Em tham khảo nha :
\(\begin{array}{l}
a)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{{3,36}}{{22,4}} = 0,15mol\\
{n_{Fe}} = {n_{{H_2}}} = 0,15mol\\
{m_{Fe}} = 0,15 \times 56 = 8,4g\\
{m_{Ag}} = 15 - 8,4 = 6,6g\\
b)\\
\% Fe = \dfrac{{8.4}}{{15}} \times 100\% = 56\% \\
\% Ag = 100 - 56 = 44\% \\
c)\\
{n_{HCl}} = 2{n_{Fe}} = 0,3mol\\
{m_{HCl}} = 0,3 \times 36,5 = 10,95g\\
{m_{{\rm{dd}}HCl}} = \dfrac{{10,95 \times 100}}{{15,6}} = 70,2g
\end{array}\)