Hướng dẫn:

${{n}_{KCl{{O}_{3}}}}=\frac{{{m}_{KCl{{O}_{3}}}}}{{{M}_{KCl{{O}_{3}}}}}=\frac{24,5}{122,5}=0,2\left( mol \right)$

$2KCl{{O}_{3}}\xrightarrow{{{t}^{0}}}2KCl+3{{O}_{2}}\uparrow $

   0,2         $\to $                  0,3            (mol)

Theo phương trình ta có:

 ${{n}_{{{O}_{2}}}}=\frac{3.{{n}_{KClO{{ & }_{3}}}}}{2}=\frac{3.0,2}{2}=0,3\left( mol \right)$

$\Rightarrow {{V}_{{{O}_{2}}}}=22,4.{{n}_{{{O}_{2}}}}=22,4.0,3=6,72\left( l \right)$

$\Rightarrow $ Chọn đáp án A