Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
PT: \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
Theo PT: \(n_{Al_2O_3\left(LT\right)}=\dfrac{1}{2}n_{Al}=0,1\left(mol\right)\)
Mà: H% = 60%
\(\Rightarrow n_{Al_2O_3\left(TT\right)}=0,1.60\%=0,06\left(mol\right)\)
⇒ mAl2O3 (TT) = 0,06.102 = 6,12 (g)