LIMITING
REAGENT
Just a Quick Note
• Gas occupies 22.4L/mol at STP (0oC)
• Gas occupies 24.5L/mol at NTP (25oC)
Limiting Reactant
In a chemical reaction where arbitrary amounts
of reactants are mixed and allowed to react, the
one that is used up first is the limiting reactant.
A portion of the other reactants remains.
There is a systematic procedure for finding the
limiting reagent based on the reactant ratio (RR)
defined as the ratio of the number of moles of a
reactant to its coefficient in a balanced chemical
equation. The reagent with the smallest reactant
ratio is the limiting reactant.
The Limiting Reagent
In most chemical reactions, not all of the reactants
are used up because they are not present in the
exact proportions required of the reaction
o
•
•
o
o
Eg. If Pb(NO3)2 + CuSO4  PbSO4 + Cu(NO3)2;
you would need exactly 313g (1mol) of Pb(NO3)2 and
303g (1mol) of CuSO4 to fully use each of the reactants
The limiting reagent is the reactant which is used up
first in a chemical reaction (i.e. it being used up,
stops the rest of the reactants from being able to fully
react)
The limiting reagent can be determined
mathematically using the Mole
Balanced
reaction!
Defines
stoichiometric
ratios!
Unbalanced (i.e., non-stoichiometric)
mixture!
Limited by syrup!
For a Reaction of the Form:
aA + bB = cC + dD
If compounds A and B are present in the mole
amounts called for in the balanced reaction, then the
following equation is valid:
moles of A a

moles of B b
or
moles of A moles of B

 RR
a
b
To proceed, first calculate the reactant ratios for all of the
reactants:
aA + bB = cC + dD
starting moles A
RR A 
,
a
starting moles B
RR B 
, ..
b
From among these, choose (RR)min, the smallest
reactant ratio. This identifies the limiting reactant.
Limiting Reactant EXAMPLE
• 2Al(s) + 6HCl(g)
2AlCl3(s) + 3H2(g)
Consider the reaction above. If we react 30.0 g Al and 20.0 g
HCl, how many moles of aluminum chloride will be
formed?
• 30.0 g Al
20.0g HCl
MOLES = M/MM
= 30 / 27
= 1.1111
MOLES = M/MM
= 20/ (1 + 35.5)
= 0.5479
RRa = (Moles of Al) / a
= 1.1111 / 2
= 0.5555
RRb = (Moles of HCl) / b
= 0.5479 / 6
= 0.09132
• Limiting reactant = reactant with RRmin =
Finding amount of Product from
Limiting Reagent
2Al(s) + 6HCl(g)
2AlCl3(s) + 3H2(g)
Calculate the Moles of AlCl3 from:
Moles of Limiting Reagent x (need/have)
= 0.5479 x
2/ 6
= 0.18264
1. 4KO2 + 2H2O  4KOH + 3O2:
0.15mol KO2 and 0.10mol H2O
a. Which is the limiting reagent?
b. How many moles of O2 can be
produced?
c. What is the volume of O2 is
produced at STP?
2. CO + H2  CH3OH: 35.4g CO
and 10.2g H2
a. Balance the equation
b. Which is the limiting reagent?
c. How many grams of the excess
reagent are left at the end of
reaction?
( Hint – [Need / have] x sentence) use this to calculate the
moles of H2 used
1. .
a. KO2 is the
limiting reagent
b. 0.11mol of O2
c. 2.52 L
2. .
a. CO + 2H2 
CH3OH
b. CO is the
limiting reagent
c. 5.1g of H2
Work on Worksheet