnO since Zn = +2 and O =
-2.)
(4) Write the left over element as the second product. Write it as diatomic if it is. (In this
case all you write is S, since sulfur is not diatomic. Now, if you knew a bit more
chemistry, you'd write sulfur as S8. We'll just leave it as S for now.)
So the final answer looks like this:
ZnS + O2 ---> ZnO + S
Example #2
How to figure out the right (or product side):
(1) Decide if the reactant element (K in this case) is normally positive or negative. (It is a
positive 1.)
(2) Identify the opposite charged (from step one) portion of the compound and its charge.
(OH has a charge of negative 1.)
Special note on water - the +1 cations, the +2 cations and Al3+ will kick out only ONE
hydrogen, leaving hydroxide (OH¯) behind.
(3) Write a formula using information from step one & two. (KOH since K = +1 and OH
= -1.) Keep in mind that OH¯ is involved, NOT O2(4) Write the left over element as the second product. Write it as diatomic if it is. (In this
case since hydrogen is diatomic, we write H2.)
So the final answer looks like this:
K + H2O ---> KOH + H2
Example #3
How to figure out the right (or product side):
(1) Decide if the reactant element (Fe in this case) is normally positive or negative. (It
can be a +2 or a +3. The ion produced depends on conditions such as concentration or
temperature. Check with your teacher on how to deal with multiple charge cations. Keep
in mind, that to be safe - especially if your teacher is unclear on what to do - you may
want to do all possible answers. I'll use +3 in this example.)
(2) Identify the opposite charged (from step one) portion of the compound and its charge.
(Cl has a charge of negative 1.)
(3) Write a formula using information from step one & two. (FeCl3 since Fe = +3 and Cl
= -1.)
(4) Write the left over element as the second product. Write it as diatomic if it is. (In this
case since hydrogen is diatomic, we write H2.)
So the final answer looks like this:
Fe + HCl ---> FeCl3 + H2
Example #4
How to figure out the right (or product side):
(1) Decide if the reactant element (Br in this case) is normally positive or negative. (It is a
negative 1.)
(2) Identify the opposite charged (from step one) portion of the compound and its charge.
(Na has a charge of +1.)
(3) Write a formula using information from step one & two. (NaBr since Na = +1 and Br
= -1.)
(4) Write the left over element as the second product. Write it as diatomic if it is. (In this
case since iodine is diatomic, we write I2.)
So the final answer looks like this:
NaI + Br2 ---> NaBr + I2
Practice Problems
Note that none of the example problems above are balanced. Write correct formulas for
the products in these single replacement reactions.
1) Al + Pb(NO3)2 --->
2) Cl2 + NaI --->
3) Fe + AgC2H3O2 --->
4) Al + CuCl2 --->
5) Br2 + CaI2 --->
6) Al + HCl --->
7) Mg + HCl --->
8) Zn + H2SO4 --->
9) Fe + CuSO4 --->
10) Cl2 + MgI2 --->
Unbalanced Answers
1) Al + Pb(NO3)2 ---> Pb + Al(NO3)3
2) Cl2 + NaI ---> I2 + NaCl
3) Fe + AgC2H3O2 ---> Ag + Fe(C2H3O2)2
If Fe(III) is used, then Fe(C2H3O2)3 would result.
4) Al + CuCl2 ---> Cu + AlCl3
5) Br2 + CaI2 ---> I2 + CaBr2
6) Al + HCl ---> AlCl3 + H2
7) Mg + HCl ---> MgCl2 + H2
8) Zn + H2SO4 ---> ZnSO4 + H2
9) Fe + CuSO4 ---> Cu + FeSO4
If Fe(III) is used, then Fe2(SO4)3 would result.
10) Cl2 + MgI2 ---> I2 + MgCl2
Here are examples of the +1 and +2 cations reacting with water:
Na + H2O ---> NaOH + H2
Li + H2O ---> LiOH + H2
Cs + H2O ---> CsOH + H2
Mg + H2O ---> Mg(OH)2 + H2
Ca + H2O ---> Ca(OH)2 + H2
Ba + H2O ---> Ba(OH)2 + H2
Reaction Types: Double Replacement
Important notes to remember: (1) NONE of the equations are balanced!! and (2) make
sure to write correct formulas. DO NOT just copy the subscripts from the reactants over
into the products.
During double replacement, the cations and anions of two different compounds switch
places.
