\(n_{H2}=0,06\left(mol\right)\)
\(H_2+O\rightarrow H_2O\)
0,06__0,06________
\(\Rightarrow m_M=3,48-0,96=2,52\left(g\right)\)
\(n_{H2}=0,045\left(mol\right)\)
\(2M+2nHCl\rightarrow2MCl_n+nH_2\)
0,09/n__________________0,045
\(\Rightarrow M_M=\frac{2,52n}{0,09}=28n\)
\(n=2\Rightarrow M_M=56\left(Fe\right)\Rightarrow MCl_n:FeCl_2\)
\(n_{Fe}=0,045\left(mol\right)\)
\(n_{Fe}:n_O=0,045:0,06=3:4\)
Vậy FexOy là Fe3O4