\(n_{Al}=\dfrac{8,1}{27}=0,3\left(mol\right);n_{O_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
PTHH: 4Al + 3O2 ---to→ 2Al2O3
Mol: 0,2 0,15 0,1
Ta có: \(\dfrac{0,3}{4}>\dfrac{0,15}{3}\) ⇒ Al dư, O2 hết
\(m_{Aldư}=\left(0,3-0,2\right).27=2,7\left(g\right)\)
\(m_{Al_2O_3}=0,1.102=10,2\left(g\right)\)