The concept of limiting reactants
FOR THE MIDTERM EXAM
This should have been included on the study guide...
For a given chemical reaction, know how to write:
• molecular (overall) equation
• complete ionic equation
• net ionic equation
In some chemical reactions, all reagents are present in the exact
amounts required to completely react with one another.
Example: In a lab experiment, ammonia is produced by reacting
6.05 g of hydrogen gas (3 moles) with 28.02 g of nitrogen gas (1 mole)
3 H2 + N2
2 NH3
In this case, hydrogen and nitrogen are said to react in
stoichiometric amounts.
The concept of limiting reactants
The concept of limiting reactants
But in many cases, a chemical reaction will take place under conditions
where one (or more) of the reactants is present in excess
But in many cases, a chemical reaction will take place under conditions
where one (or more) of the reactants is present in excess
-- i.e., there is more than enough of that reactant available for the
reaction to proceed
Example: Combustion of 85.0 g of propane in air
C3H8(g) + 5 O2(g)
-- i.e., there is more than enough of that reactant available for the
reaction to proceed
The limiting reactant is the reactant that is not present in excess
-- the limiting reactant will be used up first (the reaction will stop
when the limiting reactant is depleted)
3 CO2(g) + 4 H2O(g)
There is more than enough oxygen available in the air to react
with all of the propane
-- the reaction will proceed until all of the 85.0 g of propane
has been consumed
-- the limiting reactant therefore limits the amount of product that
can be formed by the reaction
In the previous example, propane was the limiting reactant
(oxygen was present in excess)
C3H8(g) + 5 O2(g)
Another food analogy…
3 CO2(g) + 4 H2O(g)
Another food analogy…
Recipe for a grilled cheese sandwich:
Recipe for a grilled cheese sandwich:
Two slices bread and one slice of cheese gives one sandwich
Two slices bread and one slice of cheese gives one sandwich
Balanced equation:
Balanced equation:
2
+
Δ
If you have 10 slices of bread and 4 slices of cheese, how many
sandwiches can you make?
• enough bread for ( 10 / 2 ) = 5 sandwiches
• enough cheese for ( 4 / 1 ) = 4 sandwiches
• you can only make 4 sandwiches before the cheese is used up
• cheese is the limiting reactant
2
+
Δ
If you have 8 slices of bread and 6 slices of cheese, how many
sandwiches can you make?
• enough bread for ( 8 / 2 ) = 4 sandwiches
• enough cheese for ( 6 / 1 ) = 6 sandwiches
• you can only make 4 sandwiches before the bread is used up
• bread is the limiting reactant
Limiting reactants
Limiting reactants
Chemistry example:
Chemistry example:
Hydrogen and chlorine gas combine to form hydrogen chloride:
H2
+
Cl2
Hydrogen and chlorine gas combine to form hydrogen chloride:
2 HCl
H2
If 4 moles of hydrogen reacts with 3 moles of chlorine, how many moles
of HCl will be formed?
How much
HCl can be
formed from
4 mol H2 ?
How much
HCl can be
formed from
3 mol Cl2 ?
( 4 mol H2 )
n
4 mol H2
2 mol HCl
=
1 mol H2
n
3 mol Cl2
1 mol Cl2
• At this point, the Cl2 will have been completely consumed and the
reaction stops (chlorine is the limiting reactant)
( 3 mol Cl2 )
• 1 mole of H2 will remain unreacted (hydrogen is present in excess)
• 6 moles of HCl will have been formed
n = 6 mol HCl
Procedure for identifying the limiting reactant
Chemistry example:
1. Calculate the amounts of product that can be formed from
each of the reactants
Hydrogen and chlorine gas combine to form hydrogen chloride:
+
Cl2
2 HCl
If 4 moles of hydrogen reacts with 3 moles of chlorine, how many moles
of HCl will be formed?
H
H
Cl
Cl
H
Cl
H
Cl
H
H
Cl
Cl
H
Cl
H
Cl
H
H
Cl
Cl
H
If 4 moles of hydrogen reacts with 3 moles of chlorine, how many moles
of HCl will be formed?
