a) nh2=\(\dfrac{8.96}{22.4}\)=0.4 (mol)
Fe + 2HCl \(\rightarrow\) FeCl2 + H2
x 2x x x
Mg + 2HCl \(\rightarrow\) MgCl2 + H2
y 2y y y
\(\begin{cases} x+y=0.4\\ 56x+24y=16 \end{cases}\)
\(\Leftrightarrow\) \(\begin{cases} x=0.2\\ y=0.2 \end{cases}\)
%mFe=\(\dfrac{0.2 * 56}{16}\)*100=70%
%mMg= 100-70=30%
b) \(\sum\)nHCl=0.4+0.4=0.8 (mol)
Vdd HCl=\(\dfrac{0.8}{2}\)= 0.4 (l)