$ZnC{{O}_{3}}+2HCl\to ZnC{{l}_{2}}+C{{O}_{2}}+{{H}_{2}}O$

0,05   $\to $   0,1                                         (mol)

${{n}_{ZnC{{O}_{3}}}}=\frac{{{m}_{ZnC{{O}_{3}}}}}{{{M}_{ZnC{{O}_{3}}}}}=\frac{6,25}{125}=0,05\left( mol \right)$

Theo phương trình ta có:

${{n}_{HCl}}=2.{{n}_{ZnC{{O}_{3}}}}=2.0,05=0,1\left( mol \right)$

${{V}_{HCl}}=\frac{{{n}_{HCl}}}{{{C}_{{{M}_{HCl}}}}}=\frac{0,1}{0,2}=0,5\left( l \right)$

$\Rightarrow $Chọn đáp án A