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Balancing Reactions
--for unbalanced students
Write the reaction
• Butane gas(C4H10) burns in oxygen gas to
form carbon dioxide gas and water vapor
Write the reaction
• Butane gas(C4H10) burns in oxygen gas to
form carbon dioxide gas and water vapor
C4H10(g) + O2(g)  CO2(g) + H2O(g)
“forms”
“and”
Write the reaction
• Aqueous sodium bromide and chlorine gas
form aqueous sodium chloride and liquid
bromine
• Silver (I) oxide decomposes to form solid
silver and oxygen gas
• Solid phosphorus burns in oxygen gas to
form diphosphorus pentoxide.
Writing a reaction
• You must include formulas for all substances.
• You may include states [(s), (l), (g), (aq)]
• You should balance the reaction
Do not include masses—even if they are part
of the problem, they are not part of the
reaction
Write the reaction
• Aqueous sodium bromide and chlorine gas
NaBr
(aq) + sodium
Cl2 (g) NaCl
Br2(l)
form
aqueous
chloride(aq)
and+liquid
bromine
• Silver (I) oxide decomposes to form solid
Ag2O(s)  Ag (s) + O2(g)
silver and oxygen gas
• Solid phosphorus burns in oxygen gas to
P4 (s) + O2pentoxide.
(g) P2O5 (s)
for diphosphorus
But…
NaBr (aq) + Cl2 (g) NaCl (aq) + Br2(l)
+
+
But…
NaBr (aq) + Cl2 (g) NaCl (aq) + Br2(l)
+
+
Seems to contradict Dalton’s Atomic Theory
But…
2NaBr(aq) +Cl2 (g)2NaCl(aq) + Br2(l)
+
+
That’s better.
It’s balanced!
But…
2NaBr(aq) +Cl2 (g)2NaCl(aq) + Br2(l)
These are
coefficients
But…
2NaBr(aq) +Cl2 (g)2NaCl(aq) + Br2(l)
Coefficients
indicate moles or
particles of the
substance
But…
2NaBr(aq) +Cl2 (g)2NaCl(aq) + Br2(l)
No coefficient? It
means there is a 1
(not written in)
Count your atoms:
• H20
• As2O3
• Ca(NO3)2
• 3 Al2(SO4)3
• 2 NH4NO3
The T diagram
NaBr(aq) +Cl2 (g)NaCl(aq) + Br2(l)
• Na
• Br
• Cl
Na
Br
Cl
Never change a good formula!
1) Count all atoms in reactants and
products
2) Fix a count by changing a coefficient
3) Repeat steps 1-2 as needed
The T diagram
NaBr(aq) +Cl2 (g)NaCl(aq) + Br2(l)
• Na
• Br
• Cl
1
1
2
Na 1
Br 2
Cl 1
You need to double the bromine on
the left.
You need to double the bromine on
the left.
• Do not try this at home or anywhere!
NaBr2
Na2Br
• Sodium bromide is NaBr
NaBr2
You need to double the bromine on
the left.
• Do not try this at home or anywhere!
NaBr2
Na2Br
NaBr2
• Sodium bromide is NaBr
• Two Sodium bromide is 2 NaBr
The T diagram
2NaBr(aq) +Cl2 (g)NaCl(aq) + Br2(l)
• Na
• Br
• Cl
12
12
2
Recount
Na 1
Br 2
Cl 1
The T diagram
2NaBr(aq) +Cl2 (g)2NaCl(aq) + Br2(l)
• Na
• Br
• Cl
12
12
2
Na 1 2
Br 2
Cl 1 2
Recount
Easy
•
__N2 + __O2  __ NO
•
__N2 + __ H2 __ NH3
•
__S8 + __ F2  __ SF4
•
__ NaI + __Br2  __I2 + __ NaBr
If it gets tough:
• Start with elements that only show up
twice.
• If one side is even, and the other is odd—
double the odd one.
• Leave hydrogen, oxygen and any pure
elements to the very end.
Medium
•
__NH3+_NaOCl_N2H4+_NaCl+_H2O
•
__Ca2C +__ H2O __CH4+__ Ca(OH)2
•
__ NH3 + __ O2 __ NO + __ H2O
•
___C4H10O + ___O2___CO2+___H2O
“Am I being punished?”
• (Later)
Types of reactions
Synthesis or composition
A+BAB
Only 1 product!
Decomposition
AB  A+B
Only 1 reactant!
Single replacement
A+BC  AC+B
or
D+BC  BD+C
A pure element in
reactants and products!
Double replacement
AB+CD  AD+CB
Two ionic compounds
switch partners!
Combustion (burning or oxidation)
AB+O2  AOx+BOy
Oxygen is a reactant!
Other Oxidation/Reduction
AB+CD  a whole
bunch of other
things.