3 moles of H2 will react with 3 moles of Cl2
Limiting reactants
H2
2 HCl
6 mol HCl can be formed from 3 mol of Cl2
2 mol HCl
=
Cl2
8 mol HCl can be formed from 4 mol of H2
( 4 mol H2 )
n = 8 mol HCl
( 3 mol Cl2 )
+
2. Determine which reactant gives the least amount of product
-- this is the limiting reactant
Do not just compare the numbers of moles of reactants -- you
must also account for the ratios in which the reactants combine
3. To find the amount of the non-limiting reactant remaining
after the reaction:
-- calculate the amount of the non-limiting reactant required
to react completely with the limiting reactant
Cl
H
H
Cl
H
-- subtract this amount from the starting quantity of the
non-limiting reactant
Limiting reactant
Limiting reactant
How many grams of silver bromide (AgBr) can be formed when solutions
containing 50.0 g of MgBr2 and 100. g of AgNO3 are mixed together?
Equation:
MgBr2 + 2 AgNO3
2 AgBr + Mg(NO3)2
Convert the amounts of reactants from grams to moles
50.0 g MgBr2 ( 1 mol MgBr2 / 184.11 g MgBr2 )
=
0.272 mol MgBr2
How many grams of silver bromide (AgBr) can be formed when solutions
containing 50.0 g of MgBr2 and 100. g of AgNO3 are mixed together?
Equation:
MgBr2 + 2 AgNO3
Amount of AgBr that can be produced by 0.272 mol of MgBr2:
n
( 0.272 mol MgBr2 )
0.272 mol MgBr2
n
100. g AgNO3 ( 1 mol AgNO3 / 169.91 g AgNO3 )
=
0.589 mol AgNO3
2 AgBr + Mg(NO3)2
=
=
0.544 mol AgBr
2 mol AgBr
1 mol MgBr2
( 0.272 mol MgBr2 )
MgBr2 is the limiting reactant
Amount of AgBr that can be produced by 0.589 mol of AgNO3 :
( 0.589 mol AgNO3 )
n
n
0.589 mol AgNO3
=
=
0.589 mol AgBr
2 mol AgBr
2 mol AgNO3
( 0.589 mol AgNO3 )
Limiting reactant
Limiting reactant
How many grams of silver bromide (AgBr) can be formed when solutions
containing 50.0 g of MgBr2 and 100. g of AgNO3 are mixed together?
Equation:
MgBr2 + 2 AgNO3
2 AgBr + Mg(NO3)2
Amount of AgBr that can be produced by 0.272 mol of MgBr2:
( 0.272 mol MgBr2 )
n
0.272 mol MgBr2
n
=
=
How many grams of silver bromide (AgBr) can be formed when solutions
containing 50.0 g of MgBr2 and 100. g of AgNO3 are mixed together?
Equation:
MgBr2 + 2 AgNO3
How many grams of the excess reactant remain unreacted?
2 mol AgBr
1 mol MgBr2
( 0.272 mol MgBr2 )
MgBr2 is the limiting reactant
0.544 mol AgBr
Convert moles of AgBr to grams of AgBr:
0.544 mol AgBr ( 187.8 g AgBr / 1 mol AgBr )
= 102 g AgBr
Since MgBr2 is the limiting reactant, all 50.0 g of MgBr2 will be used up.
Calculate how much AgNO3 is required to react with 50.0 g MgBr2:
Remember that in the previous steps, we calculated that 50.0 g of MgBr2 is
equal to 0.272 moles of MgBr2.
We then need to determine how many moles of AgNO3 are required to
react with 0.272 moles of MgBr2, then convert from moles to grams.
Yields
Limiting reactant
How many grams of silver bromide (AgBr) can be formed when solutions
containing 50.0 g of MgBr2 and 100. g of AgNO3 are mixed together?
Equation:
2 AgBr + Mg(NO3)2
MgBr2 + 2 AgNO3
Theoretical yield -- the calculated amount (mass) of product that can be
obtained from a given amount of reactant based on the balanced chemical
equation for a reaction
2 AgBr + Mg(NO3)2
Actual yield -- the amount of product actually obtained from a reaction
How many grams of the excess reactant remain unreacted?
( 0.272 mol MgBr2 )
n
0.272 mol MgBr2
=
2 mol AgNO3
1 mol MgBr2
The actual yield observed for a reaction is almost always less than the
theoretical yield due to:
( 0.272 mol MgBr2 )
n = 0.544 mol AgNO3 ( 169.9 g AgNO3 / 1 mol AgNO3 )
= 92.4 g AgNO3
• side reactions that form other products
• incomplete / reversible reactions
• loss of material during handling and transfer from one vessel to another
The actual yield can never be greater than the theoretical yield
When all 50.0 g of MgBr2 has reacted, 92.4 g of AgNO3 will have been
consumed. The amount of unreacted AgNO3 left over is given by:
actual yield
Percent yield
= 100 x
theoretical yield
100.0 g – 92.4 g = 7.6 g AgNO3
Yields
Yields
Silver bromide was prepared by reacting 200.0 g of magnesium
bromide and an excess amount of silver nitrate.
