Hướng dẫn:

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

${{n}_{Cu}}=\frac{{{m}_{Cu}}}{{{M}_{Cu}}}=\frac{3,2}{64}=0,05\left( mol \right)$

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

0,05  0,075                             (mol)

$Cu+C{{l}_{2}}\xrightarrow{{{t}^{0}}}CuC{{l}_{2}}$     (2)

0,05  0,05                             (mol)

Theo phương trình (1) ta có: ${{n}_{C{{l}_{2}}}}=\frac{3.0,05}{2}=0,075\left( mol \right)$

Theo phương trình (2) ta có: ${{{n}'}_{C{{l}_{2}}}}={{n}_{Cu}}=0,05\left( mol \right)$

$\Rightarrow \sum{{{n}_{C{{l}_{2}}}}={{n}_{C{{l}_{2}}}}+{{{{n}'}}_{C{{l}_{2}}}}=0,075+0,05=0,125\left( mol \right)}$

$\Rightarrow {{V}_{C{{l}_{2}}}}=22,4.\sum{{{n}_{C{{l}_{2}}}}=22,4.0,125=2,8\left( l \right)}$

Chọn đáp án C