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The concept of limiting reactants
The concept of limiting reactants
In some chemical reactions, all reagents are present in the exact
amounts required to completely react with one another.
But in many cases, a chemical reaction will take place under conditions
where one (or more) of the reactants is present in excess
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
-- 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
2 NH3
C3H8(g) + 5 O2(g)
In this case, hydrogen and nitrogen are said to react in
stoichiometric amounts.
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 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
-- i.e., there is more than enough of that reactant available for the
reaction to proceed
Another food analogy…
Recipe for a grilled cheese sandwich:
Two slices bread and one slice of cheese gives one sandwich
Balanced equation:
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)
-- 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)
3 CO2(g) + 4 H2O(g)
2
+
!
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
Limiting reactants
Another food analogy…
Recipe for a grilled cheese sandwich:
Chemistry example:
Two slices bread and one slice of cheese gives one sandwich
Hydrogen and chlorine gas combine to form hydrogen chloride:
H2
Balanced equation:
2
!
+
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
+
Cl2
2 HCl
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
( 4 mol H2 )
n = 8 mol HCl
( 3 mol Cl2 )
n
3 mol Cl2
2 mol HCl
=
1 mol Cl2
( 3 mol Cl2 )
n = 6 mol HCl
Limiting reactants
Limiting reactants
Chemistry example:
Chemistry example:
Hydrogen and chlorine gas combine to form hydrogen chloride:
Hydrogen and chlorine gas combine to form hydrogen chloride:
H2
+
Cl2
2 HCl
If 4 moles of hydrogen reacts with 3 moles of chlorine, how many moles
of HCl will be formed?
8 mol HCl can be formed from 4 mol of H2
H2
+
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
Cl
H
Cl
H
H
6 mol HCl can be formed from 3 mol of Cl2
3 moles of H2 will react with 3 moles of Cl2
• At this point, the Cl2 will have been completely consumed and the
reaction stops (chlorine is the limiting reactant)
• 1 mole of H2 will remain unreacted (hydrogen is present in excess)
• 6 moles of HCl will have been formed
Limiting reactant
Procedure for identifying the limiting reactant
1. Calculate the amounts of product that can be formed from
each of the reactants
2. Determine which reactant gives the least amount of product
-- this is the 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
Do not just compare the numbers of moles of reactants -- you
must also account for the ratios in which the reactants combine
50.0 g MgBr2 ( 1 mol MgBr2 / 184.11 g MgBr2 )
=
0.272 mol MgBr2
100. g AgNO3 ( 1 mol AgNO3 / 169.91 g AgNO3 )
=
0.589 mol AgNO3
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
-- 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
Amount of AgBr that can be produced by 0.272 mol of MgBr2:
n
( 0.272 mol MgBr2 )
0.272 mol MgBr2
n
=
=
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
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
=
=
0.544 mol AgBr
2 mol AgBr
1 mol MgBr2
MgBr2 is the limiting reactant
Convert moles of AgBr to grams of AgBr:
2 mol AgBr
2 mol AgNO3
( 0.589 mol AgNO3 )
( 0.272 mol MgBr2 )
0.544 mol AgBr ( 187.8 g AgBr / 1 mol AgBr )
= 102 g AgBr
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
How many grams of the excess reactant remain unreacted?
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:
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
How many grams of the excess reactant remain unreacted?
( 0.272 mol MgBr2 )
n
0.272 mol MgBr2
=
2 mol AgNO3
1 mol MgBr2
( 0.272 mol MgBr2 )
n = 0.544 mol AgNO3 ( 169.9 g AgNO3 / 1 mol AgNO3 )
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.
MgBr2 + 2 AgNO3
= 92.4 g AgNO3
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:
100.0 g – 92.4 g = 7.6 g AgNO3
Yields
Yields
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
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
Actual yield -- the amount of product actually obtained from a reaction
The actual yield observed for a reaction is almost always less than the
theoretical yield due to:
• 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
actual yield
Percent yield
= 100 x
theoretical yield
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 )
= 1.086 mol MgBr2
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?
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
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
( 1.086 mol MgBr2 )
percent yield
= 100 x
actual yield
theoretical yield
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
percent yield = 100 x
= 407.9 g AgBr
Chemical Reactions
375.0 g
407.9 g
=
91.93 %
Classes of chemical reactions
1. Combination reactions
2. Decomposition reactions
3. Precipitation reactions
4. Acid-base neutralization reactions
5. Oxidation-reduction reactions
Types of chemical reactions
Combination reaction -- two reactants combine to give one product
Types of chemical reactions
Decomposition reaction -- a single substance is broken down into two or more
products
General form:
A + B
General form:
AB
AB
A + B
Example: Burning magnesium in the presence of oxygen
2 Mg(s) + O2(g)
!
Example: Decomposition of mercury (II) oxide
2 MgO(s)
2 HgO(s)
!
2 Hg(l) + O2(g)
O2
Hg
HgO
Decomposition reactions
Heating compounds containing oxygen often results in decomposition reactions
Example: Metal oxides
2 HgO(s)
2 PbO2(s)
!
2 Hg(l) + O2(g)
!
2 PbO(s) + O2(g)
Carbonates and hydrogen carbonates decompose to yield CO2
CaCO3(s)
2 NaHCO3(s)
!
CaO(s) + CO2(g)
!
