\(n_{H_2}=\dfrac{3,36}{22,4}=0,15(mol)\\ n_{NaOH}=2.0,2=0,4(mol)\\ Al+NaOH+H_2O\to NaAlO_2+\dfrac{3}{2}H_2(1)\\ Al_2O_3+2NaOH\to 2NaAlO_2+H_2O(2)\\ \Rightarrow n_{Al}=n_{NaOH(1)}=0,1(mol)\Rightarrow m_{Al}=0,1.27=2,7(g)\\ \Rightarrow n_{NaOH(2)}=0,4-0,1=0,3(mol)\\ \Rightarrow n_{Al_2O_3}=0,15(mol)\Rightarrow m_{Al_2O_3}=0,15.102=15,3(g)\\ \Rightarrow \%_{Al}=\dfrac{2,7}{2,7+15,3}.100\%=10\%\\ \Rightarrow \%_{Al_2O_3}=100\%-15\%=85\%\)