Giải thích các bước giải:
3,
\(\begin{array}{l}
Mg + 2HCl \to MgC{l_2} + {H_2}\\
{n_{Mg}} = 0,3mol\\
\to {n_{{H_2}}} = {n_{Mg}} = 0,3mol\\
\to {V_{{H_2}}} = 6,72l\\
{n_{MgC{l_2}}} = {n_{Mg}} = 0,3mol\\
\to {m_{MgC{l_2}}} = 28,5g
\end{array}\)
3,
\(\begin{array}{l}
F{e_2}{O_3} + 3{H_2} \to 2Fe + 3{H_2}O\\
{n_{{H_2}}} = 0,3mol\\
\to {n_{F{e_2}{O_3}}} = \dfrac{1}{3}{n_{{H_2}}} = 0,1mol\\
\to {m_{F{e_2}{O_3}}} = 16g\\
\to {n_{Fe}} = \dfrac{2}{3}{n_{{H_2}}} = 0,2mol\\
\to {m_{Fe}} = 11,2g
\end{array}\)