$n_{H_2}=\dfrac{3,36}{22,4}=0,15 mol$
Gọi a, b là mol $Al$, $Al_2O_3$.
$\Rightarrow 27a+102b=9$ (1)
$2Al+2NaOH+2H_2O\to 2NaAlO_2+3H_2$
$Al_2O_3+2NaOH\to 2NaAlO_2+H_2O$
$\Rightarrow 1,5a=0,15$ (2)
(1)(2)$\Rightarrow a=0,1; b=0,062$
$n_{NaOH}=a+2b=0,224 mol$