$n_{H_2}=\dfrac{13,44}{22,4}=0,6(mol)$

$PTHH:Fe+2HCl\to FeCl_2+H_2(1)$

$Mg+2HCl\to MgCl_2+H_2(2)$

$Zn+2HCl\to ZnCl_2+H_2(3)$

Theo PT(1) và PT(3): $n_{Zn}=n_{H_2(3)};n_{Fe}=n_{H_2(1)}$

Mà $V_{H_2(3)}=2V_{H_2(1)}\Rightarrow n_{H_2(3)}=2n_{H_2(1)}$

$\Rightarrow n_{Zn}=2n_{Fe}$

Mà $n_{Zn}+n_{Fe}+n_{Mg}=0,6$

Và $65n_{Zn}+56n_{Fe}+24n_{Mg}=25,8$

Giải được: $n_{Zn}=0,2(mol);n_{Fe}=0,1(mol);n_{Mg}=0,3(mol)$

$\Rightarrow m_{Zn}=0,2.65=13(g);m_{Fe}=0,1.56=5,6(g)$

$\Rightarrow \%m_{Zn}=\dfrac{13}{25,8}.100\%=50,39\%$

$\Rightarrow \%m_{Fe}=\dfrac{5,6}{25,8}.100\%=21,71\%$

$\Rightarrow \%m_{Mg}=100-21,71-50,39=27,9\%$

$b)$ Từ 3 PT: $n_{HCl}=2n_{H_2}=1,2(mol)$

$\Rightarrow V_{dd_{HCl}}=\dfrac{1,2}{0,2}=6(l)=6000(ml)$

Em tham khảo nha:

\(\begin{array}{l}
a)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{{13,44}}{{22,4}} = 0,6\,mol\\
hh:Fe(a\,mol),Mg(b\,mol),Zn(c\,mol)\\
\left\{ \begin{array}{l}
56a + 24b + 65c = 25,8\\
a + b + c = 0,6\\
2a - c = 0
\end{array} \right.\\
 \Rightarrow a = 0,1;b = 0,3;c = 0,2\\
\% {m_{Fe}} = \dfrac{{0,1 \times 56}}{{25,8}} \times 100\%  = 21,71\% \\
\% {m_{Mg}} = \dfrac{{0,3 \times 24}}{{25,8}} \times 100\%  = 27,91\% \\
\% {m_{Zn}} = 100 - 21,71 - 27,91 = 50,38\% \\
b)\\
{n_{HCl}} = 2{n_{{H_2}}} = 1,2\,mol\\
{V_{{\rm{dd}}HCl}} = \dfrac{{1,2}}{{0,2}} = 6l = 6000ml
\end{array}\)