Limiting Reagents
Caution: this stuff is difficult to follow at first.
Be patient.
Podcast 8.3
Limiting Reagents and
Percent Yield
Limiting Reactants
• Cake example:
– Recipe:
2 cups flour
1½ TBSP baking powder
2 eggs
1 cup water
1 cup sugar
1/3 cup oil
Suppose in your kitchen you have 14 cups flour,
4 eggs, 9 cups sugar, 15 TBSP baking powder,
10 cups of water, and 3 1/3 cups oil. How
many cakes can be baked?
•
•
•
•
Limiting Reactants
Flour... need 2 cups, have 14 cups = ________
7 cakes
2 cakes
Eggs…need 2 eggs, have 4 eggs =_________
9 cakes
Sugar… need 1 cup, have 9 cups = _________
Baking powder…need 1½ TBSP,
10 cakes
have 15 TBSP = __________
cakes
• Water…need 1 cup, have 10 cups = 10
________
10 cakes
• Oil…need 1/3 cup, have 3 1/3 = __________
Limiting Reactants
• Back to the question, how many cakes can be
baked?
• Answer is _______
– Only 4 eggs, so we have extra of all the other
ingredients
– Eggs, in this case, is the ____________ reactant
Example One
Given: 4NH3 + 5O2  6H2O + 4NO
How many moles of NO are produced if
__ mol NH3 are burned in __ mol O2?
4 mol NH3, 5 mol O2 4 mol NO, works out exactly
4 mol NH3, 20 mol O2 4 mol NO, with leftover O2
8 mol NH3, 20 mol O2 8 mol NO, with leftover O2
• Here, NH3 limits the production of NO; if there
was more NH3, more NO would be produced
• Thus, NH3 is called the “limiting reagent”
Example Two
Given: 4NH3 + 5O2  6H2O + 4NO
How many moles of NO are produced if
__ mol NH3 are burned in __ mol O2?
•4 mol NH3, 2.5 mol O2 2 mol NO, leftover NH3
•In limiting reagent questions we use the
limiting reagent as the “given quantity” and
ignore the reagent that is in excess
Example Three
4NH3 + 5O2  6H2O + 4NO
How many grams of NO are produced if
4 moles NH3 are burned in 20 mol O2?
Since NH3 is the limiting reagent we will use this
as our “given quantity” in the calculation
# g NO=
4 mol NH3 x 4 mol NO x 30.0 g NO= 120 g NO
4 mol NH3 1 mol NO
Example Four
4NH3 + 5O2  6H2O + 4NO
How many g NO are produced if 20 g NH3 is
burned in 30 g O2?
Step 1: Calculate the number of moles of
EACH reactant
# mol NH3= 20 g NH3 x 1 mol NH3 = 1.176
mol NH3
17.0 g NH3
# mol O2=
30 g O2
1 mol O2
0.9375
x
=
mol O2
32.0 g O2
Step 2: determine the limiting reagent using a
comparison chart
Comparison chart
NH3
O2
What we
have
1.176 mol
0.937 mol
What we
need
0.75 mol NH3
1.5 mol O2
1.176molNH
1.47mol
There is more
NH3 (what
we= have)
thanO2
3 x __5molO
2__
needed (what we
need).
4 mol
NH3 Thus NH3 is in
0.937O
__4molNH
= 0.7496mol
excess,
O2 is the
reagent. NH3
2 xand
3 limiting
5 mol O2
.
Step 3: Perform the stoichiometry using the
____________ ____________ as the
“given” quantity (the starting place)
How many g NO are produced if 20 g NH3 is
burned in 30 g O2?
4NH3 + 5O2  6H2O + 4NO
# g NO=
30 g O2 x 1 mol O2 x 4 mol NO x 30.0 g NO
32.0 g O2
5 mol O2 1 mol NO
=
22.5 g NO
Example Five: The Shortcut
• Do _________ separate calculations using
both given quantities. The ___________
answer is correct.
How many g NO are produced if 20 g NH3 is
burned in 30 g O2? 4NH3 + 5O2 6H2O+ 4NO
# g NO=
20 g NH3 x 1 mol NH3 x 4 mol NO x 30.0 g NO
17.0 g NH3 4 mol NH3 1 mol NO
=
35.3 g NO
30 g O2 x 1 mol O2 x 4 mol NO x 30.0 g NO
32.0 g O2
5 mol O2
1 mol NO
=
22.5 g NO
Example Six: Use the Shortcut
2Al + 6HCl  2AlCl3 + 3H2
If 25 g aluminum was added to 90 g HCl, what
mass of H2 will be produced?
# g H2= 25 g Al x 1 mol Al x 3 mol H2 x 2.0 g H2 = 2.78 g H2
27.0 g Al 2 mol Al 1 mol H2
# g H2 =90 g HCl x1 mol HClx 3 mol H2 x 2.0 g H2 = 2.47 g H2
36.5 g HCl 6 mol HCl 1 mol H2
Percent Yield
• _________________ is the amount of product
made in a chemical reaction.
• Three Types:
– _______________ Yield –
• What you actually get in the lab
– Theoretical Yield –
• What the balanced equation tells
_____________ be made
– ___________________ Yield –
•
Percent Yield
• Percent yield tells us how ________________ a
reaction is.
• Percent yield CANNOT be bigger than _________
unless contamination occurs
• Theoretical yield will always be larger than actual
yield.
– Why?
– Impure reactants, competing side reactions, loss
of product in filtering or transferring between
containers
Example Seven
6.78 g of copper is produced when 3.92 g of Al
are reacted with excess copper (II) sulfate.
• What is the actual yield?
–6.78 g Cu
• What is the theoretical yield?
–13.8 g Cu
• What is the percent yield?
–49.1%