$n_{Al}=4,05/27=0,15mol$
$a/$
$4Al+3O_2\overset{t^o}{\longrightarrow}2Al_2O_3$
$b/$
Theo pt :
$n_{Al_2O_3}=1/2.n_{Al}=1/2.0,17=0,075mol$
$⇒m_{Al_2O_3}=0,075.102=7,65g$
$c/$
Theo pt:
$n_{O_2}=3/4.n_{Al}=3/4.0,15=0,1125mol$
$⇒V_{kk}=5.0,1125.22,4=12,6l$