Đáp án B
${{n}_{{{H}_{2}}}}=0,1\,\,mol$
Ta có:
$\begin{array}{l} F{\rm{e}} \to \mathop {F{\rm{e}}}\limits^{ + 2} + 2{\rm{e}}\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{0,2}}\\ 2\mathop H\limits^{ + 1} + 2{\rm{e}} \to {\mathop H\limits^0 _2}\\ \,\,\,\,\,\,\,\,\,\,\,0,2\,\,\,\,\,\,0,1 \end{array}$
Bte: ${{n}_{e\left( + \right)}}={{n}_{e\left( - \right)}}=0,1.2=0,2\,\,mol$
Nên ${{n}_{F\text{e}}}=\frac{0,2}{2}=0,1\,mol$
Vậy $m=0,1.56=5,6\left( g \right)$
