Hướng dẫn:

$\begin{align}  & 2NaOH+{{H}_{2}}S{{O}_{4}}\to N{{a}_{2}}S{{O}_{4}}+2{{H}_{2}}O \\ & {{n}_{NaOH}}={{V}_{NaOH}}.{{C}_{{{M}_{NaOH}}}}=0,1.1=0,1mol \\ & {{n}_{{{H}_{2}}S{{O}_{4}}}}={{V}_{{{H}_{2}}S{{O}_{4}}}}.{{C}_{{{M}_{{{H}_{2}}S{{O}_{4}}}}}}=0,5.0,15=0,075mol \\\end{align}$

Theo phương trình ta nhận thấy: NaOH phản ứng hết, ${{H}_{2}}S{{O}_{4}}$dư.

$\begin{align}  & {{n}_{N{{a}_{2}}S{{O}_{4}}}}=\frac{1}{2}{{n}_{NaOH}}=\frac{1}{2}.0,1=0,05mol \\ & {{m}_{N{{a}_{2}}S{{O}_{4}}}}={{n}_{N{{a}_{2}}S{{O}_{4}}}}.{{M}_{N{{a}_{2}}S{{O}_{4}}}}=0,05.142=7,1g \\\end{align}$

                   $\Rightarrow $ Chọn đáp án D.