Written using generic symbols, it is:
AB + XY ---> AY + XB
A and X are the cations (postively-charged ions) in this example, with B and Y being the
anions (negatively-charged ions).
Here is another way to look at the above generic example:
a) the outside portions (the cation A and anion Y) combine to make a formula called AY.
b) The inside portions (the anion B and the cation X) switch order so that X (postively
charged) goes first and B (negatively charged) goes second making a formula called XB.
Keep in mind that, when it comes to writing actual formulas, you MUST write
chemically correct formulas. Please do not assume from the AY and XB examples that
the product formulas will always be one-to-one in terms of positive and negative.
Some examples of actual reactions are:
KOH + H2SO4 ---> K2SO4 + H2O
FeS + HCl ---> FeCl2 + H2S
NaCl + H2SO4 ---> Na2SO4 + HCl
AgNO3 + NaCl ---> AgCl + NaNO3
Note that none of them are balanced yet.
These three are also examples of double replacement, but there is something special
about them:
CaCO3 + HCl ---> CaCl2 + CO2 + H2O
K2SO3 + HNO3 ---> KNO3 + SO2 + H2O
NH4Cl + NaOH ---> NaCl + NH3 + H2O
Notice how, one of the two product compounds decomposes. Whenever H2CO3, H2SO3,
or NH4OH is a product formula, the correct technique is to write the products as done in
the examples. Don't forget that!!
In other words,
CaCO3 + HCl ---> CaCl2 + H2CO3
is incorrect.
One additional comment on the above - a acid or a base is one of the two substances
involved in the reactants. If no acid or base, the decomposition does not take place. See
practice problem #6 for an example. There is another example in the 10 problems, but
you'll have to figure out which one!!
In double replacement, both reactants are compounds, each with a cation part and an
anion part. Diatomic elements do not count; they are included in the single replacement
category.
Typically, you will be given the left-hand (reactant side) and asked to provide the
products to the reaction. You need to be able to recognize double replacement reactions
AND be able to break a formula apart into proper cations and anions as well as write
correct formulas
Here are several examples which are solved below:
1) Ca(OH)2 + HCl --->
2) Al(NO3)3 + H2SO4 --->
3) Pb(NO3)2 + K2S --->
4) Pb(NO3)2 + CuSO4 --->
Example #1
How to figure out the right (or product side):
(1) Identify the cations and anions in each compound:
Ca(OH)2 has Ca2+ and OH¯
HCl has H+ and Cl¯
All you have to do is identify each, you need not worry about amounts yet.
(2) Pair up each cation with the anion from the OTHER compound:
Ca2+ pairs with Cl¯
H+ pairs with OH¯
(3) Write two new (CORRECT!!) formulas using the pairs from step two.
CaCl2 since Ca is positive 2 and Cl is minus one
H2O since H is plus one and OH is negative one
So the final answer looks like this:
Ca(OH)2 + HCl ---> CaCl2 + H2O
One warning - take a look at that word "CORRECT" just above. In these types of
problems, many kids (that's you, most likely) tend to just pair up the cation and anion in a
one-to-one ratio to get the formulas. In the example above, they would get CaCl
(wrong!!) and H2O (correct, but maybe just lucky!!). Be careful to write correct formulas.
Example #2
How to figure out the right (or product side):
(1) Identify the cations and anions in each compound:
Al(NO3)3 has Al3+ and NO3¯
H2SO4 has H+ and SO42¯
All you have to do is identify each, you need not worry about amounts yet.
(2) Pair up each cation with the anion from the OTHER compound:
Al3+ pairs with SO42¯
H+ pairs with NO3¯
(3) Write two new (CORRECT!!) formulas using the pairs from step two.
Al2(SO43) since Al is positive 3 and sulfate is minus 2
HNO3 since H is plus one and nitrate is negative one
So the final answer looks like this:
Al(NO3)3 + H2SO4 ---> Al2(SO4)3 + HNO3
Make sure that you keep polyatomic ions together (except for the ones that decompose!!)
Example #3
How to figure out the right (or product side):
(1) Identify the cations and anions in each compound:
Pb(NO3)2 has Pb2+ and NO3¯
K2S has K+ and S2¯
All you have to do is identify each, you need not worry about amounts yet.
(2) Pair up each cation with the anion from the OTHER compound:
Pb2+ pairs with S2¯
K+ pairs with NO3¯
(3) Write two new (CORRECT!!) formulas using the pairs from step two.
