Đáp án:
$2,52(l)$
Giải thích các bước giải:
${m_{ddgiam}} = {m_{CaC{O_3}}} - {m_{C{O_2}}} - {m_{{H_2}O}} = 12 \Rightarrow {m_{C{O_2}}} + {m_{{H_2}O}} = 25,5$
${n_{C{O_2}}} = {n_{CaC{O_3}}} = 0,375mol \Rightarrow {m_{C{O_2}}} = 16,5g$
$ \Rightarrow {m_{{H_2}O}} = 25,5 - 16,5 = 9g \Rightarrow {n_{{H_2}O}} = 0,5mol$
$ \Rightarrow {n_{ancol}} = {n_{{H_2}O}} - {n_{C{O_2}}} = 0,5 - 0,375 = 0,125mol$
${m_{ancol}} = {m_C} + {m_H} + {m_O} \Rightarrow {m_O} = 9,1 - 0,375.12 - 0,5.2 = 3,6g$
$ \Rightarrow {n_O} = \dfrac{{3,6}}{{16}} = 0,225mol$
$ \Rightarrow {n_{OH}} = {n_O} = 0,225mol$
$R{(OH)_x} + xNa \to R{(ONa)_x} + \dfrac{x}{2}{H_2}$
Ta có:
$\begin{gathered}
{n_{{H_2}}} = \dfrac{1}{2}{n_{OH}} = \dfrac{1}{2}.0,225 = 0,1125mol \hfill \\
\Rightarrow {V_{{H_2}}} = 0,1125.22,4 = 2,52(l) \hfill \\
\end{gathered} $