a, $4Al+3O_{2} \buildrel{{t^o}}\over\to 2Al_{2}O_{3}$
b, $n_{Al}=\dfrac{10,8}{27}=0,4(mol)$
$\to n_{Al_{2}O_{3}}=\dfrac{0,4.2}{4}=0,2(mol)$
$\to m_{Al_{2}O_{3}}=0,2.102=20,4g$
c,$n_{O_{2}}=\dfrac{0,4.3}{4}=0,3(mol)$
$V_{O_{2}}=n.22,4=0,3.22,4=6,72l$
$\to V_{kk}=5.V_{O_{2}}=5.6,72=33,6l$