$2Al+Fe_2O_3\rightarrow
2Fe+Al_2O_3 $$2Al+2NaOH+2H_2O\rightarrow 2NaAlO_2+3H_2O$ $n_{Al}=\frac{6,48}{27}=0,0888888889 mol;n_{Fe_2O_3}=0,1 mol\Rightarrow H=100\%\Rightarrow Al:$Hết$\Rightarrow n_{H_2}=0,06 mol\Rightarrow n_{Al dư}=0,04 mol\Rightarrow H=\frac{6,48-0,04.27}{6,48}.100=83,33\%$
$2Al+Fe_2O_3\rightarrow
2Fe+Al_2O_3 $$2Al+2NaOH+2H_2O\rightarrow 2NaAlO_2+3H_2O$ $n_{Al}=\frac{6,48}{27}=0,0888888889 mol;n_{Fe_2O_3}=0,1 mol\Rightarrow H=100\%\Rightarrow Al:$Hết$\Rightarrow n_{H_2}=0,06 mol\Rightarrow n_{Al đư}=0,04 mol\Rightarrow H=\frac{6,48-0,04.27}{6,48}.100=83,33\%$
$2Al+Fe_2O_3\rightarrow
2Fe+Al_2O_3 $$2Al+2NaOH+2H_2O\rightarrow 2NaAlO_2+3H_2O$ $n_{Al}=\frac{6,48}{27}=0,0888888889 mol;n_{Fe_2O_3}=0,1 mol\Rightarrow H=100\%\Rightarrow Al:$Hết$\Rightarrow n_{H_2}=0,06 mol\Rightarrow n_{Al
dư}=0,04 mol\Rightarrow H=\frac{6,48-0,04.27}{6,48}.100=83,33\%$