\(\begin{array}{l}
Fe+2HCl\to FeCl_2+H_2\\
2Al+6HCl\to 2AlCl_3+3H_2\\
n_{HCl}=\frac{21,9}{36,5}=0,6(mol)\\
a)\\
Gia\,su\,hon\,hop\,toan\,Fe\\
\to n_{hh(min)}=n_{Fe}=\dfrac{22}{56}≈0,4(mol)\\
Theo\,PT:\,0,6=n_{HCl}<2n_{Fe}=2n_{hh(min)}≈0,8(mol)\\
\to Axit\,het,\,hon\,hop\,KL\,du\\
b)\\
Theo\,PT:\,n_{H_2}=\frac{1}{2}n_{HCl}=0,3(mol)\\
\to V_{H_2(đktc)}=0,3.22,4=6,72(l)=6720(cm^3)\\
\to V=6720
\end{array}\)

Giải thích các bước giải:

$a)$

Giả sử hỗn hợp chỉ có $Fe.$ 

$n_{X\ (min)}=n_{Fe}=\dfrac{22}{56}=0,4(mol)$

$n_{HCl}=\dfrac{21,9}{36,5}=0,6(mol)$

$Fe+2HCl\to FeCl_2+H_2$

$n_{HCl\ (X)}=2n_{Fe}=0,8(mol)>n_{HCl}=0,6(mol)$

Suy ra X không tan hết.

$b)$

Bảo toàn nguyên tố H:

$n_{H_2}=\dfrac{1}{2}n_{HCl}=0,3(mol)\\V_{H_2}=0,3×22,4=6,72(l)$