What is the reaction between Mg and NaOH?

Usually, one would not expect an earth alkali metal to reduce an alkali metal salt, as the alkali metal is the stronger reductant, i.e., it is the less noble metal. This holds true if, e.g., you tried to reduce sodium chloride with Mg: This won’t work, as the reaction would be endothermic:
2NaCl + Mg → MgCl2 + 2 Na ΔHr=+180.4 kJ/mol.

But the above reaction is different, as sodium hydroxide is used. There are two reductions going on at the same time: the endothermic reduction of Na+ ions and the exothermic reduction of protons H+. The latter reaction helps the first one and makes the overall reaction exothermic:
2 NaOH + 2 Mg → 2 MgO + H2 + 2 Na ΔHr=-264.9 kJ/mol

Magnesium hydroxide cannot be and is not the oxidation product, as it would lose water at these high tempereature. Again, the simple anion swap from Na to Mg would not work for itself, as this would be an endothermic process like in the chloride case:
Mg + 2 NaOH → 2 Na + Mg(OH)2 ΔHr=+13.76 kJ/mol

In conclusion: The reduction of sodium hydroxide by magnesium to sodium only works because the hydroxide is reduced to hydrogen at the same time.

when you mix Mg and NaOH powder you would form sodium metal

Mg + 2NaOH →Mg(OH)2 + 2Na

A rather spectacular reaction:

Well, following the pattern of a normal synthesis reaction, one would assume that the reaction of Mg + NaOH would go as follows:

Mg + NaOH → Na + Mg(OH)2 (the 2 on the OH to balance out Mg’s 2+ charge)…

However,

This is not the case, as in order for magnesium (Mg) to replace sodium (Na), which is already bonded to hydroxide (OH) in this reaction, it would have to be more reactive then sodium (to be able to “kick sodium out”), and, according to the reactivity chart, it is less reactive

Thus, there is No Reaction.

For reference, see “Activity Series” chart below, where sodium (#4) is above magnesium (#5):

IN SHORT: No reaction will happen.

This is a single replacement reaction. In a single replacement reaction, a more active element replaces an ion in an ionic compound. What does that mean? Well, we have to look at the activity series. This is a list or ranking of how reactive the metals are.

Let’s say you have an ionic compound AX (in which A is the cation and X is the anion) and a metal B in elemental state. If B is more active (higher up on the activity series), it will “steal” the anion X from the cation A, and bond ionically with X to form the compound BX and the elemental metal A.

An analogy would be basketball players — player B is more skilled than player A, and player A currently has the ball. Most likely, player B will end up stealing the ball, but A wouldn’t be able to get it back since A is not skilled enough.

In this reaction (Mg + NaOH), the compound AX is NaOH, and B is Mg. As you can see in the activity series, Na (sodium) is higher up on the activity series (a “better player”) than Mg. Therefore, Mg wouldn’t be able to steal the “ball” (the anion OH) from Na.

Na(0) is obnoxiously unstable due to having 1 electron in its outer shell. that unfilled orbital desperately wants electrons. You can't easily make an unstable atom from a stable one- it just defies our physical law of tending toward a state of rest. Mg is in a more more stable state because it's 3s and 2p orbitals are full.

Magnesium reduces the sodium hydroxide to elemental sodium. This is a bit surprising because Mg is lower than Na on the reactivity series. However, the reactivity series only gives general trends and not universally valid rules. For a detailed explanation, please see my answer to What is the reaction between Mg and NaOH? .

The exothermic reaction between Mg and NaOH is the reaction

Mg + NaOH -> MgO + Na + 1/2 H2 ,

but the simple displacement reaction

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

would already be exothermic, though less exothermic than the first reaction, which is why the first one happens. In addition to my theoretical explanation and the thermochemical evidence given by others and me on What is the reaction between Mg and NaOH?, there are YouTube videos that apparently show the real reaction, and (at least) one of them (the one by NurdRage) also explains how that can be.

