Adjusting To Reality This is not the entire story In reality you never have the exact amounts of both reactants you need At the end of the reaction one reactant has been ID: 748173
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Slide1
Stoichiometry
Adjusting To RealitySlide2
Adjusting To Reality
This
is not the
entire
story. In reality, you
never
have the
exact
amounts of
both
reactants you need. At the
end
of the reaction, one reactant has been
completely consumed
and there is some
“left over”
of the other reactant.
Let’s take a second look at the reaction between PbS and O
2
. Slide3
2PbS + 3O2
2PbO + 2SO
2
Since
oxygen
is not costly (free in our atmosphere), it’s usually the reactant there is
plenty
of and only a
certain amount
of
lead II sulfide
would have been purchased for this reaction
.
What
would our BCA table look like if we had 0.40 moles of lead (II) sulfide reacting with an abundance (
excess
) of oxygen?Slide4
Adjusting To Reality
Equation:
2
PbS +
3
O
2
2
PbO +
2
SO
2
Before
:
.40
mol
xs
mol
0
mol
0
mol
Change
-
.40
mol
-
xs
mol
+
.
40
mol
+
.
40
mol
____________________________________
After
0
mol
xs
mol
.
40
mol
.
40
mol
Excess
is written (
xs
) and indicates there is some
reactant
remaining
. Here,
PbS
is
completely
consumed
and some O
2
remains
after the reaction is complete.Slide5
Further Reality
What if only a certain
amount
of each reactant were available?
25.50 g of oxygen reacts with 114.85 g lead (II) sulfide producing lead (II) oxide and sulfur dioxide. What
mass
of lead (II) oxide
would be
produced?
We cannot directly
measure
moles
,
so the
reactant
amounts are given in
grams
.
In order to use our BCA table (for
mole
ratios
) we need the amounts in moles. Using
molar mass
,
we convert
the
mass
of the
reactants
to
moles
of
reactants
.Slide6
Mass To Moles, “Molar Mass”
25.50 g O
2
x
1 mol O2
_
= .80 mol O
2
32.0 g O
2
114.85 g PbS
x _
1
mol
PbS_
= .
48 mol
PbS
239.27g
PbSSlide7
A Second Look
Equation:
2
PbS +
3
O
2
2
PbO +
2
SO
2
B
efore:
.
48
mol
.80
mol
0
mol
0
mol
C
hange
-
__
mol
-
__
mol +__mol+
__
mol
_______________________________
A
fter
__
mol
__mol
__
mol __ molSlide8
Limiting & Excess Reactants
Our first task is to find out which reactant will be completely
consumed
(
limiting
reactant
) and
which reactant will have some
remaining
after the reaction is complete (
reactant
in
excess
).
We
will use
mole
ratios
from the BCA table for
this task.Slide9
Limiting & Excess Reactants
.80 mol O
2
x
2 mol PbS
= .53 mol PbS
3 mol O
2
We need .53 mol of PbS to
completely
react .80 mol of O
2
.48 mol PbS x
3 mol O
2
= .72 mol O
2
2 mol PbS
We need .72 mol of O
2
to
completely
react .48 mol of PbSSlide10
Evaluating Our Answers
We
need 0.53 mol PbS to completely burn 0.80 mol O
2
. We only have 0.48 mol PbS. Not all of the O
2
will be
“consumed”
.
The reaction will
stop
when the PbS has run out.
This tells us the PbS will
limit
the reaction (
limiting reactant
) and some oxygen will
remain
after the reaction is complete (
reactant in excess
).Slide11
Amount of R
eactant Remaining
We need 0.72 mol O
2
to
completely
react
with 0.48
mol PbS. We have 0.80 mol
O
2
.
0.08
mol O
2
will
remain
after
all of the PbS
has been
consumed
.Slide12
Importance of Limiting Reactant
The PbS
limits
the reaction. PbS
“
runs out
”
before all the O
2
is consumed. PbS is the
reactant
that
determines
how much product
will
be
produced
.Slide13
Putting It All Together!
Equation: 2PbS
+
3O
2
2PbO + 2SO
2
B
efore:
.
48
mol
.80
mol
0
mol
0
mol
C
hange
-
.
48
mol
-
.
72
mol
+
.
48
mol +
.
48
mol
___________________________________
A
fter:
0
mol
.
08
mol
.48
mol
.48
mol