Hướng dẫn giải

Ta có: n_Fe= \[\frac{{40,32}}\]= 0,72 mol;

n_O= \[\frac{{50,56 - 40,32}}\]= 0,64 mol

Ta có:

\[\begin{array}{*{20}{l}}\begin{array}{l}\mathop {Fe}\limits^0 {\rm{ }} \to {\rm{ }}\mathop {Fe}\limits^{ + 3} \; + 3e\\0,72\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2,16\,\,\,\,\,\,\,\,\,\,\,mol\end{array}\\\begin{array}{l}\mathop O\limits^0 \,\,\,\, + {\rm{ }}2e{\rm{ }} \to \mathop O\limits^{ - 2} \;\;\\0,64\,\,\,\,\,\,1,28\;\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,mol\;\;\;\;\;\;\;\end{array}\\\begin{array}{l}\mathop S\limits^{ + 6} + 2e \to \mathop S\limits^{ + 4} \\x\,\,\,\,\,\,2x\,\,\,\,\,\,\,\,\,\,mol\end{array}\end{array}\]

Áp dụng định luật bảo toàn e

\[{n_{S{O_2}}} = \frac{{2,16 - 1,28}}{2}\] = 0,44 (mol)

Vậy \[{V_{S{O_2}}}\]= 0,44 .22,4 = 9,856(lít)