Em tham khảo nha:

\(\begin{array}{l}
F{e_3}{O_4} + 8HCl \to 2FeC{l_3} + FeC{l_2} + 4{H_2}O(1)\\
FeC{l_3} + 3NaOH \to 3NaCl + Fe{(OH)_3}(2)\\
FeC{l_2} + 2NaOH \to 2NaCl + Fe{(OH)_2}(3)\\
4Fe{(OH)_2} + {O_2} + 2{H_2}O \to 4Fe{(OH)_3}(4)\\
2Fe{(OH)_3} \to F{e_2}{O_3} + 3{H_2}O(5)\\
{n_{F{e_3}{O_4}}} = \dfrac{m}{{232}}\,mol\\
 \Rightarrow {n_{FeC{l_2}}} = {n_{F{e_3}{O_4}}} = \dfrac{m}{{232}}\,mol\\
{n_{FeC{l_3}}} = 2{n_{F{e_3}{O_4}}} = \dfrac{m}{{116}}\,mol\\
{n_{Fe{{(OH)}_3}}}(2) = {n_{FeC{l_3}}} = \dfrac{m}{{116}}\,mol\\
{n_{Fe{{(OH)}_2}}}(3) = {n_{FeC{l_2}}} = \dfrac{m}{{232}}\,mol\\
{n_{Fe{{(OH)}_3}}}(4) = {n_{Fe{{(OH)}_2}}}(3) = \dfrac{m}{{232}}\,mol\\
{n_{Fe{{(OH)}_3}}}(5) = \dfrac{m}{{116}} + \dfrac{m}{{232}} = \dfrac{{3m}}{{232}}\,mol\\
{n_{F{e_2}{O_3}}} = \dfrac{{3m}}{{232}} \times \dfrac{1}{2} = \dfrac{{3m}}{{464}}\,mol\\
{m_{F{e_2}{O_3}}} = \dfrac{{3m}}{{464}} \times 160 = \dfrac{{30m}}{{29}}(g)
\end{array}\)

$n_{Fe_3O_4}=\dfrac{m}{232} (mol)$ 

Quá trình biến đổi chất:

$Fe_3O_4\buildrel{{HCl, -H_2O}}\over\longrightarrow (FeCl_2, FeCl_3)\buildrel{{+NaOH, -NaCl}}\over\longrightarrow (Fe(OH)_2, Fe(OH)_3)\buildrel{{t^o kk, -H_2O}}\over\longrightarrow Fe_2O_3$ 

Bảo toàn Fe: $3n_{Fe_3O_4}=2n_{Fe_2O_3}$

Chất rắn sau khi nung là $Fe_2O_3$

$n_{Fe_2O_3}=\dfrac{3m}{464}(mol)$

$\to m_{Fe_2O_3}=160.\dfrac{3m}{464}\approx 1,034m(g)$