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 Click here to return to the Chemistry Archives

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