It is true that Na is higher on the reactivity series than Mg, but we shouldn’t forget that the reactivity series only gives (over-)generalized trends that are not universally applicable. Saying that Na is more reactive than Mg is like saying that Jack Smith is more attractive than John Smith - most ladies like Jack more than John, but not all. Similarly, most non-metals like Na more than Mg, but there are exceptions. One such exception is oxygen. Franco Bolli says that the reaction Mg + 2 NaOH → 2 Na + Mg(OH)2 would be endothermic, but according to what I know, the opposite is true:

ΔHf( NaOH (s) ) = −427 kJ/mol (Sodium hydroxide - Wikipedia)
ΔHf( NaOH (s) ) = −425.93 kJ/mol (Sodium hydroxide)
ΔHf( Mg(OH)2 (s) ) = −924.7 kJ/mol (Magnesium hydroxide - Wikipedia)
ΔHf( Mg(OH)2 (s) ) = −924.66 kJ/mol (magnesium hydroxide)

Therefore, magnesium should reduce sodium hydroxide exothermically to elemental sodium metal:

Mg (s) + 2 NaOH (s) → 2 Na (s) + Mg(OH)2 (s) ΔHr = 2*(−ΔHf( NaOH (s) )) + ΔHf( Mg(OH)2 (s) ) = −70.7 kJ/mol (according to the data from Wikipedia)
Mg (s) + 2 NaOH (s) → 2 Na (s) + Mg(OH)2 (s) ΔHr = −72.8 kJ/mol (according to the data from NIST)
Mg (s) + 2 NaOH (s) → 2 Na (s) + Mg(OH)2 (s) ΔHr = −70.66 kJ/mol (according to the data selection most disadvantageous to Mg)

We see that the simple anion swap from Na to Mg would in fact work* by itself and be an exothermic process. Franco Bolli is right in saying that the reaction
2 NaOH + 2 Mg → 2 MgO + H2 + 2 Na
would happen (although the enthalpy change for
2 NaOH (s) + 2 Mg (s) → 2 MgO (s) + H2 (g) + 2 Na (s)
would actually be at least as exothermic as −348.48 kJ/mol according to the data above and not just −264.9 kJ/mol) and that the exothermic reduction of H+ helps the reaction, but contrary to what he says, the reduction of Na+ is also exothermic according to the above data. So, a helping reaction does actually occur, but even if it didn’t, the Mg would still reduce sodium hydroxide to sodium metal. The reaction of Mg with NaOH does not depend on helping reactions and would happen anyway.*

But there is more:

ΔHf( Na2O (s) ) = −416 kJ/mol (Sodium hydroxide - Wikipedia)
ΔHf( Na2O (s) ) = −417.98 kJ/mol (Sodium hydroxide)
ΔHf( Na2O2 (s) ) = −515 kJ/mol (Sodium peroxide - Wikipedia)
ΔHf( Na2O2 (s) ) = −513.21 kJ/mol (disodium peroxide)
ΔHf( MgO (s) ) = −601.6 kJ/mol (Magnesium hydroxide - Wikipedia)
ΔHf( MgO (s) ) = −601.24 kJ/mol (magnesium hydroxide, Chase)

Mg (s) + Na2O (s) → 2 Na (s) + MgO (s) ΔHr = −ΔHf( Na2O (s) ) + ΔHf( MgO (s) ) = −185.6 kJ/mol (according to the data from Wikipedia)
Mg (s) + Na2O (s) → 2 Na (s) + MgO (s) ΔHr = −183.26 kJ/mol (according to the data from NIST, Chase)

2 Mg (s) + Na2O2 (s) → 2 Na (s) + 2 MgO (s) ΔHr = −ΔHf( Na2O2 (s) ) + 2*ΔHf( MgO (s) ) = −688.2 kJ/mol (according to the data from Wikipedia)
Mg (s) + Na2O (s) → 2 Na (s) + MgO (s) ΔHr = −689.27 kJ/mol (according to the data from NIST)

As we can see, magnesium wins* the fight for oxygen against sodium in every way fair and square, i.e. without the need for any “tricks” such as helping reactions. Even in a simple displacement reaction without any boosting side reactions, magnesium metal exothermically reduces sodium ions to give sodium metal and magnesium ions if the anions are oxide or hydroxide ions.*

Why is that so?

