\(a,PTHH:4Al+3O_2\rightarrow2Al_2O_3\\ TL:....4.....3.....2\left(mol\right)\\ BR:....0,4......0,3......0,2\left(mol\right)\\ b,n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right);n_{Al_2O_3}=\dfrac{20,4}{102}=0,2\left(mol\right)\)
Vì \(\dfrac{n_{Al}}{4}=\dfrac{n_{Al_2O_3}}{2}\) nên phản ứng xảy ra hoàn toàn
\(\Rightarrow m_{O_2}=0,3\cdot32=9,6\left(g\right)\)