2Al+3H2SO4--->Al2(SO4)3+3H2
Al2O3+3H2SO4---->Al2(SO4)3+3H2O
n\(_{H2}=\frac{3,36}{22,4}=0,15\left(mol\right)\)
Theo pthh1
n\(_{Al}=\frac{2}{3}n_{H2}=0,1\left(mol\right)\)
%m\(_{Al}=\frac{0,1.27}{15.6}.100\%=17,31\%\%=\%=\)
%m\(_{Al2O3}=100-17,31=82,69\%\)