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Topic: Al + NaOH --> Al(OH)4 not Al(OH)3?  (Read 87494 times)

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Offline zeoblade

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Al + NaOH --> Al(OH)4 not Al(OH)3?
« on: February 27, 2010, 01:37:28 AM »
When reacting Al metal and NaOH(aq), I thought initially I would get the following reaction:
Al(s) + 3NaOH(aq) --> Al(OH)3(aq) + 3Na(s)

But I feel this is not correct and wondered if it should be:
2Al(s) + 2NaOH(aq) + 3H2O(l) --> 2Al(OH)4-(aq) + 2Na+(aq) + 3H2(g)

In explaining this reaction, how does hydrogen gas combine together to be evolved in this reaction?

Is there a possibility it could also be like this and why?
2Al(s) + 2NaOH(aq) + 2H2O(l)  2NaAlO2(s) + 3H2(g)

Offline Schrödinger

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Re: Al + NaOH --> Al(OH)4 not Al(OH)3?
« Reply #1 on: February 27, 2010, 04:12:43 AM »
But I feel this is not correct and wondered if it should be:
2Al(s) + 2NaOH(aq) + 3H2O(l) --> 2Al(OH)4-(aq) + 2Na+(aq) + 3H2(g)
Is the equation balanced?

Is there a possibility it could also be like this and why?
2Al(s) + 2NaOH(aq) + 2H2O(l)  2NaAlO2(s) + 3H2(g)
This one is correct. It is actually one of the commercial methods to produce H2 gas(pure), if I'm not wrong.
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Online Borek

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Re: Al + NaOH --> Al(OH)4 not Al(OH)3?
« Reply #2 on: February 27, 2010, 04:21:28 AM »
But I feel this is not correct and wondered if it should be:
2Al(s) + 2NaOH(aq) + 3H2O(l) --> 2Al(OH)4-(aq) + 2Na+(aq) + 3H2(g)
Is the equation balanced?

Is there a possibility it could also be like this and why?
2Al(s) + 2NaOH(aq) + 2H2O(l)  2NaAlO2(s) + 3H2(g)
This one is correct. It is actually one of the commercial methods to produce H2 gas(pure), if I'm not wrong.

Note: once balanced they both describe the same process, I would not call one "correct" and the other "incorrect".
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Offline Schrödinger

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Re: Al + NaOH --> Al(OH)4 not Al(OH)3?
« Reply #3 on: February 27, 2010, 04:37:55 AM »
Note: once balanced they both describe the same process, I would not call one "correct" and the other "incorrect".
But what about the aluminium hydroxide idea? Is that correct?
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Offline zeoblade

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Re: Al + NaOH --> Al(OH)4 not Al(OH)3?
« Reply #4 on: February 27, 2010, 04:38:48 AM »
Al(s) + 2NaOH(aq) + 2H2O(l) --> Al(OH)4-(aq) + 2Na+(aq) + H2(g)

Sorry it should be this

2Al(s) + 2NaOH(aq) + 2H2O(l) --> 2NaAlO2(s) + 3H2(g)

So both describing the same process, how are Al(OH)4-(aq) and 2NaAlO2(s) related?

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Re: Al + NaOH --> Al(OH)4 not Al(OH)3?
« Reply #5 on: March 24, 2014, 10:24:02 AM »
So both describing the same process, how are Al(OH)4-(aq) and 2NaAlO2(s) related?

This is basically the same substance:

Na+ + Al(OH)4- :rarrow: NaAlO2 + 2H2O
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Offline bubblegumpi

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Re: Al + NaOH --> Al(OH)4 not Al(OH)3?
« Reply #6 on: August 09, 2017, 10:59:33 AM »
Note: once balanced they both describe the same process, I would not call one "correct" and the other "incorrect".
But what about the aluminium hydroxide idea? Is that correct?

If you think about how the energy would flow or entropy the sodium wouldn't go into solution.

First the sodium doesn't have anything to balance it out; like in the case of NaCl the chloride ions keep the sodium ions from reacting with the water.
Second if the sodium was made it would react with the water forming first sodium oxide then sodium hydroxide, which is what you started with so there would be an energy/entropy "imbalance". All reactions go to to the lowest energy state.
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