\(n_{H_2SO_4}=\dfrac{147.20\%}{98}=0,3\left(mol\right)\\ Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ n_{H_2}=n_{axit}=0,3\left(mol\right)\\ Đặt:A_2O_3\\ A_2O_3+3H_2\rightarrow\left(t^o\right)2A+3H_2O\\ n_{oxit}=\dfrac{n_{H_2}}{3}=\dfrac{0,3}{3}=0,1\left(mol\right)\\ M_{oxit}=\dfrac{16}{0,1}=160\left(\dfrac{g}{mol}\right)=2M_A+48\\ \Rightarrow M_A=56\left(\dfrac{g}{mol}\right)\\ \Rightarrow A:Sắt\left(Fe=56\right)\\ \Rightarrow Oxit:Fe_2O_3\)