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Review of Classifying and Balancing
Chemical Equations
• Lets take a moment… sit back…
relax… and review some previously
learned concepts…
5 main types of chemical
reactions:
1. Formation/Synthesis
2. Decomposition/De-formation
3. Single Replacement
4. Double Replacement
5. Hydrocarbon Combustion
6.
Other Combustions
1. Formation/Synthesis Reactions
Reaction in which 2 or more pure elements combine to form a
new compound.
General Equation:
A + B  AB
Element + Element  Compound
Example:
Na + Cl  NaCl
(Sodium) (Chlorine)
(Sodium Chloride) SALT
2. Decomposition/De-Formation
Reaction in which a compound breaks down into the base
elements that formed it.
General Equation:
AB  A + B
Compound  Element + Element
Example:
H2O(l)  H2(g) + O2(g)
(Water)
(Hydrogen)
(Oxygen)
3. Single Replacement
Reaction in which an element and a compound react. The
“partners” are traded and the elements that make up the
compound and lone element are substituted.
General Equation:
AB +C  AC + B
Compound + Element  Compound + Element
Example:
NaCl(s) + Li(s) LiCl(s) + Na(s)
(Salt)
(Lithium)
(Lithium Chloride)
(Soduim)
4. Double Replacement
Reaction in which two compounds react and “switch partners”.
The elements that make up each compound are swapped to form
two new compounds.
General Equation:
AB +CD  AD + CB
Compound + Compound Compound + Compound
Example:
NaCl(s) + LiBr(s) LiCl(s) + NaBr(s)
(Salt)
(Lithium
Bromide)
(Lithium
Chloride)
(Sodium
Bromide)
5. Hydrocarbon Combustion
Reaction in which a compound containing Hydrogen and Carbon
are burned to produce H2O(g) and CO2(g).
General Equation:
CXHX +O2(g)  H2O(g) + CO2(g)
CxHx + Oxygen Water + Carbon Dioxide
Example:
CH4(g) + O2(g) H2O(g) + CO2(g)
(Methane)
(Oxygen)
(Water
Vapour)
(Carbon
Dioxide)
6. Other Combustions
Reaction in which a compound OR element is burned forming its
most common oxide.
General Equation:
X +O2(g)  XO(g)
X + Oxygen XO
Example:
X+ O2(g) XO
Steps to Balance a Chemical
Equation
1) Write out the correct chemical formulae
for the products and the reactants. Be sure
to include all states of matter!
2) Balance the atoms or ion present in the
greatest number. You may do this by finding
the lowest common multiple of the two.
3) Continue to systematically balance the rest
of the atoms or ions.
4) Check the final equation. Make sure every
type of atom or ion balances.
• The coefficients that are placed in front of the
compounds and elements represent the mole-to-mole
ratio of the reactants and products.
Ex. 2NaCl(aq) + MgSO4(aq) → Na2SO4(aq) + MgCl2(s)
– 2 moles of sodium chloride reacts with one mole of
magnesium sulfate to produce one mole of sodium sulfate and
one mole of magnesium chloride
• When you are making quantitative calculations, please
use the mole-to-mole ratio in your calculations
:
RXN Limitations
• RXNS don’t communicate temp. and pressure
• RXNS don’t communicate progress or process
• RXNS don’t communicate measurable
quantities of substances (can’t really weigh or
calculate moles, atoms, molecules, ions, etc.
without a calculation first)
RXN Assumptions
• RXNS are spontaneous
• RXNS are fast
• RXNS are quantitative (more than 99%
complete)
• RXNS are stoichiometric (simple, whole
number ratio of chemical amounts of
reactants and products
Net Ionic Equations
• Net ionic equations are useful in that they
show only those chemical species
participating in a chemical reaction
• The key to being able to write net ionic
equations is the ability to recognize
monoatomic and polyatomic ions, and the
solubility rules.
Net Ionic Equations
Step 1:
Pb(NO3)2(aq) + 2HCl(aq)  PbCl2(s) + 2 HNO3(aq)
Step 2:
Dissociate (split up) and soluble ionic compounds AND ionize (split up)
all Strong Acids.
Pb2+(aq) + 2NO3-(aq) + 2 H3O+(aq) + 2 Cl-(aq)
 PbCl2(s) + 2 H3O+(aq) + 2NO3-(aq)
Net Ionic Equations
Pb2+(aq) + 2NO3-(aq) + 2 H3O+(aq) + 2 Cl-(aq) PbCl2(s) + 2 H3O+(aq) + 2NO3-(aq)
Total Ionic Equation
Step 3:
Cancel out the spectator ions. (Ions that are the same on both sides)
Step 4:
Write down whatever is left over after canceling. Simplify balancing
if necessary
Pb2+(aq) + 2 Cl-(aq)  PbCl2(s)
Net Ionic Equation
BaCl2
+ Na2SO4
(aq)
(aq)
2 NaCl + BaSO4
(aq)
(s)
Complete Balanced Chemical Equation
What type of reaction?
