${{H}_{2}}S{{O}_{4}}+Ba{{\left( N{{O}_{3}} \right)}_{2}}\to BaS{{O}_{4}}\downarrow +2HN{{O}_{3}}$

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

${{n}_{{{H}_{2}}S{{O}_{4}}}}={{V}_{{{H}_{2}}S{{O}_{4}}}}.{{C}_{{{M}_{{{H}_{2}}S{{O}_{4}}}}}}=0,1.2=0,2\left( mol \right)$

Theo phương trình ta có:

 $\begin{align}  & {{n}_{BaS{{O}_{4}}}}={{n}_{{{H}_{2}}S{{O}_{4}}}}=0,2\left( mol \right) \\ & {{m}_{BaS{{O}_{4}}}}={{n}_{BaS{{O}_{4}}}}.{{M}_{BaS{{O}_{4}}}}=0,2.233=46,6\left( g \right) \\\end{align}$

$\Rightarrow $Chọn đáp án C.