Na2O + H2O --> 2NaOH, what kind of reaction is this?

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username4221054
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dont know if this posted but is this oxidation, reduction or disproportionation and could anyone tell me why it is?
i wrote out the oxidation states and i dont understand what im doing
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charco
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(Original post by earthworm206)
dont know if this posted but is this oxidation, reduction or disproportionation and could anyone tell me why it is?
i wrote out the oxidation states and i dont understand what im doing
There are no electron transfers, so nothing to do with REDOX.
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P.Ree
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synthesis reaction my friend
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Kallisto
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(Original post by earthworm206)
dont know if this posted but is this oxidation, reduction or disproportionation and could anyone tell me why it is?
i wrote out the oxidation states and i dont understand what im doing
Synthesis reaction, as P.Ree said it before.

In your case, Na2O and H2O forming a new product by balancing the equation: The O-Atom in Na2O is double negative charged and is attracted by one of the partial charges of the H-Atoms in H2O. It comes to two negative charged hydroxyl groups.

Na2O -> 2 Na(+) + O(-2)
O(-2) + H2O -> 2 OH(-)

Those hydroxyl groups reacting with the positive charged Na-ions and complete the equation:

2 Na(+) + 2 OH(-) -> 2 NaOH

So at the end you got: Na2O + H2O -> 2 NaOH
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charco
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(Original post by Kallisto)
Synthesis reaction, as P.Ree said it before.

In your case, Na2O and H2O forming a new product by balancing the equation: The O-Atom in Na2O is double negative charged and is attracted by one of the partial charges of the H-Atoms in H2O. It comes to two negative charged hydroxyl groups.

Na2O -> 2 Na(+) + O(-2)
O(-2) + H2O -> 2 OH(-)

Those hydroxyl groups reacting with the positive charged Na-ions and complete the equation:

2 Na(+) + 2 OH(-) -> 2 NaOH

So at the end you got: Na2O + H2O -> 2 NaOH
The sodium ions are spectator so your equation boils down to :

O2- + H2O --> 2OH-
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Kallisto
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(Original post by charco)
The sodium ions are spectator so your equation boils down to :

O2- + H2O --> 2OH-
I am afraid I don't understand completely. Is that the way to say that I did a step too many? was the sodium equation above unneccesary to write down?
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charco
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(Original post by Kallisto)
I am afraid I don't understand completely. Is that the way to say that I did a step too many? was the sodium equation above unneccesary to write down?
yes, there is no "reaction" between sodium ions and hydroxide ions.
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username4221054
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thanks everyone
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