\(\begin{align} & {{n}_{NaHC{{O}_{3}}}}=2\text{a}\,\,\,mol \\ & {{n}_{NaOH}}=x\,\,\,\,mol \\ & HCO_{3}^{-}\,\,\,\,+\,\,\,\,O{{H}^{-}}\,\,\,\to \,\,\,CO_{3}^{2-}\,\,\,\,+\,\,\,\,{{H}_{2}}O \\ \end{align}\)

\(T{{H}_{1}}:\,\,x\,<\,2\text{a}\)

\(dd\,\,X\left| \begin{align} & CO_{3}^{2-}:x \\ & HCO_{3\,(d)}^{-}:2a-x \\ & N{{a}^{+}} \\ \end{align} \right.\)

2 lít dd X + BaCl₂ (dư) -> 2.0,06 mol \(\begin{align}BaC{{O}_{3}}\downarrow \end{align}\)

\(\begin{align} & B{{a}^{2+}}\,\,\,\,\,+\,\,\,\,\,\,\,CO_{3}^{2-}\,\,\,\,\,\to \,\,\,\,BaC{{O}_{3}}\downarrow \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(mol) \\ & x=0,12\,\,\,mol \\ & m=4,8\,\,\,gam \\ \end{align}\)

2 lít dd X + CaCl₂ \(\xrightarrow{{{t}^{0}}}\) 2.0,07 mol \(\begin{align}CaC{{O}_{3}}\downarrow \end{align}\)

\(\begin{align} & C{{a}^{2+}}\,\,\,\,\,+\,\,\,\,\,CO_{3}^{2-}\,\,\,\,\to \,\,\,\,CaC{{\text{O}}_{3}}\downarrow \\ & C{{a}^{2+}}\,\,\,\,+\,\,\,\,2HCO_{3}^{-}\,\,\to \,\,\,Ca{{(HC{{O}_{3}})}_{2}} \\ & Ca{{(HC{{O}_{3}})}_{2}}\,\xrightarrow{{{t}^{0}}}\,\,CaC{{\text{O}}_{3}}\downarrow \,\,\,\,\,\,\,+\,\,\,\,C{{O}_{2}}\,\,\,\,+\,\,\,{{H}_{2}}O \\ \end{align}\)

\(\begin{align} & {{n}_{CaC{{\text{O}}_{3}}}}={{n}_{CO_{3}^{2-}}}+\frac{1}{2}.{{n}_{HCO_{3}^{-}}} \\ & \Leftrightarrow 0,14=x+\frac{1}{2}.(2\text{a}-x) \\ & \Leftrightarrow a=0,08 \\ \end{align}\)