Hỗn hợp E gồm axit cacboxylic đơn chức X, ancol no đa chức Y và Z

- TN1: x = \({{n}_{C{{O}_{2}}}}\)= 0,15 mol \(\Rightarrow \) y + z = 0,5 – 0,15 = 0,35 mol

- TN2: n_{ancol Y thu được} = y + z = 0,35 mol \(\Rightarrow \) M_Y = \(\frac{32,2}{0,35}=92\) \(\Rightarrow \) Y phù hợp là C₃H₅(OH)₃

n_NaOH = 0,15 + m.z = 0,65 

\(\,\left[ \begin{array}{l} m = 2\, \Rightarrow \,z = \,0,25\,\, \to \,y = 0,1\, < \,x\, \to \,thoa\,man\\ m = 3 \Rightarrow \,z = \frac{{0,5}}{3}\, \to y = 0,18 > x\, \to \,loai \end{array} \right.\)

\({\mathop{\rm E}\nolimits} \left\{ \begin{array}{l} {C_a}{H_b}COOH:\,0,15\,mol\\ {C_3}{H_5}{(OH)_3}:0,1\,mol\\ {({C_a}{H_b}COO)_2}{C_3}{H_5}OH:\,0,25\,mol \end{array} \right.\, \)\( + {O_{2du}} \to \,C{O_2}\,3,65mol + {H_2}O\,2,85mol\)

\(\left\{ {\begin{array}{*{20}{l}} {{\rm{n\;}}}&{{\rm{BT}}({\rm{C}}):{\mkern 1mu} ({\rm{a + 1}}).{\rm{0}},{\rm{15}}{\mkern 1mu} {\rm{ + }}{\mkern 1mu} {\rm{3}}.{\rm{0}},{\rm{1}}{\mkern 1mu} {\rm{ + }}{\mkern 1mu} ({\rm{2a + 5}}).{\rm{0}},{\rm{25 = 3}},{\rm{65}}}\\ {{\rm{n\;}}}&{{\rm{BT}}({\rm{H}}):{\mkern 1mu} ({\rm{b + 1}}).{\rm{0}},{\rm{15 + 8}}.{\rm{0}},{\rm{1 + }}({\rm{2b + 6}}).{\rm{0}},{\rm{25 = 2}}.{\rm{2}},{\rm{85}}} \end{array}} \right. \to \left\{ {\begin{array}{*{20}{l}} {{\rm{n\;}}}&{{\rm{a = 3}}}\\ {{\rm{n\;}}}&{{\rm{b = 5}}} \end{array}} \right.{\mkern 1mu} \to {\rm{E}}\left\{ {\begin{array}{*{20}{l}} {{\rm{n\;}}}&{{{\rm{C}}_{\rm{3}}}{{\rm{H}}_{\rm{5}}}{\rm{COOH}}:{\mkern 1mu} {\rm{0}},{\rm{15}}{\mkern 1mu} {\rm{mol}}}\\ {{\rm{n\;}}}&{{{\rm{C}}_{\rm{3}}}{{\rm{H}}_{\rm{5}}}{{({\rm{OH}})}_{\rm{3}}}:{\mkern 1mu} {\rm{0}},{\rm{1}}{\mkern 1mu} {\rm{mol}}}\\ {{\rm{n\;}}}&{{{({{\rm{C}}_{\rm{3}}}{{\rm{H}}_{\rm{5}}}{\rm{COO}})}_{\rm{2}}}{{\rm{C}}_{\rm{3}}}{{\rm{H}}_{\rm{5}}}{\rm{OH}}:{\mkern 1mu} {\rm{0}},{\rm{25}}{\mkern 1mu} {\rm{mol}}} \end{array}} \right.{\mkern 1mu} \to {\mkern 1mu} \% {{\rm{m}}_{\rm{Z}}}{\rm{ = 72}},{\rm{06}}\% \)