$4Al+3O_2\xrightarrow{t^o}2Al_2O_3$
$4$ : $3$ : $2$ (mol)
$0,2$ : $0,15$ : $0,1$
$n_{Al}=\dfrac{5,4}{27}=0,2(mol)$
$⇒V_{O_2}=0,15.22,4=0,15.22=3,36(l)=\dfrac{1}{5}V_{kk}$
$⇒V_{kk}=5V_{O_2}=5.3,36=16,8(l)$
$m_{Al_2O_3}=0,1.(27.2+16.3)=0,1.102=10,2(g)$