1. Caution: this stuff is difficult to follow at first.
5/2/2019
Limiting Reagents
Caution: this stuff is difficult to follow at first.
Be patient.

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

3. 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 NH3, 20 mol O2
8 mol NH3, 20 mol O2
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
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 …

4. Limiting reagents in stoichiometry
4NH3 + 5O2  6H2O + 4NO
E.g. How many grams of NO are produced if 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 NO
4 mol NH3 x
30.0 g NO
1 mol NO x
4 mol NH3
= 120 g 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 …

5. 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
1 mol NH3
17.0 g NH3 x
1.176 mol NH3 =
# mol NH3=
20 g NH3
1 mol O2
32.0 g O2 x
mol O2 =
# mol O2=
30 g O2
A – Once the number of moles of each is calculated we can determine the limiting reagent via a chart …

6. 2: Comparison chart 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 4 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.

7. 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=
1 mol O2
32.0 g O2 x
4 mol NO
5 mol O2 x
30.0 g NO
1 mol NO x
30 g O2
22.5 g NO =

8. 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=
1 mol NH3
17.0 g NH3 x
4 mol NO
4 mol NH3 x
30.0 g NO
1 mol NO x
20 g NH3
35.3 g NO =
1 mol O2
32.0 g O2 x
4 mol NO
5 mol O2 x
30.0 g NO
1 mol NO x
30 g O2
22.5 g NO =

9. Practice questions 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?

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

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

12. N2 is the limiting reagent
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=
1 mol N2
28.0 g N2 x
2 mol NH3
1 mol N2 x
17.0 g NH3
1 mol NH3 x
20 g N2
= 24.3 g H2
# g NH3 =
1 mol H2
2.0 g H2 x
2 mol NH3
3 mol H2 x
17.0 g NH3
1 mol NH3 x
5.0 g H2
= 28.3 g H2
N2 is the limiting reagent

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

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

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

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

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

18. 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=
1 mol MgCl2
95.21 g MgCl2 x
2 mol AgCl
1 mol MgCl2 x
143.3 g AgCl
1 mol AgCl x
25 g MgCl2
75.25 g AgCl =
# g AgCl=
1 mol AgNO3
g AgNO3 x
2 mol AgCl
2 mol AgNO3 x
143.3 g AgCl
1 mol AgCl x
68 g AgNO3
57.36 g AgCl =
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