Đáp án:

\(\begin{array}{l}
{V_{{H_2}}} = 4,48l\\
\% {m_{Mg}} = 15,58\% \\
\% {m_{Zn}} = 84,42\% 
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
Mg + {H_2}S{O_4} \to MgS{O_4} + {H_2}\\
Zn + {H_2}S{O_4} \to ZnS{O_4} + {H_2}\\
{P_2}:\\
hh:Mg(a\,mol),Zn(b\,mol)\\
{n_{{H_2}}} = \dfrac{{13,44}}{{22,4}} = 0,6\,mol\\
{m_{hh({P_2})}} = \dfrac{{46,2}}{3} \times 2 = 30,4g\\
\left\{ \begin{array}{l}
24a + 65b = 30,4\\
a + b = 0,6
\end{array} \right.\\
 \Rightarrow a = 0,2;b = 0,4\\
{P_2}:\\
{n_{{H_2}S{O_4}}} = 0,2 \times 1 = 0,2\,mol\\
{n_{hh}} = \dfrac{{0,2 + 0,4}}{2} = 0,3\,mol\\
 \Rightarrow \text{ Hỗn hợp dư tính theo $H_2SO_4$ } \\
{n_{{H_2}}} = {n_{{H_2}S{O_4}}} = 0,2\,mol\\
{V_{{H_2}}} = 0,2 \times 22,4 = 4,48l\\
\% {m_{Mg}} = \dfrac{{0,3 \times 24}}{{46,2}} \times 100\%  = 15,58\% \\
\% {m_{Zn}} = 100 - 15,58 = 84,42\% 
\end{array}\)