Đáp án đúng: A
24,625 gam
TN1:
${{n}_{C{{O}_{2}}}}\,=\,0,2\,mol;\,{{n}_{O{{H}^{-}}}}\,=\,0,3\,mol$ ⟹ tạo 2 muối
$\left\{ \begin{array}{l}2{{n}_{CO_{3}^{2-}}}\,+\,{{n}_{HCO_{3}^{-}}}\,=\,{{n}_{N{{a}^{+}}}}\,+\,{{n}_{{{K}^{+}}}}\,=\,0,3\\{{n}_{CO_{3}^{2-}}}\,+\,{{n}_{HCO_{3}^{-}}}\,=\,{{n}_{C{{O}_{2}}}}\,=\,0,2\end{array} \right.\,\Rightarrow \,\left\{ \begin{array}{l}{{n}_{CO_{3}^{2-}}}\,=\,0,1\\{{n}_{HCO_{3}^{-}}}\,=\,0,1\end{array} \right.$
TN2:
$\begin{array}{l}\,HCO_{3}^{-}\,\,\,+\,\,\,\,O{{H}^{-}}\,\xrightarrow{{}}CO_{3}^{2-}\,+\,O{{H}^{-}}\\0,025\,\,\leftarrow \,\,\,0,025\,\,\,\,\to \,\,\,0,025\end{array}$
$\begin{array}{l}\,\,B{{a}^{2+}}\,\,+\,\,\,CO_{3}^{2-}\,\to \,BaC{{O}_{3}}\downarrow \\0,125\,\leftarrow \,0,125\,\,\to \,\,0,125\end{array}$
$\Rightarrow {{m}_{BaC{{O}_{3}}}}\,=\,24,635\,gam$