Silver bromide was prepared by reacting 200.0 g of magnesium
bromide and an excess amount of silver nitrate.
Equation:
Equation:
MgBr2 + 2 AgNO3
Mg(NO3)3 + 2 AgBr
a) What is the theoretical yield of silver bromide?
Mg(NO3)3 + 2 AgBr
a) What is the theoretical yield of silver bromide?
Step 1: Convert the amount of MgBr2 from grams to moles
200.0 g MgBr2 ( 1 mol MgBr2 / 184.1 g MgBr2 )
MgBr2 + 2 AgNO3
= 1.086 mol MgBr2
Step 2: Determine how many moles of AgBr can be formed from
this amount of MgBr2 ( i.e., 1.086 moles)
( 1.086 mol MgBr2 )
n
1.086 mol MgBr2
n
=
=
2 mol AgBr
1 mol MgBr2
( 1.086 mol MgBr2 )
2.172 mol AgBr
Step 3: Convert from moles to grams
2.172 mol AgBr ( 187.8 g AgBr / 1 mol AgBr )
This is the theoretical yield
= 407.9 g AgBr
Yields
SOLUTIONS
Silver bromide was prepared by reacting 200.0 g of magnesium
bromide and an excess amount of silver nitrate.
Equation:
MgBr2 + 2 AgNO3
Mg(NO3)3 + 2 AgBr
b) Calculate the percent yield if 375.0 g of silver bromide was
obtained from the reaction.
theoretical yield = 407.9 g AgBr
percent yield
= 100 x
percent yield = 100 x
actual yield
theoretical yield
375.0 g
407.9 g
=
91.93 %
Concentration of solutions
Concentration units based on
the number of moles of solute
concentration -- the amount of solute dissolved in a given
quantity of solvent or solution
There are many different types of concentration units:
molarity
molarity -- number of moles of solute per liter of solution
we will focus on this
mass %
volume %
mass/volume %
Molarity (M ) =
moles of solute
liters solution
parts per million (ppm)
parts per billion (ppb)
mole fraction
molality (not to be confused with molarity)
Preparation of a 1 molar solution
What is the molarity of a solution made by dissolving 2.00 g of
potassium chlorate in enough water to make 150. ml of solution?
Step 1: Start with the definition of molarity:
Molarity (M ) =
moles of solute
liters solution
Step 2: Determine the number of moles of solute
Molar mass of KClO3 = 39.10 + 35.45 + 3(16.00) = 122.6 g / mol
2.00 g KClO3
1 mole KClO3
122.6 g KClO3
= 0.0163 moles KClO3
What is the molarity of a solution made by dissolving 2.00 g of
potassium chlorate in enough water to make 150. ml of solution?
moles of solute
Molarity (M ) =
Molarity (M ) =
liters solution
Molarity (M ) =
1 liter
=
1000 ml
moles of solute
liters solution
Step 4: Plug values into molarity equation
Step 3: Determine the number of liters of solution
150. ml
What is the molarity of a solution made by dissolving 2.00 g of
potassium chlorate in enough water to make 150. ml of solution?
0.0163 moles KClO3
0.150 L
0.150 L
Molarity (M ) = 0.109 moles KClO3 / L = 0.109 M KClO3
How many grams of potassium hydroxide are required to
prepare 600. ml of 0.450 M KOH solution?
How many grams of potassium hydroxide are required to
prepare 600. ml of 0.450 M KOH solution?
