Hướng dẫn:

Gọi a, b lần lượt là số mol của clo và brom

${{n}_{Fe}}=\frac{{{m}_{Fe}}}{{{M}_{Fe}}}=\frac{2,8}{56}=0,05\left( mol \right)$

$2Fe+3C{{l}_{2}}\xrightarrow{{{t}^{0}}}2FeC{{l}_{3}}$                (1)

$\frac{2.a}{3}$  a      $\to $$\frac{2.a}{3}$   (mol)

$2Fe+3B{{r}_{2}}\xrightarrow{{{t}^{0}}}2FeB{{r}_{3}}$         (2)

$\frac{2.b}{3}$  b      $\frac{2.b}{3}$   (mol)

Theo 2 phương trình ta có:

${{n}_{Fe}}=\frac{2.{{n}_{C{{l}_{2}}}}}{3}+\frac{2.{{n}_{C{{l}_{2}}}}}{3}=\frac{2.a}{3}+\frac{2.b}{3}\left( mol \right)$

$\Rightarrow \frac{2.a}{3}+\frac{2.b}{3}=0,05\left( mol \right)$                             (*)

Theo phương trình (1) ta có: ${{n}_{FeC{{l}_{3}}}}=\frac{2.{{n}_{C{{l}_{2}}}}}{3}=\frac{2.a}{3}\left( mol \right)$

Theo phương trình (2) ta có: ${{n}_{FeB{{r}_{3}}}}=\frac{2.{{n}_{B{{r}_{2}}}}}{3}=\frac{2.b}{3}\left( mol \right)$

${{m}_{muoi}}={{m}_{FeC{{l}_{3}}}}+{{m}_{FeB{{r}_{3}}}}={{n}_{FeC{{l}_{3}}}}.{{M}_{FeC{{l}_{3}}}}+{{n}_{FeB{{r}_{3}}}}.{{M}_{FeB{{r}_{3}}}}$

$\Rightarrow 162,5.a+296.b=15,525$                           (**)

Từ (*) và (**)  có: $a=0,05\left( mol \right)$

                               b=0,025(mol)

$\Rightarrow {{m}_{C{{l}_{2}}}}={{n}_{C{{l}_{2}}}}.{{M}_{C{{l}_{2}}}}=0,05.71=3,55\left( g \right)$

${{m}_{B{{r}_{2}}}}={{n}_{B{{r}_{2}}}}.{{M}_{B{{r}_{2}}}}=0,025.160=4\left( g \right)$

$\Rightarrow $Chọn đáp án D