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Same Reactants Different Products
Name: Katie
Status: student
Grade: 9-12
Location:N/A
Country: N/A
Date: 5/31/2005
Question:
While working on some practice stoichiometry questions I
encountered an equation that I could balance two entirely different ways.
6HNO3 + 2Al(OH)3 ---> 2Al + 3O2 + 6HOH + 6NO2
and
HNO3 + Al(OH)3 ---> Al + O2 + 2HOH + NO2
Same reactants, same products, so why do I get two entirely different
equations? I asked my teacher and she was not sure why I was getting two
equations either. Is this possible with many reactions?
Replies:
Katie-
These two reactions are an example of competing reactions. In the first
reaction, there is much more HNO3 than Al(OH)3 -- the mole ratio is
3:1. While in the second reaction the mole ratio is 1:1. A chemist or
chemical engineer needs to know which product he or she wants to emphasize
-- get lots of AL, use equation 2; get lots of NO2 -- use equation 1.
Warren Young
Chemist
Glenbrook North High School
Katie,
The problem is that the equation you were given is actually two different
reactions rolled into one. Note that while there is only one element that
is being oxidized (N), there are two elements that are being reduced (Al
and O). While such simultaneous oxidation-reduction reactions do occur,
normally it is better to split apart the equation so that there is only
one pair of oxidation-reduction. This will make balancing easier.
Moreover, the H+ of the nitric acid will react with the OH- of the
aluminum hydroxide (acid base reaction) to give water. So you actually
have three reactions rolled into one here.
I suspect there is a typo in the problem as well.
You wrote:
6HNO3 + 2Al(OH)3 ---> 2Al + 3O2 + 6HOH + 6NO2
and
HNO3 + Al(OH)3 ---> Al + O2 + 2HOH + NO2
I do not have data on the NO2, but what I found was this:
(1) NO3- + 4H+ 3e- = NO + 2H2O +0.96V
(2) O2 + 2H2O + 4e- = 4OH- +0.40V
(3) Al3+ + 3e- = Al - 1.66V
So let us assume for the moment that there is a typo and the product is NO
and not, as written, NO2. This would mean that you would have to reverse
equation 2 (so that when all three equations are added together, you end
up with something similar to what was given). This will produce a negative
voltage - suggesting that this reaction is not spontaneous or is done
under electrolytic conditions (or is a poorly written question
Anyway, if we do that, add equation 1 and 3 together, we get:
(4) NO3- + 4H+ + Al3+ + 6e- = NO + 2H2O + Al
(5) 4OH- = O2 + 2H2O + 4e-
and then to balance the electrons on both sides when we add up equations 4
and 5, we need to multiply equation 4 by 2 and equation 5 by 3, and then
when we add these together we get:
(6) 2NO3- + 8H+ + 2Al3+ + 12OH- = 2NO + 10H2O + 2Al + 3O2.
But wait, the H+ and the OH- are going to react with each other to give
H2O, so we reduce the waters and get:
(7) 2NO3- + 2Al3+ + 4OH- = 2NO + 4H2O + 2Al + 3O2
Thus, equation 7 is what I would expect as the balanced equation.
I would say, that whoever gave you this equation to balance should
carefully figure out whether there is any point in trying to do this. The
problem is made complex because three different reactions (2oxidReduc, and
1 acid-base) were mushed into one. Balancing each reaction independently
should give the final balanced equation, but that would require that you
know that the equation can be split into those three.
Greg (Roberto Gregorius)
Ignore for the moment (even without going through the thermochemical
calculation) that it is highly unlikely that nitric acid (a strong
oxidizing agent) is going to reduce Al(+3) to produce elemental Al(0).
Even so, these two reactions are not the same: In the first reaction the
ratio of HNO3 / Al(OH)3 = 6 / 2 or 3 / 1. In the second reaction the ratio
of HNO3 / Al(OH)3 = 1 / 1. That makes the two reactions different from one
another, since the ratio of reactants is different. Put another way: If
the two reactions were identical, you should be able to multiply the
second reaction by '2' and subtract it from the first to yield "zero" that
is:
'no reaction', just like algebraic equations: However, doing this gives:
(6 - 2) HNO3 + (2 - 2) Al(OH)3 ---> (2 - 2) Al + (3 - 2) O2 + (6 - 4) HOH +
(6 - 2) NO2 or:
4 HNO3 ---> O2 + 2 HOH + 4 NO2
Obviously, there is a "net" reaction. The point you illustrate, however, has
merit. It was observant on your part, and important, to recognize that
"there was a problem", and that is half the battle of learning -- knowing
there is a problem.
Vince Calder
Katie,
I am sorry I could not respond sooner and I am not sure if you received
a response from anyone else. I do have concerns with the equation that
you presented in your e-mail noted below.
Aluminum Hydroxide is a light fluffy solid which has a very slimy feel
to it. In your equation, we are adding nitric acid to this material.
What will happen as the pH is depressed is that the aluminum will
dissolve in the acid and the aluminum hydroxide will disappear. The
actual equation will be something like:
3HNO3 + Al(OH)3 ---> Al(NO3)3 + 3H2O
In your reaction, you are producing a pure metal by the addition of an
acid to a metal compound. This does not really happen (although many
manufacturers would wish for it!)
When dealing with stoichiometry, the reaction and reaction products must
first make sense before the equation can be balanced. I hope I have
understood correctly what you were trying to say. If not, feel free to
contact me again.
Bob Hartwell
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Update: June 2012
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