Transcript Stoichiometry PowerPoint

Chemistry 11
Chapter 6
STOICHIOMETRY OF EXCESS
QUANTITIES
Introduction: So far ...
 we
have
assumed that a
given reactant is
completely used
up during the
reaction
In reality...

reactions are often
carried out in such
a way that one or
more of the
second reactants
actually are
present in EXCESS
amounts.
Definitions



EXCESS REACTANT = the reactant in excess
LIMITING REACTANT = the reactant that
completely reacts
THE LIMITING REACTANT determines the yield of
the product (how much product(s) will form)
A Simple Analogy
Imagine you work
at McDonalds™ …
 You have 10
hamburger buns
and 5 beef patties
 How many regular
hamburgers can
you make?

Da Answer ...

Indeed, you would
get 5 regular
hamburgers!

And what was left
over?
Connecting the Lingo …


There would be 5
hamburger buns in
EXCESS!
Therefore, the beef
patty is known as the
LIMITING ingredient
since it “limits” or
determines how many
regular buns can be
made!
Note that
 we
do not predict based on the
number of hamburger buns
Example 1

If 20.0 g of
hydrogen gas
react with 100.0 g
of oxygen, which
reactant is present
in excess and by
how many grams?
Step 1 …
 The
balanced equation:
2H2(g) + O2(g)
 2H2O(l)
Step 2 …

First PREDICT which
reactant is limiting
(it’s ok if you predict
wrong)

USUALLY the reactant
with the least number
of moles is limiting
(but not always)
Convert masses to moles
Number of moles of H2 present
= 20.0 g x 1 mol H2
=
10 mol H2
2.0 g H2
Number of moles of O2 present
= 100.0 g x 1 mol O2
= 3.125 mol O2
32.0 g O2
Let’s make a prediction ...
Prediction: O2 is limiting
Mass of H2 that reacts with 100.0 g O2
= 100.0 g O2 x 1 mol O2 x 2 mol H2 x 2.0 g H2
32.0 g O2 1 mol O2
1 mol H2
= 12.5 g H2
Analysing the numbers …




What we have: 100.0 g O2 and 20.0 g H2
We predict O2 is limiting (i.e. all 100.0 g reacted)
We calculated that we would need 12.5 g H2
Is the prediction correct ?
Da Answer (again!)



Yes!! Prediction is
correct
only 12.5 g H2 is
required, so we have
an excess of 7.5 g H2
(20.0 g - 12.5 g)
so H2 is in EXCESS of
7.5 g.
The other side of the coin
So what if we predicted that H2 was
limiting?
Mass of O2 that reacts with 20.0 g H2
= 20.0 g H2 x 1 mol H2 x 1 mol O2 x 32.0 g O2
2.0 g H2
2 mol H2 1 mol O2
= 160.0 g O2
Therefore …

If ALL 20.0 g of H2 were to completely react we
would need 160.0 g of O2

BUT we only have 100.0 g of O2

So the prediction that H2 limiting is INCORRECT!
Example 2.
If 79.1 g of Zn reacts
with 1.05 L of 2.00 M
HCl,
a) Which reactant is
in excess and by
how much?
b) What is the mass
of each product?
a) which reactant is excess?
The balanced equation:
Zn + 2HCl
 ZnCl2 + H2
79.1 g 1.05 L, 2.00 M


1.21 mol 2.10 mol
(what we HAVE)
xg
yg
Prediction: Zn is limiting
Moles of HCl required
= 1.21 mol Zn x 2 mol HCl = 2.43 mol HCl
1 mol Zn
Therefore 2.42 mol HCl would be required to react
with 1.21 mol Zn.
We ONLY have 2.10 mol HCl
So is our prediction correct?
Uh Oh! You’re wrong!


We would need more
HCl (2.42 mol) than
what we have (2.10
mol) if all the Zn were
to react
Thus: Zn is in excess,
and HCl is limiting!
To find how much in
excess:
We must find how many
moles of Zn is required
to react with 2.10 molHCl
Mol of Zn
= 2.10 mol HCl x 1 mol Zn
2 mol HCl
= 1.05 mol Zn
Excess Zn = 1.21 - 1.05
= 0.16 mol Zn
b) mass of products?
Since HCl is limiting we MUST use this amount to
calculate the mass of products
x g ZnCl2 = 2.10 mol HCl x 1 mol ZnCl2 x 136.4 g
2 mol HCl
1 mol ZnCl2
= 143 g ZnCl2
y g H2
= 2.10 mol HCl x 1 mol H2
x 2.0 g
2 mol HCl
1 mol H2
= 2.1 g H2
Example 3:

3.00 L of 0.1 M NaCl reacts with 2.50 L of 0.125 M
AgNO3. Calculate the yield of solid AgCl (in
grams) that will be produced.

