$n_{Al}=\dfrac{5,4}{27}=0,2mol \\a.PTHH : \\4Al+3O_2\overset{t^o}\to 2Al_2O_3 \\b.Theo\ pt : \\n_{Al_2O_3}=\dfrac{1}{2}.n_{Al}=\dfrac{1}{2}.0,2=0,1mol \\⇒m_{Al_2O_3}=0,1.102=10,2g \\c.Theo\ pt : \\n_{O_2}=\dfrac{3}{4}.n_{Al}=\dfrac{3}{4}.0,2=0,15mol \\⇒V_{O_2}=0,15.22,4=3,36l \\d.V_{kk}=3,36.5=16,8l$