Đáp án:
4,
a, \({m_{Fe}} = 11,2g\)
b, \({m_{F{e_2}{O_3}}} = 16g\)
c, \(C{M_{HCl}} = \dfrac{{0,4}}{{0,8}} = 0,5M\)
Giải thích các bước giải:
\(\begin{array}{l}
F{e_2}{O_3} + 3{H_2} \to 2Fe + 3{H_2}O\\
{n_{{H_2}}} = 0,3mol\\
\to {n_{Fe}} = \dfrac{2}{3}{n_{{H_2}}} = 0,2mol\\
\to {m_{Fe}} = 11,2g\\
{n_{F{e_2}{O_3}}} = \dfrac{1}{3}{n_{{H_2}}} = 0,1mol\\
\to {m_{F{e_2}{O_3}}} = 16g\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
\to {n_{HCl}} = 2{n_{Fe}} = 0,4mol\\
\to C{M_{HCl}} = \dfrac{{0,4}}{{0,8}} = 0,5M
\end{array}\)