$n_{Al}=\dfrac{2,4.10²²}{6.10²³}=0,04mol$
$4Al+3O_2\overset{t^o}{\longrightarrow}2Al_2O_3$
$a/$
Theo pt :
$n_{O_2}=3/4.n_{Al}=0,03mol$
$⇒V_{O_2}=0,03.22,4=0,672l$
$⇒V_{kk}=0,672.5=3,36l$
$b/$
Theo pt:
$n_{Al_2O_3}=1/2.n_{Al}=1/2.0,04=0,02mol$
$⇒m_{Al_2O_3}=0,02.102=2,04g$