$a) PTPƯ: 4Al + 3O_{2} → 2Al_{2}O_{3}$
$\text{- Phản ứng tổng hợp. }$
$\text{b) -Theo bài ra, ta có: }$
$n_{Al} = \dfrac{5,4}{27} = 0,2$ $(mol)$
$n_{O_{2}} = \dfrac{2,688}{22,4} = 0,12$ $(mol)$
$\text{Ta thấy:}$ $\dfrac{n_{Al}}{4} = \dfrac{0,2}{4} > \dfrac{n_{O_{2}}}{3} = \dfrac{0,12}{3}$
$\text{⇒ Al dư, $O_{2}$ hết}$
$\text{-Theo PTPƯ, ta có: }$
$n_{Al_{2}O_{3}} = \dfrac{2}{3}.n_{O_{2}} = \dfrac{2}{3}.0,12 = 0,08$ $(mol)$
$⇒ m_{Al_{2}O_{3}} = n.M = 0,08.102 = 8,16$ $(g)$