PbS since Pb is positive 2 and sulfide is minus 2
KNO3 since K is plus one and nitrate is negative one
So the final answer looks like this:
Pb(NO3)2 + K2S ---> PbS + KNO3
Example #4
How to figure out the right (or product side):
(1) Identify the cations and anions in each compound:
Pb(NO3)2 has Pb2+ and NO3¯
CuSO4 has Cu2+ and SO42¯
All you have to do is identify each, you need not worry about amounts yet.
(2) Pair up each cation with the anion from the OTHER compound:
Pb2+ pairs with SO42¯
Cu2+ pairs with NO3¯
(3) Write two new (CORRECT!!) formulas using the pairs from step two.
PbSO4 since Pb is positive 2 and sulfate is minus 2
Cu(NO3)2 since Cu is plus two and nitrate is negative one
So the final answer looks like this:
Pb(NO3)2 + CuSO4 ---> PbSO4 + Cu(NO3)2
Practice Problems
Note that none of the example problems above are balanced. Write correct formulas for
the products in these double replacement reactions.
1) Ca(OH)2 + H3PO4 --->
2) K2CO3 + BaCl2 --->
3) Cd3(PO4)2 + (NH4)2S --->
4) Co(OH)3 + HNO3 --->
5) AgNO3 + KCl --->
6) Na2CO3 + H2SO4 --->
7) Al(OH)3 + HC2H3O2 --->
8) Al2(SO4)3 + Ca3(PO4)2 --->
9) Cr2(SO3)3 + H2SO4 --->
10) AgC2H3O2 + K2CrO4 --->
Practice Problem Answers
Write correct formulas for the products in these double replacement reactions.
1) Ca(OH)2 + H3PO4 ---> Ca3(PO4)2 + H2O
2) K2CO3 + BaCl2 ---> KCl + BaCO3
3) Cd3(PO4)2 + (NH4)2S ---> CdS + (NH4)3PO4
4) Co(OH)3 + HNO3 ---> Co(NO3)3 + H2O
5) AgNO3 + KCl ---> AgCl + KNO3
6) Na2CO3 + H2SO4 ---> Na2SO4 + CO2 + H2O
7) Al(OH)3 + HC2H3O2 ---> Al(C2H3O2)3 + H2O
8) Al2(SO4)3 + Ca3(PO4)2 ---> AlPO4 + CaSO4
9) Cr2(SO3)3 + H2SO4 ---> Cr2(SO4)3 + SO2 + H2O
10) AgC2H3O2 + K2CrO4 ---> Ag2CrO4 + KC2H3O2
Reaction Types: Decomposition
Important notes to remember: (1) NONE of the equations are balanced!! and (2) make
sure to write correct formulas. DO NOT just copy the subscripts from the reactants over
into the products.
During decomposition, one compound splits apart into two (or more pieces). These pieces
can be elements or simpler compounds
Written using generic symbols, it is usually shown as:
AB ---> A + B
However, that really only works for splitting apart into the elements, like these examples.
HgO ---> Hg + O2
H2O ---> H2 + O2
MgCl2 ---> Mg + Cl2
FeS ---> Fe + S
Decomposition can also split one compound into two simpler compounds (or compound
and an element) as in these examples:
CaCO3 ---> CaO + CO2
Na2CO3 ---> Na2O + CO2
KClO3 ---> KCl + O2
Ba(ClO3)2 ---> BaCl2 + O2
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. Don't forget that!!
Figuring out what the products are in decomposition is harder (maybe you'll think it's
easier!!) because you will have to recognize several categories of decomposition
reactions. Here are your first (yes, there's more!) three:
1) All binary compounds (like the four in the first example set above) will break down
into their elements.
2) All carbonates (like the first two in the second example set above) break down to the
oxide and carbon dioxide.
3. Chlorates (like KClO3 and Ba(ClO3)2 in the example) will break down to the binary
salt and oxygen.
Here is one more category of decomposition reactions:
Ca(OH)2 ---> CaO + H2O
NaOH ---> Na2O + H2O
HNO3 ---> N2O5 + H2O
H3PO4 ---> P2O5 + H2O
The first two substances are bases and the last two are acids. In each case, the acid or
base breaks down into the oxide of the metal (in the case of bases) or the oxide of the
nonmetal (in the case of acids) plus water.
