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} $