Hỗn hợp E gồm ba este no, mạch hở X (đơn chức), Y (hai chức), Z (ba chức) đều được tạo thành từ axit cacboxylic và ancol. Khối lượng của X trong m gam E là?

* Hướng dẫn giải

\(\to \,{{n}_{COONa}}={{n}_{NaOH}}=2{{n}_{N{{a}_{2}}C{{O}_{3}}}}=0,7\,mol\to \,{{n}_{C{{O}_{2}}}}=0,35\,mol\)

\(\to \,{{n}_{C(T)}}={{n}_{N{{a}_{2}}C{{O}_{3}}}}+{{n}_{C{{O}_{2}}}}=0,7\,mol={{n}_{COONa}}\,\Rightarrow \,\left\{ \begin{align} & HCOONa:0,3\,(BTH) \\ & {{(COONa)}_{2}}:0,2\,(BTNa) \\ \end{align} \right. \to {{m}_{T}}=47,2\,gam\,\to {{m}_{E}}=47,2+28,6-40.0,7=47,8\,gam\)

\( \to \left\{ \begin{array}{l} C{O_2}:\,x\,mol\\ {H_2}O:\,y\,mol \end{array} \right.\, \Rightarrow \left\{ \begin{array}{l} {m_E} = 12x + 2y + 32.0,7 = 47,8\\ x - y = 0,425 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} x = 1,875\,mol\\ y = 1,45\,mol \end{array} \right.\)

\(Ancol\left\{ \begin{align} & {{n}_{C(ancol)}}=1,875-0,7=1,175\,mol={{n}_{C{{O}_{2}}}} \\ & {{n}_{O(ancol)}}={{n}_{NaOH}}=0,7\,mol \\ & {{n}_{H(ancol)}}=\frac{28,6-12.1,175-16.0,7}{1}=3,3\,mol\Rightarrow {{n}_{{{H}_{2}}O}}=1,65\,mol \\ \end{align} \right.\)

\(\begin{array}{l} \to \left\{ \begin{array}{l} {C_n}{H_{2n + 2}}O:\,a\,mol\\ {C_m}{H_{2m + 2}}{O_2}:\,b\,mol \end{array} \right. \Rightarrow \left\{ \begin{array}{l} {n_O} = a + 2b = 0,7\\ a + b = 1,65 - 1,175 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} a = 0,25\,mol\\ b = 0,225\,\,mol \end{array} \right.\\ \to \,0,25n + 0,225m = 1,175 \Rightarrow 10n + 9m = 47 \to \,\left\{ \begin{array}{l} n = 2:\,{C_2}{H_5}OH:\,0,25\,mol\\ m = 3:\,{C_3}{H_6}{(OH)_2}:\,0,225\,mol \end{array} \right. \end{array}\)

⇒ hhE \(\left\{ \begin{align} & X:HC\text{OO}{{\text{C}}_{2}}{{H}_{5}}:0,05\,mol \\ & Y:{{(HC\text{OO})}_{2}}{{C}_{3}}{{H}_{6}}:0,025\,mol \\ & Z:\,{{C}_{2}}{{H}_{5}}\text{OO}C-C\text{OO}{{\text{C}}_{3}}{{H}_{6}}\text{OO}CH:\,0,2mol \\ \end{align} \right.\Rightarrow {{m}_{X}}=3,7\,gam.\)