Đáp án:
$m = 10,4\,\,gam$
Giải thích các bước giải:
${n_{S{O_2}}} = 0,1\,\,mol,\,\,{n_{NaOH}} = 0,1\,\,mol$
Nhận thấy: $\dfrac{{{n_{NaOH}}}}{{{n_{S{O_2}}}}} = \dfrac{{0,1}}{{0,1}} = 1 \to $ tạo muối $NaHSO_3$
PTHH:
$\begin{gathered} S{O_2} + NaOH \to NaHS{O_3}\,\,\,\,\,\,\,\,\,\,\,\, \hfill \\ 0,1\,\,\,\,\,\,\,\,\,\,0,1\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,0,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,mol \hfill \\ \to {m_{muối}} = 0,1.104 = 10,4\,\,gam\,\, \hfill \\ \end{gathered} $