Em tham khảo nha :

\(\begin{array}{l}
1)\\
a)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{{2,24}}{{22,4}} = 0,1mol\\
{n_{Fe}} = {n_{{H_2}}} = 0,1mol\\
{m_{Fe}} = 0,1 \times 56 = 5,6g\\
{m_{Ag}} = 16,4 - 56 = 10,8g\\
b)\\
{n_{HCl}} = 2{n_{{H_2}}} = 0,2mol\\
{C_{{M_{HCl}}}} = \dfrac{{0,2}}{{0,2}} = 1M\\
c)\\
4HCl + Mn{O_2} \to MnC{l_2} + C{l_2} + 2{H_2}O\\
{n_{C{l_2}}} = \dfrac{{{n_{HCl}}}}{4} = \dfrac{{0,2}}{4} = 0,05mol\\
{V_{C{l_2}}} = 0,05 \times 22,4 = 1,12l\\
2)\\
a)\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{{3,36}}{{22,4}} = 0,15mol\\
{n_{Mg}} = {n_{{H_2}}} = 0,15mol\\
{m_{Mg}} = 0,15 \times 24 = 3,6g\\
\% Mg = \dfrac{{3,6}}{{10}} \times 100\%  = 36\% \\
\% Cu = 100 - 36 = 64\% \\
b)\\
{n_{HCl}} = 2{n_{{H_2}}} = 0,3mol\\
{C_{{M_{HCl}}}} = \dfrac{{0,3}}{{0,15}} = 2M\\
c)\\
{n_{MgC{l_2}}} = {n_{Mg}} = 0,15mol\\
{m_{MgC{l_2}}} = 0,15 \times 95 =  = 14,25g
\end{array}\)