nFeO=14,4/72=0,2(mol)
PTHH: FeO + 2 HCl -> FeCl2 + H2O
nHCl=2.0,2=0,4(mol)
=>VddHCl=0,4/2=0,2(l)=200(ml)
=>V=200(ml)
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$n_{FeO} = \dfrac{14,4}{72} = 0,2(mol)$
$FeO + 2HCl \to FeCl_2 + H_2O$
$n_{HCl} = 2n_{FeO} = 0,2.2 = 0,4(mol)$
$V_{dd\ HCl} = \dfrac{0,4}{2} = 0,2(lít) = 200(ml)$