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This question already has an answer here:

I would like to know how to complete the reaction $\ce{SnO}+\ce{NaOH ->}$, how can I form a salt and water from this?

I've seen that the complete reaction is:

$$ \ce{SnO + 2NaOH ->Na2SnO2 + H2O} $$

How to know what will be formed?

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marked as duplicate by Todd Minehardt, airhuff, Jan, ringo, Jon Custer Mar 20 '17 at 13:30

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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I've discovered the answer, at least I guess...

The given reaction is

$$ \ce{SnO + 2NaOH -> Sn^{2+} + O^{2-} + 2Na+ + 2OH-} $$

The hidroxy breaks in $\ce{H2O + O^{2-}}$. This released oxigen will get together with the oxigen from the $\ce{SnO}$, to form $\ce{O2^{2-}}$.

$$ \ce{SnO + 2NaOH -> Sn^{2+} + \underbrace{O2^{-2}}_{\textrm{nox = -4}} + 2Na+ + H2O} $$

The oxidation state of this molecule is -4. Performing the double replacement reaction we get

$$ \ce{SnO + 2NaOH -> SnO2^{-2} + 2Na+ + H2O} $$

And finally,

$$ \ce{SnO + 2NaOH -> Na2SnO2 + H2O} $$

What is already balanced.

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Apart from your reaction that gives sodium stannate by reacting tin(II) oxide and sodium hydroxide at 400°C, there is an another reaction but that involves aqueous medium:

$$\ce{SnO + NaOH <=>[H2O] Na[Sn(OH)3]}$$

Tin(II) oxide react with sodium hydroxide and water to produce sodium trihydroxostannate(II). Sodium hydroxide - concentrated solution. The reaction proceeds at room temperature. (Chemiday)

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