Đáp án:
\(\begin{array}{l}
a/{V_{HCl}} = 0,4l\\
b/{m_{FeC{l_2}}} = 25,4g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
Fe + 2HCl \to FeC{l_2} + {H_2}\\
FeO + 2HCl \to FeC{l_2} + {H_2}O\\
{n_{{H_2}}} = 0,1mol\\
\to {n_{Fe}} = {n_{{H_2}}} = 0,1mol\\
\to {m_{Fe}} = 5,6g\\
\to {m_{FeO}} = 7,2g \to {n_{FeO}} = 0,1mol\\
a/\\
{n_{HCl}} = 2{n_{Fe}} + 2{n_{FeO}} = 0,4mol\\
\to {V_{HCl}} = 0,4l\\
b/\\
{n_{FeC{l_2}}} = {n_{Fe}} + {n_{FeO}} = 0,2mol\\
\to {m_{FeC{l_2}}} = 25,4g
\end{array}\)