Hướng dẫn:

${{n}_{CO}}=\frac{{{V}_{CO}}}{22,4}=\frac{2,24}{22,4}=0,1\left( mol \right)$

${{n}_{{{H}_{2}}}}=\frac{{{V}_{{{H}_{2}}}}}{22,4}=\frac{3,36}{22,4}=0,15\left( mol \right)$

$CO+FeO\xrightarrow{{{t}^{0}}}C{{O}_{2}}+Fe$         (1)

0,1            $\to $                0,1              (mol)

${{H}_{2}}+FeO\xrightarrow{{{t}^{0}}}{{H}_{2}}O+Fe$          (2)

0,15           $\to $              0,15             (mol)

Theo phương trình (1) và (2) có:

${{n}_{Fe}}={{n}_{CO}}+{{n}_{{{H}_{2}}}}=0,1+0,15=0,25\left( mol \right)$

${{m}_{Fe}}={{n}_{Fe}}.{{M}_{Fe}}=0,25.56=14\left( g \right)$

$\Rightarrow $ Chọn đáp án B