Đáp án:

$1)\ Fe_3O_4\\ 2)\ Fe\\Fe_3O_4$

Giải thích các bước giải:

$1)\\ n_{H_2}=\dfrac{2,352}{22,4}=0,105(mol)\\ Fe_xO_y+yCO\xrightarrow{t^0} xFe+yCO_2\\ Fe+H_2SO_4\to FeSO_4+H_2\\ \xrightarrow{\text{Từ PTHH}}\ n_{Fe}=n_{H_2}=0,105(mol)\\ \xrightarrow{\text{Từ PTHH}}\ n_{Fe_xO_y}=\dfrac{1}{x}.n_{Fe}=\dfrac{0,105}{x}(mol)\\ \Rightarrow M_{Fe_xO_y}=8,12:\dfrac{0,105}{x}=\dfrac{232}{3}x(g/mol)\\ \Rightarrow 56x+16y=\dfrac{232}{3}x\\ \Rightarrow \dfrac{64}{3}x=16y\\ \Rightarrow \dfrac{x}{y}=\dfrac{3}{4}\\ \to CTHH:\ Fe_3O_4$

$2)\\ Oxit:\ M_2O_n\\ M_2O_n+nH_2\xrightarrow{t^0} 2M+nH_2O\ (1)\\ 2M+2mHCl\to 2MCl_m+mH_2O\\ n_{H_2\ (1)}=\dfrac{1,344}{22,4}=0,06(mol)\\ n_{H_2\ (2)}=\dfrac{1,008}{22,4}=0,045(mol)\\ n_{M_2O_n}=\dfrac{1}{n}.n_{H_2\ (1)}=\dfrac{0,06}{n}(mol)\\ \Rightarrow M_{M_2O_n}=3,48:\dfrac{0,06}{n}=58n(g/mol)\\ \Rightarrow 2M+16n=58n\\ \Rightarrow M=21n\\ \Rightarrow n=\dfrac{8}{3};\ M=56(Fe)\\ \to M:\ Fe\\ Oxit:\ Fe_3O_4$