- Chọn D.

- Ta có: ${{n}_{{{O}_{2}}}}=\frac{3,36}{22,4}=0,15\left( mol \right)$.

- PTHH: $2KCl{{O}_{3}}\xrightarrow{{{t}^{o}}}2KCl+3{{O}_{2}}$

                2(mol)                      3 (mol)

                x(mol)                     0,15(mol)

$x=\frac{2\times 0,15}{3}\times \frac{80}{100}=0,08\left( mol \right).$$\to {{m}_{KCl{{O}_{3}}}}=0,08\times 122,5=9,8\left( gam \right)$.