Limiting Reagents
Caution: this stuff is difficult to follow at first.
Be patient.
Balloon & Flask Demonstration
1
2
3
g of
NaHCO3
10
10
10
mL of 3M
HCl
25
50
100
How can we prove that our conclusions
about limiting reagents is correct?
Limiting reagent defined
Given: 4NH3 + 5O2  6H2O + 4NO
Q - 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”
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 …
Limiting reagents in stoichiometry
4NH3 + 5O2  6H2O + 4NO
E.g. 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
• 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 1: g to mol
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
mol NH3
17.0 g NH3
1 mol O2
0.9375
x
=
mol O2
32.0 g O2
A – Once the number of moles of each is
calculated we can determine the limiting
reagent via a chart …
# mol O2=
30 g O2
2: Comparison chart NH
3
What we have
What we need
O2
1.176
1.176/0.937
= 1.25 mol
4
0.937
0.937/0.937
= 1 mol
5
4/5 = 0.8 mol
5/5 = 1 mol
*Choose the smallest value to divide each by
** You should have “1 mol” in the same column
twice in order to make a comparison
A - There is more NH3 (what we have) than
needed (what we need). Thus NH3 is in
excess, and O2 is the limiting reagent.
3: Stoichiometry (given = limiting)
So far we have followed two steps …
1) Expressed all chemical quantities as moles
2) Determined the limiting reagent via a chart
Finally we need to …
3) Perform the stoichiometry using the limiting
reagent as the “given” quantity
Q - 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
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. 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)?
2. N2 + 3H2  2NH3: If you have 20 g of N2 and
5.0 g of H2, which is the limiting reagent?
3. What mass of aluminum oxide is formed
when 10.0 g of Al is burned in 20.0 g of O2?
4. 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?
5. How can you tell if a question is a limiting
reagent question vs. typical stoichiometry?
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 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
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
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
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 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.
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
=
18.87 g Al O
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 g CO2
4
Limiting Reagents: shortcut
MgCl2 + 2AgNO3  Mg(NO3)2 + 2AgCl
If 25 g magnesium chloride was added to 68 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
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