Stoichiometry Unit V

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StoichiometryUnit V
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I. The Mole

• a) Definition An arbitrary unit used in
  chemistry to unite three measured properties of
  matter
• Mass
• Volume
• of particles

• Similar to a dozen, except instead of 12, its
• 6.02 X 1023 (in scientific notation)

This number is named in honor of Amedeo Avogadro
(1776 1856)
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II. Molar Mass

• a) Definition the mass of 1 mole (in grams)
• Equal to the numerical value of the average
  atomic mass (get from periodic table), or add the
  atoms together for a molecule
• 1 mole of C atoms 12.0 g
• 1 mole of Mg atoms 24.3 g
• 1 mole of O2 molecules 32.0 g

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b) Molar Mass of Compounds

• The molar mass (MM) of a compound is determined
  the same way, except now you add up all the
  atomic masses for the molecule (or compound)
• Ex. Molar mass of CaCl2
• Element of atoms Atomic Mass
• Calcium 1 X 40.08g 40.08g
• Chlorine 2 X 35.45g 70.90g
• Molar Mass of calcium chloride 110.98 g/mol

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Practice

• Calculate the Molar Mass of calcium phosphate
• Formula
• Gram Atomic Mass
• Ca 3 X 40.1
• P 2 X 31.0
• O 8 X 16.0
• Molar Mass

Ca3(PO4)2
120.3 g
62.0 g
128.0 g
120.3g 62.0g 128.0g
310.3 g/mol
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1. Converting Moles to Mass

• Find the molar mass of the substance.

• Multiply the molar mass by the number of moles
  of the substance.

Ex Find the mass of 4.5 moles of Calcium Sulfate
Chemical Name Calcium Sulfate Chemical
Formula CaSO4
Ca 1 X 40.08 40.08 S 1 X 32.06 32.06
O 4 X 15.9994 63.9976
Molar Mass 136.1 g/mol
136.1 g/mol X 4.5 mol 612.6 g
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2. Converting Mass to Moles

• Find the molar mass of the substance.

• Divide the given mass by the molar mass to
  determine the number of moles.

Ex Find the number of moles in 450 g of glucose
(C6H12O6 ).
C 6 X 12.0111 72.0666 H 12 X 1.00794 12.
09528 O 6 X 15.9994 95.9964
Molar Mass 180.16 g/mol
Given Mass/Molecular Mass 450 g 180.16 g/mol
2.5 moles
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III. Molar Volume

• a) Definition the volume of 1 mole of a
  substance (in liters)
• One mole of any gas at STP occupies a volume of
  22.4 liters.
• 1 mole of CO2 (g) molecules 22.4 liters
• 1 mole of N2 (g) molecules 22.4 liters
• 1 mole of O2 (g) molecules 22.4 liters

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1. Converting moles to volume

• Multiply the number of moles of the chemical
  (given) by 22.4.

Practice 2.0 moles of CO2 (g)
____________liters 0.5 moles of O2 (g)
____________liters 4.0 moles of CH4 (g)
________________liters
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2. Converting volume to moles

• Divide the volume of the chemical (given) by
  22.4.

Practice
_____moles of N2 (g) 33.6
liters _____moles of H2 (g)
5.6 liters _____moles of CCl4 (g) 56.0 liters
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IV. Molar Particles

• Definition the of particles
  (atoms/molecules/ions/electrons/etc.) in 1 mole.
• 1 mole of C atoms 6.02 X 1023 atoms
• 2 mole of Mg atoms 12.04 X 1023 atoms
• 2 mole of O2 molecules 1.204 X 1024
  molecules
• 1 mole of C6H12O6 6.02 X 1023 molecules of
  glucose
• 6 moles of C atoms
• 12 moles of H atoms
• 6 moles of O atoms
• (6) 6.02 X 1023 atoms of carbon
• 36.12 X 1023 atoms of carbon
• (12) 6.02 X 1023 atoms of hydrogen
• 72.24 X 1023 atoms of hydrogen

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b) Conversions
1. Moles to Particles

• Multiply the number of moles (given) by 6.02 x
  1023

Ex 1 2.5 moles of CO2 __________________mole
cules
6.02 X 1023 molecules x molecules
1 mole
2.5 moles
X 15.05 x 1023 molecules of CO2
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2. Particles to Moles

• Divide the number of particles (given) by 6.02 x
  1023.

