limiting reagent

1. Limiting Reagents
Caution: this stuff is difficult to follow at first.
Be patient.

2. g of
NaHCO3
mL of 3M
HCl
1 10 25
2 10 50
3 10 100
How can we prove that our conclusions
about limiting reagents is correct?
Balloon & Flask Demonstration

3. Limiting reagent definedLimiting reagent defined
Q - How many moles of NO are produced if
__ mol NH3 are burned in __ mol O2?
4 mol NH3, 5 mol O2
4 mol NH3, 20 mol O2
8 mol NH3, 20 mol O2
Given: 4NH3 + 5O2 → 6H2O + 4NO
4 mol NO, works out exactly
4 mol NO, with leftover 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
• In limiting reagent questions we use the limiting
reagent as the “given quantity” and ignore the
reagent that is in excess …
2 mol NO, leftover NH3

4. Limiting reagents in stoichiometryLimiting reagents in stoichiometry
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
4NH3 + 5O2 → 6H2O + 4NO
4 mol NO
4 mol NH3
x
# g NO=
4 mol NH3 = 120 g NO
30.0 g NO
1 mol NO
x
• 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 …

5. Solving Limiting reagents 1: g to molSolving Limiting reagents 1: g to mol
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
4NH3 + 5O2 → 6H2O + 4NO
1 mol NH3
17.0 g NH3
x# mol NH3= 20 g NH3
1.176
mol NH3
=
1 mol O2
32.0 g O2
x# mol O2= 30 g O2
0.9375
mol O2
=
A – Once the number of moles of each is
calculated we can determine the limiting
reagent via a chart …

6. NH3 O2
What we have
What we need
1.176 0.937
1.176/0.937
= 1.25 mol
0.937/0.937
= 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
4 5
4/5 = 0.8 mol 5/5 = 1 mol
A - There is more NH3 (what we have) than
needed (what we need). Thus NH3 is in
excess, and O2 is the limiting reagent.
2: Comparison chart2: Comparison chart

7. 3: Stoichiometry (given = limiting)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
4 mol NO
5 mol O2
x
# g NO=
30 g O2
22.5 g NO=
30.0 g NO
1 mol NO
x1 mol O2
32.0 g O2
x

8. Limiting Reagents: shortcutLimiting 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
4 mol NO
5 mol O2
x30 g O2
22.5 g NO=
30.0 g NO
1 mol NO
x1 mol O2
32.0 g O2
x
4 mol NO
4 mol NH3
x
# g NO=
20 g NH3
35.3 g NO=
30.0 g NO
1 mol NO
x1 mol NH3
17.0 g NH3
x

9. Practice questionsPractice 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?

10. 11 1 mol Al
27.0g Al
x#mol Al =25 g Al = 0.926 mol
# mol HCl= 90 g HCl 1 mol HCl
36.5g HCl
x = 2.466 mol
Al HCl
What we
have
What we
need
0.926 2.466
0.926/0.926
= 1 mol
2.466/0.926
= 2.7 mol
2 6
2/2 = 1 mol 6/2 = 3 mol
HCl is
limiting.
3 mol H2
6 mol HCl
x
# g H2 =
90 g HCl
2.0 g H2
1 mol H2
x1 mol HCl
36.5 g HCl
x = 2.47 g H2

11. Question 1: shortcutQuestion 1: shortcut
2Al + 6HCl → 2AlCl3 + 3H2
If 25 g aluminum was added to 90 g HCl, what
mass of H2 will be produced?
3 mol H2
2 mol Al
x# g H2= 25 g Al = 2.78 g H2
2.0 g H2
1 mol H2
x1 mol Al
27.0 g Al
x
3 mol H2
6 mol HCl
x# g H2= 90 g HCl = 2.47 g H2
2.0 g H2
1 mol H2
x1 mol HCl
36.5 g HCl
x

