Hỗn hợp X gồm Ba, BaO, Na, Na2O và K. Cho m gam hỗn hợp X vào nước...

$m(gam)X\left\{ \begin{aligned}
& Ba \\
& Na \\
& K \\
& O \\
\end{aligned} \right.\xrightarrow{{{H}_{2}}O}\left\langle \begin{aligned}
& {{H}_{2}}:0,14(mol) \\
& Y\left\{ \begin{aligned}
& Ba{{(OH)}_{2}}:0,93m(g) \\
& NaOH:0,18(mol) \\
& KOH:0,044m(g) \\
\end{aligned} \right.\xrightarrow{C{{O}_{2}}:0,348(mol)} \\
\end{aligned} \right.$
$BTKL:{{m}_{X}}+{{m}_{{{H}_{2}}O}}={{m}_{Y}}+{{m}_{{{H}_{2}}}}$
$\to {{m}_{{{H}_{2}}O}}=0,93m+7,2+0,044m+0,14.2-m=7,48-0,026m$
$\to {{n}_{{{H}_{2}}O}}=\dfrac{7,48-0,026m}{18}(mol)$
$BTNT(H):\dfrac{7,48-0,026m}{18}=\dfrac{0,93m}{171}+\dfrac{1}{2}.0,18+\dfrac{1}{2}.\dfrac{0,044m}{56}+0,14$
$\to m=25,5(gam)$
$\to {{n}_{O{{H}^{-}}(Y)}}=2{{n}_{Ba{{(OH)}_{2}}}}+{{n}_{NaOH}}+{{n}_{KOH}}=0,4774(mol)$
$O{{H}^{-}}\xrightarrow{C{{O}_{2}}}\downarrow $
$T=\dfrac{{{n}_{O{{H}^{-}}}}}{{{n}_{C{{O}_{2}}}}}=1,37\to \left\{ \begin{aligned}
& CO_{3}^{2-}:a \\
& HCO_{3}^{-}:b \\
\end{aligned} \right.\to \left\{ \begin{aligned}
& a+b=0,348 \\
& 2a+b=0,4774 \\
\end{aligned} \right.\to \left\{ \begin{aligned}
& a=0,1294 \\
& b=0,2186 \\
\end{aligned} \right.$
$B{{a}^{2+}}+CO_{3}^{2-}\to BaC{{O}_{3}}$
${{n}_{B{{a}^{2+}}}}=0,1387<{{n}_{CO_{3}^{2-}}}=0,1294$
$\to {{n}_{BaC{{O}_{3}}}}=0,1294(mol)$
$\to m=25,5(gam)$
Chọn đáp án A.