Here is one example of each category which are then solved below:
1) NaClO3 --->
2) Li2CO3 --->
3) KOH --->
4) NaCl --->
Example #1
How to figure out the right (or product side):
(1) Identify the type of compound decomposing:
NaClO3 is a chlorate
Notice that you have to be able to "read" a formula and identifiy the parts (cation and
anion) that make it up.
(2) Apply the rule for that type:
chlorates decompose to the binary salt and oxygen gas
(3) Write two new (CORRECT!!) formulas using the rule from step two.
NaCl since Na is positive 1 and Cl is minus one
O2 since oxygen is a diatomic gas
So the final answer looks like this:
NaClO3 ---> NaCl + O2
Example #2
How to figure out the right (or product side):
(1) Identify the type of compound decomposing:
Li2CO3 is a carbonate
(2) Apply the rule for that type:
carbonates decompose to the binary oxide and carbon dioxide gas
(3) Write two new (CORRECT!!) formulas using the rule from step two.
Li2O since Li is positive 1 and O is minus two
CO2 is the formula for carbon dioxide gas
So the final answer looks like this:
Li2CO3 ---> Li2O + CO2
Example #3
How to figure out the right (or product side):
(1) Identify the type of compound decomposing:
KOH is a base
(2) Apply the rule for that type:
bases decompose to the binary oxide and water
(3) Write two new (CORRECT!!) formulas using the rule from step two.
K2O since K is positive 1 and O is minus two
H2O is the formula for water
So the final answer looks like this:
KOH ---> K2O + H2O
Example #4
How to figure out the right (or product side):
(1) Identify the type of compound decomposing:
NaCl is a binary compound (that is not an acid or a base. I left this point until now.)
(2) Apply the rule for that type:
binary compounds decompose to the elements
(3) Write two new (CORRECT!!) formulas using the rule from step two.
Na is the proper symbol
Cl2 is the proper symbol for chlorine since it is diatomic
So the final answer looks like this:
NaCl ---> Na + Cl2
Example #5
There is another type of acid which does not have oxygen in it. HCl, HBr and HI are
examples. These acids simply decompose into their elements:
HCl ---> H2 + Cl2
Practice Problems
Note that none of the example problems above are balanced. Write correct formulas for
the products in these decomposition reactions. #3 might be tough - remember to preserve
nitrogen's oxidation number.
1) Ni(ClO3)2 --->
2) Ag2O --->
3) HNO2 --->
4) Fe(OH)3 --->
5) ZnCO3 --->
6) Cs2CO3 --->
7) Al(OH)3 --->
8) H2SO4 --->
9) RbClO3 --->
10) RaCl2 --->
Practice Problem Answers
Write correct formulas for the products in these decomposition reactions. #3 might be
tough - remember to preserve nitrogen's oxidation number.
1) Ni(ClO3)2 ---> NiCl2 + O2
2) Ag2O ---> Ag + O2
3) HNO2 ---> N2O3 + H2O
4) Fe(OH)3 ---> Fe2O3 + H2O
5) Zn(CO3) ---> ZnO + CO2
6) Cs2CO3 ---> Cs2O + CO2
7) Al(OH)3 ---> Al2O3 + H2O
8) H2SO4 ---> SO3 + H2O
9) RbClO3 ---> RbCl + O2
10) RaCl2 ---> Ra + Cl2
Reaction Types: Synthesis
Important notes to remember: (1) NONE of the equations are balanced!! and (2) make
sure to write correct formulas. DO NOT just copy the subscripts from the reactants over
into the products.
Synthesis are, at this introductory level, almost always the reverse of a decomposition
reaction. That means that two pieces join together to produce one, more complex
compound. These pieces can be elements or simpler compounds. Complex simply means
that the product compound has more atoms than the reactant molecules. Usually!!
Written using generic symbols, it is usually shown as:
A + B ---> AB
These are some examples:
Mg + O2 ---> MgO
H2 + O2 ---> H2O
K + Cl2 ---> KCl
Fe + O2 ---> Fe2O3
Notice that two elements are combining in each example. Synthesis can also be two
compounds making a more complex compound (or a compound and an element joining
together) as in these examples:
CaO + CO2 ---> CaCO3
Na2O + CO2 ---> Na2CO3
KCl + O2 ---> KClO3
Ba(ClO3)2 ---> BaCl2 + O2
Notice how, in every case so far, there is only one substance on the right-hand (product)
side. This is not always the case in a synthesis reaction. Sometimes there will be two
products. Here's an example:
CO2 + H2O ---> C6H12O6 + O2
You might recognize this as the photosynthesis equation. However, the majority of what
you will see at this level will be two substances combining to make one product.
