Transcript 1 mol - RHS

Stoichiometry
Unit: Stoichiometry
Topic: Intro to Stoichiometry
Objectives: Day 1 of 4
• To learn how we go from moles of a
reactant to moles of a product
• To learn how to calculate between moles of
reactants to moles of products
Quickwrite
Answer one of the questions below 1-2
sentences:
• A sandwich recipe requires 2 pieces of bread, 2
slices of meat and 1 slice of cheese; you need to
make 50 sandwiches, how many slices of bread,
meat and cheese do you need????
Stoichiometry
• Chemistry is really all about reactions
• Reactions involve the rearrangement of atoms
• The calculation of the quantities of chemical
elements or compounds involved in chemical
reactions is called Stoichiometry
• It is the coefficients in the balanced chemical
equation that enables us to determine just how
much product forms
• What we once called coefficients are now called
moles!!!!
What is Stoichiometry?
quantities of
• The calculation of ________
chemical elements or
compounds involved in
reactions
chemical ________
• What we once called
coefficients are now called
_____
moles
Answer Bank
Reactants
conversion
moles
Quantities
Reactions
products
Stoichiometry
• To explore this idea, consider a nonchemical analogy
• A particular type of sandwich requires 2
pieces of bread, 3 slices of meat, and 1
slice of cheese
• Or, you might represent this by:
2 pieces of bread + 3 slices of meat + 1 slice of cheese →1 sandwich
Stoichiometry
• Now, your boss sends you to the store to get
enough ingredients to make 50 sandwiches
• How do you figure out how much of each
ingredient you need to buy?
• You could multiply the previous equation by
50:
50(2 pieces of bread) + 50(3 slices of meat) + 50(1 slice of cheese) → 50(1 sandwich)
100 pieces of bread + 150 slices of meat + 50 slice of cheese
→ 50 sandwich
Stoichiometry
• Notice that the numbers 100:150:50
correspond to the ratio 2:3:1, which
represents the original numbers of bread,
meat and cheese
• The equation for chemical reaction gives
you the same type of information
• It indicates the relative numbers of
reactant and product molecules required
for the reaction to take place
Stoichiometry
• To illustrate how this idea works, consider
the reaction between gaseous carbon
monoxide and hydrogen to produce liquid
methanol
• The reactants and products are:
• Unbalanced: CO(g) + H2(g) → CH3OH(l)
• Because atoms are just rearranged (not
created or destroyed) in a chemical
reaction, we must always balance the
chemical equation
Stoichiometry
Balanced: CO(g) + 2H2(g) → CH3OH(l)
• It is important to recognize that the coefficients in
a balanced equation give the relative number of
molecules
• That is, we could multiply this balanced equation
by any number and still have a balanced equation
• For example, we could multiply by 12,
• 12 [ CO(g) + 2H2(g) → CH3OH(l) ] to obtain
• 12 CO(g) + 24H2(g) → 12CH3OH(l)
Stoichiometry
• 12 CO(g) + 24H2(g) → 12CH3OH(l)
• This is still a balanced equation
• Because 12 represents a dozen, we could even
describe the reaction in terms of dozens:
1 dozen CO(g)+ 2 dozen H2(g) →1 dozen CH3OH(l)
• We could also multiply the original equation by
a very large number, such as 6.022 x 1023
Stoichiometry
• 6.022 x 1023 [ CO(g) + 2H2(g) → CH3OH(l) ]
• Which leads to the equation below:
6.022 x 1023 CO(g) + 2(6.022 x 1023 )H2(g) → 6.022 x 1023 CH3OH(l)
• We also know this number 6.022 x 1023
is equal to What?????
• Our equation can be written in terms of
moles:
1 mol CO(g) + 2 mol H2(g) → 1 mol CH3OH(l)
Mole – Mole Relationships
• Now that we have discussed the meaning of a
balanced chemical reaction in terms of moles of
reactants and products, we can use an equation
to predict the moles of products that a given
number of moles of reactants will yield
• For example, consider the reaction of the
decomposition of water:
• 2H2O(l) → 2H2(g) + O2(g)
• The equation tells us that 2 mol of H2O yields 2
mol of H2 and 1 mol of O2
Mole – Mole Relationships
• Now suppose we decompose 5.8 mol of water
• What number of moles of products are formed
in this process?
