$1/$
$n_{Al}=2,16/27=0,08mol$
$4Al + 3O_2\overset{t^o}{\longrightarrow}2Al_2O_3$
$0,08→0,06$
$⇒V_{O_2}=0,06.22,4=1,344l$
$2/$
$n_{Na}=2,3/23=0,1mol$
$4Na+O_2\overset{t^o}{\longrightarrow}2Na_2O$
$0,1→0,025$
$⇒V_{O_2}=0,025.22,4=0,56l$
$3/$
$n_{Mg}=4,32/24=0,18mol$
$2Mg+O_2\overset{t^o}{\longrightarrow}2MgO$
$0,18→0,09$
$⇒V_{O_2}=0,09.22,4=2,016l$