\(n_{Al}=\dfrac{162}{27}=6\left(mol\right)\)
\(2Al_2O_3\underrightarrow{dpnc,criolit}4Al+3O_2\)
\(3.............................6\)
\(m_{Al_2O_3\left(lt\right)}=3\cdot102=306\left(g\right)\)
\(m_{Al_2O_3\left(tt\right)}=\dfrac{306}{80\%}=382.5\left(g\right)\)