Here's another example of a synthesis reaction:
H2 + O2 ---> H2O2
This happens to be a reaction that can never take place. Hydrogen peroxide is made in
other ways, NOT by direct union of the elements. Nonetheless, it is a valid synthesis
reaction and useful in contexts otherthan how H2O2 is made.
Since synthesis reactions are the reverse of decomposition, you might ask if the decomp.
categories apply, just in reverse. The answer is yes!
1) Direct union of two elements will produce a binary compound.
2) Metallic oxides and carbon dioxide react to produce carbonates.
3. Binary salts and oxygen react to produce a chlorate.
Here is one more category of decomposition reactions:
CaO + H2O ---> Ca(OH)2
Na2O + H2O ---> NaOH
N2O5 + H2O ---> HNO3
P2O5 + H2O ---> H3PO4
The first two substances are metallic oxides and the last two are nonmetallic oxides. In
each case, the oxide plus water will produce a base (in the case of the metallic oxide) or
an acid (in the case of the nonmetallic oxide).
Here is one example of each category which are then solved below:
1) LiCl + O2 --->
2) Na2O + CO2 --->
3) SO3 + H2O --->
4) N2 + H2 --->
Example #1
How to figure out the right (or product side):
(1) Ask yourself what type of decomposition produces these products:
LiCl + O2 are the products of a chlorate decomposing.
Notice that you have to be able to "read" a formula and identifiy the parts (cation and
anion) that make it up.
(2) Write the reactant formula using the compounds from step one.
Chlorate is always ClO3¯
Li is plus one
So the final answer looks like this:
LiCl + O2 ---> LiClO3
Example #2
How to figure out the right (or product side):
(1) Ask yourself what type of decomposition produces these products:
Na2O + CO2 are the products of a carbonate decomposing.
Notice that you have to be able to "read" a formula and identifiy the parts (cation and
anion) that make it up.
(2) Write the reactant formula using the compounds from step one.
Carbonate is always CO32¯
Na is plus one
So the final answer looks like this:
Na2O + CO2 ---> Na2CO3
Example #3
How to figure out the right (or product side):
(1) Ask yourself what type of decomposition produces these products:
SO3 + H2O are the products of an acid decomposing.
Notice that you have to be able to "read" a formula and identifiy the parts (cation and
anion) that make it up.
(2) Write the reactant formula using the compounds from step one. With special regard to
acids and bases, you must preserve the oxidation number of the nonmetal (acid) or metal
(base).
In SO3 the S has an oxidation number of +6 H has its usual value of +1 and O has its
usual value of -2
So the final answer looks like this:
SO3 + H2O ---> H2SO4
If this oxidation number business is unknown to you, the only other way is to memorize a
set of reactions. For example, SO2 produces H2SO3, SO3 produces H2SO4 and so on.
Example #4
How to figure out the right (or product side):
(1) Ask yourself what type of decomposition produces these products:
N2 + H2 are the products of a binary compound decomposing.
Notice that you have to be able to "read" a formula and identifiy the parts (cation and
anion) that make it up.
(2) Write the reactant formula using the compounds from step one.
N has a charge of -3 H has its usual value of +1
So the final answer looks like this:
N2 + H2 ---> NH3
Positives are written first in formula, so why is NH3 reversed? Good question. For the
time being, just do it!
Practice Problems
Note that none of the example problems above are balanced. Write correct formulas for
the products in these synthesis reactions.
1) MgCl2 + O2 --->
2) Na + O2 --->
3) P2O3 + H2O --->
4) K2O + H2O --->
5) BaO + CO2 --->
6) BeO + CO2 --->
7) Al2O3 + H2O --->
8) N2O5 + H2O --->
9) NaCl + O2 --->
10) Ra + Cl2 --->
Practice Problem Answers
Write correct formulas for the products in these synthesis reactions.
1) MgCl2 + O2 ---> Mg(ClO3)2
2) Na + O2 ---> Na2O
3) P2O3 + H2O ---> H3PO3
4) K2O + H2O ---> KOH
5) BaO + CO2 ---> BaCO3
6) BeO + CO2 ---> BeCO3
7) Al2O3 + H2O ---> Al(OH)3
8) N2O5 + H2O ---> HNO3
9) NaCl + O2 ---> NaClO3
10) Ra + Cl2 ---> RaCl2