$P_1:\begin{cases}n_{Fe}=0,0675 mol \\n_{Al_2O_3}=0,03 mol\\ n_{Al}=0,0525 mol \end{cases}\Rightarrow P_2:\begin{cases}n_{Fe}=0,0675t \\ n_{Al_2O_3}=0,03t\\n_{Al}=0,0525t \end{cases}$$\Rightarrow 0,0675t+1,5.0,0525t=\frac{9,828}{22,4}=0,43875\Rightarrow t=3$
$\Rightarrow m_{P_2}=3m_{P_1}\Rightarrow m=4m_{P_1}=33,03 gam$