2.10 Limiting Reagents
 in a chemical reaction, one reactant always limits how
much product will be produced
o ex: oxygen limits how much water can be produced
when hydrogen & oxygen gases combine
 ….H2 (g) + ….O2 (g)  ….H2O (l)
m = n × mm
Balanced Eq’n
2 H2 (g) +
O2 (g)
Moles, n
2 mol
1 mol
Molar mass, mm
2.02 g/mol
 2 H2O (l)
2 mol
32.00 g/mol 18.02 g/mol
Mass, m
conservation of mass!!
limiting reagent = the reactant that limits the amount of
product that can be formed in a
chemical reaction
o the reactant runs out first
excess reagent = any reactant that is not completely
consumed in the chemical reaction
o any reactant that’s leftover
o a reactant that doesn’t limit the amount of product
 in wine making, the alcohol content is controlled by the
sugar added to the mixture, because sugar is the limiting
reagent
o when the sugar is used up, the fermentation process
stops
 alcohol is a byproduct of yeast fermentation
 salicyclic acid is the limiting reagent in the production of
Aspirin
 in baking, sodium bicarbonate (baking soda) is often the
limiting reagent
o baking soda controls the texture of the food, since its
concentration in the batter controls how much CO2 is
produced
 the quantity of the limiting reagent in a reaction must be
carefully calculated because it ultimately determines how
much product will be made
o this is why you shouldn’t just guess how much
baking soda to put into a cake / muffin / cookie mix!
 recall the peanut brittle lab where you put the
baking soda in last
 besides baking, where else do you need to have an
accurate understanding of proportions?
o agriculture (measuring fertilizer, herbicides, etc.)
o putting gas in a chainsaw, weed trimmer (gas-to-oil
mix)
o adding chlorine to water at a purification plant
o mixing paint colours
o pharmaceutical industry (creating drugs of a certain
dosage)
o esthetics (colouring hair)
Analogies
 burger analogy
o what molar ratio exists between the bread & meat?
 2 bread, 1 meat
 2 : 1 molar ratio
o what is the limiting reagent? The excess reagent?
 dish-washing analogy on p. 149
 S’mores demo
o what type of reaction is this?
o what is the limiting reagent in this reaction?
How do you tell what the
limiting reagent is?
 whichever reactant has the lowest molar ratio is the
limiting reagent
o think back to the synthesis of water
 2 H2 (g) + O2 (g)  2 H2O (l)
oxygen is the limiting reagent here
 if a word problem states that one reactant is “excess”, it
is the excess reagent
o the other reactant is the limiting reagent
o e.g., aqueous potassium chromate combines with
excess silver nitrate solution to produce…
 potassium chromate is the limiting reagent
 in a mass-mass stoichiometry question:
o if you’re given the mass of a reactant, that’s the
limiting reagent
 you have to find how much precipitate (ppte) is
produced
o if you’re given the mass of the ppte and are required
to find the mass of a reactant, that’s the limiting
reagent
 scientists are able to determine which reactant in a
chemical reaction is the limiting reagent
o not necessary for you to know in this course
(…anymore!)
Calculating the
Limiting Reagent
4.80 mol sodium metal reacts with 2.7 mol chlorine gas to
form solid sodium chloride.
a) Write a balanced stoichiometric equation that represents
this reaction.
b) Determine the limiting reagent in this reaction.
c) Calculate the amount of NaCl that is produced.
a)
2 Na (s)
b)
2 mol
n = 4.80 mol
+

Cl2 (g)
1 mol
n = 2.70 mol
2 NaCl (s)
2 mol
Calculate the moles you need from the moles you have
need
2 mol of Na : 1 mol of Cl2
4.80 mol : x
2 (x) = 4.80 (1)
x = 2.40 mol Cl2
NEED 2.40 mol Cl2
HAVE 2.70 mol Cl2
 You have 2.70 mol of Cl2 but you only need 2.40
mol for the reaction, therefore Na is the limiting
reagent, and Cl2 is in excess.
OR
need
2 mol of Na : 1 mol of Cl2
x
: 2.70 mol
2 (2.70) = 1(x)
x = 5.40 mol Na
NEED 5.40 mol Na
HAVE 4.80 mol Na
 You have 4.80 mol of Na but you need 5.40 mol for
the reaction, therefore Na is the limiting reagent, and
Cl2 is in excess.
c)
amount of NaCl formed from the limiting reagent:
2 mol of Na
4.80 mol
2 (x)
x
m
:
:
=
=
2 mol NaCl
x
2 (4.80)
4.80 mol NaCl
= n * mm
= 4.80 mol *(22.99 g/mol + 35.45 g/mol)
= 280.8 g
 280.0 g of NaCl is formed in this reaction.