LIMITING AND EXCESS REAGENTS
“Reagent” is another name for “reactant”
Limiting reagent is the one that is COMPLETELY
consumed in a reaction.
The other reactants that are NOT COMPLETELY consumed in
a reaction called excess reagents.
If the limiting reagent runs out, the reaction will stop.
Example. Synthesis of ammonia
N2(g) + 3H2(g) 2NH3(g).
N2 H2 NH3
1 molecule 3 molecules 2 molecules
(or 1 mole) (or 3 moles) (or 2 moles)
No reactant is limiting. All quantities used up.
This is rare. Usually you have one quantity consumed and the other in excess.
Limiting/Excess reagent Scenario:
N2 H2 NH3
2 molecules 3 molecule ? molecules
(or 1 mole) (or 3 moles) (or ? moles)
How many molecules of NH3 would be produced?
Which is the limiting reagent?
Which is the excess reagent? By how much?
Before Reaction Before the reaction has
occurred, there are 0 molecules
N2 N2 H2 H2 H2
of NH3 since they have not been
Balanced equation N2(g) + 3H2(g) 2NH3(g)
After reaction 0 molecules of
H2 left. NH3 NH3
N2 in excess by 1
Excess Reagent Limiting
1 molecule of N2 reacted with 3 molecules of H2. There is still one more
molecule of N2 left. However, the molecules of H2 have all been used up. The
reaction can no longer continue since H2 is used up. H2 is the limiting reagent
and N2 is the excess reagent.
The quantity of the limiting reagent therefore determines ALL the quantities of
the products formed in the reactions
In limiting reagent questions we use the limiting reagent as the “given
quantity” and ignore the reagent that is in excess …
If we use the excess reagent, we will overestimate the amount of product that
can be produced.
Identifying the limiting reagent in a chemical reaction
This is because the amount of limiting reactant that is available
for a chemical reaction determines the amount of product that is
formed and the amount of excess reactant that is left over.
The limiting reagent yields the smaller amount of product in a
The limiting reactant forms less product
- Limiting reactant is not necessarily the reactant this is present in the smaller
amount (or smaller mass). It is the reactant that forms the smaller amount of
e.g. 2H2(g) + O2(g)2H2O (g)
If oxygen is present in excess and 2 mol of hydrogen is available, then 2
mol of water is produced (stoichiometric amounts)
If hydrogen is in excess and 2 mol of oxygen is available then 4 mol of
water is produced
The moles of products are always determined by the starting moles of the
Example 1: Sodium metal reacts with chlorine gas to form solid sodium
chloride according to the following balanced equation:
2Na(s) + Cl2(g) 2NaCl
If 4.80 mol Na reacts with 2.70 mol Cl2, determine the following:
a) The limiting reagent
b) The mass of excess reagent that remains unreacted.
c) The amount of NaCl that is produced
Step 1: write a balanced equation:
Already done for you: 2Na(s) + Cl2(g) 2NaCl
Step 2: Analyze the mole ratios/proportions.
The equation tells us that 2 moles of Na will react with 1 mole of Cl2 to
produce 2 moles of NaCl.
Step 3: Check the reagent ratios to see if they are correct according to the
balanced equation ratio.
4.80 mol Na will react with how many mole Cl2?
4.80 mol Na x 1 mole Cl2 = 2.40 mole of Cl2
2 mole of Na
The question tells us that there are 2.70 mole of Cl2.
Which means that Cl2 is in excess since 2.70 > 2.40 (the actual amount that
4.80 mole of Na would react with).
Therefore, Na(s) is the limiting reagent.
Excess mol Cl2= 2.70 -2.40 mol = 0.30 mol
Excess mass of Cl2 that remains unreacted:
mcl2= 0.30 mol x 71 g/mol=21.3 g
1. If Mole stated in question > calculated mole, then that is your excess
2. If Mole stated in question < calculated mole, then that is your limiting
To calculate the amount of NaCl produced:
Use the mole amount of the limiting reagent in order to calculate amounts of
any other substance in the equation.
Calculate the amount of ammonia that is produced using the moles of the
4.80 mole Na x 2 mole NaCl = 4.80 mole of NaCl produced.
2 mole Na
If you used the other value?
2.70 mole Cl2 x 2 mole NaCl = 5.40 mole of NaCl produced.
1 mole Cl2
If you calculated using excess reagent, the amount of product
would mistakenly be too high!
Method 2: a second method to determine limiting reagents
Example 2: The chemical compound ammonia is prepared from its elements
according to the following chemical equation:N2(g) + 3H2(g) 2NH3(g)
If 4.20 g of nitrogen gas reacts with 0.750 g of hydrogen gas, which is the
Plan your strategy Act on your strategy
Calculate the molar MN2 = 2 x MN = 2 (14.01 g/mol) = 28.02 g/mol
masses, M, of nitrogen MH2 = 2 x MH = 2(1.01 g/mol) = 2.02 g/mol
Convert the masses of nN2 = mN2/MN2 = 4.20 g/28.02 g/mol = 0.14989 mol of N2
nitrogen and hydrogen nH2 = mH2/MN2 = 0.750g/2.02 g/mol = 0.37129 mol of H2
into amounts (in moles)
using n = m/M
Calculate the amount of nNH3 = 0.14989 mol N2 X 2 Mol NH3 = 0.300 mol NH3
ammonia that is produced 1 mol N2
by the given amount of nNH3 = 0.37129 mol H2 X 2 mol NH3 = 0.248 mol NH3
nitrogen and the given 3 mol H2
amount of hydrogen
(using mole ratios).
Compare the amounts of The calculation with H2 gas produces less NH3 than N2.
ammonia that are Therefore Hydrogen gas is the limiting reagent despite
produced by nitrogen and there being more moles of Hydrogen gas present!
hydrogen to determine the DO NOT ASSUME THAT THE LIMITING
limiting reactant. REAGENT IS THE REAGENT PRESENT IN THE
SMALLEST QUANTITY! IT IS ALL ABOUT THE
Adjusted Approach for Limiting Reagent Problems
KEY: If 2 masses are given (ie. the mass of both reactants), then
the question is a limiting
1) Balance the given equation
2) a) Convert the quantity of the given substance to moles of the
b) Determine limiting reagent
3) Using the moles of the limiting reagent to determine the moles
of the unknown substance (use
mole ratio derived from balanced equation)
4) Convert the moles of the unknown to the desired quantity of
**In any of the above cases, the mass calculated will be a
theoretical mass. This mass may then be used to determine the