• We could answer this question by re-balancing
the equation as follows:
• First, we divide all coefficients of the balanced
equation by 2,
• 2H2O(l) → 2H2(g) + O2(g) This gives:
• H2O(l) → H2(g) + ½ O2(g)
Mole – Mole Relationships
• Now because we have 5.8 mol of water,
we multiply this equation by 5.8
• 5.8 [H2O(l) → H2(g) + ½ O2(g) ] which gives,
• 5.8H2O(l) → 5.8H2(g) + 5.8(½) O2(g)
• 5.8H2O(l) → 5.8H2(g) + 2.9 O2(g)
• So, 5.8 mol of water yields 2.9 mol of of
Oxygen and 5.8 mol of hydrogen
Mole Ratio
• Take a look at the equation below:
2H2O(l) → 2H2(g) + 1O2(g)
• The above equation tells us that 2 moles of H2O
will produce 2 moles of H2 and 1 mol of O2
2 moles H2O or __2 moles H2___
• We can write this as a ratio:
2 moles H2
2 moles H2O
• Or, we can use this ratio:
2 moles H2O
1 moles O2
or
1 moles O2___
2 moles H2O
__
• This ratio is important because it allows us to go from moles
of reactants to moles of products
• The mole ratio is conversion factor that allows us to go from
moles of reactants to moles of products
moles of A on the reactant side moles B on the product
What is the mole ratio?
conversion factor that takes
• A __________
us from moles of ________
reactants to
moles of ________
products
moles of A on the reactant side moles B on the product
Answer Bank
Reactants
conversion
moles
Quantities
Reactions
products
Practice: First balance the equations
below and determine the mole ratios
• 3NO2(g) + H2O (l) → 2 HNO3(aq) + NO (g)
• Determine the mole ratio for NO2 and HNO3
3 moles of NO2
2 moles of HNO3
or
2 moles of HNO3
3 moles of NO2
• C6H6 (g) + 3 H2 (g) → C6H12(g)
• Determine the mole ratio for H2 and C6H12
3 moles of H2
1 moles of C6H12
or
1 moles of C6H12
3 moles of H2
Practice:
• Using the equation below, what number of
moles of O2 will be produced by the
decomposition of 6.4 mol of water
• 2H2O(l) → 2H2(g) + O2(g)
6.4 mol H2O
1 mol O2
= 3.2 mol of O2
2 mol H2O
Practice:
• Using the equation below, calculate the
number of moles of NH3 that can be made
from 1.3 mol H2 reacting with excess N2
• N2(g)+ 3H2(g) → 2NH3(g)
1.3 mol H2
2 mol NH3
3 mol H2
= 0.867 mol of NH3
Practice:
• How many moles of N2 are needed to
produce 8.5 moles of NH3?
• N2(g)+ 3H2(g) → 2NH3(g)
8.5 mol NH3
1 mol N2
2 mol NH3
= 4.25 mol of N2
Practice:
• Using the equation below, calculate the
number of grams of NH3 that can be made
from 3.9 mol H2 reacting with excess N2
• N2(g)+ 3H2(g) → 2NH3(g)
3.9 mol H2 2 mol NH3
3 mol H2
17.0 g of NH3
1 mol NH3
= 44.2 grams
of NH3
Summarize:
• Stoichiometry is……..
• The ____ ____is a conversion factor that takes us
from moles of _____ to moles of ____
• When performing stoichiometry problems always
make sure that your _____ cancel
• Review: A solid that dissolves in water is ______ & a
solid that does not dissolve in water is _______
Unit: Stoichiometry
Topic: Mass Calculations
Objectives: Day 2 of 4
• To learn how to perform mass calculations
(convert grams of a product into grams of a
reactant)
Quickwrite
Answer one of the questions below 1-2
sentences:
• In chemistry, we count by weighing moles; We
can’t measure moles in a lab, BUT what unit of
measurement can we use in the lab to count
atoms or molecules???
• Review: Using the equation below, 2 moles of
H2O will produce how many moles of O2 and H2?
