Em tham khảo nha :

\(\begin{array}{l}
F{e_3}{O_4} + 8HCl \to FeC{l_2} + 2FeC{l_3} + 4{H_2}O\\
{n_{F{e_3}{O_4}}} = \dfrac{{23,2}}{{232}} = 0,1mol\\
{n_{HCl}} = 0,4 \times 2 = 0,8mol\\
\dfrac{{0,1}}{1} = \dfrac{{0,8}}{8} \Rightarrow\text{Phản ứng xảy ra hoàn toàn} \\
{n_{FeC{l_2}}} = {n_{F{e_3}{O_4}}} = 0,1mol\\
{n_{FeC{l_3}}} = 2{n_{F{e_3}{O_4}}} = 0,2mol\\
{n_{Cu}} = \dfrac{{25,6}}{{64}} = 0,4mol\\
\end{array}\)

\(\begin{array}{l}
Cu + 2FeC{l_3} \to CuC{l_2} + 2FeC{l_2}\\
{n_{Cu}} > \dfrac{{{n_{FeC{l_3}}}}}{2} \Rightarrow \text{ Cu dư }\\
{n_{CuC{l_2}}} = \dfrac{{0,2}}{2} = 0,1\,mol\\
{n_{FeC{l_2}}} = {n_{FeC{l_3}}} = 0,2\,mol\\
CuC{l_2} + 2AgN{O_3} \to Cu{(N{O_3})_2} + 2AgCl\\
FeC{l_2} + 2AgN{O_3} \to Fe{(N{O_3})_2} + 2AgCl\\
Fe{(N{O_3})_2} + AgN{O_3} \to Fe{(N{O_3})_3} + Ag\\
{n_{AgCl}} = 2{n_{CuC{l_2}}} + 2{n_{FeC{l_2}}} = 0,8\,mol\\
{n_{Fe{{(N{O_3})}_2}}} = {n_{FeC{l_2}}} = 0,3\,mol\\
{n_{Ag}} = {n_{Fe{{(N{O_3})}_2}}} = 0,3\,mol\\
m = 0,8 \times 143,5 + 0,3 \times 108 = 147,2g
\end{array}\)