a,
$Fe+2HCl\to FeCl_2+H_2$
$FeO+2HCl\to FeCl_2+H_2O$
b,
Gọi x, y là số mol $Fe$, $FeO$.
$\Rightarrow 56x+72y=19,4$ $(1)$
$n_{H_2}=\dfrac{4,48}{22,4}=0,2(mol)$
$\Rightarrow x=0,2$ $(2)$
$(1)(2) \Rightarrow y=0,1139$
$\%m_{Fe}=\dfrac{56x.100}{19,4}=57,73\%$
$\%m_{FeO}=42,27\%$
c,
$n_{HCl}=2x+2y=0,3139(mol)$
$\Rightarrow m_{dd HCl}=0,3139.36,5:7,3\%=156,95g$
d,
$m_{dd \text{spứ}}=19,4+156,95-0,2.2=175,95g$
$n_{FeCl_2}=x+y=0,15695(mol)$
$\Rightarrow C\%_{FeCl_2}=\dfac{0,15695.127.100}{175,95}=11,33\%$