a) n_h2=\(\dfrac{8.96}{22.4}\)=0.4 (mol)

Fe + 2HCl \(\rightarrow\) FeCl₂ + H₂

x 2x x x

Mg + 2HCl \(\rightarrow\) MgCl₂ + H₂

y 2y y y

\(\begin{cases} x+y=0.4\\ 56x+24y=16 \end{cases}\)

\(\Leftrightarrow\) \(\begin{cases} x=0.2\\ y=0.2 \end{cases}\)

%m_Fe=\(\dfrac{0.2 * 56}{16}\)*100=70%

%m_Mg= 100-70=30%

b) \(\sum\)n_HCl=0.4+0.4=0.8 (mol)

V_{dd HCl}=\(\dfrac{0.8}{2}\)= 0.4 (l)