Hỗn hợp X gồm Na, Ba, Na2O và BaO

Chọn đáp án D

Giải thích:$Taco:n_{Ba{\left(OH\right)}_2}=\mathrm{0,72};n_{C{O}_2}=\mathrm{1,8};n_{H_2}=\mathrm{0,3.}$

$\underset{\mathrm{131,4}gam}{\underbrace{\left\{\begin{array}{l}Ba:\mathrm{0,72}\\ Na:x\\ O:y\end{array}\right\}}}\stackrel{quydoi}{\leftarrow }\underset{\mathrm{131,4}gam}{\underbrace{\left\{\begin{array}{l}Na,N{a}_{2}O\\ Ba,BaO\end{array}\right\}}}\stackrel{H_{2}O}{\to }\left\{\begin{array}{l}NaOH:x\\ Ba{\left(OH\right)}_2:\mathrm{0,72}\end{array}\right\}\underset{\mathrm{1,8}mol}{\overset{C{O}_2}{\to }}\underset{m=?}{\underbrace{BaC{O}_3\downarrow }}$

$\begin{array}{l}+\left\{\begin{array}{l}{m}_{honhop}=m_{Ba}+m_{Na}+m_O\\ BTE:2n_{Ba}+n_{Na}=2n_O+2n_{H_2}\end{array}\right.\Rightarrow \left\{\begin{array}{l}{m}_{honhop}=\mathrm{0,72.137}+23x+16y=\mathrm{131,4}\\ BTE:\mathrm{0,72.2}+x=2y+\mathrm{0,3.2}\end{array}\right.\\ \Rightarrow \left\{\begin{array}{l}x=\mathrm{0,84}\\ y=\mathrm{0,84}\end{array}\right.\Rightarrow \underset{taoraC{O_3}^{2-}vaøHC{O_3}^{-}}{\underbrace{1<\frac{n_{O{H}^{-}}}{n_{C{O}_2}}=\frac{\mathrm{2,28}}{\mathrm{1,8}}<2}}\Rightarrow \left\{\begin{array}{l}{n}_{C{O_3}^{2-}}=n_{O{H}^{-}}-n_{C{O}_2}=\mathrm{0,48}<n_{B{a}^{2+}}\\ n_{BaC{O}_3}=n_{C{O_3}^{2-}}=\mathrm{0,48}\Rightarrow m_{BaC{O}_3}=\mathrm{94,56}gam.\end{array}\right.\end{array}$