Đáp án đúng: A
19,7 gam và 20,6 gam
${{n}_{C{{O}_{2}}}}\,=\,0,2\,mol;\,\,{{n}_{O{{H}^{-}}}}\,=\,0,3\,mol$ ⟹ tạo 2 muối
$\left\{ \begin{array}{l}{{n}_{CO_{3}^{2-}}}\,+\,{{n}_{HCO_{3}^{-}}}\,=\,{{n}_{C{{O}_{2}}}}\,=\,0,2\\2{{n}_{CO_{3}^{2-}}}\,+\,{{n}_{HCO_{3}^{-}}}\,=\,{{n}_{N{{a}^{+}}}}\,+\,{{n}_{{{K}^{+}}}}\,=\,0,3\end{array} \right.$$\Rightarrow \left\{ \begin{array}{l}{{n}_{CO_{3}^{2-}}}\,=\,0,1\,mol\\{{n}_{HCO_{3}^{-}}}\,=\,0,1\,mol\end{array} \right.$
Cô cạn X có${{m}_{m.khan}}\,=\,{{m}_{N{{a}^{+}}}}\,+\,{{m}_{{{K}^{+}}}}\,+\,{{m}_{CO_{3}^{2-}}}\,+\,{{m}_{HCO_{3}^{-}}}=20,6\,gam$
½ X có 0,05 mol$CO_{3}^{2-}$ và 0,05 mol$HCO_{3}^{-}$, tác dụng với Ba(OH)2 dư
⟹${{n}_{BaC{{O}_{3}}}}\,=\,{{n}_{CO_{3}^{2-}}}\,+\,{{n}_{HCO_{3}^{-}}}\,=\,0,1\,mol\,\Rightarrow \,m\,=\,19,7\,gam$
