Đáp án:

 1. 435,24 kg

2. 

a) muối: 143,7 kg; glixerol: 15,5 kg

b) 20,67 kg 

Giải thích các bước giải:

 1. ${({C_{15}}{H_{31}}COO)_3}{C_3}{H_5} + 3NaOH \to 3{C_{15}}{H_{31}}COONa + {C_3}{H_5}{(OH)_3}$

$\begin{gathered}
  {m_{{C_{15}}{H_{31}}COONa}} = \dfrac{{500.72}}{{100}} = 360kg \hfill \\
   \Rightarrow {n_{{C_{15}}{H_{31}}COONa}} = \dfrac{{360}}{{278}} = 1,295kmol \hfill \\ 
\end{gathered} $

Theo PTHH: 

$\begin{gathered}
  {n_{este(pu)}} = \dfrac{1}{3}{n_{{C_{15}}{H_{31}}COONa}} = 0,4317kmol \hfill \\
  H = 80\%  \Rightarrow {n_{este(tt)}} = \dfrac{{0,4317}}{{80\% }} = 0,54kmol \hfill \\
   \Rightarrow {m_{este}} = 0,54.806 = 435,24kg \hfill \\ 
\end{gathered} $

2. 

a) ${({C_{17}}{H_{35}}COO)_3}{C_3}{H_5} + 3{H_2}O\underset{{{t^o}}}{\overset{{{H_2}S{O_4}}}{\longleftrightarrow}}3{C_{17}}{H_{35}}COOH + {C_3}{H_5}{(OH)_3}$

$\begin{gathered}
  H = 75\%  \Rightarrow {m_{este(pu)}} = 200.\dfrac{{75}}{{100}} = 150kg \hfill \\
   \Rightarrow {n_{{C_{17}}{H_{35}}COOH}} = 3{n_{este(pu)}} = 3.\dfrac{{150}}{{890}} = 0,506kmol \hfill \\
  {n_{{C_3}{H_5}{{(OH)}_3}}} = {n_{este(pu)}} = 0,1685kmol \hfill \\
   \Rightarrow {m_{{C_{17}}{H_{35}}COOH}} = 0,506.284 = 143,7kg \hfill \\
  {m_{{C_3}{H_5}{{(OH)}_3}}} = 0,1685.92 = 15,5kg \hfill \\ 
\end{gathered} $

b) 

$\begin{gathered}
  {n_{este}} = \dfrac{{200}}{{890}} = 0,2247kmol \hfill \\
   \Rightarrow {n_{{C_3}{H_5}{{(OH)}_3}}} = {n_{este}} = 0,2247kmol \hfill \\
   \Rightarrow {m_{{C_3}{H_5}{{(OH)}_3}}} = 0,2247.92 = 20,67kg \hfill \\ 
\end{gathered} $