This problem requires us to determine how much
product (AgCl) will form, so we will need to first
determine which reactant is limiting.
The balanced equation:
NaCl(aq) + AgNO3(aq) 
3.00 L
2.50 L
0.1M
0.125 M


0.300 mol 0.325 mol
Limiting
Excess
NaNO3(aq) + AgCl(s)
?g
(since 1:1 ratio)
NaCl limiting ...
Therefore: mol NaCl = mol AgCl = 0.300 mol
(also 1:1 ratio)
Mass of AgCl = 0.300 mol AgCl x 143.5 g AgCl
1 mol AgCl
= 43.1 g AgCl
Percent Yield


Often 100% of the
expected amount of
product cannot be
obtained from a
reaction
The term “Percent
Yield” is used to
describe the amount of
product actually
obtained as a
percentage of the
expected amount
Reasons for reduced yields
A) the reactants may not all
react because:
i) not all of the pure
material
actually reacts
ii) the reactants may be
impure
B) Some of the products are
lost during procedures
such as solvent extraction,
filtration etc
The equation:
Percent Yield



= ACTUAL YIELD
THEORETICAL YIELD
x 100%
Actual yield = amount of product obtained
(determined experimentally)
Theoretical yield = amount of product expected
(determined from calculations based on the
stoichiometry of the reaction)
The amounts may be expressed in g, mol,
molecules
Types of calculations
A) Find the percentage yield, given the mass of
reactant used and mass of product formed
B) Find the mass of product formed, given the mass
of reactant used and the percentage yield
C) Find the mass of reactant used, given the mass
of product formed and percentage yield
 Note that the percentage yield must be less than
100%
 But when calculating the theoretical yield assume
a 100% yield
Example 1
When 15.0 g of CH4 is
reacted with an excess
of Cl2 according to the
reaction:
CH4 + Cl2  CH3Cl + HCl
a total of 29.7 g of CH3Cl
is formed. Calculate the
percentage yield.
The solution ...
The actual yield of CH3Cl = 29.7 g
To find the theoretical yield of CH3Cl: (assuming a
100% yield)
g of CH3Cl = 15.0 g CH4 x 1 mol CH4 x 1 mol CH3Cl x 50.5 g
16.0g CH4 1 mol CH4
1 mol CH3Cl
= 47.34 g
Then:
Percentage yield = actual yield
x 100%
theoretical yield
= 29.7 g x 100%
= 62.7 %
47.34 g
Example deux!
What mass of K2CO3 is produced when 1.50 g of
KO2 is reacted with an excess of CO2 if the reaction
has a 76.0% yield? The reaction is:
4KO2(s) + 2 CO2(g)  2K2CO3(s) + 3O2(g)
The solution:
We are looking for the actual yield (some idiot
forgot to weigh and record the mass of product!)
First calculate the mass of K2CO3 produced
(assuming a 100% yield) i.e. the theoretical yield
g of K2CO3 = 1.50 g KO2 x 1 mol KO2 x 2 mol K2CO3 x 138.2 g
71.1 g KO2 4 mol KO2 1 mol K2CO3
= 1.458 g
actual yield = 76.0 % x 1.458 g = 0.760 x 1.458 = 1.11 g K2CO3
Last (but not least) example
...
What mass of CuO is required to make 10.0 g of Cu
according to the reaction
2NH3 + 3CuO  N2 + 3Cu + 3H2O
if the reaction has 58.0 % yield?
Here we go again …
Actual yield = 10.0 g Cu
From the percentage yield equation, calculate the
theoretical yield of Cu.
Theoretical yield of Cu = Actual yield x 100%
Percentage yield
= 10.0 g x 100 %
58.0 %
= 17.24 g
Now find the mass of CuO:
Use this theoretical yield and find the mass of CuO
that would be needed:
g CuO = 17.24 g Cu x 1 mol Cu x 3 mol CuO x 79.5g
63.5 g Cu 3 mol Cu
1mol CuO
= 21.6 g CuO