Double replacement
AB+CD  AD+BC
If both AD and BC are ionic
compounds that dissolve in water—
the product never comes together
Two types of double replacement
AB+CD  AD+BC
In a neutralization
reaction, A=H+, D=OH-,
so AD=water
Write and balance:
•
•
•
•
HCl +NaOH
H2SO4 + KOH
H3PO4 + Al(OH)3
HClO4 + Ba(OH)2 
A neutralization reaction forms a salt and water
Write and balance:
•
•
•
•
HCl +NaOH H2O +NaCl
H2SO4 + KOH K2SO4 + H2O
H3PO4 + Al(OH)3 AlPO4 + 3H2O
2HClO4 + Ba(OH)2  2H2O+ Ba(ClO4)2
A neutralization reaction forms a salt and water
Two types of double replacement
AB+CD  AD+BC
In a precipitation
reaction, one of the
products won’t
dissolve.
For example:
NaCl(aq)+AgNO3(aq) NaNO3(aq)+AgCl(s)
NaNO3 dissolves in water, AgCl doesn’t
AgCl is the precipitate
No precipitate=no reaction
NaNO3(aq)+KCl(aq)
All of these dissolve in water
Before
Contents: aqueous
Na+ Cl- K+ NO3-
No precipitate=no reaction
KNO3(aq)+NaCl(aq)
All of these dissolve in water
Before
After
Contents: aqueous
Na+ Cl- K+ NO3-
Better yet:
NaCl(aq)+AgNO3(aq) NaNO3(aq)+AgCl(s)
Ag+(aq) + Cl-(aq) AgCl(s)
The sodium and nitrate ions
(not changed in the reaction)
are spectator ions, omit them
Do you want to save some trouble?
NaCl(aq)+AgNO3(aq) NaNO3(aq)+AgCl(s)
is
Ag+(aq) + Cl-(aq) AgCl(s)
This is the net
ionic equation
Review Questions:
• Your lab.
• It’s not a question.
Mole conversions review:
•
•
•
•
How many moles of Mg are in 36.0 g Mg?
What is the formula mass of Mg3N2?
What is the mass of .49 moles Mg3N2
Write and balance: Magnesium metal and
nitrogen gas form solid magnesium nitride
• In this reaction: how many moles of
magnesium nitride can be made from 1.48 mol
of magnesium?
The mass to mass problem
g
1 mol
mol
g
mol
g
1 mol
g
The mass to mass problem
g
1 mol
mol
g
Given
Value
mol
g
1 mol
g
The mass to mass problem
g
1 mol
mol
g
Formula
Mass
mol
g
1 mol
g
The mass to mass problem
g
1 mol
mol
g
Mole ratio
(from a
balanced
reaction)
mol
g
1 mol
g
The mass to mass problem
g
1 mol
mol
g
Formula
Mass
mol
g
1 mol
g
The mass to mass problem
g
1 mol
mol
g
Answer
(probably
wrong)
mol
g
1 mol
g
The mass to mass problem
g
1 mol
mol
g
mol
g
1 mol
g
What mass of liquid bromine is
formed by the action of chlorine
gas on 12.5 g NaBr?
g
1 mol
mol
g
mol
g
1 mol
g
What mass of liquid bromine is
formed by the action of chlorine
gas
on
12.5
g
NaBr?
Step 1
2NaBr(aq) +Cl2 (g)2NaCl(aq) + Br2(l)
g
1 mol
mol
g
mol
g
1 mol
g
What mass of liquid bromine is
formed by the action of chlorine
gas
on
12.5
g
NaBr?
Step 2
2NaBr(aq) +Cl2 (g)2NaCl(aq) + Br2(l)
102.89 g/mol
g
159.81g/mol
1 mol
mol
g
mol
g
1 mol
g
What mass of liquid bromine is
formed by the action of chlorine
gas
on
12.5
g
NaBr?
Step 3
12.5 g
1 mol NaBr
1 mol Br2
102.89 g NaBr
2 mol NaBr
159.81g Br2
1 mol Br2
9.71 g
Name the reactants and products,
1.
2.
3.
4.
5.
Fe2O3(s) + CO (g)  FeO (s) + CO2 (g)
FeO (s) + CO (g)  Fe (s) + CO2 (g)
C12H22O11(s)+O2(g)  CO2 (g) + H2O (g)
Fe (s) + O2 (g)  Fe2O3 (s)
Ca (s) + H2O (l)  Ca(OH)2 (aq) + H2 (g)
Balance the equations
1.
2.
3.
4.
5.
Fe2O3(s) + CO (g)  FeO (s) + CO2 (g)
FeO (s) + CO (g)  Fe (s) + CO2 (g)
C12H22O11(s)+O2(g)  CO2 (g) + H2O (g)
Fe (s) + O2 (g)  Fe2O3 (s)
Ca (s) + H2O (l)  Ca(OH)2 (aq) + H2 (g)
Calculate the molecular weight of
the 1st reactants and 1st products.
