The reaction of aluminium with a solution of sodium hydroxide produces hydrogen (the gas formed) and sodium tetrahyroxoaluminate $Na^+[Al^{3+}(OH^-)_4]^-$ which dissolves in water:
First we write down the redox reaction where aluminium metal reduces the hydrogen ion producing elementary hydrogen gas and aluminium ion:
$Al$ $-$ $3e^-$ $\longrightarrow$ $Al^{3+}|\cdot 2$ $2H^+$ $+$ $2e^-$ $\longrightarrow$ $H_2|\cdot 3$
_ _ _ _
$6H^+$ $+$ $2Al$ $\longrightarrow$ $2Al^{3+}$ $+$ $3H_2 $
Then we neutralize the 6 hydrogen ions with 6 hydroxyde ions
$6H_2O$ $+$ $2Al$ $\longrightarrow$ $2Al^{3+}$ $+$ $3H_2$ $+$ $6OH^-$
We add enough hydoxide ions needed to form tétrahydroxoaluminate:
$6H_2O$ $+$ $2Al$ $+$ $8OH^-$ $\longrightarrow$ $2[Al^{3+}(OH^-)_4]^-$ $+$ $3H_2$ $+$ $6OH^-$
We simplify:
$6H_2O$ $+$ $2Al$ $+$ $2OH^-$ $\longrightarrow$ $2[Al^{3+}(OH^-)_4]^-$ $+$ $3H_2$
We add the sodium ions which have not really participated in the reaction ("spectator" ions )
$6H_2O$ $+$ $2Al$ $+$ $2Na^+$ $+$ $2OH^-$ $\longrightarrow$ $2Na^+$ $+$ $2[Al^{3+}(OH^-)_4]^-$ $+$ $3H_2 $