Limiting & Excess Reactants
How do you know which one is
which?
What does limiting & excess
mean?
 Limiting Reactant - the reactant that runs out first
in a chemical reaction, thus determining the
amount of product produced
 Excess Reactant - the reactant that there is a
quantity of left over after a chemical reaction
 *The excess reactant should be the cheaper
reactant since we do not like to waste unused
chemical
Vinegar (5% by mass acetic acid) and Baking
Soda Demo
 NaHCO3 (s) + HC2H3O2 (aq)  NaC2H3O2 (aq) +
H2O (l) + CO2 (g)
Mass of
Baking
Soda
Mass of
Vinegar
Flask #1
Flask #2
Flask #3
1.50 g
3.50 g
5.50 g
50.00 g
(2.5 g
HC2H3O2)
50.00 g
(2.5 g
HC2H3O2)
50.00 g
(2.5 g HC2H3O2)
Outcome from Demo
 Flask #1 showed the smallest amount of CO2 produced in
the balloon with a clear solution in the flask.
 Flask #2 showed the same amount of CO2 produced in the
balloon as in flask #3, which was a larger amount
compared to flask #1. The solution in the flask was still
clear like in flask #1.
 Flask #3 showed the same amount of CO2 produced in the
balloon as in flask #2. However, the solution in the flask
was cloudy.
Which reactant was limiting and which
reactant was excess in each flask based off
observations?
Flask #1
Flask #2
Flask #3
Baking Soda
limiting
theoretical
amount
excess
Vinegar
excess
theoretical
amount
limiting
The proof is in the stoichiometry
50.0 g
5g
1 mol
1 mol
84.01 g
3.50 g
vinegar HC2H3O2 HC2H3O2 NaHCO3 NaHCO3 NaHCO3
100 g
vinegar
60.06 g
1 mol
1 mol
HC2H3O2 HC2H3O2 NaHCO3
The theoretical amount of baking soda needed to react
completely with the 50.0 grams of vinegar is 3.50 g.
Practice Problem
2NaI + Cl2  2NaCl + I2
 1. You are given 22.1 g of NaI and 4.13 g of Cl2.
What is the limiting reactant?
 Pick one of the values given for your reactants and
through stoichiometry find out how much you
need of the other reactant.
22.1 g NaI
1 mol NaI
1 mol Cl2
70.90 g Cl2
149.89 g
NaI
2 mol NaI
1 mol Cl2
5.23 g Cl2
You have 4.13 g Cl2 available
You need 5.23 g to react completely with 22.1 g NaI
Therefore, Cl2 is the limiting reactant which will run out first
in the reaction and determine the amount of product produced.
2. How much NaCl is produced?
*Remember that you must always start with
the limiting reactant quantity that you
have because this is what determines the
amount of product produced.
4.13 g Cl2
1 mol Cl2
70.90 g Cl2
2 mol NaCl 58.44 g NaCl
1 mol Cl2
1 mol NaCl
6.81 g
NaCl
3. How much excess reactant will be left over?
*Remember that you must always start with the
limiting reactant quantity that you have because
this is what determines the amount of the excess
reactant you will need.
4.13 g Cl2
1 mol
Cl2
2 mol
NaI
149.89 g
NaI
70.90 g
Cl2
1 mol
Cl2
1 mol
NaI
17.5 g NaI
You have 22.1 g available
You need 17.5 g
22.1 g NaI – 17.5 g NaI = 4.6 g NaI will be left over
after the reaction is complete
% Yield
 Chemists like to know how right we are 
 % yield = (actual/theoretical) x 100
 Actual  lab results
 Theoretical  mathematical prediction
through stoichiometry
% error = |theoretical – experimental|/theoretical
x 100
% error + % yield = 100