Na2CO3(s) + H2O(g) + CO2(g)
Sample problems
Which of these are combination reactions? Which of these are
decomposition reactions?
H2 + Br2
Ba(ClO3)2
2 HBr
!
BaCl2 + 3 O2
2 SO2 + O2
2 SO3
4 Al + 3 O2
2 Al2O3
2 LiAlH4
!
2 LiH + 2 Al + 3 H2
Precipitation reactions
Sample problems
Predict the products of the following reactions (don’t forget to
balance the equations):
2 Ba
+
O2
BaCl2 (aq) + 2 AgNO3 (aq)
!
2 HgO
2 Al
2 BaO
2 Hg
+
!
MgO
+
insoluble precipitate
indicated by (s) after its formula
CO2
Solubility rules
Precipitation reactions
Most precipitation reactions occur when the anions and cations of two
inorganic compounds switch partners
General form:
AB + CD
AD + BC
To predict whether a precipitation reaction will occur:
• look at the potential products of the reaction (i.e., make the anions and
cations switch partners)
• determine whether either product is an insoluble solid
Example: Will a precipitation reaction occur when aqueous zinc chloride
and potassium hydroxide are mixed?
ZnCl2 (aq) + 2 KOH (aq)
2 AgCl (s) + Ba(NO3)2 (aq)
O2
2 AlCl3
+ 3 Cl2
MgCO3
In a precipitation reaction, an insoluble solid (called a
precipitate) is formed when reactants in aqueous solution
(i.e., dissolved in water) are combined.
Zn(OH)2
+
2 KCl
How can you tell if either of these is an insoluble solid?
AN IONIC COMPOUND
IS SOLUBLE IN WATER
IF IT CONTAINS THE
FOLLOWING IONS:
EXCEPTIONS
Ammonium ion (NH4+)
none
Alkali metal (Group IA) ions
(Li+, Na+, K+, Rb+, Cs+)
none
Nitrate (NO3-)
Acetate (C2H3O2-)
none
Halides (Cl-, Br-, I-)
Sulfate (SO42-)
Compounds containing
Ag+, Pb2+, Hg22+
Compounds containing
Ag+, Pb2+, Hg22+, Ca2+, Sr2+, Ba2+
Solubility rules
AN IONIC COMPOUND IS
NOT SOLUBLE IN WATER
IF IT CONTAINS THE
FOLLOWING IONS:
Carbonate (CO32-)
Phosphate (PO43-)
Hydroxide (OH-)
Oxide (O2-)
Sulfide (S2-)
Sample problems
Are the following compounds soluble or insoluble in water?
EXCEPTIONS
Compounds containing
NH4+, Alkali metal (Group IA) ions
Compounds containing
NH4+, Alkali metal (Group IA) ions
Compounds containing
metal (Group IA) ions
Compounds containing Group IIA ions are
slightly soluble (decompose in water)
soluble
CaCO3
insoluble
MgSO4
soluble
BaSO4
insoluble
Zn(OH)2 (s) + 2 KCl (aq)
Acid-base neutralization reactions
Classical definition of acids and bases (Arrhenius)
acid -- a substance that increases the concentration of
hydrogen ions (H+) in aqueous solutions
Example:
2 NaCl (aq) + (NH4)2CO3 (aq)
no reaction
HCl
H+ (aq)
+
Cl- (aq)
hydrochloric
acid
CuS (s) + H2SO4 (aq)
insoluble precipitate
Na2CO3 (aq) + 2 NH4Cl (aq)
NH4PO4
NH4+, Alkali
insoluble precipitate
CuSO4 (aq) + H2S (aq)
soluble
Compounds containing
Ca2+, Ba2+, Sr2+ are slightly soluble
Will the following reactions take place?
ZnCl2 (aq) + 2 KOH (aq)
NaCl
base -- a substance that increases the concentration of
hydroxide ions (OH-) in aqueous solutions
Example:
NaOH
sodium
hydroxide
Na+ (aq)
+
OH- (aq)
Acid-base neutralization reactions
Acid-base neutralization reactions
In a neutralization reaction, an acid reacts with a base
to form water plus an ionic compound (called a salt)
As is the case for precipitation reactions, neutralization
reactions occur when the anions and cations of two
inorganic compounds switch partners
acid
+
base
water
salt
• but water is formed as one of the products
Note: This can be any ionic compound,
not just “table salt” (NaCl)
Example:
HCl (aq) + KOH (aq)
Net result
combines with OHto form water
(a neutral compound)
+
H2O (l) + KCl (aq)
instead of an insoluble solid
Example:
HCl (aq) + KOH (aq)
H2O (l) + KCl (aq)
H–O–H
H+
H+
+
OH-
H2O
Types of acid-base neutralization reactions
Note: Heat is given off during an acid-base neutralization reaction
Homework assignment
Acid + metal hydroxide
Chapter 6 Problems:
Example:
HBr (aq) + NaOH (aq)
H2O (l) + NaBr (aq)
Acid + metal oxide
Example:
2 HNO3 (aq) + CuO (s)
H2O (l) + Cu(NO3)2 (aq)
Acid + carbonate (or bicarbonate)
Example:
2 HCl (aq) + Na2CO3 (s)
H2CO3(aq) spontaneously
decomposes to form water
and carbon dioxide
2 NaCl (aq) + H2O (l) + CO2 (g)
6.72, 6.74, 6.80, 6.82, 6.99, 6.102