Double Replacement
Step 1:
DONE
****Have to use solubility
table to determine the state
(aq) or (s)****
BaCl2
+ Na2SO4
(aq)
(aq)
2 NaCl + BaSO4
(aq)
(s)
Complete Balanced Chemical Equation
Ba2+(aq)
2Cl-(aq) 2Na+(aq)
Step 2:
SO42-(aq)
2Na+(aq)
2Cl-(aq)
BaSO4(s)
Dissociate soluble ionic and ionize strong acids.
Ba2+(aq) + 2Cl-(aq) + 2Na+(aq) + SO4+(aq) 2Na+(aq) + 2Cl-(aq) + BaSO4(s)
Step 2: DONE
DONE out the spectator ions. (Ions that are the same on
Step 3: Cancel
both sides)
Step 4: Write
DONEdown whatever is left over after canceling. Simplify
balancing if necessary.
Ba2+(aq) + SO4+(aq)  BaSO4(s)
Gravimetric Stoichiometry:
The process of calculating the masses involved in a chemical reaction using the balancing
numbers and (Need/Got).
Steps for Starting stoichiometry:
1. Balanced Chemical Equation.
2. Write information…mols….masse under the chemical/element
that it relates to.
3. ***Convert information given in question into mols.***
4. Find the chemical species the question wants to know about
and put a N for (need to find) above it.
5. Put a G for (got information about this) over the species you
have information about from the question.
Step 1:
Write a Balanced Chemical Equation for the reaction.
n=m
M
1 Fe2O3(s) +3 CO 2 Fe
3
(g)
(s) + CO2(g)
m=?
m = 100.0g
n =1.79
= ? mol
Step 2:
Write the information from the question under the correct
element/compound
Step 3:
Convert information to mols.
n = 100.0 g
55.85 g/mol
n =1.79 mol
Step 4:
Put an N over the species you NEED to find out about.
N
G
1 Fe2O3(s) +3 CO 2 Fe
3
(g)
(s) + CO2(g)
m=?
m = 100.0g
n =1.79 mol
Step 5: Put a G over the compound you know information about.
n=m
M
n = 100.0 g
55.85 g/mol
n = 1.79 mol
Hydrogen gas and Oxygen gas react to produce water. How much
hydrogen gas would be needed to produce 12.56 g of water?
Step 1:
Write a Balanced Chemical Equation for the reaction.
N
G
m=?
m = 12.56 g
2 H2(g) + 1 O2(g)  2 H20(l)
n = 0.697 mol
Step 2:
Write the information from the question under the correct
element/compound
Convert information to mols.
Step 3:
Step 4: Put an N over the species you NEED to find out about.
Step 5: Put a G over the compound you know information about.
Gravimetric Stoichiometry:
The process of calculating the masses involved in a chemical reaction using the balancing
numbers and (Need/Got).
Steps for FINISHING stoichiometry:
6. Convert mols of got (G) to mols of need (N) by multiple by the
conversion factor (N/G).
7. Convert the mols of need (N) into a mass by multiplying by the
molar mass (M) of the needed compound.
Step 6:
Convert mols of got (G) to mols of need (N) by multiple by the conversion
factor (N/G).
N
G
1 Fe2O3(s) +3 CO 2 Fe
3
(g)
(s) + CO2(g)
m = 100.0g
m=?
n =0.895 mol
(
N1
G2
n =1.79 mol
)
nFE x (1/2) = nFe2O3
(1.79mol) X (1/2) = nFe2O3
0.895 mol = nFe2O3
Step 7:
Convert mols of need (N) into mass of Need (N) by multiplying by N’s
Molar Mass.
N
G
1 Fe2O3(s) +3 CO 2 Fe
3
(g)
(s) + CO2(g)
m = 142.93
g
?
n =0.895 mol
(
n=m
M
m = 100.0g
m = nM
m = (0.895 mol)(
n =0.8953 mol
1
2
)
)
159.70 g/mol
MFe2O3(s) = 2 x 55.85 g/mol
3 x 16.00 g/mol
159.70 g/mol
m = 142.93 g
Hydrogen gas and Oxygen gas react to produce water. How
much hydrogen gas would be needed to produce 12.56 g of
water?
N
G
2 H2(g) + 1 O2(g)  2 H20(g)
m = 1.4079
?
g
m = 12.56 g
n = 0.697 mol
Step 6: Convert mols of got (G) to mols of need (N) by multiple by the conversion
factor (N/G).
Step 7: Convert the mols of need (N) into a mass by multiplying by the molar mass
(M) of the needed compound.
Excess and Limiting Reagents
Excess and limiting reagents refer to the reactant that will run out first and stop more
product from forming.
Limiting Reagent
Excess Reagent
+
24 Crackers
=
7 Pieces of Cheese
Which will run out
first?
Excess and Limiting Reagents
12.4 g of Hydrogen gas reacts with 12.4 g of Oxygen gas to form liquid
water. Determine the excess and limiting reagent. Calculate the
maximum amount of water that could be produced by reacting these
two gases together.
