[hóa 10] giúp

K

kakashi_hatake

$n_{FeO}=n_{Fe_2O_3}$ nên coi hỗn hợp chất này là $Fe_3O_4$
$n_{Fe_3O_4}=0.01 \ mol \\ Fe_3O_4+8HCl \rightarrow FeCl_2+2FeCl_3+4H_2O \\ 0.01 \ \ \ \ \ \ \ \ \ \ \ \ 0.08 \\ V_{HCl}=\dfrac{n}{C_M}=\dfrac{0.08}{1}=0.08l=80ml$
 
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