Download Step by Step Stoichiometry

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Step by Step Stoichiometry!
Mole Ratios:
A mole ratio is a conversion factor that relates the amounts in moles of any two substances
involved in a chemical reaction.
equation.
The information is found directly from the balance chemical
The coefficients in a chemical equation represent the relative amounts of moles in the
reactants and products.
Example:
2Al2O3  4Al + 3O2
The above balanced equation tells us that for every 2 moles of aluminum oxide there are 4
moles of aluminum and 3 moles of oxygen.
Now you try!
Practice:
Write the correct balanced equation for the following reaction, solid iron reacts with
oxygen to produce iron (III) oxide.
Using the balanced equation above write all the possible mole ratios (there are 6 possibilities):
Practice:
Write the correct balanced equation for the following reaction, sulfur and oxygen react
to form sulfur dioxide.
Using the balanced equation above, write all the possible mole ratios (there are 6 possibilities):
Mole to Mole:
These problems will ask you to calculate the amount in moles of one substance that will react
with or be produced from the given amount of moles in another substance.
use the mole ratio from the balanced equation to answer these problems.
You will have to
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Example:
2Al2O3  4Al + 3O2
If you are given 20 moles of Al2O3, how many moles of aluminum will be produced in a
decomposition reaction?
Set up the problem using dimensional analysis:
Things to ID before you
start:
Given: 20 moles Al2O3
Unknown: _____moles of
20 Mol
Al2O3
Al
Mole ratio of Al2O3 to Al:
4 mol Al
2 mol
= 40 moles of Al
Al2O3
2:4
Now you try!
Practice: 2.5 moles of solid iron reacts with oxygen to produce iron (III) oxide. How many
moles of iron (III) oxide will be produced in this reaction?
(You should have the balanced
reaction and mole ratios from the previous practice problems)
Practice: Sulfur and oxygen react to form sulfur dioxide.
How many moles of oxygen are
needed to react with .84 moles of sulfur? (You should have the balanced reaction and mole
ratios from the previous practice problems)
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Mass to Mass:
These problems will ask you to calculate the amount of one substance, in grams, that will be
needed to react or be produced from a given mass for another substance.
The plan for solving these kinds of problems:
Mass of given  amount of given in moles  amount of unknown in moles mass of
unknown
Example:
2Al2O3  4Al + 3O2
If you are given 6.25g of Al2O3, how many grams of aluminum will be produced in a
decomposition reaction?
Things to ID before you
start:
Given: 6.25 g Al2O3
Unknown: _____ g Al
Set up the problem using dimensional analysis:
Mole ratio of Al2O3 to Al:
2:4
Atomic mass of Al2O3:
102g
6.25 g Al2O3
1 mol Al2O3
4 mol Al
27 g Al
102 g Al2O3
2 mol Al2O3
1 mol Al
= 3.31 g Al
Atomic Mass of Al: 27 g
Note: 1 mol = atomic mass
of the element or
compound
Now you try!
Practice: 22g of solid iron reacts with oxygen to produce iron (III) oxide.
iron (III) oxide will be produced in this reaction?
How many grams of
(You should have the balanced reaction and
mole ratios from the previous practice problems)
Practice: Sulfur and oxygen react to form sulfur dioxide.
How many grams of oxygen are
needed to react with 1.23 grams of sulfur? (You should have the balanced reaction and mole
ratios from the previous practice problems)
Limiting Reactant (sometimes called limiting reagent):
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In some problems more than one amount is given.
which amount is going to be the limiting reactant.
In these cases you must first determine
The limiting reactant is the reactant that limits
the amount of product that can be produced in a reaction. Let’s look at a practical example
before we move on to a chemistry example:
Mr. Potato Head
List the 12 parts needed to make one complete Mr. Potato Head:
If I had 15 parts, how many complete Mr. Potato Heads can I make?
What about 35 parts?
As you can see I’m limited in making a complete Mr. Potato Head by the parts available to me.
Even though I have 15 parts, I can only make 1 Mr. Potato Had in the first example, leaving 3
parts in unused or in excess.
If I have 35 parts, I am one part shy of making a 3rd Mr. Potato
Head meaning I can only make 2 complete Mr. Potato Heads, leaving 11 parts unused or in
excess.
Now our chemistry example: CO + 2H2  CH3OH
A mixture of 1.20 g of H2 and 7.45 g of CO are allowed to react.
How much CH3OH will be
produced in this reaction?
Since I am given two amounts in my problem, I know this is a limiting reactant problem.
I must
first figure out which amount I am limited by, prior to solving the problem using stoichiometry.
The way to do this is to do two stoichiometry problems, one with each given to find out how
much product can be made with each amount.
The amount of reactant that makes the least
amount of product (the smaller number) is your limiting reactant and is therefore the amount of
product that can be produced in the reaction.
1.20 g H2
1 mol H2
2g H2
1 mol
34 g
CH3OH
CH3OH
2 mol H2
1 mol
CH3OH
= 10.2 g
CH3OH
CO is your
limiting
reactant
because
with 7.45 g
you can
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7.45 g CO
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1 mol CO
28 g CO
1 mol
34 g
CH3OH
CH3OH
1 mol CO
1 mol
= 9.05 g CH3OH
CH3OH
Now you try!
Practice: How many grams of nitric acid can be prepared form the reaction of 92.0 g of NO2
with 36.0 g of H2O? (Start by writing a balanced equation and go from there)
Practice:
How many grams of H2O will form when 16.0 g of H2 is allowed to react with 16.0 g
of O2?
Percent Yield:
In the lab, sometimes experiments do not go exactly as planned.
You may end up with more or
less product than you should due to experimental errors or if the experiment does not go to
completion.
You can determine how well the experiment went by calculating the percent yield.
The percent yield is the actual yield (amount of actual product measured in the lab) divided by
the theoretical yield (amount of product that should have been made in the lab based on the
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numbers you started with) multiplied by 100.
For our purposes the actual yield will be given to
you but the theoretical yield will need to be calculated using stoichiometry!
Example:
3H2 + N2  2 NH3
If 55.6 g of H2 reacts with excess N2, what is the theoretical yield of NH3?
What is the
percent yield if the actual yield was 159g?
Let’s solve the first question first then use that amount to solve the second part of the question.
First off, don’t let the word excess hold you up, it only means that H2 is the limiting reactant
and that amount is given to you so go ahead and work this problem like the typical mass to
mass stoichiometry problem it is!
55.6 g H2
1 mol H2
2 mol NH3
17 g NH3
2 g H2
3 mol H2
1 mol NH3
= 315.1 g NH3
Based on our calculations, we should have made 315.1 g of product but the second part of the
question tells us that we only made 159 g of product when we conducted this experiment in the
lab. We will use this information to calculate the percent yield.
Actual yield
Theoretical
X 100
159 g NH3
X
315.1 g NH3 100
yield
Now you try!
= 50.46
%
Our percent yield
indicates that we
only made about half
the product that we
should have in the
lab
Practice:
The reaction of 25.0 g of benzene, C6H6, with excess HNO3 resulted in 21.4 g
C6H5NO2.
What is the percent yield?