Đáp án:

 \(\begin{array}{l}
a)\\
\% {m_{Fe}} = 43,58\% \\
\% {m_{Cu}} = 24,9\% \\
\% {m_{Al}} = 31,52\% \\
b)\\
x = 2,6\\
c)\\
{V_{{\rm{dd}}{H_2}S{O_4}}}80\,ml\\
d)\\
{m_{Ag}} = 21,6g
\end{array}\)

Giải thích các bước giải:

 \(\begin{array}{l}
a)\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{{14,56}}{{22,4}} = 0,65mol\\
{m_{cr}} = {m_{Cu}} = 6,4g\\
hh:Al(a\,mol),Fe(b\,mol)\\
27a + 56b = 25,7 - 6,4\\
1,5a + b = 0,65\\
 \Rightarrow a = 0,3;b = 0,2\\
\% {m_{Fe}} = \dfrac{{0,2 \times 56}}{{25,7}} \times 100\%  = 43,58\% \\
\% {m_{Cu}} = \dfrac{{6,4}}{{25,7}} \times 100\%  = 24,9\% \\
\% {m_{Al}} = 100 - 43,58 - 24,9 = 31,52\% \\
b)\\
{n_{HCl}} = 2{n_{{H_2}}} = 0,65 \times 2 = 1,3\,mol\\
x = {C_M}HCl = \dfrac{{1,3}}{{0,5}} = 2,6M\\
c)\\
Cu + 2{H_2}S{O_4} \to CuS{O_4} + S{O_2} + 2{H_2}O\\
{n_{Cu}} = \dfrac{{6,4}}{{64}} = 0,1\,mol\\
{n_{{H_2}S{O_4}}} = 2{n_{Cu}} = 0,2\,mol\\
{V_{{\rm{dd}}{H_2}S{O_4}}} = \dfrac{{0,2}}{{2,5}} = 0,08l = 80\,ml\\
d)\\
Cu + 2AgN{O_3} \to Cu{(N{O_3})_2} + 2Ag\\
{n_{AgN{O_3}}} = 0,2 \times 1,5 = 0,3\,mol\\
\dfrac{{0,1}}{1} < \dfrac{{0,3}}{2} \Rightarrow \text{ $AgNO_3$ dư}\\
{n_{Ag}} = 2{n_{Cu}} = 0,2\,mol\\
{m_{Ag}} = 0,2 \times 108 = 21,6g
\end{array}\)