# How to predict (know) the products of the reaction SnO+NaOH? [duplicate]

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?

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,

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

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)