Trong 22,36 (g) E có n_NaOH = 0,16/2 = 0,08 mol  → n_axit + 3n_este = 0,08 (1)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVGI8VfYFOqpC0xXdHaVhbbf9v8qqaqFr0xc9pk0xbb % a9q8WqFfea0-yr0RYxir-Jbba9q8aq0-yq-He9q8qqQ8frFve9Fve9 % Ff0dmeaabaqaaiaacaGaaeqabaWaaeaaeaaakeaadaGabaabaeqaba % Waa4ajaSqaaiaadkeacaWGubGaam4saiaadYeaaeqakiaawkziaiaa % isdacaaI0aGaamOBamaaBaaaleaacaWGdbGaam4tamaaBaaameaaca % aIYaaabeaaaSqabaGccqGHRaWkcaaIXaGaaGioaiaad6gadaWgaaWc % baGaamisamaaBaaameaacaaIYaaabeaaliaad+eaaeqaaOGaeyypa0 % JaaGOmaiaaikdacaGGSaGaaG4maiaaiAdacqGHRaWkcaaIYaGaaiil % aiaaicdacaaIXaGaaiOlaiaaiodacaaIYaaabaWaa4ajaSqaaiaadk % eacaWGubGaaiOoaiaaysW7caWGpbaabeGccaGLsgcacaaIYaGaaiOl % aiaacIcacaGGUbWaaSbaaSqaaiaabggacaqG4bGaamyAaiaadshaae % qaaOGaey4kaSIaaG4maiaac6gadaWgaaWcbaGaamyzaiaadohacaWG % 0bGaamyzaaqabaGccaGGPaGaey4kaSIaaGOmaiaacYcacaaIWaGaaG % ymaiaac6cacaaIYaGaeyypa0JaaGOmaiaad6gadaWgaaWcbaGaam4q % aiaad+eadaWgaaadbaGaaGOmaaqabaaaleqaaOGaey4kaSIaamOBam % aaBaaaleaacaWGibWaaSbaaWqaaiaaikdaaeqaaSGaam4taaqabaaa % aOGaay5EaaGaeyO0H49aaiqaaqaabeqaaiaad6gadaWgaaWcbaGaam % 4qaiaad+eadaWgaaadbaGaaGOmaaqabaaaleqaaOGaeyypa0JaaGym % aiaacYcacaaI0aGaaG4maaqaaiaad6gadaWgaaWcbaGaamisamaaBa % aameaacaaIYaaabeaaliaad+eaaeqaaOGaeyypa0JaaGymaiaacYca % caaIZaGaaGOmaaaacaGL7baaaaa!892E! \left\{ \begin{gathered} \xrightarrow{{BTKL}}44{n_{C{O_2}}} + 18{n_{{H_2}O}} = 22,36 + 2,01.32 \hfill \\ \xrightarrow{{BT:\;O}}2.({n_{{\text{ax}}it}} + 3{n_{este}}) + 2,01.2 = 2{n_{C{O_2}}} + {n_{{H_2}O}} \hfill \\ \end{gathered} \right. \Rightarrow \left\{ \begin{gathered} {n_{C{O_2}}} = 1,43 \hfill \\ {n_{{H_2}O}} = 1,32 \hfill \\ \end{gathered} \right.\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVGI8VfYFOqpC0xXdHaVhbbf9v8qqaqFr0xc9pk0xbb % a9q8WqFfea0-yr0RYxir-Jbba9q8aq0-yq-He9q8qqQ8frFve9Fve9 % Ff0dmeaabaqaaiaacaGaaeqabaWaaeaaeaaakeaacaWGUbWaaSbaaS % qaaiaadoeacaWGpbWaaSbaaWqaaiaaikdaaeqaaaWcbeaakiabgkHi % Tiaad6gadaWgaaWcbaGaamisamaaBaaameaacaaIYaaabeaaliaad+ % eaaeqaaOGaeyypa0JaaiikaiaadUgadaWgaaWcbaGaaGymaaqabaGc % cqGHRaWkcaaIXaGaeyOeI0IaaGymaiaacMcacaGGUaGaaiOBamaaBa % aaleaacaqGHbGaaeiEaiaadMgacaWG0baabeaakiabgUcaRiaacIca % caWGRbWaaSbaaSqaaiaaikdaaeqaaOGaey4kaSIaaG4maiabgkHiTi % aaigdacaGGPaGaaiOlaiaad6gadaWgaaWcbaGaamyzaiaadohacaWG % 0bGaamyzaaqabaGccqGHshI3daagaaqaamaabmaabaGaam4AamaaBa % aaleaacaaIXaaabeaakiaac6cacaGGUbWaaSbaaSqaaiaabggacaqG % 4bGaamyAaiaadshaaeqaaOGaey4kaSIaam4AamaaBaaaleaacaaIYa % aabeaakiaac6cacaWGUbWaaSbaaSqaaiaadwgacaWGZbGaamiDaiaa % dwgaaeqaaaGccaGLOaGaayzkaaaaleaacaWGUbWaaSbaaWqaaiaadk % eacaWGYbWaaSbaaeaacaaIYaaabeaaaeqaaaGccaGL44pacqGHRaWk % caaIYaGaamOBamaaBaaaleaacaWGLbGaam4CaiaadshacaWGLbaabe % aakiabg2da9iaaicdacaGGSaGaaGymaiaaigdaaaa!7D1F! {n_{C{O_2}}} - {n_{{H_2}O}} = ({k_1} + 1 - 1).{n_{{\text{ax}}it}} + ({k_2} + 3 - 1).{n_{este}} \Rightarrow \underbrace {\left( {{k_1}.{n_{{\text{ax}}it}} + {k_2}.{n_{este}}} \right)}_{{n_{B{r_2}}}} + 2{n_{este}} = 0,11\)

→ n_este = 0,01

Từ (1) suy ra: n_axit = 0,05 mol.

Xét 44,72 gam E (gấp 2 lần ở phần trên) có C₃H₅(OH)₃: 0,02 mol và H₂O: 0,1 mol → m = 47,48 (g)