Ex 1.806 X 1024 molecules of CO2 is equal to
_____________________moles.
6.02 X 1023 molecules 1.806 x 1024 molecules
1 mole
x moles
X 3 moles of CO2
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V. Conversions
a) Mass to Volume/Volume to Mass

• One must first convert the given measurement to
  the mole.

• After converting to the mole, convert to the
  desire measurement.

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Ex At STP, 88 grams of CO2 will occupy a volume
of________liters.
1. Find the molar mass of CO2.
2. Find the number of moles represented by the
given mass.
3. Determine the number of liters represented by
the of moles.
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1. Find the molar mass of CO2 .
CO2
C 1 X 12 12 O 2 X 16 32 44
grams/mole
2. Find the number of moles represented by the
given mass.
88 g
2 moles
Given Mass


of moles
Molecular Mass
44 g
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3. Determine the number of liters represented by
the of moles.
22.4 liters x liters
1 mole 2 moles
X 44.8 liters of CO2
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Ex Find the mass of 56.0 liters of SO2 (g) at
STP.
1. Determine the molar mass of SO2.
2. Convert from liters to moles.
3. Calculate the mass from the determined of
moles and the molar mass.
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Ex Find the mass of 56.0 liters of SO2 (g) at
STP.
1. Determine the molar mass of SO2.
SO2
S 1 X 32 32 O 2 X 16 32 64 grams
2. Convert from liters to moles.
1 mole
known
56.0 liters
22.4 liters

X 2.5 moles
1 mole
X moles
3. Calculate the mass from the determined of
moles and the molar mass.
64 grams x grams
X 160 grams
1 mole 2.5 moles
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b) Converting Mass to Particles/Particles to Mass

• One must first convert the given measurement to
  the mole.

• After converting to the mole, convert to the
  desire measurement.

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Ex 1 88g of carbon dioxide
________________molecules
1. Find the gram formula mass of CO2
2. Find the number of moles represented by the
determined mass
3. Determine the number of particles (molecules)
based upon the calculated number of moles and the
known ratio of moles to particles
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Ex 1 88g of carbon dioxide
________________molecules
1. Find the gram formula mass of CO2 C
1 X 12 12 O 2 X 16 32 44g/mol
CO2 44 g/mol
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2. Find the number of moles represented by the
determined mass
88g of CO2 44g of CO2 x moles 1 mole
X 2 moles of carbon dioxide
3. Determine the number of particles (molecules)
based upon the calculated number of moles and the
known ratio of moles to particles
6.02 X 1023 x molecules 1 mole
2 moles
X 12.04 X 1023 molecules or 1.204 X 1024
molecules
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Ex 2 9.03 X 1023 molecules of SO2
________________grams
1. Find the gram formula mass of SO2
2. Find the number of moles represented by the
given number of particles using the ratio 6.02 x
10 23 particles is to one mole.
3. Determine the number of grams be multiplying
the number of moles (calculated in step 2) by the
gram formula mass
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Ex 2 9.03 X 1023 molecules of SO2
________________grams
1. Find the gram formula mass of SO2 S
1 X 32 32 O 2 X 16 32 64g/mol
SO2 64 g/mol
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2. Find the number of moles represented by the
given number of particles using the ratio 6.02 x
10 23 particles is to one mole
6.02 X 1023 9.03 x 10 23 molecules 1 mole
x moles
X 1.5 moles of sulfur dioxide
3. Determine the number of grams be multiplying
the number of moles (calculated in step 2) by the
gram formula mass
64 g of SO2 x g of SO2 1 mole 1.5
moles
X 96 grams of SO2
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Gram Atomic (Formula) Mass
Multiply by atomic/molar mass from periodic table
Divide by atomic/molar mass from periodic table
Moles
Multiply By 22.4
Multiply by 6.02 X 1023
Divide By 22.4
Divide by 6.02 X 1023
Volume
Atoms or Molecules
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• 1. How many atoms of Cu are present in 35.4 g of
  Cu?

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• 2. What is the mass (in grams) of
• 1.20 X 1024 molecules of glucose (C6H12O6)?

