Hướng dẫn:

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

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

$4{{H}_{2}}+F{{e}_{3}}{{O}_{4}}\xrightarrow{{{t}^{0}}}4{{H}_{2}}O+3Fe$                       (2)

0,4 $\leftarrow $ 0,1                                         (mol)
Theo phương trình (1) có: ${{n}_{CO}}={{n}_{FeO}}=0,15\left( mol \right)$

Theo phương trình (2) có: ${{n}_{{{H}_{2}}}}=4.{{n}_{F{{e}_{3}}{{O}_{4}}}}=4.0,1=0,4\left( mol \right)$
$\Rightarrow {{V}_{CO}}=22,4.0,15=3,36\left( l \right)$

     ${{V}_{{{H}_{2}}}}=22,4.{{n}_{{{H}_{2}}}}=22,4.0,4=8,96\left( l \right)$

$\Rightarrow $ Chọn đáp án A