Lab
“Oreo Lab”
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6.3 Limiting Reagents
Caution: this stuff can be difficult to follow
at first. Be patient.
Limiting Reagents and
Reagents in Excess
• So far we have assumed that all of the
reactants taking part in the reaction
have been transformed into products.
(No left-overs)
• Not usually the case.
3
Ex. Wood burning
– Unlimited oxygen in the air
\ for the reaction
available
(oxygen = reagent in excess)
– When all wood is burned, the
reaction stops, no longer using
oxygen from the air, and making
no more water or
CO2.
-The amount of
wood limits the
amount of products,
therefor, it is the
limiting reagent
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Ex. Cake baking
-The recipe calls for 2 eggs to combine with 1
cup of flour and this make 1 cake.
-How many cakes could you bake if you have 4
eggs in your fridge, and 4 cups of flour in your
pantry?
-To use up all 4 eggs, you would only need 2
cups of flour. (2 cups left over).
-Flour = reagent in excess
-Eggs = limiting reagent
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In Chemistry:
Given: 4NH3 + 5O2  6H2O + 4NO
Q - How many moles of NO are produced if
__ mol NH3 are burned in __ mol O2?
4 mol NO, works out exactly
a) 4 mol NH3, 5 mol O2
b) 4 mol NH3, 20 mol O2 4 mol NO, with leftover O2
c) 8 mol NH3, 20 mol O2 8 mol NO, with leftover O2
•
In all of these cases, NH3 limits the production
of NO; if there was more NH3, more NO would
be produced
•
Thus, NH3 is called the “limiting reagent”
In Chemistry:
Given: 4NH3 + 5O2  6H2O + 4NO
Q - How many moles of NO are produced if
__ mol NH3 are burned in __ mol O2?
(d) 4 mol NH3, 2.5 mol O2
•
•
•
2 mol NO, 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”
Sometimes the question is more complicated. For
example, if grams of the two reactants are given
instead of moles.
Procedure for solving
limiting reagent problems
1.If not already, balance the chemical equation.
2.Convert information given into units of moles.
3.Divide the mole amount of each reactant by its
stoichiometric coefficient from the balanced
chemical equation.
4.The reactant with the smallest quotient from
step #3 will be the limiting reagent.
5.Work as typical stoichiometry problem using
the limiting reagent.
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Example #1
2Ca + O2 → 2CaO
How many moles of calcium oxide can you form starting with 15
moles of calcium and 12 moles of oxygen?
1.If not already, balance the chemical equation. Balanced.
2.Convert information given into units of moles.
Already in units of moles.
3.Divide the mole amount of each reactant by its stoichiometric
ratio from the balanced chemical equation.
15 moles Ca = 7.5
12 moles O2 = 12
2 moles Ca
1 mol O2
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Example #1 (con’t)
2Ca + O2 → 2CaO
How many moles of calcium oxide can you form starting with 15
moles of calcium and 12 moles of oxygen?
4.The reactant with the smallest quotient from step #3 will be the
limiting reagent. Calcium will be the limiting reagent in this
problem.
5.Work as typical stoichiometry problem using the limiting reagent.
15mol Ca x 2 mol CaO = 15mol CaO
2 mol Ca
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Example #2
What mass of CO2 can be produced by the reaction of 8.0 grams
of CH4 with 48 grams of O2 according to the equation below?
CH4 + 2O2 → CO2 + 2H2O
1.If not already, balance the chemical equation. Balanced.
2.Convert information given into units of moles. Information given
is in units of grams. We need to use the molar mass of each
reactant to find moles of each reactant.
8.0g CH4 x 1mol CH4 = 0.50mol CH4
16.04g CH4
48g O2 x 1mol O2 = 1.50mol O2
32.0g O2
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Example #2 (con’t)
What mass of CO2 can be produced by the reaction of 8.0 grams
of CH4 with 48 grams of O2 according to the equation below?
CH4 + 2O2 → CO2 + 2H2O
3.Divide the mole amount of each reactant by its stoichiometric
ratio from the balanced chemical equation.
