Stoichiometric balancing coefficients
Chemistry 6A F2007
They are the numbers in front of the chemical
formulas.
Dr. J.A. Mack
They give the ratio of reactants and products.
Conversion
factors
ratio
The balancing coefficients allow us to convert between
numbers of reactants and products.
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Stoichiometric ratio: The ratio of any two species
(reactants or products) in a balanced chemical reaction.
A2B3
2A
2 mols A
3B
3 mols B
The molar ratios allow us to relate the amounts of reactants
and products in a chemical equation.
2 A’s combine with 3B’s
CONVERSION FACTORS!!!
2A
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i.e.
3B
3B
2
Since the individual ratios must scale to moles, we can
write:
Stoichiometry: The branch of chemistry that deals with
the mole proportions of chemical reactions.
2A + 3B
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3 mols B ×
2 mols A
= 2 mols A
3 mols B
2A
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Quantitative calculations: Mass and moles
2HCl (aq) + Ba(OH)2 (aq) Æ 2H2O (l) + BaCl2(aq)
Consider the following reaction:
HCl (aq) + Ba(OH)2 (aq) Æ H2O (l) + This
BaCl2means
(aq) that
there is more
than enough
2 HCl (aq) + Ba(OH)2 (aq) Æ 2 H2O (l) + BaCl
Ba(OH)
2(aq) 2 to react
all of the HCl.
Balacing:
How many moles of HCl are consumed if 1.50 g of BaCl2 are
produced assuming that Ba(OH)2 is in excess?
Solution:
convert
g BaCl2
to
then convert
Æ
mol BaCl2
using the Molar mass of BaCl2
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the calculation follows the sequence…
g BaCl2 Æ mol BaCl2
1.50 g BaCl 2 ×
1 mol BaCl 2
2 mol HCl
×
= 0.0144 mol HCl
208.24 g BaCl 2 1 mol BaCl 2
to
Æ
Molar mass BaCl2
mol HCl
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Step 1: Put your calculator down!
Step 2: Write the balanced chemical equation.
1 CO2(g)
+
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mols product
stoichometric
molar ratios
CaO(s)
Step 3: use the molar ratios to convert the moles
1mol CO 2
1mol CaCO3
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We can also determine the number of grams of each
product that will be produced:
mols reactant
3.55mol CaCO3 ×
Molar ratio for equation
using the Molar ratio for equation
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Calcium carbonate decomposes to form carbon dioxide and calcium
oxide
If 3.55 mol of calcium carbonate decomposes, how many mols of
carbon dioxide will form?
1 CaCO3(s)
Æ mol HCl
3.55mol CaCO3 ×
= 3.55mol CO 2
mass product
molar mass
44.01g CO 2
1mol CO 2
×
=
1mol CO 2
1mol CaCO3
= 156g CO2
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2
If 3.50g of calcium oxide remain after the decomposition
of calcium carbonate, how many moles of calcium
carbonate must have been initially present?
(a) How many grams of nitrogen will be needed to
produce 0.384 mols of ammonia?
N2(g) + 3 H2(g)
Step 1: Write the balanced chemical equation
CaCO3 (s) → CaO(s) + CO2 (g)
0.384 mols NH3 ×
Step 2: Outline your conversion
g CaO
3.50g CaO ×
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mols CaO
mols CaCO3
1mol CaO 1mol CaCO3
×
56.08g CaO 1mol CaO
= 0.0624mol CaCO3
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How many grams of nitrogen are needed to completely
react with 0.525 g of hydrogen in the formation of
ammonia?
1mol N 2 28.02g N 2
×
2mol NH 3 1mol N 2
= 5.38g N2
(b) How many moles of hydrogen are needed to combine
with 5.84g of nitrogen?
1mol N 2
3mol H 2
×
5.38g N 2 ×
= 0.576 mols H 2
28.02 gN 2
1mol N 2
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How many grams of nitrogen are needed to completely
react with 0.525 g of hydrogen in the formation of
ammonia?
To answer this question, one must go through moles.
N2(g) + 3 H2(g)
2 NH3(g)
N2(g) + 3 H2(g)
2 NH3(g)
0.525gH 2×
The equation relates moles, not mass.
2 NH3(g)
1 mol H 2 1mol N 2
28.02g N 2
×
×
= 2.43g N 2
2.02g H 2 3mol H 2
1mol N 2
How many grams of ammonia form?
g H2
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mols H2
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mols N2
g N2
0.525gH 2×
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1 mol H 2 2 mol NH 3 17.04g NH 3
×
×
= 2.95g NH 3
1mol NH 3
2.02g H 2
3mol H 2
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Stoichiometry
When propane (C3H8) burns the products are carbon dioxide and
water. How many grams of water will result form the burning of
5.05g propane in excess oxygen.?