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VI. Chemical Equations

• A reaction indicates the concentration of
  reactants that are needed to get a product.
• The reactions must be balanced to establish
  ratios.
• Example
• 2 Na Cl2 ? 2 NaCl
• This reaction tells us that by mixing 2 moles of
  sodium atoms with 1 mole of chlorine molecules we
  will get 2 moles of sodium chloride
• What if we wanted 4 moles of NaCl? 10 moles? 50
  moles?

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Practice

• Write the balanced reaction for hydrogen gas
  reacting with oxygen gas.
• 2 H2 O2 ? 2 H2O
• How many moles of reactants are needed?
• What if we wanted 4 moles of water?
• What if we had 3 moles of oxygen, how much
  hydrogen would we need to react and how much
  water would we get?
• What if we had 50 moles of hydrogen, how much
  oxygen would we need and how much water produced?

2 mol H2 1 mol O2
4 mol H22 mol O2
6 mol H2, 6 mol H2O
25 mol O2, 50 mol H2O
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a) Mole Ratios

• These mole ratios can be used to calculate the
  moles of one chemical from the given amount of a
  different chemical
• Example How many moles of chlorine is needed to
  react with 5 moles of sodium (without any sodium
  left over)?
• 2 Na Cl2 ? 2 NaCl

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1. Mole-Mole Conversions

• How many moles of sodium chloride will be
  produced if you react 2.6 moles of chlorine gas
  with an excess (more than you need) of sodium
  metal?
• 2 Na Cl2 ? 2 NaCl

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2. Mole-Mass Conversions

• Most of the time in chemistry, the amounts are
  given in grams instead of moles
• We still go through moles and use the mole ratio,
  but now we also use molar mass to get to grams
• Example How many grams of chlorine are required
  to react completely with 5.00 moles of sodium to
  produce sodium chloride?
• 2 Na Cl2 ? 2 NaCl

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Practice

• Calculate the mass in grams of Iodine required to
  react completely with 0.50 moles of aluminum.
• 2 Al 3 I2 ? 2 AlI3

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3. Mass-Mole

• We can also start with mass and convert to moles
  of product or another reactant
• We use molar mass and the mole ratio to get to
  moles of the compound of interest
• Calculate the number of moles of ethane (C2H6)
  needed to produce 10.0 g of water
• 2 C2H6 7 O2 ? 4 CO2 6 H20

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Practice
Calculate how many moles of oxygen are required
to make 10.0 g of aluminum oxide 4 Al 3 O2 ? 2
Al2O3
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4. Mass-Mass Conversions

• Most often we are given a starting mass and want
  to find out the mass of a product we will get
  (called theoretical yield) or how much of another
  reactant we need to completely react with it.
• Now we must go from grams to moles, mole ratio,
  and back to grams of compound we are interested in

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Mass-Mass Conversion

• Ex. Calculate how many grams of ammonia are
  produced when you react 2.00g of nitrogen with
  excess hydrogen.
• N2 3 H2 ? 2 NH3

2.00g N2 1 mol N2 2 mol NH3 17.06g NH3
28.02g N2 1 mol N2 1 mol
NH3
2.4 g NH3
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Practice

• How many grams of calcium nitride are produced
  when 2.00 g of calcium reacts with an excess of
  nitrogen?

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5. Mass-Volume Conversions

• In calculating mass-volume questions, one must
  first convert the given mass to moles (calculate
  molar mass).

• Then, establish a ratio knowing the number of
  liters one mole of a gas occupies at STP and the
  number of calculated moles.

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Ex Mass-Volume Conversions
2 C8H18 25 O2 ? 16 CO2 18 H2O

• How many liters of CO2 at STP are produced by
  the combustion of 342 grams of octane according
  the above equation?

• How many grams of O2 at STP are required to
  completely combust 89.6 liters of octane
  according the above equation?

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VIII. Percent Composition, Molecular and
Empirical Formulas
a) Percent Composition the composition of a
compound in terms of the percentage of each
component present with respect to the whole.
Ex 1 What is the percent composition by mass of
the elements in sodium sulfate?

• Find the formula of sodium sulfate.

• Find the molar mass of sodium sulfate.

• Using the formula for composition by mass
  (reference table T), calculate the composition
  of each element in the compound.

