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CHAPTER 9: STOICHIOMETRY
9.1 Interpreting a chemical Equation
H2 (g) +
1 molecule
N2
____molecule(s)
Cl2 (g)
1 molecule
+
2 HCl (g)
2 molecules
3 H2 (g)
____molecule(s)
2 NH3 (g)
____molecule(s)
It follows that any multiples of these coefficients will be in same ratio!
2 H2 (g)
O2 (g) +
X 1000 ____molecule(s)
XN
____molecule(s)
2 H2O (g)
____molecule(s) ____ molecule(s)
____molecule(s) ____ molecule(s)
Since N=Avogadro’s # = 6.02x1023 molecules = 1 mol
2 H2 (g)
+
____mol (s)
O2 (g) 2 H2O (g)
____mol (s)
____ mol (s)
Thus, the coefficients in a chemical equation give the mole ratios of reactants and products
in a reaction.
Give the mole ratios for each of the following:
1.
H2 (g) +
____mol(s)
2.
C3H8 (g)
____mol(s)
Cl2 (g) 2 HCl (g)
____mol(s)
+
5 O2 (g)
____mol(s)
____mol(s)
3 CO2 (g)
____mol(s)
+ 4 H2O (g)
____mol(s)
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9.2
MOLE-MOLE RELATIONSHIPS
Consider the following reaction:
C3H8 (g)
+
5 O2 (g)
3 CO2 (g)
+
4 H2O (g)
1. Use unit factors to determine how many moles of O2 are needed to completely react
with 2.25 moles of C3H8.
2. How many moles of CO2 form when 3.50 moles of O2 react?
3. How many moles of H2O form when 4.75 moles of CO2 form?
4. How many moles of C3H8 are required to produce 1.50 moles of H2O?
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9.3 Stoichiometry Problems
stoichiometry (STOY-key-OM-etry):
-refers to the amounts of reactants and products in a chemical reaction
-a stoichiometry problem generally involves relating amounts of reactants and/or
products to each other in terms of moles
9.4 Mass-Mass (Stoichiometry) Problems
Molar
MASS OF
KNOWN
Mole-Mole
MOLES OF
KNOWN
Mass
Molar
MOLES OF
UNKNOWN
Ratio
MASS OF
UNKNOWN
Mass
Example: Consider the mass of CO needed to react completely with 50.0 g of Fe2O3.
Fe2O3 (s) + 3 CO 2 Fe (s) + 3 CO2 (g)
1. Calculate the mass of CO needed to react completely with 50.0 g of Fe2O3.
2. Calculate the mass of iron produced when 123 g of CO reacts completely.
3. Calculate the mass of CO2 produced when 75.0 g of iron is produced.
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9.5 The Limiting Reactant Concept (LIMITING REAGENT)
In practice, reactants will not always be present in the exact amounts necessary for all
reactants to be converted completely into products.
Some reactants (usually the least expensive) are present in larger amounts and are never
completely used up
“reactant(s) in excess”
Only in a limited supply of the other reactants (usually the more expensive) are present, so
these are completely used up
“limiting reactant” since it limits the amount of product that can be made
MAKING BICYCLES
-Parts needed:
- 1 bicycle frame
- 1 seat
- 2 pedals
- 2 wheels (rims + tires)
Example: How many bicycles can be made with 5 frames, 6 seats, 15 pedals and 8 wheels?
(Indicate the limiting reactant and reactants in excess.)
4
MAKING BISQUICK
TM
PANCAKES
2 cups Bisquick
1 cup milk
2 eggs
14 pancakes
2 cups of BisquickTM + 1 Cup milk + 2 eggs 14 pancakes
Example: If you have 10 cups of BisquickTM, 10 cups of milk, and 12 eggs, how many
pancakes can you make? (Indicate the limiting reagent(s) and reagent(s) in excess.)
9.6 Limiting Reactant Problems
GUIDELINES FOR SOLVING LIMITING REAGENT PROBLEMS:
1. Calculate the amount (moles or mass) of product formed using the amount of each
reactant given
-Use mass-mass conversions
Smallest amount = amount of product formed!
2. Whichever reactant produces the smaller amount of product
limiting reagent
3. All other reactant(s) in excess
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Ex. 1: Consider the reaction between aluminum metal and hydrochloric acid to produce
hydrogen gas:
2 Al(s) + 6 HCl (aq) 2 AlCl3 (aq) + 3 H2 (g)
Calculate the number of moles of hydrogen gas produced when 5.00 moles of
aluminum metal react with 5.00 moles of HCl.
Limiting reactant = ____________________ Reactant in excess =___________________
Ex. 2: Consider the reaction fro propane (C3H8) burning:
C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (g)
Calculate the number of moles of Carbon dioxide gas produced. When 1.50 moles of
propane react with 5.00 moles of oxygen.
Limiting reactant = ____________________ Reactant in excess =___________________
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actual yield
9.7 Percent yield =
x 100%
theoretical yield
theoretical yield: Amount of product one should get based on the chemical equation and
the amount of reactants present
-One generally calculates this in grams from info given
Actual yield: Amount of produce one actually obtains
-Generally smaller than the theoretical yield because of impurities
and other adverse conditions in the lab
-This value has to be provided in a problem
Ex.: Calculate the percent yield for a reaction with a theoretical yield of 75.0 g of
carbon dioxide if the actual amount of carbon dioxide produced was 59.2 g.
