$PTPƯ:$
$Mg+2HCl\xrightarrow{} MgCl_2+H_2↑$ $(1)$
$Fe+2HCl\xrightarrow{} FeCl_2+H_2↑$ $(2)$
$n_{H_2}=\dfrac{4,48}{22,4}=0,2mol.$
$⇒m_{H_2}=0,2.2,=0,4g.$
$Theo$ $pt$ $(1),(2):$ $n_{HCl}=2n_{H_2}=0,4mol.$
$⇒m_{HCl}=0,4.36,5=14,6g.$
$\text{Bảo toàn khối lượng:}$
$⇒m_{muối}=m_{hỗn\ hợp}+m_{HCl}-m_{H_2}=8+14,6-0,4=22,2g.$
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