G
2
N
H2(g) + 1 O2(g)  2 H2O(g)
m = 12.4 g
n=m
M
n = 6.138….mol
m = 12.4 g
m=?
n=m
M
n = 0.3875 mol
***From here on it’s like a double stoichiometry
to see which can make more H2O(l)***
Excess and Limiting Reagents
12.4 g of Hydrogen gas reacts with 12.4 g of Oxygen gas to form liquid
water. Determine the excess and limiting reagent. Calculate the
maximum amount of water that could be produced by reacting these
two gases together.
G
2
N
H2(g) + 1 O2(g)  2 H2O(g)
m = 12.4 g
n = 6.138….mol
m = 12.4 g
m=?
n = 0.3875 mol
(
N
2
G
2
)
H2 n = 6.138…
Excess and Limiting Reagents
12.4 g of Hydrogen gas reacts with 12.4 g of Oxygen gas to form liquid
water. Determine the excess and limiting reagent. Calculate the
maximum amount of water that could be produced by reacting these
two gases together.
Excess
2
Limiting
N
G
H2(g) + 1 O2(g)  2 H2O(g)
m = 12.4 g
n = 6.138….mol
m=?
m = 12.4 g
O2 n = 0.775 mol
n = 0.3875 mol
(
2
N
1
G
)
H2 n = 6.138…
Percent Yield
In every reaction you can use stoichiometry to calculate the theoretical amount of
product that could be made. (Maximum or Total).
When you actually do an experiment, the actual amount that you are able to make
is called the actual amount.
Actual Amount
% Yield =
Theoretical Amount
X 100%
Stoichiometry ALWAYS involves mols and a balanced chemical
equation.
2
H2(g) + 1 O2(g)  2 H2O(g)
What information can get us n?
What formula’s do we know that have n in them?
2
H2(g) + 1 O2(g)  2 H2O(g)
n=m
M
STP
22.4 mol/L
n = 1 mol
V = 22.4 L
ALREADY DID THIS.
Use this when we turn a mass into mols
Called Gravimetric Stoichiometry.
and do stoichiometry.
SATP
24.8 mol/L
n = 1 mol
V = 24.8 L
***Because we can calculate mols
we can then do stoichiometry just
like before.***
1 C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(g)
6
4
10
Even Number
N
G
1 C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(g)
m = 275g
n = 31.2 mol
n=m
M
n = 275g
44.11 g/mol
VV == 698
? L
(
N
5
)
G1
n = 6.23 mol
nO2(g) @ STP =
=
22.4 L
1 mol
X
31.2 mol
X = 698 L
(YOU TRY)
G
N
2 Na(s) +2H2O(l)  1H2(g) + 2NaOH(s)
m=
m =37.1
? g
n=m
M
n = 1.61…mol
V = 20.0 L
m= nM
m=
(1.61mol)(22.99g/mol)
m= 37.1 g
( 21 )
nH2(g) @ SATP =
n = 0.806…mol
24.8 L
1 mol
= 20.0 L
X
X = 0.80645 mol
N
1 N2(g) + 3 H2(g)

2 NH3(g)
V=?
P = 450 kPa
T = 333.15
80°C K
G
1 N2(g) + 3 H2(g)
N

m = 7500
7.5 Kgg
2 NH3(g)
V=?
P = 450 kPa
T = 333.15 K
G
1 N2(g) + 3 H2(g)
n=m
M
n = 7500g
2.02 g/mol
N

m = 7500 g
n = 3700 mol
( 2N3G )
2 NH3(g)
V=?
P = 450 kPa
T = 333.15 K
R = 8.314
n = 2466.6.. mol
Have to calculate mols to do stoich.
V=?
P = 450 kPa
T = 333.15 K
R = 8.314
n = 2466.6.. mol
PV = nRT
V = (2466.6 mol)(8.314)(333.15)
(450 kPa)
V = 16000 L of NH3
V = 16 KL of NH3
Stoichiometry ALWAYS involves mols and a balanced chemical
equation.
Solution Stoichiometry:
Using the principles that govern concentrations to solve for n (mols).
Once you have moles the stoichiometry is the same.
C=n
V
***Once you get mols…the rest is the
same.***
H2SO4(aq) + KOH(aq)  H2O(aq) + K SO
2
4(aq)
1 H2SO4(aq) +2 KOH(aq) 2H2O(aq) + 1 K SO
2
4(aq)
N
G
1 H2SO4(aq) +2 KOH(aq) 2H2O(aq) + 1 K SO
2
4(aq)
V = 0.010 L
C=?
n = 0.00119
n = 0.00119
V = 0.0159 L
C = 0.150 mol/L
1
N
2
G
( )
n = CV
n = (0.150 mol/L)(0.0159 L)
n = 0.00239 mol
C=n
V
N
G
1 H2SO4(aq) +2 KOH(aq) 2H2O(aq) + 1 K SO
2
4(aq)
V = 0.010 L
V = 0.0159 L
C = ?0.119 mol/L
C = 0.150 mol/L
n = 0.00119
C = 0.00119 mol
0.010L
C = 0.119 mol/L
C=n
V