$\{a\}\\4Al + 3O_2 \xrightarrow{t^o}2Al_2O_3\\\{b\}\\n_{Al}=\dfrac{13,5}{27}=0,5\,(\text{mol})\\\to n_{Al_2O_3}=\dfrac{1}{2}n_{Al}=0,25\,(\text{mol})\\\to m_{Al_2O_3}=0,25\times102=25,5\,(\text{g})\\\{c\}\\n_{O_2}=\dfrac{3}{4}n_{Al}=0,375\,(\text{mol})\\\to V_{O_2}=0,375\times22,4=8,4\,(\text{l})$