Hấp thụ hoàn toàn 896 ml khí CO2 (đktc)...

C

- Hướng dẫn giải

Phương pháp giải:

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