$M_xO_y+ yCO\to xM+ yCO_2$

$CO_2+Ca(OH)_2\to CaCO_3+H_2O$

$Al_2O_3+2NaOH\to 2NaAlO_2+H_2O$

$2M+nH_2SO_4\to M_2(SO_4)_n+ nH_2$

Gọi x là mol $CO_2$ tạo thành.

$\Delta m= m_{CaCO_3}-m_{CO_2}$

$\Rightarrow 100x-44x=3,36$

$\Leftrightarrow x=0,06= n_{O(M_xO_y)}$

$\Rightarrow m_Z=m_X-m_O=7,305-0,06.16=6,345g$

$n_{NaOH}=0,075 mol$

$\Rightarrow n_{Al_2O_3}=0,0375mol$

$\Rightarrow m_T= m_Z-m_{Al_2O_3}=6,345-0,0375.102=2,52g$

$n_{H_2}=0,045 mol$

$\Rightarrow n_M=\frac{0,09}{n}$

$M_M=\frac{2,52n}{0,09}=28n$

$n=2\Rightarrow M_M=56(Fe)$ 

$n_{Fe}=\frac{0,09}{2}=0,045 mol$ 

$n_O= 0,06 mol$

$n_{Fe} : n_O = 0,045:0,06=3:4$

$\Rightarrow$ Oxit là $Fe_3O_4$

$n_O= 4n_{Fe_3O_4}$

$\Rightarrow n_{Fe_3O_4}=0,015 mol$

$\%m_{Fe_3O_4}=\frac{0,015.232.100}{7,305}=47,64\%$

$\%m_{Al_2O_3}=52,36\%$