Hướng dẫn:

${{n}_{Fe{{\left( N{{O}_{3}} \right)}_{2}}}}=\frac{{{m}_{Fe{{\left( N{{O}_{3}} \right)}_{2}}}}}{{{M}_{Fe{{\left( N{{O}_{3}} \right)}_{2}}}}}=\frac{27}{180}=0,15\left( g \right)$ 

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

      0,15           $\to $                0,075   0,3           (mol)

Theo phương trình ta có:

${{n}_{{{O}_{2}}}}=\frac{{{n}_{Fe{{\left( N{{O}_{3}} \right)}_{2}}}}}{2}=\frac{0,15}{2}=0,075\left( mol \right)$ 

${{n}_{N{{O}_{2}}}}=\frac{4.{{n}_{Fe{{\left( N{{O}_{3}} \right)}_{2}}}}}{2}=\frac{4.0,15}{2}=0,3\left( mol \right)$ 

$\Rightarrow {{n}_{hh}}={{n}_{{{O}_{2}}}}+{{n}_{N{{O}_{2}}}}=0,075+0,3=0,375\left( mol \right)$ 

${{V}_{hh}}=22,4.{{n}_{hh}}=22,4.0,375=8,4\left( l \right)$ 

Chọn đáp án D