Stoichiometry
Chapter 11
REVIEW
Chemical Reactions/Equations
Reactant
+ Reactant → Product + Product
Balanced chemical equations are written to
show chemical changes taking place
Balanced chemical equations reflect the
Law of Conservation of Mass
Coefficients balance chemical equations
and indicate the relative amounts of
substances
REVIEW
Conversion Factors
1.
Avogadro’s number and moles
1 mol = 6.02 x 1023 representative particles
Representative particles:
Atoms – element from periodic table
Ions – charged atom (due to loss/gain of e–)
Formula units – ionic compound
Molecules – covalent compound
2.
Molar mass
3.
1 mol = g of substance
where = (atomic mass × subscript)
Mole ratio of element to compound
mol element = 1 mol compound
where = subscript of element in compound
PREVIEW
Unit Objectives
Interpret
chemical equations in terms of
particles, moles, and mass
Write mole ratios from balanced equations
Calculate number of moles and mass of a
reactant or product when given number of
moles or mass of another reactant or
product
Identify limiting reactants in chemical
reactions
Determine percent yield of chemical
reactions
U8-5
Stoichiometry
Defined:
The
study of quantitative relationships
between amounts of reactants used and
products formed by a chemical reaction
Involves calculating quantities of reactants
or products in a reaction using relationships
found in balanced chemical equation
Based
Mass:
Mass
on the Law of Conservation of
is neither created nor destroyed in
any process; it is conserved
U8-5
Interpreting Equations
Write
the balanced chemical equation for
the synthesis of ammonia from nitrogen and
hydrogen gases.
N2
Interpret
+ 3 H2
→
2 NH3
the equation in terms of particles,
moles, and mass.
Then, show the Law of Conservation of
Mass is obeyed.
U8-5
Interpreting Equations
Interpret the equation in terms of particles.
N2 + 3 H2 → 2 NH3
Number
of particles for each substance is
indicated by the coefficients
Four particle types: atoms, ions, formula units,
or molecules
1 molecule N2 + 3 molecules H2 → 2 molecules NH3
U8-5
Interpreting Equations
Interpret the equation in terms of moles.
N2 + 3 H2 → 2 NH3
Number
of moles for each substance is
indicated by the coefficients
1 mole N2 + 3 moles H2 → 2 moles NH3
U8-5
Interpreting Equations
Interpret the equation in terms of mass.
N2 + 3 H2 → 2 NH3
Calculate
the molar mass for each reactant
and product
N2 :
2 mol x 14.0 g/mol
28.0 g/mol N2
H 2:
2 mol x 1.0 g/mol
6.0 g/mol H2
NH3:
1 mol x 14.0 g/mol = 14.0 g
3 mol x 1.0 g/mol = 3.0 g
17.0 g/mol NH3
U8-5
Interpreting Equations
Interpret the equation in terms of mass.
N2 + 3 H2 → 2 NH3
28.0
g/mol
2.0
g/mol
17.0
g/mol
Multiply
number of moles of each reactant
and product by molar mass
g N2 =
1 mol N2
28.0 g N2
1 mol N2
= 28.0 g N2
U8-5
Interpreting Equations
Interpret the equation in terms of mass.
N2 + 3 H2 → 2 NH3
28.0
g/mol
2.0
g/mol
17.0
g/mol
Multiply
number of moles of each reactant
and product by molar mass
g H2 =
3 mol H2
2.0 g H2
1 mol H2
= 6.0 g H2
U8-5
Interpreting Equations
Interpret the equation in terms of mass.
N2 + 3 H2 → 2 NH3
28.0
g/mol
2.0
g/mol
17.0
g/mol
Multiply
number of moles of each reactant
and product by molar mass
g NH3 =
2 mol NH3
17.0 g NH3
1 mol NH3
= 34.0 g NH3
U8-5
Interpreting Equations
Interpret the equation in terms of mass.
N2 + 3 H2 → 2 NH3
28.0 g
6.0 g
34.0 g
28.0 g N2 + 6.0 g H2 → 34.0 g NH3
U8-5
Interpreting Equations
Show that the Law of Conservation of Mass is
observed.
