$Mg + 2HCl \rightarrow MgCl_{2} + H_{2}\uparrow$
$2Al + 6HCl \rightarrow 2AlCl_{3} + 3H_{2}\uparrow$
$Fe + 2HCl \rightarrow FeCl_{2} + H_{2}\uparrow$
$\text{Ta có: }n_{H_{2}}=\frac{11,2}{22,4}=0,5 (mol)$
$\text{Theo ptpư: }n_{HCl}=2n_{H_{2}}=1 (mol)$
$\Rightarrow m_{HCl}=1.36,5=36,5 (g)$
$m_\text{muối}=m_\text{kim loại}+m_{HCl}-m_{H_{2}}$
$\Rightarrow m_\text{muối}=20+36,5-0,5.2=55,5 (g)$