0.5mol CH4 = 0.5
1.5mol O2 = 0.75
1 mol CH4
2mol O2
4.The reactant with the smallest quotient from step #3 will be the
limiting reagent. CH4
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Example #2 (con’t)
What mass of CO2 can be produced by the reaction of 8.0 grams
of CH4 with 48 grams of O2 according to the equation below?
CH4 + 2O2 → CO2 + 2H2O
4.Work as typical stoichiometry problem using the limiting reagent.
0.5mol CH4 x 1mol CO2 x 44g CO2 = 22g CO2
1mol CH4 1mol CO2
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Limiting Reagents: shortcut
• Do two separate calculations using both given
quantities. The smaller answer is correct.
Ex3. 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
17.0 g NH3 4 mol NH3
=
30 g O2 x 1 mol O2 x 4 mol NO
32.0 g O2
5 mol O2
=
30.0 g NO
x
1 mol NO
35.3 g NO
30.0 g NO
x
1 mol NO
22.5 g NO
Assignment
1.
2.
3.
4.
5.
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
the steps & 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?
2
# mol Al = 25 g Al x 1 mol Al = 0.926 mol
27.0 g Al
2mol
= 0.463
# mol HCl = 90 g HCl x 1 mol HCl = 2.466 mol
36.5 g HCl
6mol
=0.411
HCl is limiting.
# 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 2: shortcut
2Al + 6HCl  2AlCl3 + 3H2
If 25 g aluminum was added to 90 g HCl, what
mass of H2 will be produced?
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
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 3: 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 4: 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 5: 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)
6.4 Assignment Con’t
6. How many grams of NH3 can be produced from the reaction of 28g of N2
and 25g of H2?
34g NH3
7. How much of the excess reagent in Problem 6 is left over?
19g H2 excess
8. What volume of hydrogen at STP is produced from the reaction of 50.0g of
Mg and the equivalent of 75g of HCl?
23.0L H2
9. How much of the excess reagent in Problem 3 is left over?
25g Mg excess
10.Silver nitrate and sodium phosphate are reacted in equal amounts of 200g
each. How many grams of silver phosphate are produced?
164.3g Ag3PO4
11.How much of the excess reagent in problem 5 is left?
135.6g21 Na3PO4
6.4
Assignment
12. Iron reacts with sulfur to form iron(II) sulfide. You have 32.0g of sulfur
and 100g of iron. Calculate the number of grams of iron(II) sulfide that
will form.
87.9g FeS
13.Zinc reacts with sulfuric acid in a single replacement reaction. You have
40g of zinc and 57g of sulfuric acid. What volume of hydrogen gas will
form?
13.0L H2
14.Silver nitrate and sodium bromide will react in a double replacement
reaction. You have 24g of silver nitrate and 39g of sodium bromide. How
many grams of sodium nitrate will form?
11.9g NaNO3
15.When hydrochloric acid reacts with iron(II) sulfide, a double replacement
reaction is to be expected. You have 67g of hydrochloric acid and 58 of
iron(II) sulfide. How many grams of hydrogen sulfide will form?
22.5g H2S
16.When aluminum is heated in oxygen, aluminum oxide is formed. You
have 384g of each reactant. How many moles of aluminum oxide will
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form?
6.45mol Al2O3
6.4 Limiting Reagents C.P.
1. A mixture of 49.1g of tin(II) nitrate and 81.2g of
sodium chloride react. How many grams of
tin(II) chloride are produced? How much excess
is left over?
2. Silicon monocarbide, commonly known as
carborundum, is prepared by heating silicon
dioxide in the presence of carbon. How many
grams of silicon monocarbide can be formed
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from 100.0g of graphite and 100.0g of silicon
6.4 Limiting Reagents C.P.
1. Silver nitrate and sodium phosphate are reacted
in equal amounts of 137g each. How many
grams of silver phosphate are produced?
2. How much of the excess reagent in #1 is left?
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What volume of hydrogen at STP is produced from the
reaction of 50.0g of Mg and the equivalent of 75g of HCl?
How much of the excess reagent in Problem 3 is left over?
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Check Point #2
A mixture of 49.1g of tin(II) nitrate and
81.2g of sodium chloride react. How
many grams of tin(II) chloride are
produced? How much excess is left over?
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