N2 + 3 H2 → 2 NH3
28.0 g
Add
6.0 g
34.0 g
the masses of the reactants.
28.0 g N2 + 6.0 g H2 = 34.0 g reactants
Add
the masses of the products.
34.0 g NH3 = 34.0 g products
When
the mass of the reactants equals the
mass of the products, the Law of Conservation
of Mass is observed.
34.0 g reactants = 34.0 g products
HW due 01/28
Textbook Practice, p. 371
#1, #2
U8-5
Mole Ratio
Ratio between the numbers of moles of
any two substances in a balanced
chemical equation
U8-5
Mole Ratios
Determine
all possible mole ratios for the
balanced equation showing the synthesis of
ammonia.
N2 + 3 H2 → 2 NH3
How
many mole ratios are possible?
This reaction has three participating species.
Multiply the number of species present by the
next lower number. 3 x 2 = 6 mole ratios
For the synthesis of ammonia, six mole ratios
are possible.
U8-5
Mole Ratios
Determine
all possible mole ratios for the
balanced equation showing the synthesis of
ammonia.
N2 + 3 H2 → 2 NH3
1 mol N2
3 mol H2
3 mol H2
2 mol NH3
2 mol NH3
1 mol N2
1 mol N2
2 mol NH3
3 mol H2
1 mol N2
2 mol NH3
3 mol H2
U8-5
Mole Ratios
Which
mole ratio should be used?
The
needed mole ratio is the one involving
the UNKNOWN and the GIVEN.
UNKNOWN
GIVEN
HW due 01/29
Textbook Practice, p. 372
#3
U8-5
Stoichiometric Calculations
Write the balanced chemical equation for the
reaction.
Identify the UNKNOWN and the GIVEN; draw the
bridge/grid.
The GIVEN must be in moles or converted to moles.
Identify the conversion factor that will cancel the
unit of the GIVEN.
Set up conversion factors and cancel units until the
only unit left standing matches the UNKNOWN.
Do the math* and express the answer to the
correct number of significant figures.
Mole-to-Mole
Mole-to-Mole Conversions
Both
the UNKNOWN and the GIVEN are in
moles
Example:
How many moles of ammonia are produced
when 10.0 moles of hydrogen react with
excess nitrogen?
excess
10.0 mol
? mol
N2 + 3 H2 → 2 NH3
Mole-to-Mole
10.0 mol
? mol
N2 + 3 H2 → 2 NH3
UNKNOWN
mol NH3 =
=
GIVEN
10.0 mol H2
g→mol (G)
Mole Ratio
2 mol NH3
3 mol H2
6.67 mol NH3
mol→g (U)
Mole-to-Mole
Mole-to-Mole Conversions
Practice 1
How many moles of zinc chloride will be formed
when 17.0 moles of hydrochloric acid react with
excess zinc metal?
17.0 mol
2 HCl +
? mol
excess
Zn
→
ZnCl2 +
H2
How many mole ratios are possible for this
equation? 12
Mole-to-Mole
Mole-to-Mole Conversions
Practice 1
2 HCl + Zn → ZnCl2 + H2
Write all possible mole ratios for the balanced
equation.
2 mol HCl
1 mol Zn
1 mol Zn
1 mol ZnCl2
1 mol ZnCl2
1 mol H2
1 mol H2
2 mol HCl
2 mol HCl
1 mol ZnCl2
1 mol Zn
1 mol H2
1 mol ZnCl2
2 mol HCl
1 mol H2
1 mol Zn
2 mol HCl
1 mol H2
1 mol Zn
2 mol HCl
1 mol ZnCl2
1 mol Zn
1 mol H2
1 mol ZnCl2
Mole-to-Mole
? mol
17.0 mol
2 HCl + Zn → ZnCl2 + H2
UNKNOWN
mol ZnCl2 =
=
GIVEN
g→mol (G)
17.0 mol HCl
Mole Ratio
1 mol ZnCl2
2 mol HCl
8.50 mol ZnCl2
mol→g (U)
Mole-to-Mole
Mole-to-Mole Conversions
Practice 2
Potassium chlorate decomposes into potassium
chloride and oxygen. How many moles of oxygen
are formed when 3.20 moles KClO3 decompose?
