PRODUCTS
WHY CAN’T I USE ALL
MY BUILDING BLOCKS?
(REACTANTS)
• WE DON’T HAVE ENOUGH OF
EACH OF THE BLOCKS TO BUILD
FULL PRODUCTS
•SOME REACTANTS ARE LEFT
OVER BUT ONE REACTANT
GETS COMPLETELY USED UP!
Limiting Reagents
STOICHIOMETRY
Limiting Reagent
• the reactant which is totally
consumed when the chemical
reaction is complete
“the amount of product formed is
limited by this reactant”
Limiting reagent
Given: 4NH3 + 5O2  6H2O + 4NO
Q - How many moles of NO are produced
if 4 mol NH3 are burned in 5 mol O2?
4 mol NO, works out
PERFECTLY – both reactants
are completely used up
Given: 4NH3 + 5O2  6H2O + 4NO
Q - How many moles of NO are
produced if 4 mol NH3 are burned in 20
mol O2?
4 mol NO, with leftover O2
Here, NH3 limits the production of NO;
If there was more NH3, more NO would
be produced
NH3 is called the “limiting reagent”
and O2 is in “excess
On your worksheet:
How many moles of NO are produced if
4 mol NH3 are burned in 2.5 mol O2?
2 mol NO, with leftover NH3
•
Here, O2 limits the production of NO; if
there was more O2, more NO would be
produced
• Thus, O2 is called the “limiting reagent”
and NH3 is in excess!
How can I tell which
reactant is the limiting
reagent?
Use a comparison chart between what we
have and what the balanced equation says
we need…
What we have
From the
question
What we need
NH3
O2
Mole given
Mole given
Mole ratio calculated
Ratio in balanced equation
Ok… let’s try it!!
4NH3 + 5O2  6H2O + 4NO
3.2 mol NH3 reacts with 1.6 mol O2
-which reactant will limit the production of
the reactants?
Comparison chart
NH3
O2
3.2
1.6
What we have
3.2/1.6 = 2
1.6/1.6 = 1
mol
mol
4/5 = 0.8
5/5 = 1
What we need
There is more NH3 than needed
to react all the O2. So O2 is the
limiting reagent which makes
NH3 the excess reagent!
• Now you can use the limiting reagent
moles to calculate how much product
you can make!
Limiting reagents in stoichiometry
4NH3 + 5O2  6H2O + 4NO
How many moles of NO are produced if
0.25 moles NH3 are burned in 0.56 mol
O2? (make a chart)
0.25 mol NH3 is the limiting reagent
NO
4 mol NO
mol= 0.25 mol NH3 x 4 mol NH =0.25 mol NO
3
Complete questions 3 and 4
on the worksheet!
4 Al(s) + 3 MnO2(aq)  2 Al2O3(aq) + 3 Mn(s)
3. What is the limiting reactant when 0.1372
mol of aluminum reacts with 0.1264 mol of
MnO2? How many moles of the aluminum
product should be yielded from this reaction?
What we have
Al
0.1372
MnO2
0.1264
0.1372/0.1264 0.1264/0.1264
1.085
1
What we need
4/3 = 1.33
3/3 = 1
So, based on the balanced equation,
to use all the MnO2 I would need
more Aluminum than I have?
Aluminum is the LIMITING REACTANT!
Mol Al2O3 made = 0.1372 mol Al x 2 mol Al2O3
4 mol Al
= 0.0686 mol
4. If 0.434 mol of both reactants are
combined, which will be the limiting
reagent?
What is the theoretical (predicted)
maximum mass of manganese that can
be yielded from this reaction?
LIMITING REACTANT IS aluminum
Theoretical maximum of Mn is 0.3255
mol which (using mm of Mn) converts
to 17.88 g
• Sometimes
the question is more
complicated. For example, if
grams of the two reactants are
given instead of moles we must
first determine moles, then decide
which is limiting …
Solving Limiting reagents mass to mole
4NH3 + 5O2  6H2O + 4NO
Q - How many g NO are produced if 20 g
NH3 is burned in 30 g O2?
A - First we need to calculate the number
of moles of each reactant
# mol NH3= 20 g NH3 x 1 mol NH3 1.176
=
17.0 g NH3 mol NH3
# mol O2=
30 g O2 x 1 mol O2 = 0.9375
32.0 g O2
mol O2
A – Once the number of moles of
each is calculated-find the LR…
Comparison chart
What we have
What we need
NH3
1.176
1.176/0.937
= 1.25 mol
4
O2
0.937
0.937/0.937
= 1 mol
5
*Choose the smallest value to divide each by
A - There is more NH3
(what we have) than
needed (what we need).
Thus NH3 is in excess,
and O2 is the limiting
reagent.
Stoichiometry
1) Expressed all chemical quantities
as moles
2) Determined the limiting reagent via
a chart
3) Use the limiting reagent to
determine how much product can
be made
Limiting Reagents: “shortcut”
• Limiting reagent problems can be solved
another way (without using a chart)…
• Do two separate calculations using both given
quantities. The smaller answer is correct.
Q - 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
Practice questions
1.
2.
3.
4.
5.
6.
2Al + 6HCl  2AlCl3 + 3H2
If 25 g of aluminum was added to 90 g of HCl, what mass of
H2 will be produced (try this two ways – with a chart & using
the shortcut)?
N2 + 3H2  2NH3: If you have 20 g of N2 and 5.0 g of H2,
which is the limiting reagent?
What mass of aluminum oxide is formed when 10.0 g of Al is
burned in 20.0 g of O2?
