$n_{Al}=\dfrac{5,}4{27}=0,2mol$
$a.PTHH :$
$Al+3O_2\overset{t^o}\to 2Al_2O_3$ 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:
$n_{O_2}=\dfrac{3}{4}.n_{Al}=\dfrac{3}{4}.0,2=0,15mol$
$=>V_{O_2}=0,15.22,4=3,36l$