3.20 mol
2 KClO3 →
? mol
2 KCl + 3 O2
Write the mole ratios involving the UNKNOWN and
the GIVEN in the problem.
3 mol O2
2 mol KClO3
2 mol KClO3
3 mol O2
Mole-to-Mole
3.20 mol
? mol
2 KClO3 → 2 KCl + 3 O2
UNKNOWN
mol O2 =
=
GIVEN
3.20 mol KClO3
g→mol (G)
Mole Ratio
3 mol O2
2 mol KClO3
4.80 mol O2
mol→g (U)
HW due 02/04
Textbook Practice, p. 375
#11
Mole-to-Mass
Mole-to-Mass Conversions
p. 4 of Stoichiometry for Students Notes Packet
Mole-to-Mass
Mole-to-Mass Conversions
GIVEN
in moles and UNKNOWN in grams
Example: Balance the following equation for the
combustion of propane.
3 CO2 + ____
4 H 2O
5 O2 → ____
____ C3H8 + ____
44.0
g/mol
32.0
g/mol
44.0
g/mol
18.0
g/mol
Calculate the molar mass for each substance.
Mole-to-Mass
Mole-to-Mass Conversions
Example: If 10.0 moles of propane are used, how
many grams of water are formed?
10.0 mol
?g
C3H8 + 5 O2 → 3 CO2 + 4 H2O
Write the mole ratios involving the UNKNOWN
and the GIVEN.
1 mol C3H8
4 mol H2O
4 mol H2O
1 mol C3H8
Mole-to-Mass
10.0 mol
?g
C3H8 + 5 O2 → 3 CO2 + 4 H2O
UNKNOWN
g H2O =
=
GIVEN
10.0 mol C3H8
720. g H2O
g→mol (G)
Mole Ratio
mol→g (U)
4 mol H2O
18.0 g H2O
1 mol C3H8
1 mol H2O
Mole-to-Mass
Mole-to-Mass Conversions
Practice
1: Sulfuric acid is produced when sulfur
dioxide reacts with oxygen and water. Balance
the equation for the synthesis reaction.
2 H2O → ____
2 H2SO4
2 SO2 + ____ O2 + ____
____
64.1
g/mol
32.0
g/mol
18.0
g/mol
98.1
g/mol
Calculate the molar mass for each substance.
Mole-to-Mass
Mole-to-Mass Conversions
Practice 1: How many grams of sulfuric acid are
produced when 1.50 moles of sulfur dioxide
completely reacts with oxygen and water ?
1.50 mol
?g
2 SO2 + O2 + 2 H2O → 2 H2SO4
Write the mole ratios involving the UNKNOWN
and the GIVEN.
2 mol H2SO4
2 mol SO2
2 mol SO2
2 mol H2SO4
Mole-to-Mass
1.50 mol
?g
2 SO2 + O2 + 2 H2O → 2 H2SO4
UNKNOWN
g H2SO4 =
=
GIVEN
1.50 mol SO2
147 g H2SO4
g→mol (G)
Mole Ratio
mol→g (U)
2 mol H2SO4
98.1 g H2SO4
2 mol SO2
1 mol H2SO4
Mass-to-Mole
Mass-to-Mole Conversions
GIVEN
in grams and UNKNOWN in moles
Example: Methane and sulfur produce carbon
disulfide and hydrogen sulfide gas, as
indicated by the following chemical equation.
2 CS2 + ____
2 CH4 + ____ S8 → ____
4 H 2S
____
16.0
g/mol
256.8
g/mol
76.2
g/mol
34.1
g/mol
Calculate the molar mass for each substance.
Mass-to-Mole
Mass-to-Mole Conversions
Example: Suppose that 19.75 g sulfur react with
an excess of methane. How many moles of
carbon disulfide will be produced?
