Hướng dẫn:

$\begin{align}  & Ba{{(OH)}_{2}}+2HCl\to BaC{{l}_{2}}+{{H}_{2}}O(1) \\ & 0,02\to 0,04mol \\ & Ba{{(OH)}_{2}}+N{{a}_{2}}S{{O}_{4}}\to BaS{{O}_{4}}\downarrow +2NaOH(2) \\ & 0,01\leftarrow 0,01\leftarrow 0,01mol \\ & {{n}_{BaS{{O}_{4}}}}=\frac{{{m}_{BaS{{O}_{4}}}}}{{{M}_{BaS{{O}_{4}}}}}=\frac{2,33}{233}=0,01mol \\ &  \\\end{align}$

Theo phương trình(2):

$\begin{align} & {{n}_{Ba{{(OH)}_{2}}(2)}}={{n}_{BaS{{O}_{4}}}}=0,01mol \\ & \Rightarrow {{n}_{Ba{{(OH)}_{2}}(1)}}=0,03-0,01=0,02mol \\ & {{n}_{HCl}}=2{{n}_{Ba{{(OH)}_{2}}(1)}}=0,02.2=0,04mol \\\end{align}$

                   $\Rightarrow $ Chọn đáp án D.