2H2O(l) → 2H2(g) + 1O2(g)
Mass Calculations
• We just saw how to use balanced equations for
a reaction to calculate the numbers of moles
• Remember, moles represent numbers of
molecules and we cannot count molecules
directly
• In chemistry, we count by weighing
• When we weigh we use the gram, therefore we
need to learn how to convert moles to mass
Mass Calculations
• Let’s consider an unbalanced combustion
reaction in which propane reacts with
oxygen to produce carbon dioxide and
water
• C3H8(g) + O2(g) → CO2(g) + H2O(g)
• What mass of oxygen will be required to
react exactly with 44.1 grams of propane?
Mass Calculations
• First, we need to balance the equation:
• C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
• Let’s summarize what we know and what we
want to find
• What we know:
– The balanced equation for the reaction
– The mass or amount of propane availible(44.1g)
• What we want to calculate:
– The mass of oxygen required to react exactly
with all the propane
Mass Calculations
44.1 g propane
requires
? Grams of O2
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
Our overall plan of attack is as follows:
1. We are given the number of grams of propane,
so we must convert to moles of propane
(C3H8), because the balanced equation deals
in moles not grams
2. Next, we can use the coefficients in the
balanced equation to determine the moles of
oxygen(O2) required
3. Finally, we will use the molar mass of O2 to
calculate grams of oxygen
Our Plan of Attack!
We are
given grams
of propane
We have to
convert grams
of propane
into moles
44.1 g propane
? moles of propane
Use mole ratio
2 convert moles
of propane into
moles of O2
We have to
convert moles
into grams
of O2
? moles of O2
? Grams of O2
44.1 g C3H8 requires ?? g O2
C3H8(g)
+
5O2(g) →
3CO2(g) + 4H2O(g)
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
••• Next,
wegiven
can
use
the
the
We are
the
number
of grams
of in
propane,
Finally,
we
will
use
thecoefficients
molar
mass
of
so to
wecalculate
must
convert
to
moles
of propane
(C3H8),
balanced
equation
to
determine
the
moles
O
grams
of
oxygen
2
the balanced
equation deals in moles
ofbecause
oxygen(O
2) required
not grams
44.1 g C3H8
1 mol C3H8
5 mol O2
32.0 g O2
44.09 g C3H8 1 mol C3H8 1 mol O2
= 160 g
of O2
Practice:
2Al(s) + 3I2(g) → 2AlI3(s)
• Consider the above reaction
• Calculate how many grams of the product aluminum
Iodide (AlI3)would be produced by the complete reaction
of 35.0 grams of Aluminum (Al)
Our Plan of Attack!
We are
given grams
of Al
We have to
convert grams
of Al into
moles of Al
Use mole ratio
2 convert moles
of Al into
moles of AlI3
35.0 g of Al
? moles of Al
? moles of AlI3
2Al(s) + 3I2(g) → 2AlI3(s)
We have to
convert moles
into grams
of AlI3
? Grams of AlI3
2Al(s) + 3I2(g) → 2AlI3(s)
• Back to the problem!!!!
• Calculate how many grams of the product aluminum
Iodide (AlI3)would be produced by the complete reaction
of 35.0 grams of Aluminum (Al)
35.0 g Al
1 mol Al
26.98 g Al
2 mol AlI3 407.68 g AlI3
2 mol Al
1 mol AlI3
= 528.87 g
of AlI3
Practice:
Br2(l) + 2NaCl(aq) → 2NaBr(aq) + Cl2(g)
• Consider the above balanced reaction
• Calculate how many grams of the product
chlorine(Cl2) be produced by the complete
reaction of 5.0 grams of sodium chloride (NaCl)
5.0 g NaCl
1 mol NaCl
1 mol Cl2
70.9 g Cl2
58.43 g NaCl 2 mol NaCl 1 mol Cl2
= 3.03 g of Cl2
Practice:
•
•
How many grams of chlorine gas are
needed to produce 10.0 g of sodium
chloride? Remember to balance first!!!!!
