Hỗn hợp X gồm Ba, BaO, Na, Na2O và K.

* Hướng dẫn giải

Chọn đáp án A

$Qui\text{\hspace{0.33em}X ve }\left\{\begin{array}{l}Na\\ Ba:x\text{\hspace{0.33em}mol}\\ K:y\text{\hspace{0.33em}mol}\\ O:z\text{\hspace{0.33em}mol}\end{array}\right.\stackrel{+H_{2}O}{\to }\left|\begin{array}{l}Y\left\{\begin{array}{l}NaOH:\mathrm{0,18}\text{\hspace{0.33em}mol}\\ Ba{\left(OH\right)}_2:x\text{\hspace{0.33em}mol}\\ KOH:y\text{\hspace{0.33em}mol}\end{array}\right.\stackrel{+C{O}_2:\mathrm{0,348}\text{\hspace{0.33em}mol}}{\to }\downarrow BaC{O}_3\\ H_2:\mathrm{0,14}\text{\hspace{0.33em}mol}\end{array}\right.$

$\begin{array}{l}\stackrel{BTE}{\to }\mathrm{0,18}+2x+y=2z+\mathrm{2.0,14}\text{ (1)}\\ \stackrel{BTKL}{\to }\mathrm{23.0,18}+137x+39y+16z=m\end{array}$

$\begin{array}{l}{m}_{Ba{\left(OH\right)}_2}=171x=\mathrm{0,93}m\to \mathrm{0,93}(\mathrm{23.0,18}+137x+39y+16z)=171x\text{ (2)}\\ \frac{{\text{m}}_{KOH}}{m_{Ba{\left(OH\right)}_2}}=\frac{\mathrm{0,044}m}{\mathrm{0,93}m}\to \frac{56y}{171x}=\frac{\mathrm{0,044}}{\mathrm{0,93}}\to \mathrm{7,524}x-\mathrm{52,08}y=0\text{ (3)}\end{array}$

$Tu\text{ (1), (2) va (3) }\to \left\{\begin{array}{l}x=\mathrm{0,1387}\\ y=\mathrm{0,02}\\ z=\mathrm{0,0987}\end{array}\right.$

$\begin{array}{l}\to n_{O{H}^{-}\left(Y\right)}=n_{NaOH}+2n_{Ba{\left(OH\right)}_2}+n_{KOH}=\mathrm{0,18}+\mathrm{2.0,1387}+\mathrm{0,02}=\mathrm{0,4774}\text{\hspace{0.33em}mol}\\ \to {\text{n}}_{C{O}_3^{2-}}=n_{O{H}^{-}\left(Y\right)}-n_{C{O}_2}=\mathrm{0,4774}-\mathrm{0,348}=\mathrm{0,1294}{\text{\hspace{0.33em}mol < n}}_{B{a}^{2+}}\\ \to {\text{\hspace{0.33em}ket tua tinh theo CO}}_3^{2-}\to m_{BaC{O}_3}=\mathrm{197.0,1294}\simeq \mathrm{25,5}\text{ gam}\end{array}$