Em tham khảo nha:

\(\begin{array}{l}
a)\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
\text{ Giả sử hỗn hợp chỉ gồm Mg }\\
{n_{Mg}} = \dfrac{{8,4}}{{24}} = 0,35\,mol\\
{n_{HCl}} = \dfrac{{36,5}}{{36,5}} = 1\,mol\\
{n_{Mg}} < \dfrac{{{n_{HCl}}}}{2} \Rightarrow HCl \text{ dư }\\
{n_{hh}} < {n_{Mg}} \Rightarrow \text{ HCl dư } \\
b)\\
{n_{{H_2}}} = \dfrac{{4,48}}{{22,4}} = 0,2\,mol\\
hh:Zn(a\,mol),Mg(b\,mol)\\
\left\{ \begin{array}{l}
a + b = 0,2\\
65a + 24b = 8,4
\end{array} \right.\\
 \Rightarrow a = 0,0878;b = 0,1122\\
{m_{Zn}} = 0,0878 \times 65 = 5,707g\\
{m_{Mg}} = 8,4 - 5,707 = 2,693g
\end{array}\)