Excess reagent (reactant): The reactant that is not
completely consumed during a chemical reaction.
Where do you start?
Burning = combustion
combustion = reacting with oxygen
Limiting reagent (reactant): The reactant that is
completely consumed in a chemical reaction, while one
or more other reactants is not consumed.
Write and balance the equation!
C3H8(g) + 5 O2(g)
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When propane (C3H8) burns the products are carbon dioxide and
water. How many grams of water will result form the burning of
5.05g propane in excess oxygen.?
C3H8(g) + 5 O2(g)
The answer to that question is:
3 CO2(g) + 4 H2O(l)
The amount of water produced is determined by the amount of
propane that reacts.
5.05gC3 H 8 ×
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mols C3H8
mols H2O
g H2O
1mol C3 H 8
4mol H 2 O 18.02g H 2 O
×
×
= 8.25g H 2 O
44.11g C3 H 8 1mol C3 H 8 1mol H 2 O
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What happens when we we don’t have an excess of oxygen?
Will the same amount of water be formed?
Excess oxygen means more than enough to react all of the propane.
g C3H8
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3 CO2(g) + 4 H2O(l)
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Depends…
It depends upon the relative amounts of each reactant.
If one reactant is used up before the other is completely consumed,
Then it limits the reaction: it is the limiting reactant!
When propane (C3H8) burns the products are carbon dioxide and
water. How many grams of water will result form the burning of
5.05g propane with 10.1g of oxygen.?
Which one limits, the propane or the oxygen!
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When propane (C3H8) burns the products are carbon dioxide and
water. How many grams of water will result form the burning of
5.05g propane with 10.1g of oxygen.?
When propane (C3H8) burns the products are carbon dioxide and
water. How many grams of water will result form the burning of
5.05g propane with 10.1g of oxygen.?
The limiting reactant will be that reactant which yields the smallest
amount of an in common product.
C3H8(g) + 5 O2(g)
5.05gC3 H 8 ×
3 CO2(g) + 4 H2O(l)
Convert:
g C3H8
mols or grams H2O or CO2
then convert:
g O2
mols or grams H2O or CO2
C3H8(g) + 5 O2(g)
10.1g O2 ×
Dr. Mack. CSUS
1mol C3 H 8
4mol H 2 O 18.02g H 2 O
×
×
= 8.25g H 2 O
44.11g C3 H 8 1mol C3 H 8 1mol H 2 O
1mol O 2
4mol H 2 O 18.02g H 2 O
×
×
= 4.55 gH 2 O
32.00g O2
5mol O2
1mol H 2 O
Even though we start with more O2 by mass, it is used up first by moles!
Which produces the least is the limiting reactant.
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3 CO2(g) + 4 H2O(l)
Oxygen is the limiting reactant.
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4Al(s) + 3O2(g) → 2Al2O3(s)
Consider the reaction of aluminum and oxygen:
Which is the limiting reactant if we start with 50.0 g Al and 50.0 g O2?
4Al(s) + 3O2(g) → 2Al2O3(s)
Which is the limiting reactant if we start with 50.0 g Al and 50.0 g O2?
How much O2 remains?
How much aluminum oxide will form?
Solution:
Solution: since Al limits, the amount of O2 left over and the amount
of product formed are determined by the moles of Al.
1 mol Al
3mol O 2 32.00 g O 2
×
50.0g Al ×
×
= 44.5g O2 needed
26.98g Al 4 mol Al 1mol O 2
g Al Æ mol Al Æ mol O2 Æ g O2 reacted
Since we start with 50.0 g of O2 and we need 44.5g of O2,
one can conclude that:
g O2 initial
Al limits
g AlÆ mol Al Æ mol Al2O3 Æ
i.e. there is more than enough O2 to react with all of the Al.
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– g O2 reacted
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= g O2 left
g Al2O3 produced
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4Al(s) + 3O2(g) → 2Al2O3(s)
50.0 g Al ×
1 mol Al 3 mol O 2 32.00 g O 2
×
×
26.98 g Al 4 mol Al 1 mol O 2
50.0 g O 2 − 44.5 g O 2
50.0 g Al ×
= 5.5 g O 2 left over
1 mol Al 2 mol Al2 O3 101.96 g Al 2 O3
×
= 94.5 g Al 2 O3 forms
×
1 mol Al2 O3
26.98 g Al 4 mol Al
note that 94.5g of Al 2 O3 + 5.5 g O 2 = 100.0 g
= 50.0 g of Al + 50.0g O2 = 100.0g
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= 44.5 g O 2 reacted
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×
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