Đáp án:
$11,2$ lít
Giải thích các bước giải:
$ 60\,gam\left\{ \begin{array}{l} F\text{e}\,\text{(x}\,\text{mol)} \\ S\,(y\,mol) \end{array} \right.\xrightarrow{{{t}^{o}}}\left\{ \begin{array}{l} \text{FeS}\,\text{(x}\,\text{mol)} \\ S \end{array} \right.\xrightarrow{{{H}_{2}}S{{O}_{4}}}\left\{ \begin{array}{l} S\,(B) \\ {{H}_{2}}S\,(D) \end{array} \right. $
$ {{m}_{S\,(du)}}=16\,(gam)\to {{n}_{S}}{{\,}_{(du)}}=0,5\,(mol) $
Bảo toàn nguyên tố Fe: $ {{n}_{F\text{e}S}}={{n}_{Fe}}=x\,(mol) $
Bảo toàn nguyên tố S: $ {{n}_{S}}={{n}_{F}}_{\text{eS}}+{{n}_{S\,(du)}} $
$ \to \left\{ \begin{array}{l} 56\text{x}+32y=60 \\ BT''S'':y=x+0,5 \end{array} \right.\to x=0,5;y=1 $
$ \begin{array}{l} F\text{e}S\,+{{H}_{2}}S{{O}_{4}}\to F\text{e}S{{O}_{4}}\,+\,{{H}_{2}}S \\ 0,5\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\to \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,5\,(mol) \\ \to V=0,5\times 22,4=11,2\,(l) \end{array} $