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Example: Consider the following equation:
2 K + Cl2 2 KCl
a. How many grams of KCl is produced from 2.50 g of K and excess Cl2.
b. From 1.00 g of Cl2 and excess K?
c. When 2.50 g of K and 1.00 g Cl2 react together, the mass of KCl produced is
_________________________, the limiting reactant is_________________,
and the reactant in excess is _________________________.
d. Calculate the percent yield if 86.7 g of KCl is actually produced when 2.50 g of
K and 1.00g Cl2 react.
8
CHAPTER 9: STOICHIOMETRY
9.6 Interpreting a chemical Equation
H2 (g) +
1 molecule
N2
_1___molecule(s)
Cl2 (g)
1 molecule
+
2 HCl (g)
2 molecules
3 H2 (g)
2 NH3 (g)
__3__molecule(s)
__2__molecule(s)
It follows that any multiples of these coefficients will be in same ratio!
2 H2 (g)
+
O2 (g) _1000___molecule(s) _2000___ molecule(s)
X 1000 _2000___molecule(s)
XN
2 H2O (g)
1.204 x 1023 molecule(s)
6.02 x 1023 molecule(s) 1.204 x 1023 molecule(s)
Since N=Avogadro’s # = 6.02x1023 molecules = 1 mol
2 H2 (g)
+
_2___mol (s)
O2 (g) 2 H2O (g)
__1__mol (s)
__2__ mol (s)
Thus, the coefficients in a chemical equation give the mole ratios of reactants and products
in a reaction.
Give the mole ratios for each of the following:
1.
H2 (g) +
__1_mol(s)
2.
C3H8 (g)
_1__mol(s)
Cl2 (g) __1_mol
+
2 HCl (g)
__2_mol(s)
5 O2 (g)
__5_mol(s)
3 CO2 (g)
_3__mol(s)
+ 4 H2O (g)
__4_mol(s)
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9.7
MOLE-MOLE RELATIONSHIPS
Consider the following reaction:
C3H8 (g)
+
5 O2 (g)
3 CO2 (g)
+
4 H2O (g)
4. Use unit factors to determine how many moles of O2 are needed to completely react
with 2.25 moles of C3H8.
2.25 mol C3H8 x 5 mol O2
1 mol C3H8
= 11.3 mol O2
5. How many moles of CO2 form when 3.50 moles of O2 react?
3.50 mol O2 x 3 mol CO2
5 mol O2
= 2.10 mol CO2
6. How many moles of H2O form when 4.75 moles of CO2 form?
4.74 mol CO2 x 4 mol H2O
3 mol CO2
= 6.33 mol H2O
4. How many moles of C3H8 are required to produce 1.50 moles of H2O?
1.50 mol H2O x 1 mol C3H8
4 mol H2O
= 0.375 mol C3H8
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9.8 Stoichiometry Problems
stoichiometry (STOY-key-OM-etry):
-refers to the amounts of reactants and products in a chemical reaction
-a stoichiometry problem generally involves relating amounts of reactants and/or
products to each other in terms of moles
9.9 Mass-Mass (Stoichiometry) Problems
Molar
MASS OF
KNOWN
Mole-Mole
MOLES OF
KNOWN
Mass
Molar
MOLES OF
UNKNOWN
Ratio
MASS OF
UNKNOWN
Mass
Example: Consider the mass of CO needed to react completely with 50.0 g of Fe2O3.
Fe2O3 (s) + 3 CO 2 Fe (s) + 3 CO2 (g)
4. Calculate the mass of CO needed to react completely with 50.0 g of Fe2O3.
Given: 50.0 g Fe2O3
Find: g CO
Format: g Fe2O3 mol Fe2O3 mol CO g CO
50.0 g Fe2O3 x 1 mol Fe2O3 x 3 mol CO x
160. g Fe2 O3
1 mol Fe2O3
28.0 g CO
1 mol CO
= 26.3 g CO
5. Calculate the mass of iron produced when 123 g of CO reacts completely.
Given: 123 g CO
Find: g Fe
Format: g CO mol CO mol Fe g Fe
123 g CO x 1 mol CO
28.0 g CO
x 2 mol Fe x
3 mol CO
55.8 g Fe
1 mol Fe
= 163 g Fe
6. Calculate the mass of CO2 produced when 75.0 g of iron is produced.
Given: 75.0 g Fe
Find: g CO2
Format: g Fe mol Fe
mol CO2 g CO2
75.0 g Fe x 1 mol Fe
55.8 g Fe
x 3 mol CO2 x
2 mol Fe
44.0 g CO2
1 mol CO2
= 88.7 g CO2
11
9.10 The Limiting Reactant Concept (LIMITING REAGENT)
In practice, reactants will not always be present in the exact amounts necessary for all
reactants to be converted completely into products.
