${{n}_{{{H}_{2}}S{{O}_{4}}}}={{V}_{{{H}_{2}}S{{O}_{4}}}}.{{C}_{{{M}_{{{H}_{2}}S{{O}_{4}}}}}}=0,2.1=0,2\left( mol \right)$

${{H}_{2}}S{{O}_{4}}+Ba{{\left( OH \right)}_{2}}\to BaS{{O}_{4}}+2{{H}_{2}}O$

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

Theo phương trình ta có:

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

$\Rightarrow $Chọn đáp án B