When C3H8 burns in oxygen, CO2 and H2O are produced. If
15.0 g of C3H8 reacts with 60.0 g of O2, how much CO2 is
produced?
How can you tell if a question is a limiting reagent question
vs. typical stoichiometry?
If 25 g magnesium chloride was added to 68 g silver nitrate,
what mass of AgCl will be produced?
MgCl2 + 2AgNO3  Mg(NO3)2 + 2AgCl
1
# mol Al = 25 g Al x 1 mol Al = 0.926 mol
27.0 g Al
# mol HCl = 90 g HCl x 1 mol HCl = 2.466 mol
36.5 g HCl
What we
have
What we
need
Al
0.926
0.926/0.926
= 1 mol
2
2/2 = 1 mol
HCl
2.466
2.466/0.926 HCl is
limiting.
= 2.7 mol
6
6/2 = 3 mol
# g H2 =
1 mol HCl 3 mol H2 2.0 g H2
90 g HCl x
x
x
= 2.47 g H2
36.5 g HCl 6 mol HCl 1 mol H2
Question 1: shortcut
2Al + 6HCl  2AlCl3 + 3H2
If 25.0 g aluminum was added to 90.0 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
Question 2
# mol N2= 20 g N2 x 1 mol N2 = 0.714 mol N2
28 g N2
# mol H2= 5.0 g H2 x 1 mol H2 = 2.5 mol H2
2 g H2
N2
H2
0.714 mol
2.5 mol
What we have
0.714/0.714
2.5/0.714
= 1 mol
= 3.5 mol
What we need
1 mol
3 mol
We have more H2 than what we need, thus H2
is in excess and N2 is the limiting factor.
Question 2: shortcut
N2 + 3H2  2NH3
If you have 20 g of N2 and 5.0 g of H2, which is
the limiting reagent?
# g NH3=
20 g N2 x1 mol N2x 2 mol NH3 x17.0 g NH3= 24.3 g H2
28.0 g N2 1 mol N2 1 mol NH3
# g NH3 =
5.0 g H2 x 1 mol H2 x 2 mol NH3 x17.0 g NH=3 28.3 g H2
2.0 g H2 3 mol H2 1 mol NH3
N2 is the limiting reagent
4Al + 3O2  2 Al2O3
# mol Al = 10 g Al x 1 mol Al = 0.37 mol Al
27 g Al
# mol O2 = 20 g O2 x 1 mol O2 = 0.625 mol O2
32 g O2
There is
Al
O2
more
0.37 mol
0.625 mol
What we
than
0.37/.37
0.625/0.37
have
enough
= 1 mol
= 1.68 mol
O
;
Al
is
2
What we
4 mol
3 mol
limiting
need
4/4 = 1 mol 3/4 = 0.75 mol
3
# g Al2O3 = 0.37 mol Al x 2 mol Al2O3 x 102 g Al2O3
4 mol Al
1 mol Al2O3
=
18.9 g Al O
Question 3: shortcut
4Al + 3O2  2 Al2O3
What mass of aluminum oxide is formed when
10.0 g of Al is burned in 20.0 g of O2?
# g Al2O3=
10.0 g Al x 1 mol Al x 2 mol Al2O3 x102.0 g Al2O3 = 18.9 g Al2O3
1 mol H2
27.0 g Al 4 mol Al
# g Al2O3=
20.0 g O2x 1 mol O2 x2 mol Al2O3x102.0 g Al2O3 = 42.5 g Al2O3
32.0 g O2 3 mol O2 1 mol H2
C3H8 + 5O2  3CO2 + 4H2O
# mol C3H8 =15 g C3H8 x 1 mol C3H8 = 0.34 mol
C 3H 8
44 g C3H8
# mol O2 = 60 g O2 x 1 mol O2 = 1.875 mol O2
32 g O2
We have
C 3H 8
O2
more
than
0.34
mol
1.875
mol
What we
enough O2,
0.34/.34
1.875/0.34
have
C3H8 is
= 1 mol
= 5.5 mol
limiting
Need
1 mol
5 mol
# g CO2 =
44
g
CO
3
mol
CO
2
2 x
0.34 mol C3H8 x
1 mol C3H8 1 mol CO2
=
45.0 g CO2
4
Question 4: shortcut
C3H8 + 5O2  3CO2 + 4H2O
When C3H8 burns in oxygen, CO2 and H2O are
produced. If 15.0 g of C3H8 reacts with 60.0 g
of O2, how much CO2 is produced?
# g CO2=
15.0 g C3H8x1 mol C3H8 x 3 mol CO2x44.0 g CO2 = 45.0 g CO2
44.0 g C3H8 1 mol C3H8 1 mol CO2
# g CO2=
60.0 g O2x 1 mol O2 x 3 mol CO2 x 44.0 g CO2 = 49.5 g CO2
32.0 g O2 5 mol O2 1 mol CO2
5. Limiting reagent questions give values for
two or more reagents (not just one)
Question 6: shortcut
MgCl2 + 2AgNO3  Mg(NO3)2 + 2AgCl
If 25.00 g magnesium chloride was added
to 68.00 g silver nitrate, what mass of AgCl
will be produced?
# g AgCl=
25 g MgCl2 x 1 mol MgCl2 x 2 mol AgCl x 143.3 g AgCl
95.21 g MgCl2 1 mol MgCl2 1 mol AgCl
=
75.25 g AgCl
# g AgCl=
68 g AgNO3 x 1 mol AgNO3 x 2 mol AgCl x 143.3 g AgCl
169.88 g AgNO3 2 mol AgNO3 1 mol AgCl
= 57.36 g AgCl