Đáp án đúng: D
Giải chi tiết:\(\begin{gathered} {n_{BaC{O_3}}} = 0,4 \hfill \\ \left\{ \begin{gathered} C{O_3}^{2 - }:0,2 \hfill \\ HC{O_3}^ - :a \hfill \\ \end{gathered} \right. \to \frac{{HC{O_3}^ - }}{{C{O_3}^{2 - }}} = 5a \hfill \\ C{O_3}^{2 - } + 2{H^ + } \to C{O_2} + {H_2}O \hfill \\ y\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2y \hfill \\ HC{O_3}^ - + {H^ + } \to C{O_2} + {H_2}O \hfill \\ 5ay\,\,\,\,\,\,\,\,\,\,\,\,\,5ay \hfill \\ \to \left\{ \begin{gathered} 2y + 5ay = {n_{{H^ + }}} = 0,5 \hfill \\ \xrightarrow{{BT:C}}0,2 + a = y + 5ay + 0,4 \hfill \\ \end{gathered} \right. \to \left\{ \begin{gathered} y = \frac{{0,5}}{{2 + 5a}} \hfill \\ 0,2 + a = y + 5ay + 0,4 \hfill \\ \end{gathered} \right. \hfill \\ \to 0,2 + a = \frac{{0,5}}{{2 + 5a}} + 5a.\frac{{0,5}}{{2 + 5a}} + 0,4 \to a = 0,6 \hfill \\ \to x = \frac{n}{V} = \frac{{0,6}}{{0,5}} = 1,2M \hfill \\ \end{gathered} \)
Đáp án D
