Phản ứng:
\(2Al_2O_3\rightarrow4Cl+3O_2\)
Ta có:
\(n_{Al2O3}=3,06.50\%=1,53\left(tan\right)\)
\(n_{Al2O3}=\frac{1,53}{27.2+16.3}=0,015\left(mol\right)\)
\(\Rightarrow n_{Al}=2n_{Al2O3}=0,03\left(mol\right)\)
\(\Rightarrow n_{Al\left(thuc.te\right)}=0,03.90\%=0,027\left(mol\right)\)
\(\Rightarrow m_{Al}=0,027.27=0,729\left(g\right)\)