Đáp án đúng: A
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