Cl2 + 2NaI  2 NaCl + I2
10.0 g NaCl
1 mol NaCl
1 mol Cl2
70.9 g Cl2
= 6.06 g Cl2
58.43 g NaCl 2 mol NaCl 1 mol Cl2
Summarize: (fill in the blank)
You are given
____ of reactant
Grams of reactant
Use Molar mass
Of reactant to
Get moles of
______
___ of reactant
Use mole _____
To get from
Moles of reactant
To moles of
product
Moles of product
Answer Bank
Product
Ratio
Mass
moles
reactant
grams
Use Molar ___
Of product to
Get grams of
product
Grams of_____
Unit: Stoichiometry
Topic: Limiting Reactant
Objectives: Day 3 of 4
• To learn what the limiting reactant is in a
chemical reaction
• To learn how to calculate the limiting
reactant using moles of reactants
Quickwrite
Answer one of the questions below 1-2
sentences:
• A sandwich recipe requires 2 pieces of bread, 3
slices of meat and 1 slice of cheese; you go into
the kitchen and realize that you have 2 pieces of
bread, 1 slice of cheese, and NO MEAT; what
limited affected your ability to make your
sandwich???
Limiting Reactants
• Earlier, we discussed making sandwiches
• Recall, that the sandwich making process
could be described as follows:
2 pieces of bread + 3 slices of meat + 1 slice of cheese →1 sandwich
• In this equation, all the products are used up,
nothing was left over
• Now assume you came to work one day and
found the following quantities of ingredients
Limiting Reactants
• Now assume you came to work one day and
found the following quantities of ingredients
• 20 slices of bread
• 24 slices of meat
• 12 slices of cheese
• How many sandwiches can you make?
• What will be left over
Limiting Reactants
• To solve this problem, let’s see how many
sandwiches we can make with each
ingredient:How many sandwiches
• Bread:
• Meat:
can
you of
make?
answer
1 sandwich
20
slices
bread The
is 8! Once you run out of meat,
2 slices of bread
you must stop making sandwiches.
The meat is the limiting ingredient!
24 slices of meat
=10 sandwiches
1 sandwich
3 slices of meat
= 8 sandwiches
• Cheese:
12 slices of cheese
1 sandwich
1 slice of cheese
=12 sandwiches
Limiting Reactants
• What do you have left over?
• Making 8 sandwiches requires 16 pieces of
bread
• You started with 20 slices, so you have 4
slices of bread left over
• You also used 8 pieces of cheese for the 8
sandwiches, so you have 4 pieces of cheese
left over
• In this example, the meat was the limiting
reactant
Limiting Reactants
• When molecules react with
each other to form
products, considerations
very similar to those making
sandwiches arise
• Consider the reaction that
occurs when Hydrogen Gas
reacts with oxygen gas to
form water
H2(gas) + O2(gas)
H2O(liquid)
Limiting Reactants
• The reaction occurs
between 10 H2
molecules and 7 O2
molecules
• Remember, each O2
molecule requires 2 H2
molecules
2H2(gas) + O2(gas)
2H2O(liquid)
Limiting Reactants
• After the reaction, 10 water
molecules formed and 2 O2
molecules are left over
• That is, the H2 molecules are
used up before the water
molecules are consumed
• We have excess (extra) O2
and H2 is the limiting reactant
because the reaction runs
out of H2 first
2H2(gas) + O2(gas)
2H2O(liquid)
What is the Limiting
Reactant?
consumed
reactant that is completely _________
• The _______
when a reaction is run to completion
• The reactant that is not completely consumed
excess
is in _____
Answer Bank
Moles
consumed
product
Limiting
reactant
excess
Practice:
N2 + 3 H2  2 NH3
You have 2 moles of N2 and 7 moles of H2
Given: 2 mole of N2
7 mol of H2
Which is the limiting reactant?
7 mol H 2
 2.3
3mol H 2
2mol N 2
2
1mol N 2
N2 is the limiting reactant because it has the lowest
ratio!!!!!!!!!!!!!!
Practice:
N2 + 3 H2  2 NH3
You have 3 moles of N2 and 6 moles of H2
Given: 3 mole of N2
6 mol of H2
Which is the limiting reactant?
6 moles of H2
3 moles of H2
3 moles of N2
1 moles of N2
H2 is the limiting reactant because it has the lowest
ratio!!!!!!!!!!!!!!
Practice:
This
reaction is25
different
the othersreacted
we have done
• Suppose
gramsfrom
of nitrogen
So
far 5in grams
that weof
arehydrogen
mixing specified
amounts
with
gas are
mixedof
Twoand
reactants
together.
