Hòa tan hết 9,19 gam hỗn hợp X gồm Ba, BaO, Na, Na...

A

- Hướng dẫn giải

Phương pháp giải:

Quy đổi hỗn hợp ban đầu thành Ba, Na, O

Giải chi tiết:

\(\begin{gathered}
\begin{array}{*{20}{l}}
{{n_{{H_2}}} = 0,02{\text{ }}mol} \\
{{n_{C{O_2}}} = 0,1{\text{ }}mol}
\end{array} \hfill \\
\left\{ \begin{gathered}
Ba:x \hfill \\
Na:y \hfill \\
O:z \hfill \\
\end{gathered} \right.\xrightarrow{{{H_2}O}}\left\{ \begin{gathered}
Ba{(OH)_2}:x \hfill \\
NaOH:y \hfill \\
\end{gathered} \right.\xrightarrow{{A{l_2}{{(S{O_4})}_3}du}}\left\{ \begin{gathered}
{\text{BaS}}{O_4}:x \hfill \\
Al{(OH)_3}:\frac{{2x + y}}{3} \hfill \\
\end{gathered} \right. \hfill \\
\to \left\{ \begin{gathered}
137x + 23y + 16z = 9,19 \hfill \\
2x + y = 2z + 0,02.2(BTe) \hfill \\
233x + 78\left( {\frac{{2x + y}}{3}} \right) = 15,81 \hfill \\
\end{gathered} \right. \to \left\{ \begin{gathered}
x = 0,05 \hfill \\
y = 0,06 \hfill \\
z = 0,06 \hfill \\
\end{gathered} \right. \hfill \\
\to {n_{O{H^ - }}} = 0,16mol \hfill \\
\frac{{{n_{O{H^ - }}}}}{{{n_{C{O_2}}}}} = 1,6 = > Tao\,C{O_3}^{2 - }\,va\,HC{O_3}^ - \hfill \\
\left\{ \begin{gathered}
{n_{C{O_3}^{2 - }}} = {n_{O{H^ - }}} - {n_{C{O_2}}} = 0,06mol \hfill \\
{n_{HC{O_3}^ - }} = 2{n_{C{O_2}}} - {n_{O{H^ - }}} = 0,04mol \hfill \\
\end{gathered} \right. \hfill \\
\to m = {m_{BaC{O_3}}} = 0,05.197 = 9,85(g) \hfill \\
Z\left\{ \begin{gathered}
N{a^ + }:0,06 \hfill \\
HC{O_3}^ - :0,04 \hfill \\
C{O_3}^{2 - }:0,01 \hfill \\
\end{gathered} \right. \hfill \\
Co\,can\,Z:2HC{O_3}^ - \xrightarrow{{{t^o}}}C{O_3}^{2 - } + C{O_2} + {H_2}O \hfill \\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,04\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,02 \hfill \\
\xrightarrow{{co\,can}}\left\{ \begin{gathered}
N{a^ + }:0,06 \hfill \\
C{O_3}^{2 - }:0,03 \hfill \\
\end{gathered} \right. \to a = 0,06.23 + 0,03.60 = 3,18(g) \hfill \\
\to m + a = 9,85 + 3,18 = 13,03(g) \hfill \\
\end{gathered} \)

Đáp án A