Reactions
Conservation of Matter,
Stoichiometry,
and
Moles
Standards
3. The conservation of atoms in chemical reactions leads to the principle of conservation of
matter and the ability to calculate the mass of products and reactants. As a basis for
understanding this concept:
a. Students know how to describe chemical reactions by writing balanced equations.
3. b. Students know the quantity one mole is set by defining one mole of carbon-12 atoms to
have a mass of exactly 12 grams.
3. c. Students know one mole equals 6.02 × 10^23 particles (atoms or molecules).
3. d. Students know how to determine the molar mass of a molecule from its chemical formula
and a table of atomic masses and how to convert the mass of a molecular substance to
moles, number of particles, or volume of gas at standard temperature and pressure.
3. e. Students know how to calculate the masses of reactants and products in a chemical reaction
from the mass of one of the reactants or products and the relevant atomic masses.
3. f.* Students know how to calculate percent yield in a chemical reaction.
3. g.* Students know how to identify reactions that involve oxidation and reduction and how to
balance oxidation-reduction reactions.
Example Reaction
Reactants  Products
what you
start out with
what you
end up with
Reaction Symbols
Symbol
Meaning
(s), (l), (g) Substance is a solid, liquid, or gas
(aq)
Aqueous, substance is dissolved in H2O
“Produces” or “makes”
“Produces” through reversible reaction
heat or Δ
Heat is added to the reactants
Pt
A catalyst is used to speed up the
reaction
Conservation of Mass and Atoms
• The total mass of the reactants is the same as
the total mass of the products. It doesn’t
change.
• The number of each type of atom for the
reactants is the same as number of each type
of atom for the products. It doesn’t change.
Example Reaction #1
2 H2 + O2  2 H2O
hydrogen
gas
oxygen
gas
+
water
Example Reaction #1
2 H2 + O2  2 H2O
hydrogen
gas
oxygen
gas
4.04 grams
water
+ 32.00 grams
36.04 grams
Basic Types of Reactions
Combination
Single Substitution
(aka Synthesis)
(aka Displacement)
A + B  AB A + BC  B + AC
Decomposition Double Substitution
AB  A + B AB + CD  CB + AD
Example Reaction #2
2 NH3 (l)  N2 (g) + 3 H2 (g)
liquid
ammonia
nitrogen
gas
hydrogen
gas
+
Example Reaction # 2
2 NH3 (l)  N2 (g) + 3 H2 (g)
liquid
ammonia
H
H
N
H
N
H
H
nitrogen
gas
hydrogen
gas
H
H
N
N + H
H
H
H
H
Combustion of Methane
CH4(g) + 2 O2(g)  CO2 (g) + 2 H2O(g)
methane
oxygen
+
carbon
dioxide
water
+
Combustion Reactions
Complete Combustion:
Fuel + O2  CO2 + H2O
Incomplete Combustion:
Fuel + O2  CO + H2O
Example Fuels: Methane (CH4), Propane
(C3H8), Octane (C8H18), Wood, etc.
How much?
???
1 cup
26 cookies
How much?
3 eggs
=
6 ounces
1.5 cups
How much?
16.05
grams
6.06
grams
CH4(g) + H2O (g)  CO (g) + 3 H2 (g)
Methane
+
water
carbon
monoxide
hydrogen
+
Stoichiometry
• Stoichiometric calculations use mole ratios
from the balanced chemical equations to
predict amounts of products or reactants.
What is the mole ratio between H2 and O2?
2 H2 + O2  2 H2O
hydrogen
gas
oxygen
gas
+
water
What is the mole ratio between H2 and O2?
2 H2 + O2  2 H2O
hydrogen
gas
oxygen
gas
water
2 mol H2
1 mol O2
What is the mole ratio between H2O and H2?
2 H2 + O2  2 H2O
hydrogen
gas
oxygen
gas
water
2 mol H2O
=
2 mol H2
1 mol H2O
1 mol H2
Stoichiometric Calculation #1
2 H2 + O2  2 H2O
How many moles of oxygen gas (O2) are
needed to react completely with 6.0 moles
of hydrogen gas (H2)?
6.0 mol H2
1 mol O2
×
1
2 mol H2
6.0
=
mol O2
2
= 3.0 mol O2
Stoichiometric Calculation #2
2 NH3 (l)  N2 (g) + 3 H2 (g)
What is the mass of ammonia (NH3) needed to
produce 9.00 moles of hydrogen gas (H2)?
9.00 mol H2 2 mol NH3 17.04 g NH3
×
=
×
1 mol NH3
1
3 mol H2
9.00 × 2 × 17.04
=
g NH3
3
= 102 g NH3
Stoichiometric Calculation #3
C3H8(g) + 5 O2(g)  3 CO2 (g) + 4 H2O(g)
How many liters of carbon dioxide (CO2) are
produced from the complete combustion of
2.00 moles of propane (C3H8)?
2.00 mol C3H8 3 mol CO2 22.4 L CO2
×
=
×
1 mol CO2
1
1 mol C3H8
2.00 × 3 × 22.4
=
L CO2
1
= 134 L CO2
The Breakfast of
Chemists!
2
+1
 16
products
reactants
Limiting and Excess Reactants
Limiting reactant – the reactant that you
will run out of.
Excess reactant – the reactant that you
have extra amounts of.
Steps for Limiting Reactant Problems
1. Divide the reactant moles by their
coefficients. Compare the results.
2.
The lower one is the limiting reactant.
The higher one is the excess reactant.
3. Only use the limiting reactant for
stoichiometric calculations.
Limiting Reactant Problem #1
Mg (s) + 2 HCl (aq)  MgCl2 (aq) + H2 (g)
3 mol
4 mol
If you have 3 moles of magnesium (Mg)
and 4 moles of hydrochloric acid (HCl),
which substance is the limiting reactant?
Which substance is the excess reactant?
3 mol
4 mol
=
2
=
3
1
2
>
excess
limiting
Limiting Reactant Problem #2
Mg (s) + 2 HCl (aq)  MgCl2 (aq) + H2 (g)
3 mol
4 mol
How many moles of magnesium chloride
(MgCl2) will be produced?
4.0 mol HCl 1 mol MgCl2 4.0
=
×
mol MgCl2
1
2 mol HCl
2
= 2.0 mol MgCl2
Add
• Breakfast of Chemists Art and intro
• Limiting Reactant (Reagent) Practice
• Cookie recipe – want to know precise amounts
– so it doesn’t turn out wrong, & so we don’t
waste ingredients
• Learn how using only the periodic table & the
chemical reaction to have the precise amount
• What is a mole, molar mass, # of particles
• Balanced chemical equations
Combustion Reactions
• Combustion is a double substitution reaction.
• For a single reaction, there are 2 reactants:
the fuel (ex. CH4) and oxygen (O2).
• With complete combustion there are 2 products:
carbon dioxide (CO2) and water (H2O).
• Incomplete combustion occurs when there is
not enough oxygen (O2) present. For this the 2
products are: carbon monoxide (CO) and
water (H2O).
H
He
Li Be B
C
N
O
F
Na Mg Al
Si
P
S
Cl Ar
K
Ca
Ne
Br Kr
I
Xe
H
He
Li Be B
C
N
O
F
Na Mg Al
Si
P
S
Cl Ar
K
Ca
Ne
Br Kr
I
Xe
4 e– in valence shell