To know how
much product
react to
form ammonia.
Calculate
the
Forms we must we must determine which reactant is
mass
of
ammonia
produced
when
this
consumed first. In other words, we must determine
reaction is The
run limiting
to completion
reactant
• N2(g) + 3H2(g) → 2NH3(g)
Practice:
• Suppose 25 grams of nitrogen reacted with 5
grams of hydrogen gas are mixed and react to form
ammonia. Calculate the mass of ammonia
produced when this reaction is run to completion
• N2(g) + 3H2(g) → 2NH3(g)
25.0 g N2
1 mol N2
28.0 g N2
5.0 g H2
1 mol H2
2.016 g H2
= .892 moles of N2
= 2.48 moles of H2
Now we must determine which
reactant Practice:
is the limiting reactant.
We have 0.892 moles of nitrogen
• Suppose
25 grams
ofmany
nitrogen
reacted
with 5
Let’s determine
how
moles
of hydrogen
grams
of hydrogen
gaswith
arethis
mixed
andnitrogen.
react to form
Are required
to react
much
ammonia.1 mol
Calculate
the mass
of with
ammonia
Because
of nitrogen
reacts
3 mol of
produced when
this reaction
is run
to completion
Hydrogen,
the number
of moles
of hydrogen
we
Need
completely
with 0.892 mol
• N2(g)
+ 3Hto2(g)react
→ 2NH
3(g)
of nitrogen is calculated as follows:
0.892 mol N2
3 mol H2
1 mol N2
= 2.68 moles of H2
Is nitrogen or hydrogen the limiting reactant?
The answer comes from the comparison:
2.48 moles of H2 available < 2.68 moles of H2 required
This means that the hydrogen will be consumed first before the nitrogen
Runs out, so hydrogen is the limiting reactant
Reflect:
Practice:
We see that 0.892 mol of nitrogen require 2.68 mol
Of hydrogen to react completely. However, only 2.48
Mol
of hydrogen
available.
This means
that the
hydrogen
• Suppose
25are
grams
of nitrogen
reacted
with
5
willgrams
be consumed
before the
runs out,
Hydrogen
of hydrogen
gasnitrogen
are mixed
andso
react
to form
Is the limiting reactant
ammonia. Calculate the mass of ammonia
producedIfwhen
this reaction
isthen
run the
to completion
the nitrogen
is excess,
• N2(g) +hydrogen
3H2(g) →
will 2NH
run out
3(g)first. Again we find
that the hydrogen limits the amount of ammonia
Formed
25.0 g N2
2 mol NH
17.0 g NH
2.48 mol Hthe
2
Because
moles of3 hydrogen 3are limiting, we must use
= the
28.1moles
g of NHof3 ammonia
Our quantity of Hydrogen to determine
3 mol H2
1 mol NH3
that
can form
Summarize:
• When determining the limiting reactant
we first convert mass into_____
• The smaller molar value of the reactants
is the _____ reactant and the larger
molar value of the reactants is in excess
• The limiting reactant will determine how
much ______ will form
Answer Bank
Moles
consumed
product
Limiting
reactant
excess
Unit: Stoichiometry
Topic: Theoretical & Percent Yield
Objectives: Day 4 of 4
• To understand what theoretical yield is
• To learn how to calculate percent yield
using the theoretical yield
Quickwrite
Answer one of the questions below 1-2
sentences:
• Chemists who work for companies like to know
how efficient a reaction is, in other word, they
would like to know how much product forms.
Why do you think they would want to know
this???
Percent Yield
• If you recall, the amount of product formed is
determined by the limiting reactant
• The amount of product calculated using
stoichiometry is called the theoretical yield
• The theoretical yield is the amount of product
predicted from the amounts of reactants used up
• Think of it as the maximum amount of product that
could be produced from 100% of the reactants
being used up
• The theoretical yield is rarely if ever actually
obtained
What is the theoretical yield?
calculated maximum
• The ________
amount of product that can be
formed when the limiting
reactant is completely used up
or ________
consumed
• It is calculated using
___________
stoichiometry
Answer Bank
actual
Consumed
percentage
stoichiometry
Calculated
theoretical
Percent Yield
• Why is the theoretical yield never reached?
