Hướng dẫn:

${{n}_{HCl}}={{V}_{HCl}}.{{C}_{{{M}_{HCl}}}}=0,3.2=0,6\left( mol \right)$

$Mn{{O}_{2}}+4HCl\xrightarrow{{{t}^{0}}}MnC{{l}_{2}}+C{{l}_{2}}+2{{H}_{2}}O$

                0,6                    0,15   (mol)

Theo phương trình ta có:${{n}_{C{{l}_{2}}}}=\frac{{{n}_{HCl}}}{4}=\frac{0,6}{4}=0,15\left( mol \right)$

$\Rightarrow {{V}_{C{{l}_{2}}}}=22,4.{{n}_{C{{l}_{2}}}}=22,4.0,15=3,36\left( l \right)$

Chọn đáp án B