Sơ đồ phản ứng:
\(X\left\{\begin{matrix} Na\\K \\ Ba \end{matrix}\right.+O_2(0,08 \ mol) \longrightarrow Y\left\{\begin{matrix} Na:x\ mol\\ K: y \ mol \\ Ba \ mol \\ O \ mol \end{matrix}\right.+H_2O \longrightarrow \left\{\begin{matrix}Z \left\{\begin{matrix} Na^+(x),K^+(y),Ba^{2+} \\ OH^- \end{matrix}\right. \\ H_2:0,14 \ mol \end{matrix}\right.\)
\(\left\{\begin{matrix}Z \left\{\begin{matrix} Na^+(x),K^+(y),Ba^{2+} \\ OH^- \end{matrix}\right. +NaHCO_3\longrightarrow BaCO_3 \ (0,2 \ mol) \\ H_2:0,14 \ mol \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \end{matrix}\right.\)
Bảo toàn nguyên tố Ba: \(n_{BaCO_3}=n_{Ba^{2+}}=n_{Ba}=0,2 \ (mol)\)
Bảo toàn nguyên tố O: \(2.n_{O_2}=n_{O}\Rightarrow n_{O}=0,16 \ (mol)\)
Bảo toàn nguyên tố H: \(n_{H_2O}=n_{H_2}+\frac{n_{OH^-}}{2} \)
Bảo toàn nguyên tố O: \(n_{O}=n_{OH^-}-n_{H_2O}=\frac{n_{OH^-}}{2}-n_{H_2}\)
\(\Leftrightarrow 2n_O=n_{OH^-}-2n_{H_2} \Leftrightarrow n_{OH^-}=2n_O+2n_{H_2}\)
\(\Leftrightarrow x+y+0,4=2.0,16+2.0,14 \Leftrightarrow x+y=0,2\)
Sục \(CO_2\) vào Z:
\(\Rightarrow n_{BaCO_3}=197.(0,6-0,45)=29,55 \ (g)\)
Chúc bạn học tốt!