# Does sodium produce water when it reacts with water?

My 10th grade chemistry book states that:

"Metals react with water and produce a metal oxide and hydrogen gas. Metal oxides that are soluble in water dissolve in it to further form metal hydroxide. Metals like potassium and sodium react violently with cold water. In case of sodium and potassium, the reaction is so violent and exothermic, that evolved hydrogen immediately catches fire."

So $\ce{Na}$ should react like this:

(i) $\ce{2Na + H2O -> Na2O + H2}$

(ii) $\ce{Na2O + H2O -> 2NaOH}$

• When hydrogen catches fire -- that is, when hydrogen reacts with $\ce{O2}$, right? -- does it form $\ce{H2O}$?

• If water is not formed, then does the water level decrease?

• Is (ii) an exothermic reaction?

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The reaction of sodium metal and water could (and usually does) indirectly produce more water, if enough heat is produced to spontaneously ignite the hydrogen gas which is produced by the initial reaction. This cyclic type of reaction is often found in explosive or extremely exothermic reactions, e.g. $\ce{K+ + I- + H2O2}$ will also react violently in a cyclic way.

Combustion of hydrogen in air is this:

$$\ce{2H2 + O2 -> 2H2O}$$

So, in a (well) closed system with excess oxygen, water could re-form and collect. It's worth noting that dissolution of $\ce{NaOH}$ in water ($\ce{NaOH -> Na+ + OH-}$) is also exothermic and adds even more heat to the system.

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– bon
Aug 7 '15 at 18:12

This does produce water, but the water level decreases due to hydrogen escaping before sodium heats up. The reaction should also speed up and create more and more hydrogen. The force of the pop from the hydrogen exploding means more water escaping. The disassociation of $\ce{NaOH}$ is exothermic. Of course, if this was a closed system nothing would be lost.