Landlord
Yellow |
Since all acetate compounds are soluble in water (one of the
wonderful advantages of using acetate compounds, of course), they're
really handy for working with metals which give mostly insoluble
compounds, like lead, Pb.
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In class, we mixed Pb(C2H3O2)2,
which all lazy chemists abbreviate as Pb(OAc)2, lead(II) acetate,
with potassium chromate, K2CrO4,
both of which are soluble (ionic) solutions to produce the insoluble
lead chromate, PbCrO4,
known disparagingly as "landlord yellow". (It makes the world's
cheapest colored paint. Fortunately, it's no longer in use since
no one wants lead on their walls! It's toxic, as are
almost all heavy metals.)
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To do so, we had to satisfy the following reaction:
WRONG!
Pb(OAc)2 + K2CrO4
PbCrO4 + K(OAc)
But that reaction shouldn't satisfy us because it is out of
atom balance! We lost a K+ and an OAc- going
from reactants (left) to products (right), so we need to correct the
stoichiometric coefficients:
RIGHT
Pb(OAc)2 + K2CrO4
PbCrO4 + 2 K(OAc)
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Stoichiometric
Calculation |
So just as we used mole-to-mole conversion ratios to analyze the
relative weights of atoms in molecules, we can do exactly the
same thing to obtain the relative weights of reactant and/or products
in balanced chemical reactions!
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Suppose we needed to know how much potassium acetate we'd be stuck
with as a by-product to the desired "landlord's yellow" when we reacted
100 kg of lead(II) acetate with sufficient potassium chromate to
use up all the lead compound?
We'd set up the same
MASS-TO-MOLES-TO-OTHER_MOLES-TO-OTHER_MASS
kind
of calculation we've done last period, but the
MOLES-TO-OTHER_MOLES
ratio will here relate to counts of different compounds which
are related by stoichiometric coefficients in unique reaction equations.
(Even the same molecular pairs may appear in different reactions with
different stoichiometric coefficients, so these molar ratios aren't
fixed for given pairs...they're fixed for given balanced reactions.)
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For the reaction equation above, then, we'd solve the problem posed
as follows:
To convert between mass and moles, we'll need the molecular weights of
both lead(II) acetate (0.3252 kg) and potassium acetate (0.0981 kg):
1 mol Pb(OAc)2 2 mol KOAc .0981 kg KOAc
100 kg Pb(OAc)2 ----------------- -------------- -------------
.3252 kg Pb(OAc)2 1 mol Pb(OAc)2 1 mol KOAc
which is 60.3 kg of KOAc, smaller than 100 kg, of course, because
potassium (39.1 g/mol) weighs far less than lead (207.2 g/mol).
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Balancing
Chemical
Equations |
As we've emphasized twice already, none of this is worth navel lint
unless the reaction equations are balanced; so we have to do valid
chemical algebra. Let's see how to go about it using another example
from the lecture, the combustion (more like
explosion)
of the anaesthetic known as "ether", actually diethylether or nowadays,
ethoxyethane,
(C2H5)2O.
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That's close to the structural formula:
H H H H
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H--C--C--O--C--C--H a highly volatile, flammable liquid!
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H H H H
but problem 3.89(e) just calls it C4H10O.
Here goes:
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WRONG!
C4H10O(l) + O2(g)
CO2(g) + H2O(l)
First look for atoms which appear in only one molecule on
each side of the reaction. The obvious candidate here is carbon.
Since C appears only once on each side, we can establish immediately
the ratio of the molecules it appears in; we may later have to scale
their coefficients when balancing other atoms but we'll never
have to change this initial ratio. In this case, we'll need a
stoichiometric coefficient of 4 for CO2 to balance
the carbons.
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fix carbon |
WRONG!
C4H10O(l) + O2(g)
4 CO2(g) + H2O(l)
We're in luck because hydrogen appears in only one molecule on each
side as well. So we can establish the ratio of those molecules
instantly too. With 10 on the left, and H coming in pairs in water
on the right, we'll need 5 as the stoichiometric coefficient
for H2O.
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fix hydrogen |
WRONG!
C4H10O(l) + O2(g)
4 CO2(g) + 5 H2O(l)
Now, if we're observant, we'll notice that the remaining unbalanced
O2 comes in pairs of O atoms. So does CO2.
But the other two molecules have an odd number of oxygens; fortunately,
C4H10O and 5 H2O
appear on opposite sides. So it's trivial to balance ether's single O
mentally with one of the water oxygens, leaving 4 water oxygens (an even
number) unbalanced on the right. Those 4 plus the 8 from 4 CO2
mean a needed 12 extra O atoms coming from O2...easily
supplied by 6 O2:
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fix oxygen |
RIGHT
C4H10O(l) + 6 O2(g)
4 CO2(g) + 5 H2O(l)
and the chemical equation is now in balance.
