Section 11.3 Limiting Reactants
• Identify the limiting reactant in a chemical equation.
• Identify the excess reactant, and calculate the
amount remaining after the reaction is complete.
• Calculate the mass of a product when the amounts
of more than one reactant are given.
molar mass: the mass in grams of one mole of any
pure substance
Section 11.3 Limiting Reactants (cont.)
limiting reactant
excess reactant
A chemical reaction stops when one of
the reactants is used up.
Why do reactions stop?
• Reactions proceed until one of the
reactants is used up and one is left in
excess.
• The limiting reactant limits the extent of the
reaction and, thereby, determines the amount
of product formed.
• The excess reactants are all the leftover
unused reactants.
Why do reactions stop? (cont.)
• Determining the limiting reactant is
important because the amount of the
product formed depends on this reactant.
Steps in Solving LR problems
When two amounts of reactants are given in a problem, we need to identify
the limiting reactant to solve for the amount of product possible!
1.
2.
3.
4.
5.
Start with a Balanced Equation
Change grams of reactants to moles – this is how
much you have.
Pick one of the reactants you just converted to mols
(how much you have) & use Stoichiometric ratio from
the balanced eqn to determine the number of mols of
other reactant needed.
Compare mols of other reactant needed to what you
have of that reactant… do you have enough?
Decide which is the limiting reactant
Example 1
Determine the limiting reactant, if 200.0 g of
S8 react with 100.0 g Cl2. Calculate the
grams of S2Cl2 produced. Calculate the
amount of excess reactant.
S8(l) + Cl2(g) → S2Cl2(l)
1. Balance Equation
2. Convert g
mols of each (What you have)
3. Pick one & convert mols
mols of other reactant. This
is what you need…Do you have enough?
4. If “yes,” then the other is your LR… if no than that one is
your LR
Example 1 (cont.)
Analyzing the excess reactant:
• Calculate mols of S8 used/needed.
0.352 mols S8
• Convert mol S8 used/needed to grams
90.4 g S8 (this is what was consumed in the
rxn.)
• Calculate excess remaining.
(Starting mass given – mass consumed = excess)
200.0g – 90.4g = 109.6 g S8 in excess
Calculating the Product when a Reactant
is Limiting (cont.)
• Using an excess reactant can speed up the
reaction.
• Using an excess reactant can drive a reaction
to completion.
Example 11.5 Determining the Limiting Reactant
The reaction between solid white
phosphorus (P4) and oxygen (O2) produces
solid tetraphosphorus decoxide (P4O10).
This compound is often called
diphosphorous pentoxide because its
empirical formula is P2O5. Determine the
mass of P4O10 formed if 25.0 g P4 and 50.0g
O2 are combined. How much excess
reactant remains after the reactions stops?
Section 11.3 Assessment
The mass of the final product in a
chemical reaction is based on what?
A. the amount of excess reactant
B. the amount of limiting reactant
D
A
0%
C
D. the amount of O2 present
A. A
B. B
C. C
0%
0%
0%
D. D
B
C. the presence of a catalyst
Section 11.3 Assessment
What is the excess reactant in the
following reaction if you start with 50.0g
of each reactant?
P4(s) + 5O2(g) → P4O10(s)
0%
A
D. unable to determine
D
C. Both are equal.
C
B. P4
A. A
B. B
C. C
0%
0%
0%
D. D
B
A. O2