Step 1: Start with the definition of molarity:
Molarity (M ) =
moles of solute
Molarity (M ) =
liters solution
moles of solute
liters solution
Step 3: Plug known values into molarity equation and solve for unknown
(moles of solute)
Step 2: Determine the number of liters of solution
0.450 M KOH =
600. ml
1 liter
=
0.450 moles KOH
1 L solution
=
x moles of KOH
0.600 L
0.600 L
1000 ml
How many grams of potassium hydroxide are required to
prepare 600. ml of 0.450 M KOH solution?
moles of solute
Molarity (M ) =
Molarity (M ) =
liters solution
Step 3: Plug known values into molarity equation and solve for unknown
(moles of solute)
(0.600 L)
0.450 moles KOH
1L
=
x moles of KOH
0.600 L
How many grams of potassium hydroxide are required to
prepare 600. ml of 0.450 M KOH solution?
moles of solute
liters solution
Step 4: Convert moles KOH to grams KOH
Molar mass of KOH = 39.10 + 16.00 + 1.008 = 56.11 g / mol
(0.600 L)
0.270 moles KOH
56.11 g KOH
1 mole KOH
0.270 moles KOH = x
= 15.1 g KOH
Calculate the number of moles of nitric acid in 325 ml of
16 M HNO3
Step 1: Start with the definition of molarity:
Molarity (M ) =
Step 1: Start with the definition of molarity:
moles of solute
Molarity (M ) =
liters solution
Step 2: Plug known values into molarity equation and solve for unknown
(moles of solute)
16 M HNO3 =
Calculate the number of moles of nitric acid in 325 ml of
16 M HNO3
16 moles HNO3
1L
=
liters solution
Step 2: Plug known values into molarity equation and solve for unknown
(moles of solute)
x moles of HNO3
0.325 L
moles of solute
(0.325 L)
16 moles HNO3
1L
=
x moles of HNO3
0.325 L
(0.325 L)
5.2 moles HNO3 = x
Dilutions
Dilution: Reducing the concentration of a solution by adding more
solvent to the solution
NO3-
• More solvent is added:
Na+
-- volume of the solution increases
• No additional solute is added
-- number of moles of solute stays the same
Net result: The molarity of the solution decreases
Molarity (M) =
moles of solute (unchanged)
liters solution
Moles = 1.0 mol
Volume = 1.0 L
Molarity =
1.0 mol
1.0 L
= 1.0 M
NaNO3 solution
Calculate the molarity of a solution prepared by diluting
125 ml of 0.400 M HCl to a final solution volume of 1.00 L.
For any dilution problem, remember that the number of moles of
solute remains the same:
NO3Na+
moles of solute (before) = moles of solute (after)
Moles = 1.0 mol
Based on the definition of molarity, this can be expressed as:
Volume = 2.0 L
Molarity =
M1 V 1 = M 2 V 2
1.0 mol
2.0 L
= 0.5 M
• Solution volume is doubled
Where M1 is the molarity of the original solution
V1 is the volume of the original solution
• Moles of solute remain the same
M2 is the molarity of the diluted solution
• Solution concentration is halved
V2 is the volume of the diluted solution
Calculate the molarity of a solution prepared by diluting
125 ml of 0.400 M HCl to a final solution volume of 1.00 L.
The
Gaseous
State of
Matter
M1 V 1 = M 2 V 2
M1 = 0.400 M
M2 = ?
V1 = 125 ml = 0.125 L
V2 = 1.00 L
(i.e., gases)
(0.400 M) (0.125 L) = (M2) (1.00 L)
1.00 L
1.00 L
0.0500 M = M2
Properties of solids, liquids and gases
liquid
gas
• indefinite shape
takes shape
of container
• indefinite shape
takes shape
of container
• definite volume
• indefinite volume
fills volume
of container
solid
• definite shape
• definite
volume
Kinetic Molecular Theory
General theory to explain the behavior and properties of gasses
Properties of solids, liquids and gases
solid
liquid
gas
• atoms/molecules are
tightly packed
• atoms/molecules are in
close contact
• atoms/molecules have
overcome attractive forces
• strong attractive
forces between
atoms/molecules
• attractive forces between
atoms/molecules
• atoms/molecules are
relatively far apart and
move independently
BUT atoms/molecules
can move freely
Homework
Assumptions
1. Gases consist of tiny particles
Homework Assignment 4: (Due Oct 14 at 9:00 pm)
2. Distance between gas particles is large
compared to size of particles (i.e., most
of gas volume is empty space)
3. Gas particles have no attraction for one
another
4. Gas particles move in straight lines in all
directions, colliding frequently with one
another and the walls of the container
5. No energy is lost during collisions
6. The average kinetic energy of gas
particles is the same for all gases at
the same temperature
An ideal gas behaves exactly
as specified by the theory
No truly ideal gases exist, but
many real gases approach
ideal behavior at non-extreme
temperatures and pressures