Đáp án:

${V_{S{O_2}}} = 0,84\,\,l$

Giải thích các bước giải:

 Lượng sắt ở hai phần là như nhau, nên số mol như nhau.

Phần 1: ${n_{{H_2}}} = 0,025\,\,mol$

$\begin{gathered} Fe\,\,\,\, + 2HCl \to FeC{l_2} + {H_2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \hfill \\ 0,025 \leftarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,025\,\,\,mol \hfill \\ \to {n_{Fe}} = 0,025\,\,mol\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \hfill \\ \end{gathered} $

Phần 2:

$\begin{gathered} 2Fe + 6{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + 3S{O_2} + 6{H_2}O \hfill \\ {n_{S{O_2}}} = \frac{3}{2}{n_{Fe}} = 0,0375\,\,mol\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \hfill \\ \to {V_{S{O_2}}} = 0,0375.22,4 = 0,84\,\,l\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \hfill \\ \end{gathered} $