Đáp án đúng: C
80 ml
${{n}_{{{H}_{2}}S}}=\frac{8,96}{22,4}=0,4\,mol\Rightarrow {{n}_{S{{O}_{2}}}}=0,4\,mol$
Giả sử tạo hoàn toàn muối NaHSO3
$\displaystyle \Rightarrow {{n}_{NaHS{{O}_{3}}}}={{n}_{S{{O}_{2}}}}=0,4\,mol\Rightarrow {{m}_{NaHS{{O}_{3}}}}=41,6\,gam$
Giả sử tạo hoàn toàn muối Na2SO3
$\Rightarrow {{n}_{N{{a}_{2}}S{{O}_{3}}}}={{n}_{S{{O}_{2}}}}=0,4\,mol\,\Rightarrow {{m}_{N{{a}_{2}}S{{O}_{3}}}}=50,4\,gam$
⇒ với 46,88 gam thì phải tạo 2 muối
$\left\{ \begin{array}{l}{{n}_{NaHS{{O}_{3}}}}+{{n}_{N{{a}_{2}}S{{O}_{3}}}}={{n}_{S{{O}_{2}}}}=0,4\\104{{n}_{NaHS{{O}_{3}}}}+126{{n}_{N{{a}_{2}}S{{O}_{3}}}}=46,88\end{array} \right.\Rightarrow \left\{ \begin{array}{l}{{n}_{NaHS{{O}_{3}}}}=0,16\\{{n}_{N{{a}_{2}}S{{O}_{3}}}}=0,24\end{array} \right.$
${{n}_{NaOH}}={{n}_{NaHS{{O}_{3}}}}+2{{n}_{N{{a}_{2}}S{{O}_{3}}}}=0,64\,mol$
$\Rightarrow {{m}_{NaOH}}=25,6\,gam\,\Rightarrow {{m}_{\text{dd}\,NaOH}}=102,4\,gam\Rightarrow {{V}_{\text{dd}\,NaOH}}=80ml$
