Đáp án:

 a) $16,8(l)$

b) $78g$

c) $\% {m_{C{H_3}OH}} = 34,13\% ;\% {m_{{C_3}{H_5}{{(OH)}_3}}} = 24,53\% ;\% {m_{{H_2}O}} = 41,34\% $

Giải thích các bước giải:

 a) $\begin{gathered}
  2C{H_3}OH + 2Na \to 2C{H_3}ONa + {H_2} \hfill \\
  {C_3}{H_5}{(OH)_3} + 3Na \to {C_3}{H_5}{(ONa)_3} + \dfrac{3}{2}{H_2} \hfill \\ 
\end{gathered} $

$2{H_2}O + 2Na \to 2NaOH + {H_2}$

$\begin{gathered}
  C{H_3}ONa + HCl \to NaCl + C{H_3}OH \hfill \\
  {C_3}{H_5}{(ONa)_3} + 3HCl \to 3NaCl + {C_3}{H_5}{(OH)_3} \hfill \\
  NaOH + HCl \to NaCl + {H_2}O \hfill \\ 
\end{gathered} $

$\begin{gathered}
  {m_{ddHCl}} = V.D = 250.1,098 = 274,5g \hfill \\
   \Rightarrow {n_{HCl}} = \dfrac{{274,5.20\% }}{{36,5}} = 1,5mol \hfill \\ 
\end{gathered} $

Theo PTHH, ta có: 

${n_{{H_2}}} = \dfrac{1}{2}\left( {{n_{C{H_3}ONa}} + 3{n_{{C_3}{H_5}{{(ONa)}_3}}} + {n_{NaOH}}} \right) = \dfrac{1}{2}{n_{HCl}} = 0,75mol$

$ \Rightarrow {V_{{H_2}}} = 0,75.22,4 = 16,8(l)$

b) Ta có: ${n_{Na}} = 2{n_{{H_2}}} = 1,5mol \Rightarrow {m_{Na}} = 1,5.23 = 34,5g$

Bảo toàn khối lượng: ${m_X} + {m_{Na}} = {m_Y} + {m_{{H_2}}}$

$ \Rightarrow {m_Y} = 45 + 34,5 - 0,75.2 = 78g$

c) Gọi x, y z là số mol $C{H_3}OH;{C_3}{H_5}{(OH)_3};{H_2}O$ trong 45 g

Ta có: $\left\{ \begin{gathered}
  \dfrac{x}{2} + \dfrac{3}{2}.y + \dfrac{z}{2} = 0,75 \hfill \\
  46x + 92y + 18z = 45 \hfill \\ 
\end{gathered}  \right.$ (1)

Trong 18 g X số mol $C{H_3}OH;{C_3}{H_5}{(OH)_3};{H_2}O$ là 0,4x; 0,4y; 0,4z

${n_{{H_2}O}} = \dfrac{{15,12}}{{18}} = 0,84mol$

$ \Rightarrow 0,4.(2x + 4y + z) = 0,84 \Rightarrow 2x + 4y + z = 2,1$ (2)

Từ (1) và (2) $ \Rightarrow \left\{ \begin{gathered}
  x = 0,48 \hfill \\
  y = 0,12 \hfill \\
  z = 0,66 \hfill \\ 
\end{gathered}  \right.$

$\begin{gathered}
   \Rightarrow \% {m_{C{H_3}OH}} = \dfrac{{0,48.32}}{{45}}.100\%  = 34,13\%  \hfill \\
  \% {m_{{C_3}{H_5}{{(OH)}_3}}} = \dfrac{{0,12.92}}{{45}}.100\%  = 24,53\%  \hfill \\
  \% {m_{{H_2}O}} = 100 - 34,13 - 24,53 = 41,34\%  \hfill \\ 
\end{gathered} $