Hướng dẫn:

${{n}_{Fe{{\left( N{{O}_{3}} \right)}_{2}}}}=\frac{{{m}_{Fe{{\left( N{{O}_{3}} \right)}_{2}}}}}{{{M}_{Fe{{\left( N{{O}_{3}} \right)}_{2}}}}}=\frac{9}{180}=0,05\left( g \right)$ 

 $2Fe{{\left( N{{O}_{3}} \right)}_{2}}\xrightarrow{{{t}^{0}}}2FeO+{{O}_{2}}+4N{{O}_{2}}$

      0,05           $\to $     0,05                                  (mol)

Theo phương trình ta có:

${{n}_{FeO}}={{n}_{Fe{{\left( N{{O}_{3}} \right)}_{2}}}}=0,05\left( mol \right)$ 

${{m}_{FeO}}={{n}_{FeO}}.{{M}_{FeO}}=0,05.72=3,6\left( g \right)$ 

Chọn đáp án D