Đáp án đúng: C
21,2 gam và 20 gam
Thoát khí CO2 ⟹ hết $CO_{3}^{2-}$
Thêm Ca(OH)2 dư vào dung dịch Z được kết tủa A ⟹ trong Z có$HCO_{3}^{-}$
$\begin{array}{l}{{H}^{+}}\,+\,CO_{3}^{2-}\,\xrightarrow{{}}\,HCO_{3}^{-}\\x\,\to \,\,\,x\,\,\,\,\,\,\,\xrightarrow{{}}\,\,\,\,\,\,\,\,x\,mol\\{{H}^{+}}\,+\,HCO_{3}^{-}\,\xrightarrow{{}}\,{{H}_{2}}O\,+\,C{{O}_{2}}\\0,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\xleftarrow{{}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,1\,mol\end{array}$
⟹ x + 0,1 = 0,8.0,5 = 0,4 ⟹ x = 0,3.
$\left\{ \begin{array}{l}{{n}_{N{{a}_{2}}C{{O}_{3}}}}\,+\,{{n}_{{{K}_{2}}C{{O}_{3}}}}\,=\,0,3\\106{{n}_{N{{a}_{2}}C{{O}_{3}}}}\,+\,138{{n}_{{{K}_{2}}C{{O}_{3}}}}\,=\,35\end{array} \right.\Rightarrow \left\{ \begin{array}{l}{{n}_{N{{a}_{2}}C{{O}_{3}}}}\,=\,0,2\\{{n}_{{{K}_{2}}C{{O}_{3}}}}\,=\,0,1\end{array} \right.$
⟹${{m}_{N{{a}_{2}}C{{O}_{3}}}}\,=\,21,2\,gam$
Trong dung dịch Z có 0,3 – 0,1 = 0,2 mol$HCO_{3}^{-}$⟹${{n}_{CaC{{O}_{3}}}}\,=\,0,2\,mol$ ⟹ m = 20 gam
