Hướng dẫn:

$\begin{align}  & BaC{{l}_{2}}+{{H}_{2}}S{{O}_{4}}\to BaS{{O}_{4}}+2HCl \\ & 0,01\leftarrow 0,01\leftarrow 0,01mol \\ & {{n}_{BaS{{O}_{4}}}}=\frac{{{m}_{BaS{{O}_{4}}}}}{{{M}_{BaS{{O}_{4}}}}}=\frac{2,33}{233}=0,01mol \\\end{align}$

Theo phương trình:

$\begin{align}  & {{n}_{{{H}_{2}}S{{O}_{4}}}}={{n}_{BaS{{O}_{4}}}}=0,01mol \\ & {{m}_{{{H}_{2}}S{{O}_{4}}}}={{n}_{{{H}_{2}}S{{O}_{4}}}}.{{M}_{{{H}_{2}}S{{O}_{4}}}}=0,01.98=0,98g \\ & {{m}_{HCl}}=1,5275-0,98=0,5475g \\\end{align}$

                     $\Rightarrow $ Chọn đáp án C.