Balancing other atoms in a redox reaction using half reaction method

I have this balanced redox reaction:
$$\ce{2H2O + 8Al + 3NaNO3 + 5NaOH -> 3NH3 + 8NaAlO2}$$
I think its unbalanced form should be (I guess water added to balanced it)
$$\ce{Al + NaNO3 + NaOH -> NH3 + NaAlO2}$$
I wrote the half reactions as
$$
\begin{align}
\ce {6H2O + 8e- + NO3- &-> NH3 + 9OH-}\label{rxn:1}\tag{1}\\
\ce {Al &-> Al^3+ + 3e-}\label{rxn:2}\tag{2}
\end{align}
$$
Then by $3\cdot\eqref{rxn:1} + 8\cdot\eqref{rxn:2}$, final half reaction is:
$$\ce{18H2O + 3NO3- + 8Al -> 3NH3 + 27OH- + 8Al^3+}$$


Now I am really confused how to get above balanced redox reaction by the final half reaction I have obtained. What is the missing idea?


$$\ce{UPDATE:- Anybody please show me how to proceed from here,}$$
$$\ce{18H2O + 3NO3- + 8Al -> 3NH3 + 27OH- + 8Al^3+}$$
$$\ce{to here}$$
$$\ce{2H2O + 8Al + 3NaNO3 + 5NaOH -> 3NH3 + 8NaAlO2}$$




Your total reaction is:


$$\ce{2H2O + 8Al + 3NaNO3 + 5NaOH⟶3NH3 + 8NaAlO2}$$


But your choice of oxidation half reaction is:


$$\ce{Al -> Al^{3+} + 3e-}$$


Notice that the right side of the half reaction doesn't represent the right side of the total reaction (once you've accounted for the sodium spectator ion).


You *must* find the corresponding oxidation half reaction, and if it's not available, then you're out of luck.


Your half reaction will have $\ce{Al}$ on the left, and $\ce{AlO2-}$ and 3 electrons on the right; with water, hydrogen ions, and hydroxides to balance.


In comments, you also asked for why this is the oxidation product. That is the wrong question to ask. All you're doing is breaking down the total reaction into component half reactions. The simple half reaction ($\ce{Al|Al^{3+}}$) is totally valid. It's just gives a different total reaction from the one you started with. There's nothing wrong with that reaction. It's just different.


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