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• Sodium Sulfate - Na2SO4

Na2SO4
Na 2 X 23 46 S 1 X 32 32 O 4 X 16 64

• Molar Mass 142 g/mol

• composition part/whole X 100

Na 46/142 X 100 32.4 S 32/142 X 100
22.5 O 64/142 X 100 45.1 100
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Ex 2 What is the percent composition by mass of
the elements in Potassium permanganate?
Ex 3 What is the percent composition by mass of
the water in Copper (II) sulfate pentahydrate?
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b) Types of Formulas

• Molecular Formulas
• Represents the type and number of elements in a
  covalent compound.
• The subscripts do not need to be reduced in a
  molecular formula.

Ex C2H4 Ethene C6H12O6 Glucose
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2. Empirical Formulas

• Represents the lowest whole number ratio of
  atoms in a compound.

• For ionic compounds, the formula is always
  empirical. This is due to the fact that one must
  reduce the subscripts in an ionic compound to the
  lowest whole number ratio.

• For covalent compounds (molecules), reducing its
  molecular formula to an empirical formula only
  shows the ratio of elements within the compound.

Ex C6H12O6 Molecular Formula
C1H2O1 Empirical Formula
Ratio CHO 121
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c) Determining empirical formulas from percent
composition

• One can determine the chemical formula for a
  substance if given the percent composition of its
  elements.

Ex What is the empirical formula of a compound
that consists of 58.80 barium, 13.75 sulfur,
and 27.45 oxygen by mass?
1. Convert the percentages to grams.
Ba 58.80 58.80g S 13.75 13.75g O 27.45 27.45
g
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2. Convert from Mass to Moles
Ba 137g 58.80g 0.43 mol 1 mol x mol S
32g 13.75g 0.43 mol 1 mol x mol O
16g 27.45g 1.71 mol 1 mol x mol
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3. Simplify the mole ratios to the lowest whole
number
Ba 0.43/0.43 1 S 0.43/0.43 1 O
1.71/0.43 4
4. The simplified ratio of elements is equal to
the subscripts of the atoms within the compounds.

Ba1S1O4 or BaSO4
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Practice Given the following compositions,
determine the formulas for the following
compounds.

• 27.3 carbon 72.7 oxygen
• 30.43 nitrogen 69.57 oxygen

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c) 37.6 carbon 12.5 hydrogen 49.9
oxygen d) 32.4 sodium 22.5 sulfur 45.1
oxygen
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c) Determining molecular formulas from percent
composition
KEY IDEA The procedures for determining the
molecular formula of a compound (given its
elements composition) is the same as finding
its empirical formula with one exception.

• YOU MUST DETERMINE THE EMPIRICAL MASS OF THE
  COMPOUND.

• YOU MUST DIVIDE THE MOLECULAR MASS BY THE
  EMPIRICAL MASS TO DETERMINE THE ACTUAL NUMBER OF
  MOLES OF ATOMS WITHIN THE COMPOUND.

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• Ex 1 In analyzing 30g of a compound, it was
  determined to consist of 80 carbon and 20
  hydrogen by mass.
• Determine the compounds empirical and molecular
  formulas.

1. Convert percentages to grams
80 C 80g C 20 H 20g H
2. Convert Mass to Moles
80g C/12g per mole 6.66 moles 20g H/1g per
mole 20 moles
3. Simplify mole ratios
C 6.66/6.66 1 H 20/6.66 3
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4. The mole ratios represent the subscripts for
the empirical formula
C1H3
5. Calculate the empirical mass
C 1 X 12 12 H 3 X 1 3 Empirical
Mass 15g/mole
6. Divide the molecular mass (given) by the
empirical mass to find the formula unit
Molecular Mass 30 2 Empirical Mass 15
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7. Distribute the formula unit
2 C1H3 C2H6
C2H6
MOLECULAR FORMULA
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IX. Balancing Equations

• To abide by the Law of Conservation of Matter
  and Energy, the number of atoms within the
  reactants of a chemical equation must be equal to
  the number of atoms found within the products.

a) Chemical Equation
- Indicates the concentration (moles) of
reactants needed to form products with a specific
concentration ( of moles).
Reactants
Products
1C1H4 (g) 2O2(g) ? 1C1O2 (g)
2H2O1 (g)
THE NUMBERS IN RED ARE CALLED COEFFICIENTS. THEY
INDICATE THE NUMBER OF MOLES OF THE CHEMICALS.
COEFFICIENTS CAN BE USED TO ESTABLISH MOLE
RATIOS.
THE NUMBERS IN BLUE ARE CALLED SUBSCRIPTS. THEY
INDICATE THE NUMBER OF ATOMS FOUND WITHIN THE
COMPOUND.
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b) Balancing Equations
The combustion of propane (C3H8) in the presence
of molecular oxygen yields carbon dioxide and
water vapor.
1. Write out the formulas for all chemicals
involved in the reaction
Propane C3H8 Molecular Oxygen O2 Carbon
Dioxide CO2 Water H2O
Reactants
Products