• One reason for this is the presence of side
reactions that consume some of the reactants
• The actual yield of product, which is the amount of
product actually obtained, is compared to the
theoretical yield
• This comparison, called the percent yield, is
expressed as a percent
• The percent yield is calculated by dividing the
actual yield / by the theoretical yield
• Percent yield =
Actual yield
x 100%
Theoretical yield
What is the percent yield?
actual (experimental) yield of a
• The ______
percentage over of the
product given as a _________
__________
theoretical yield
• Percent yield =
Actual yield
x 100%
Theoretical yield
Answer Bank
actual
Consumed
percentage
stoichiometry
Calculated
theoretical
Practice:
• Consider the balanced reaction below:
• 2H2(g) + CO (g) → CH3OH(l)
• Suppose 68.5 grams of CO is reacted with 8.6
grams of H2
• Calculate the theoretical yield of methanol
CH3OH(l)
• Your experiment actually produces 35.7 grams of
methanol CH3OH(l)
• What is the percent yield of methanol CH3OH(l) ?
Practice:
• Consider the balanced reaction below:
Step 1: Calculate the moles of reactants
• 2H2(g) + CO (g) → CH3OH(l)
• Suppose 68.5 grams of CO is reacted with
8.6 grams of H2
• Calculate the theoretical yield of methanol
CH3OH(l)
68.5 g CO
1 mol CO
28.01 g CO
8.60 g H2
1 mol H2
2.016 g H2
= 2.45 moles of CO
= 4.27 moles of H2
Practice:
Step 2:
Now we determine
reactant is reaction
limiting using
• Considerwhich
the balanced
below:
The mole ratio of between CO and H2
• 2H2(g) + CO (g) → CH3OH(l)
• Suppose 68.5 grams of CO is reacted with
8.6 grams of H2
• Calculate the theoretical yield of methanol
CH3OH(l)
2.45 mol CO
2 mol H2
1 mol CO
= 4.9 moles of H2
Is CO or H2 the limiting reactant?
The answer comes from the comparison:
4.27 moles of H2 present < 4.9 moles of H2 needed to react with all the CO
This means that the hydrogen will be consumed first before the CO
Runs out, so hydrogen is the limiting reactant
Practice:
Step 3: We must use the amount of H
and
• Consider the balanced reaction below:
the mole ratio between CH3OH and H2 to determine
• 2H
+ CO
the 2(g)
maximum
amount
methanol
that can be produced
(g) →of CH
3OH(l)
• Suppose 68.5 grams of CO is reacted with
8.6 grams of H2
• Calculate the theoretical yield of methanol
CH3OH(l)
4.27 mol H2 1 mol CH3OH(l)
2 mol H2
2
= 2.14 moles of CH3OH
2.14 represents the theoretical yield
Which if you recall, is the maximum amount of a given
product that can be formed when the limiting reactant is
completely used up or consumed
Practice:
• Consider the balanced reaction below:
Step 4: Convert moles to grams
• 2H2(g) + CO (g) → CH3OH(l)
• Suppose 68.5 grams of CO is reacted with
8.6 grams of H2
• Calculate the theoretical yield of methanol
CH3OH(l)
2.14 mol CH3OH 32.04 g CH3OH
1 mol CH3OH
= 68.6 grams of CH3OH
Practice:
• Consider
the balanced
below: the
Step 5: Calculate
percentreaction
yield by dividing
(35.7
g)
by
the
theoretical
yield
(68.6
g)
•actual
2H2(g)yield
+ CO
→
CH
OH
(g)
3
(l)
• Suppose 68.5 grams of CO is reacted with
8.6 grams of H2
• Calculate the theoretical yield of methanol
CH3OH(l)
• Your experiment actually produces 35.7
grams of methanol CH3OH(l)
• Percent yield = 35.7 g CH3OH x100 = 52%
68.6 g CH3OH
Summarize:
• The _____ ______ is the calculated maximum
amount of product that can be formed when
the limiting reactant is completely used up or
consumed
• The _____ ______ is the actual (experimental)
yield of a product given as a percentage over
of the theoretical yield
• % Yield = ___????____ x 100%
????