19.75 g ? mol
2 CH4 + S8 → 2 CS2 + 4 H2S
Mass-to-Mole
19.75 g ? mol
2 CH4 + S8 → 2 CS2 + 4 H2S
UNKNOWN
mol CS2 =
=
GIVEN
g→mol (G)
19.75 g S8
1 mol S8
2 mol CS2
256.8 g S8
1 mol S8
0.1538 mol CS2
Mole Ratio
mol→g (U)
Mass-to-Mole
Mass-to-Mole Conversions
Practice
1: Sodium fluoride is formed when
sodium metal reacts with fluorine gas.
Calculate the molar mass for each substance.
2 NaF
2 Na + ____ F2 → ____
____
23.0
g/mol
38.0
g/mol
Balance the equation.
42.0
g/mol
Mass-to-Mole
Mass-to-Mole Conversions
Practice 1: How many moles of sodium fluoride
can be formed when 4.57 grams of fluorine gas
reacts completely with excess sodium?
4.57 g
? mol
2 Na + F2 → 2 NaF
Mass-to-Mole
4.57 g
? mol
2 Na + F2 → 2 NaF
UNKNOWN
mol NaF =
=
GIVEN
g→mol (G)
4.57 g F2
1 mol F2
2 mol NaF
38.0 g F2
1 mol F2
0.241 mol NaF
Mole Ratio
mol→g (U)
HW due 02/05
Textbook Practice, p. 376
#14
Mass-to-Mass
Mass-to-Mass Conversions
GIVEN
in grams and UNKNOWN in grams
Example: The following equation shows the
combustion of butane.
Calculate the molar mass for each substance.
Balance the equation.
13 O2 → ____
8 CO2 + ____
2 C4H10 + ____
10 H2O
____
58.0
g/mol
32.0
g/mol
44.0
g/mol
18.0
g/mol
Mass-to-Mass
Mass-to-Mass Conversions
Example: If 75.5 grams of carbon dioxide are
produced, how many grams of butane were
used?
?g
75.5 g
2 C4H10 + 13 O2 → 8 CO2 + 10 H2O
Mass-to-Mass
?g
75.5 g
2 C4H10 + 13 O2 → 8 CO2 + 10 H2O
UNKNOWN
g C4H10 =
=
GIVEN
g→mol (G)
Mole Ratio
mol→g (U)
75.5 g CO2
1 mol CO2
2 mol C4H10
44.0 g CO2
58.0 g C4H10
8 mol CO2
1 mol C4H10
24.9 g C4H10
Mass-to-Mass
Mass-to-Mass Conversions
Practice 1: Use the balanced equation for the
combustion of butane. How many grams of
oxygen are necessary to react completely with
217 grams of butane?
217 g
?g
2 C4H10 + 13 O2 → 8 CO2 + 10 H2O
Mass-to-Mass
217 g
?g
2 C4H10 + 13 O2 → 8 CO2 + 10 H2O
UNKNOWN
g O2 =
=
GIVEN
g→mol (G)
Mole Ratio
mol→g (U)
217 g C4H10
1 mol C4H10
13 mol O2
58.0 g C4H10
32.0 g O2
2 mol C4H10
1 mol O2
778 g O2
Mass-to-Mass
Mass-to-Mass Conversions
Practice
2: Balance the following single
replacement reaction.
?g
75.9 g
2 Fe + ____
3 H2O → ____ Fe2O3 + ____
3 H2
____
How many grams of iron must react in order to
produce 75.9 grams of iron (III) oxide?
Because grams are involved for both UNKNOWN and
GIVEN, the molar mass for both must be calculated.
Molar mass:
55.8 g/mol
UNKNOWN
Molar mass:
159.6 g/mol
GIVEN
Mass-to-Mass
?g
75.9 g
2 Fe + 3 H2O → Fe2O3 + 3 H2
UNKNOWN
g Fe =
=
GIVEN
g→mol (G)
Mole Ratio
mol→g (U)
75.9 g Fe2O3
1 mol Fe2O3
2 mol Fe
55.8 g Fe
159.6 g Fe2O3
1 mol Fe2O3
1 mol Fe
53.1 g Fe
HW due 02/06
Textbook Practice, p. 378
#21
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