H2O Al/Hg stoich check.

[Vibrating_Lights]

How much H2O is needed to complete the Al2O3 formation in an al/Hg with 1 mole of Imine to amine And three moles Of nitromethane to MeAm. So

(3*2)+1= 7 total O to reduce.  How much water is needed here.
3MeNO2 + 6Al(6H+)+ _?H2O[1]  --> 3MeNH2 +  3Al2O3
RO-MeNH2)R +1Al(1H+)_?H2O[2] --> RMeNH2R +  AlO3

Would the H2O for [1] = 3
and [2] = 1
so a totalk of 4 moles of H20 to complete the reduction. two excess moles of MeNH2 to be collected after the Rxn.
does that sound right.  The excess of MeAm i snot a problem.
VL_
So much game I could sell a hooker some pussy
Vl_

[terbium]

I would make the assumption that the aluminum goes to the hydroxide not the oxide.

Then I would model the reduction of nitromethane as being the sum of these two reactions:

A) Al + 3H₂O ^__> Al(OH)₃ + 3H

B) CH₃NO₂ + 6H ^__> CH₃NH₂ + 2H₂O

Then, combining 2A + B, we obtain the overall reaction:

C) 2Al + 4H₂O + CH₃NO₂  ^__> CH₃NH₂ + 2Al(OH)₃

which requires 4 moles of water for every mole of nitromethane.

If, instead of reaction A, we assume that the aluminum goes to the oxide rather than the hydroxide (a bad assumption in my guestimation) then we have:

D) 2Al(OH)₃ ^__> Al₂O₃ + 3H₂O

Then, combining D + C, we have an overall reaction:

E) 2Al + H₂O + CH₃NH₂ ^__> CH₃NH₂ + Al₂O₃

which requires only 1 mole of water per mole of nitromethane but, as I said before, I would expect reaction C to be more likely thus requiring the 4 moles of water per mole of nitromethane.

[terbium]

For the reaction of the methylamine with ketone to produce imine we have:

A) R₂CO + CH₃NH₂ ^__> R₂CNCH₃ + H₂O

For the reduction of the imine we have:

B) R₂CNCH₃ + 2H ^__> R₂CHNHCH₃

And, as before, the reaction of aluminum with water:

C) Al + 3H₂O ^__> Al(OH)₃ + 3H

Then, combining 3A + 3B + 2C, we have the overall reaction:

D) 3R₂CO + 3CH₃NH₂ + 2Al + 3H₂O ^__> 3R ₂CHNHCH₃ + 2Al(OH)₃

which requires 1 mole of water per mole of ketone.

[terbium]

One other consideration that might reduce the amount of water needed in reductions performed with aluminum in methanol is that the methanolic OH is also quite reactive and could possible take the place of water thusly:

Al + 3CH₃OH ^__> Al(OCH₃)₃ + 3H