• Arrange the reactants on the left side and the
  products on the right side of the yield sign

__C3H8 (g) __O2 (g) ? __CO2 (g)
__H2O (g)
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3. Create a checklist of the types and amounts
of atoms present in the reactants and products
__C3H8 (g) __O2 (g) ? __CO2 (g)
__H2O (g)
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1
3
4
C H O
3 8 2
1 2 3
X 3
X 8
X 7
X 10
10 X
4. Manipulate the coefficient to balance the
of atoms in the reactants and products.
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X. Limiting Reactant

• Many times in chemistry the actual ratio of
  reactants is different from the theoretical one.
• This results in having an excess of a given
  reactant. (there is an overabundance of it).
• The other reactant limits how much product we
  get. Once it runs out, the reaction s.
  This is called the limiting reactant.

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Limiting Reactant

• To find the limiting reagent, one must calculate
  the number of moles of reactants and products.
• You then must calculate how much of a product can
  be produced from each of the reactants to
  determine which reactant is the limiting one.
• The lower amount of a product is the correct
  answer.
• The reactant that makes the least amount of
  product is the limiting reactant. Once you
  determine the limiting reactant, you should
  ALWAYS start with it!
• Be sure to pick a product! You cant compare to
  see which is greater and which is lower unless
  the product is the same!

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Limiting Reactant Example
LimitingReactant

• 10.0g of aluminum reacts with 35.0 grams of
  chlorine gas to produce aluminum chloride. Which
  reactant is limiting, which is in excess, and how
  much product is produced?
• 2 Al 3 Cl2 ? 2 AlCl3
• Start with Al
• Now Cl2

10.0 g Al 1 mol Al 2 mol AlCl3 133.5
g AlCl3 27.0 g Al 2 mol Al
1 mol AlCl3
49.4g AlCl3
35.0g Cl2 1 mol Cl2 2 mol AlCl3
133.5 g AlCl3 71.0 g Cl2 3
mol Cl2 1 mol AlCl3
43.9g AlCl3
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LR Example Continued

• We get 49.4g of aluminum chloride from the given
  amount of aluminum, but only 43.9g of aluminum
  chloride from the given amount of chlorine.
  Therefore, chlorine is the limiting reactant.
  Once the 35.0g of chlorine is used up, the
  reaction comes to a complete .

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Limiting Reactant Practice

• 15.0 g of potassium reacts with 15.0 g of iodine.
  Calculate which reactant is limiting and how
  much product is made.

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Finding the Amount of Excess

• By calculating the amount of the excess reactant
  needed to completely react with the limiting
  reactant, we can subtract that amount from the
  given amount to find the amount of excess.
• Can we find the amount of excess potassium in the
  previous problem?

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Finding Excess Practice

• 15.0 g of potassium reacts with 15.0 g of iodine.
  2 K I2 ? 2 KI
• We found that Iodine is the limiting reactant,
  and 19.6 g of potassium iodide are produced.

15.0 g I2 1 mol I2 2 mol K 39.1 g
K 254 g I2 1 mol I2
1 mol K
4.62 g K USED!
15.0 g K 4.62 g K 10.38 g K EXCESS
Given amount of excess reactant
Amount of excess reactant actually used
Note that we started with the limiting reactant!
Once you determine the LR, you should only start
with it!
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Limiting Reactant Recap

• You can recognize a limiting reactant problem
  because there is MORE THAN ONE GIVEN AMOUNT.
• Convert ALL of the reactants to the SAME product
  (pick any product you choose.)
• The lowest answer is the correct answer.
• The reactant that gave you the lowest answer is
  the LIMITING REACTANT.
• The other reactant(s) are in EXCESS.
• To find the amount of excess, subtract the amount
  used from the given amount.
• If you have to find more than one product, be
  sure to start with the limiting reactant. You
  dont have to determine which is the LR over and
  over again!