$n_{Al}=\dfrac{10,8}{27}=0,4\ (mol)$
$4Al+3O_2\xrightarrow{t^o} 2Al_2O_3$
a, Theo PT:
$n_{Al_2O_3}=\dfrac{1}{2}n_{Al}=\dfrac12.0,4=0,2\ (mol)$
$\to m_{Al_2O_3}=0,2.102=20,4\ (g)$
b, Theo PT:
$n_{O_2}=\dfrac34n_{Al}=\dfrac34.0,4=0,3\ (mol)$
$\to V_{O_2}=0,3.22,4=6,72\ (l)$