\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\\ n_{O_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
PTHH: 4Al + 3O2 --to--> 2Al2O3
LTL: \(\dfrac{0,2}{4}=\dfrac{0,15}{3}\)=> pư vừa đủ
=> Ko có Vdư
\(n_{Al_2O_3}=\dfrac{1}{2}.0,2=0,1\left(mol\right)\\ m_{Al_2O_3}=0,1.102=10,2\left(g\right)\)