12. Question 2: shortcutQuestion 2: shortcut
N2 + 3H2 → 2NH3
If you have 20 g of N2 and 5.0 g of H2, which is the
limiting reagent?
2 mol NH3
1 mol N2
x
# g NH3=
20 g N2 = 24.3 g H2
17.0 g NH3
1 mol NH3
x1 mol N2
28.0 g N2
x
2 mol NH3
3 mol H2
x
# g NH3=
5.0 g H2 = 28.3 g H2
17.0 g NH3
1 mol NH3
x1 mol H2
2.0 g H2
x
N2 is the limiting reagent

13. Question 3: shortcutQuestion 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?
2 mol Al2O3
4 mol Al
x
# g Al2O3=
10.0 g Al = 18.9 g Al2O3
102.0 g Al2O3
1 mol H2
x1 mol Al
27.0 g Al
x
2 mol Al2O3
3 mol O2
x
# g Al2O3=
20.0 g O2 = 42.5 g Al2O3
102.0 g Al2O3
1 mol H2
x
1 mol O2
32.0 g O2
x

14. Question 4: shortcutQuestion 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?
3 mol CO2
1 mol C3H8
x
# g CO2=
15.0 g C3H8 = 45.0 g CO2
44.0 g CO2
1 mol CO2
x1 mol C3H8
44.0 g C3H8
x
3 mol CO2
5 mol O2
x
# g CO2=
60.0 g O2 = 49.5 g CO2
44.0 g CO2
1 mol CO2
x
1 mol O2
32.0 g O2
x
5. Limiting reagent questions give values for
two or more reagents (not just one)

15. N2 H2
What we have
What we need
Question 2Question 2
0.714 mol 2.5 mol
0.714/0.714
= 1 mol
2.5/0.714
= 3.5 mol
We have more H2 than what we need, thus H2 is
in excess and N2 is the limiting factor.
1 mol 3 mol
1 mol N2
28 g N2
x# mol N2= 20 g N2 0.714 mol N2
=
1 mol H2
2 g H2
x# mol H2= 5.0 g H2 2.5 mol H2
=

16. Al O2
33 4Al + 3O2 → 2 Al2O3
1 mol Al
27 g Al
x# mol Al = 10 g Al 0.37 mol Al=
1 mol O2
32 g O2
x# mol O2 = 20 g O2
0.625 mol O2=
0.37 mol 0.625 mol
0.37/.37 = 1
mol
0.625/0.37
= 1.68 mol
4 mol 3 mol
4/4 = 1 mol 3/4 = 0.75 mol
What we
have
What we
need
There is
more
than
enough
O2; Al is
limiting
2 mol Al2O3
4 mol Al
x# g Al2O3 = 0.37 mol Al
18.87 g Al O=
102 g Al2O3
1 mol Al2O3
x

17. C3H8 O2
44 C3H8 + 5O2 → 3CO2 + 4H2O
1 mol C3H8
44 g C3H8
x# mol C3H8 = 15 g C3H8
0.34 mol
C3H8
=
1 mol O2
32 g O2
x# mol O2 = 60 g O2
1.875 mol O2=
0.34 mol 1.875 mol
0.34/.34 = 1
mol
1.875/0.34 =
5.5 mol
1 mol 5 mol
What we
have
Need
3 mol CO2
1 mol C3H8
x
# g CO2 =
0.34 mol C3H8
45 g CO2
=
44 g CO2
1 mol CO2
x
We have
more than
enough O2,
C3H8 is
limiting

18. Limiting Reagents: shortcutLimiting 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?
2 mol AgCl
1 mol MgCl2
x
# g AgCl=
25 g MgCl2
75.25 g AgCl=
143.3 g AgCl
1 mol AgCl
x1 mol MgCl2
95.21 g MgCl2
x
2 mol AgCl
2 mol AgNO3
x
# g AgCl=
68 g AgNO3
57.36 g AgCl=
143.3 g AgCl
1 mol AgCl
x1 mol AgNO3
169.88 g AgNO3
x
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