Đáp án đúng: C
Giải chi tiết:\(\begin{array}{*{20}{l}} {{n_{C{O_2}}} = 0,04{\text{ }}mol} \\ {{n_{O{H^ - }}} = 2{n_{Ba{{\left( {OH} \right)}_2}}} + {n_{NaOH}} = 0,1{\text{ }}mol} \\ {{n_{B{a^{2 + }}}} = 0,02{\text{ }}mol} \\ {\frac{{{n_{O{H^ - }}}}}{{{n_{C{O_2}}}}} = 2,5 = > O{H^ - }{\text{ }}du,\,phan\,ung\,chi\,tao{\text{ }}C{O_3}^{2 - }} \\ {{n_{C{O_3}^{2 - }}} = {n_{C{O_2}}} = 0,04{\text{ }}mol} \\ {{n_{O{H^ - }{\text{ }}du}} = {\text{ }}{n_{O{H^ - }{\text{ }}b}}-2{n_{C{O_3}^{2 - }}} = 0,1-0,04.2 = 0,02{\text{ }}mol} \\ {{n_{BaC{O_{3{\text{ }}}}}} = {n_{B{a^{2 + }}}}{\text{ }}\left( {vi{\text{ }}B{a^{2 + }}{\text{ }}het} \right) = 0,02{\text{ }}mol} \\ {Vay{\text{ }}dung{\text{ }}dich{\text{ }}X{\text{ }}gom:} \\ {\left\{ \begin{gathered} C{O_3}^{2 - }:{\text{ }}0,02{\text{ }}mol \hfill \\ O{H^ - }:{\text{ }}0,02{\text{ }}mol \hfill \\ N{a^ + }:{\text{ }}0,06{\text{ }}mol \hfill \\ \end{gathered} \right.} \\ \begin{gathered} {H^ + }\,\,\,\, + \,\,\,\,\,O{H^ - } \to {H_2}O \hfill \\ 0,02 \leftarrow 0,02 \hfill \\ \end{gathered} \\ \begin{gathered} {H^ + }\,\,\,\, + \,\,\,\,\,C{O_3}^{2 - } \to HC{O_3}^ - \hfill \\ 0,02 \leftarrow 0,02 \hfill \\ \to n{H^ + } = 0,04(mol) \to V = \frac{{0,04}}{{0,5}} = 0,08(l) = 80ml \hfill \\ \end{gathered} \end{array}\)
Đáp án C