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Limiting
Reactant |
When we spoke earlier of reacting lead(II) acetate with an excess
of potassium chromate, we wanted to make sure that all of the lead
reacted; we didn't want to run out of chromate before all the
lead was gone. In other words, we wanted the lead(II) acetate to be
the limiting reactant not the potassium chromate.
Toward that end, it would be no trick to calculate the minimum
amount of the chromate for any given amount of lead by comparing
moles to moles as we've done before. However, we should also be able
to tell which reactant among a mixture of them is there in the
shortest supply, because it, the limiting reactant, will determine
how much product the reaction produces! Reactants in relative
abundance will have excesses (waste!) left over. It's clear that
in commercial production, you really want not to have
unnecessary waste...after all you paid for what you're not using!
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So let's look at a reaction with several reactants present in
given amounts and find some simple method of picking the one that
will run out first and thus will govern the efficiency of the whole
process.
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WRONG!
KCl + MnO2 + H2SO4
K2SO4 + MnSO4 + Cl2(g) +
H2O
This reaction produces chlorine gas by "reduction" of manganese;
notice that manganese started as manganic and ended up as manganous!
This is an example of a wonderfully useful type of reaction, called
redox for short, that we'll look at in earnest later in
freshman chemistry. For now, let's just balance it as if there wasn't
anything exciting about it.
Cl appears in only one molecule on each side; so we try to balance it
first:
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fix chlorine |
WRONG!
2 KCl + MnO2 + H2SO4
2 K2SO4 + MnSO4 + Cl2(g) +
H2O
Fortuitously, that has also fixed the imbalance in K. Next Mn is already
in balance, so we move onto SO42- not S! Note
that sulfur appears only as SO42- in
three compounds; so we might as well treat SO42-
as a unit...as it if were an atom to be balanced...and not worry about
the S and its associated O atoms separately.
(We couldn't get away with this if sulfur appeared in more than one
form.) That thinking leads us
to double the sulfuric acid since there are two sulfates on the right:
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fix sulfate |
WRONG!
2 KCl + MnO2 + 2 H2SO4
K2SO4 + MnSO4 + Cl2(g) +
H2O
Now we're down to H and O (ignoring the O in the sulfates). All the O
that's left are the two in MnO2 and the one in water.
So we double the water:
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fix non-sulfate
oxygen |
RIGHT
2 KCl + MnO2 + 2 H2SO4
K2SO4 + MnSO4 + Cl2(g) +
2 H2O
and lo the hydrogens now also balance! As is the entire equation.
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Find the
Limiting
Reactant |
Suppose we're given a reaction mixture with 40 g KCl, 25 gm MnO2,
and 50 g of sulfuric acid. Which species is limiting the reaction?
One way (suggested by the text) is to methodically assume that each
reactant in turn is the limiting reactant and calculate some
product amount (say Cl2) based upon that one pretending
the rest of the reactants are there in abundance. Clearly, the one
that produces the least Cl2 is the winner (or loser,
depending upon how you think about it), e.g., the reaction quits when
it is used up and that smallest Cl2 product is all you're
going to see.
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A faster way involves comparing what you have with what you want,
given the reaction stoichiometry. Calculate how many grams of each
reactant would be perfect to make the reaction run as written. In this
case, that would require 2 mol of KCl (2×74.55 g/mol = 149.1 g),
one of MnO2 (86.9 g), and 2 of sulfuric acid
(2×98.1 g/mol = 196.2 g).
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Needed
Available |
149.1 86.9 196.2 grams
2 KCl + MnO2 + 2 H2SO4 K2SO4 + MnSO4 + Cl2(g) + 2 H2O
40 25 50 grams
All we have to do is divide Available grams by Needed
grams to tell how far (in a fractional sense) the available grams of
each reactant will allow the reaction to go! Had we supplied what was
"needed," each of those ratios would've been 1.0 which would
have meant "all the way, baby!" (There's no reason not to find
ratios greater than one. No problem; means more moles are supplied
than the stoichiometric coefficients called for.)
As it is, the ratios are
- KCl: (40 / 149.1) = 0.27
- MnO2: (25 / 86.9) = 0.29
- H2SO4: (50 / 196.2) = 0.25
SMALLEST of the 3
So sulfuric acid limits the reaction to only ¼ of what it
would have produced with the "needed" weights. Knowing that, we can
obtain the amount of Cl2 by finding ¼ of what the
reaction would have produced if the number of moles supplied were
consistent with the stoichiometric coefficients (the "needed"
amount). Clearly that would be 0.25×MW Cl2 or
¼×(2×35.45 g) = 18 g Cl2. Quick, huh?
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