1.
2.
3.
4.
5.
Fe2O3(s) + CO (g)  FeO (s) + CO2 (g)
FeO (s) + CO (g)  Fe (s) + CO2 (g)
C12H22O11(s)+O2(g)  CO2 (g) + H2O (g)
Fe (s) + O2 (g)  Fe2O3 (s)
Ca (s) + H2O (l)  Ca(OH)2 (aq) + H2 (g)
Starting with 2.0 grams of of the
first reactant, determine the mass
of the first product.
1.
2.
3.
4.
5.
Fe2O3(s) + CO (g)  FeO (s) + CO2 (g)
FeO (s) + CO (g)  Fe (s) + CO2 (g)
C12H22O11(s)+O2(g)  CO2 (g) + H2O (g)
Fe (s) + O2 (g)  Fe2O3 (s)
Ca (s) + H2O (l)  Ca(OH)2 (aq) + H2 (g)
Percent Yield
• If you are given an actual yield—compare
it to the expected yield (from the mass-tomass problem)
• Yields should be less than 100%
(measurement error otherwise)
• %yield = actual/expected x 100%
C12H22O11(s)+O2(g)CO2(g)+H2O (g)
• 25.0 g sucrose, C12H22O11 (FM=342g/mol),
is burned, but only 30.0 g CO2 is recovered.
• What is the percent yield?
C12H22O11(s)+O2(g)CO2(g)+H2O (g)
• 25.0 g sucrose, C12H22O11 (FM=342g/mol),
is burned, but only 30.0 g CO2 is recovered.
• What is the percent yield?
Do the mass-to-mass problem to find the expected yield.
Divide the actual/expected, convert to a %
Potassium iodate decomposes to form
potassium iodide and oxygen gas
• Write the reaction.
Potassium iodate decomposes to form
potassium iodide and oxygen gas
• 50 g of KIO3 decomposes and only 35 g KI
is recovered.
• What is the percent yield?
C12H22O11(s)+O2(g)CO2(g)+H2O (g)
• 25.0 g sucrose, C12H22O11 (FM=342g/mol),
is burned, but only 30.0 g CO2 is recovered.
• What is the percent yield?
Do the mass-to-mass problem to find the expected yield.
Divide the actual/expected, convert to a %
Why are yields less than 100%?
Why are yields less than 100%?
• Incomplete reaction
• Some of your product is not recovered
• A side reaction could use up some reactant
How could a yield be over 100%?
How could a yield be over 100%?
• Contamination in the product
Limiting reactant
• If you are given the amount of two reactants—
you will probably run out of one first.
• Which one?
• Do two mass-to-mass problems, the expected
yield is the lesser of the two.
• The limiting reactant gives this answer
• The excess (XS) reactant is left over. (How
much?)
Do not assume…
…that the one with more mass is in
excess—you might need more of it.
…that the one with more moles is in
excess—you might need more of it.
Do not assume…
…that the one with more mass is in
excess—you might need more of it.
If 150 g nitrogen and 40 g hydrogen make
ammonia…
…that the one with more moles is in
excess—you might need more of it.
If 5 moles oxygen and 8 moles hydrogen
make water…
Do not assume…
…that the one with more mass is in
excess—you might need more of it.
If 150 g nitrogen and 40 g hydrogen make
ammonia…hydrogen is in excess
…that the one with more moles is in
excess—you might need more of it.
If 5 moles oxygen and 8 moles hydrogen
make water…oxygen is in excess
Do not assume…
…that the one with more mass is in
excess—you might need more of it.
If 150 g nitrogen and 40 g hydrogen make
ammonia… 8g hydrogen is left over
…that the one with more moles is in
excess—you might need more of it.
If 5 moles oxygen and 8 moles hydrogen
make water… 1 mol oxygen is left over
Starting with 2.0 grams of each
reactant, identify the limiting reagent
1. Fe2O3(s) + CO (g)  FeO (s) + CO2 (g)
Starting with 2.0 grams of each
reactant, identify the limiting reagent
1.
2.
3.
4.
5.
Fe2O3(s) + CO (g)  FeO (s) + CO2 (g)
FeO (s) + CO (g)  Fe (s) + CO2 (g)
C12H22O11(s)+O2(g)  CO2 (g) + H2O (g)
Fe (s) + O2 (g)  Fe2O3 (s)
Ca (s) + H2O (l)  Ca(OH)2 (aq) + H2 (g)
The mole to mass problem
mol
mol
mol
g
1 mol
g
The mass to mole problem
g
1 mol
mol
mol
g
mol
The mass to volume problem (at STP)
g
1 mol
mol
g
mol
22.4L
1 mol
L
Etc.
g
1 mol
mol
g
mol
g
1 mol
g