In order to answer that question, we have to look at several factors that together determine whether Mg or Na wins and analyze them precisely instead of relying on over-generalizations such as the reactivity series:

1. Atomization energy (the higher, the more endothermic the reaction of the metal to form a compound): weaker metallic bond of Na than of Mg due to fewer valence electrons, therefore lower atomization energy of Na − advantage for Na
2. Ionization energy (the higher, the more endothermic the reaction of the metal to form a compound): valence electron(s) feel smaller effective nuclear charge and is/are further from nucleus in Na than in Mg & only one valence electron in Na compared to 2 in Mg (the second one has an even higher ionization energy), therefore lower ionization energy of Na − advantage for Na
3. Lattice energy (the higher, the more exothermic the reaction of the metal to form a compound): smaller ionic radius and larger charge in magnesium ion than in sodium ion, therefore stronger electrostatic attraction between magnesium ions and anions than between sodium ions and anions, therefore higher lattice energy of Mg ionic compounds than of corresponding Na compounds − advantage for Mg

We see that some of the factors oppose the others. Without quantitative analysis, such as using thermochemical data or making complicated quantum mechanical calculations (if that’s even feasible), we cannot determine which set of factors will outweigh the others. As it happens, in most cases, the first two factors outweigh the third one, making Na generally more reactive than Mg and thus putting Na above Mg on the reactivity series. By the way, that refers to thermodynamic reactivity. Kinetic reactivity is determined only by the first two factors, meaning that Na is easier to activate than Mg.

However, in the reaction with oxygen, the third factor apparently trumps the first two, making Mg (thermodynamically) more reactive with O than Na. My hypothesis for why O prefers Mg while, say, Cl prefers Na, runs as follows:

Oxide ions have a higher (negative) charge (and, according to Ionic radius - Wikipedia, also a smaller radius), than chloride ions. Since electrostatic attractions are directly proportional to each charge and inversely proportional to the square of the distance between the charges by Coulomb’s Law, the difference between electrostatic attractions is also greater if oxide ions are involved than when chloride ions are involved. Therefore, the difference in lattice energy between MgO (−3795 kJ/mol according to 8.3 Lattice Energies in Ionic Solids and Lattice energy - Wikipedia) and Na2O (−2481 kJ/mol according to 8.3 Lattice Energies in Ionic Solids) is greater than the difference in lattice energy between MgCl2 (−2526 according to lattice enthalpy (lattice energy)) and NaCl (−786 kJ/mol or 787 kJ/mol according to Lattice energy - Wikipedia or lattice enthalpy (lattice energy), respectively) per electron transferred (so you have to double the lattice energy of NaCl). This means that the advantage of Mg (factor 3) is greater in the reaction with O than in the reaction with Cl, which is probably why Mg loses the fight for Cl to Na but wins the fight for O against Na.

I hope my answer helps and I am open to any corrections.

*What actually determines whether a reaction is thermodynamically favorable is the Gibbs Free Energy change ΔG and not the enthalpy change ΔH alone. If you calculate the molar enthalpy changes and entropy changes for the reactions above using NIST (Chase) data, you get the following:

Mg (s) + Na2O (s) → 2 Na (s) + MgO (s)
ΔH = -183,260 J/mol
ΔS = +22.06 J/(mol*K)
ΔG = ΔH - TΔS = -183,260 J/mol − T*(+22.06 J/(mol*K))

2 NaOH (s) + 2 Mg (s) → 2 MgO (s) + H2 (s) + 2 Na (s)
ΔH = -350,620 J/mol
ΔS = +92.5 J/(mol*K)
ΔG = ΔH - TΔS = -350,620 J/mol − T*(+92.5 J/(mol*K))

Mg (s) + 2 NaOH (s) → 2 Na (s) + Mg(OH)2 (s)
ΔH = -72,800 J/mol
ΔS = +4.51 J/(mol*K)
ΔG = ΔH - TΔS = -72,800 J/mol − T*(+4.51 J/(mol*K))

2 Mg (s) + Na2O2 (s) → 2 Na (s) + 2 MgO (s)
ΔH = -689,270 J/mol
ΔS = -3.5 J/(mol*K)
ΔG = ΔH - TΔS = -689,270 J/mol − T*(-3.5 J/(mol*K)) = -689,270 J/mol + T*(3.5 J/(mol*K))

We see that in the first three reactions, ΔG is even a little more negative (i.e. favorable) than ΔH because the entropy change ΔS is positive, which means that these reactions are favorable at all temperatures.
In the fourth reaction, ΔS is negative, but that is far outweighed by the negative ΔH, which dominates the expression ΔH - TΔS = ΔG for all temperatures less than about 196,934 K (at that temperature, you probably won’t have any chemical bonds left).
So, all four reactions are exothermic and thermodynamically feasible at all reasonable temperatures (and the first three even at all temperatures).