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.