\(\begin{array}{l}
Dat\,oxit\,la\,M_2O_n\\
\to n_{M_2O_n}=\frac{6,4}{2M+16n}(mol)\\
M_2O_n+nH_2SO_4\to M_2(SO_4)_n+nH_2O\\
Theo\,PT:\begin{cases}n_{H_2SO_4}=\frac{6,4n}{2M+16n}(mol)\\ n_{M_2(SO_4)_n}=\frac{6,4}{2M+16n}(mol) \end{cases}\\
\to \begin{cases} m_{dd\,H_2SO_4}=\frac{627,2n}{10\%(2M+16n)}=\frac{6272n}{2M+16n}(g)\\ m_{M_2(SO_4)_n}=\frac{12,8M+614,4n}{2M+16n}(g)\end{cases}\\
\to m_{dd\,spu}=6,4+\frac{6272n}{2M+16n}=\frac{12,6M+6374,4n}{2M+16n}(g)\\
\to C\%_{M_2(SO_4)_n}=\frac{12,8M+614,4n}{12,6M+6374,4n}.100=12,9\%\\
\to \begin{cases} M=56\\ n=3 \end{cases}\\
\to M:Fe;Oxit:Fe_2O_3\\
Fe_2O_3+3H_2SO_4\to Fe_2(SO_4)_3+3H_2O\\
Theo\,PT:\,n_{Fe_2(SO_4)_3}=n_{Fe_2O_3}=\frac{6,4}{160}=0,04(mol)\\
Dat\,CT\,hidrat\,la\,Fe_2(SO_4)_3.nH_2O\\
\to n_{Fe_2(SO_4)_3.nH_2O}=n_{Fe_2(SO_4)_3}=0,04(mol)\\
\to n_{Fe_2(SO_4)_3.nH_2O(tt)}=0,04.70\%=0,028(mol)\\
\to M_{Fe_2(SO_4)_3.nH_2O}=\frac{15,736}{0,028}≈562(g/mol)\\
\to 400+18n=562\\
\to n=9\\
\to CT:Fe_2(SO_4)_3.9H_2O
\end{array}\)

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