Some reactants (usually the least expensive) are present in larger amounts and are never
completely used up
“reactant(s) in excess”
Only in a limited supply of the other reactants (usually the more expensive) are present, so
these are completely used up
“limiting reactant” since it limits the amount of product that can be made
MAKING BICYCLES
-Parts needed:
- 1 bicycle frame
- 1 seat
- 2 pedals
- 2 wheels (rims + tires)
Example: How many bicycles can be made with 5 frames, 6 seats, 15 pedals and 8 wheels?
(Indicate the limiting reactant and reactants in excess.)
1 bicycle frame + 1 seat + 2 pedals + 2 wheels = 1 bicycle
5 frames x 1 bike = 5 bikes
1 frame
6 seats x 1 bike = 6 bikes
1 seat
15 pedals x 1 bike = 7.5 bikes
2 pedals
12
8 wheels x 1 bike = 4 bikes
2 wheels
Limiting reactant – wheels
Excess reactants – Bicycle frames, seats, pedals
MAKING BISQUICK
TM
PANCAKES
2 cups Bisquick
1 cup milk
2 eggs
14 pancakes
2 cups of BisquickTM + 1 Cup milk + 2 eggs 14 pancakes
Example: If you have 10 cups of BisquickTM, 10 cups of milk, and 12 eggs, how many
pancakes can you make? (Indicate the limiting reagent(s) and reagent(s) in excess.)
2 c Bisquick + 1 C milk + 2 eggs = 14 pancakes
10 c. Bisquick x 14 pancakes = 70 pancakes
2 c. Bisquick
10 c milk x 14 pancakes = 140 pancakes
1 c milk
12 eggs x 14 pancakes = 84 pancakes
2 eggs
Limiting reagent = Bisquick
Excess reagent = milk, eggs
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9.6 Limiting Reactant Problems
GUIDELINES FOR SOLVING LIMITING REAGENT PROBLEMS:
4. Calculate the amount (moles or mass) of product formed using the amount of each
reactant given
-Use mass-mass conversions
Smallest amount = amount of product formed!
5. Whichever reactant produces the smaller amount of product
limiting reagent
6. All other reactant(s) in excess
Ex. 1: Consider the reaction between aluminum metal and hydrochloric acid to produce
hydrogen gas:
2 Al(s) + 6 HCl (aq) 2 AlCl3 (aq) + 3 H2 (g)
Calculate the number of moles of hydrogen gas produced when 5.00 moles of
aluminum metal react with 5.00 moles of HCl.
+ 6 HCl 2 AlCl3 + 3 H2
2 Al
5.00 mol Al x 3 mol H2 = 7.5 mol H2
2 mol Al
5.00 mol HCl x 3 mol H2 = 2.5 mol H2
6 mol HCl
Limiting reactant = ___HCl____________ Reactant in excess =_______Al____________
Ex. 2: Consider the reaction fro propane (C3H8) burning:
C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (g)
Calculate the number of moles of Carbon dioxide gas produced. When 1.50 moles of
propane react with 5.00 moles of oxygen.
C3H8 + 5 O2
3 CO2 + 4 H2O
1.50 mol C3H8 x 3 mol CO2 = 4.5 mol CO2
1 mol C3H8
14
5.00 mol O2 x 3 mol CO2 = 3 mol CO2
5 mol O2
Limiting reactant = _____O2_________ Reactant in excess =_________C3H8__________
actual yield
9.7 Percent yield =
x 100%
theoretical yield
theoretical yield: Amount of product one should get based on the chemical equation and
the amount of reactants present
-One generally calculates this in grams from info given
Actual yield: Amount of produce one actually obtains
-Generally smaller than the theoretical yield because of impurities
and other adverse conditions in the lab
-This value has to be provided in a problem
Ex.: Calculate the percent yield for a reaction with a theoretical yield of 75.0 g of
carbon dioxide if the actual amount of carbon dioxide produced was 59.2 g.
59.2 g CO2 x 100 = 78.9%
75.0 g CO2
15
Example: Consider the following equation:
2 K + Cl2 2 KCl
a. How many grams of KCl is produced from 2.50 g of K and excess Cl2.
Given: 2.50 g K
Find: g KCl
Format: g K mol K mol KCl g KCl
2.50 g K x 1 mol K x 2 mol KCl x 74.6 g KCl = 4.77 g KCl
39.1 g K 2 mol K
1 mol KCl
b. From 1.00 g of Cl2 and excess K?
Given: 1.00 g Cl2
Find: g KCl
Format: g Cl2 mol Cl2 mol KCl g KCl
1.00 g Cl2 x 1 mol K x 2 mol KCl x 74.6 g KCl = 2.10 g KCl
1 mol KCl
70.9 g Cl2 1 mol Cl2
c. When 2.50 g of K and 1.00 g Cl2 react together, the mass of KCl produced is
_____2.10 g_______________, the limiting reactant is____Cl2____________,
and the reactant in excess is _______K__________________.
d. Calculate the percent yield if 86.7 g of KCl is actually produced when 2.50 g of
K and 1.00g Cl2 react.
16
86.7 x 100 = 4130%
2.10
17