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I'm given this equation in word form:

$$\ce{sodium (s) + water (aq) -> unknown~products (s/g/aq)}$$

and I have to convert this to a chemical equation. I understand how to covert this normally, that equation can hence be represented by:

$$\ce{Na + H2O -> unknown~products\:(s/g/aq)}$$

This is indeed a single displacement reaction so it should become (according to what I know so far) $\ce{Na2O + H2}$ but after doing some research, I got that it should be:

$$\ce{Na + H2O -> NaOH + \frac{1}{2}H2}$$ or $$\ce{2Na + 2H2O -> 2NaOH + H2}$$

I'm lost now, where did $\ce{NaOH}$ come from? Especially the H since there is no reason for hydrogen to be present.

Please explain why and how to balance this reaction or convert this and what factors to look at when dealing with these types of equations.

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  • $\begingroup$ The Na2O is unstable so it would very quickly convert to the more stable NaOH. $\endgroup$ – Technetium Oct 31 '15 at 12:14
  • $\begingroup$ Please visit this page, this page and this ‎one on how to format your posts better.‎ Alternatively, visit this chatroom for further formatting guidance. $\endgroup$ – M.A.R. Oct 31 '15 at 12:15
  • $\begingroup$ The hydrogen is created from the exothermic reaction between the highly reactive sodium metal and water. $\endgroup$ – Technetium Oct 31 '15 at 12:16
  • $\begingroup$ @Joel But how can I identify when it does that? and where did H come from? $\endgroup$ – Imagine Dragons Oct 31 '15 at 12:25
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    $\begingroup$ The H comes from the water, think about what happened to the water molecule that reacted with the sodium ion and left an oxygen atom , hence the speculation of sodium oxide formation, where did the hydrogens go?. Sorry identify when what does what? $\endgroup$ – Technetium Oct 31 '15 at 12:29
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Water is normally a mix of hydroxide and hydrogen, at very dilute levels. For example, neutral water (pH of 7) is 10-7 of OH- and H3O+ ions. Add sodium and it combines with the OH-. The loss of electrons by sodium are available to the hydrogen ions, which combine to make hydrogen gas. Sodium in metal form is so reactive, it will sustain this reaction even at the low hydroxide levels in pure water, and even in slightly acidic seawater.

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    $\begingroup$ There are some pretty significant errors in your answer (including but not limited to the first sentence, especially regarding hydrogen). $\endgroup$ – Todd Minehardt Jul 8 '16 at 19:06
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Sodium(s)+water(aq)⟶Unknown Products(s/g/aq)

In this case you treat the hydrogen as a cation, or as a metal which is why it is displaced. Treat H2O as H(OH).

I know I have answered my own question but this is the answer I was looking for a long time ago.

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    $\begingroup$ This is unfortunately wrong. The reaction is not a displacement reaction. It is a redox reaction. One electron of sodium gets transferred to the hydrogen of water, which liberates a dihydrogen molecule. $\endgroup$ – Martin - マーチン Jul 27 '16 at 15:14

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