Đáp án:
\[n_{H_2}=\dfrac{6,72}{22,4}=0,3(mol)\]
\[2Al+2NaOH+6H_2O\to 2Na[Al(OH)_4]+3H_2\] \[0,2\qquad\qquad\qquad\qquad\qquad\qquad\qquad\leftarrow\quad \quad\quad0,3\qquad(mol)\]
\[Al_2O_3+2NaOH+3H_2O\to 2Na[Al(OH)_4]\]
\[\to m_{AL}=0,2\times 27=5,4(g)\]
\[\to \%m_{Al}=\dfrac{5,4}{10}\times 100\%=54\%\]