Hỗn hợp A gồm este đơn chức X và hai este no, hai chức, mạch hở Y và Z (M_Y < M_Z). Đốt cháy hoàn toàn 7,08 gam A cần vừa đủ 0,326 mol O₂, thu được 3,96 gam H₂O. Mặt khác 7,08 gam A tác dụng vừa đủ với 0,104 mol NaOH trong dung dịch, thu được 3,232 gam hai ancol no, đơn chức. cô cạn dung dịch thu được m gam hỗn hợp muối T. Giá trị của m là

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaacaGaaiaabeqaamaabaabaaGcbaGaamOBamaaBa % aaleaacaWGibWaaSbaaWqaaiaaikdaaeqaaSGaam4taaqabaGccqGH % 9aqpcaaIWaGaaiilaiaaikdacaaIYaaaaa!3D96! {n_{{H_2}O}} = 0,22\)

Bảo toàn khối lượng \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaacaGaaiaabeqaamaabaabaaGcbaGaeyOKH4Qaam % OBamaaBaaaleaacaWGdbGaam4tamaaBaaameaacaaIYaaabeaaaSqa % baGccqGH9aqpcaaIWaGaaiilaiaaiodacaaIWaGaaGioaaaa!403F! \to {n_{C{O_2}}} = 0,308\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaacaGaaiaabeqaamaabaabaaGcbaGaeyOKH4Qaam % OBamaaBaaaleaacaWGpbWaaeWaaeaacaWGbbaacaGLOaGaayzkaaaa % beaakiabg2da9maalaaabaGaamyBamaaBaaaleaacaWGbbaabeaaki % abgkHiTiaad2gadaWgaaWcbaGaam4qaaqabaGccqGHsislcaWGTbWa % aSbaaSqaaiaadIeaaeqaaaGcbaGaaGymaiaaiAdaaaGaeyypa0JaaG % imaiaacYcacaaIXaGaaGioaiaaisdaaaa!4B12! \to {n_{O\left( A \right)}} = \frac{{{m_A} - {m_C} - {m_H}}}{{16}} = 0,184\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaacaGaaiaabeqaamaabaabaaGcbaGaamOBamaaBa % aaleaacaWGobGaamyyaiaad+eacaWGibaabeaakiabg6da+maalaaa % baGaamOBamaaBaaaleaacaWGpbWaaeWaaeaacaWGbbaacaGLOaGaay % zkaaaabeaaaOqaaiaaikdaaaaaaa!4093! {n_{NaOH}} > \frac{{{n_{O\left( A \right)}}}}{2}\) nên X là este của phenol.

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaacaGaaiaabeqaamaabaabaaGceaqabeaacaWGUb % WaaSbaaSqaaiaad+eadaqadaqaaiaadgeaaiaawIcacaGLPaaaaeqa % aOGaeyypa0JaaGOmaiaad6gadaWgaaWcbaGaamiwaaqabaGccqGHRa % WkcaaI0aGaamOBamaaBaaaleaacaWGzbaabeaakiabgUcaRiaaisda % caWGUbWaaSbaaSqaaiaadQfaaeqaaOGaeyypa0JaaGimaiaacYcaca % aIXaGaaGioaiaaisdaaeaacaWGUbWaaSbaaSqaaiaad6eacaWGHbGa % am4taiaadIeaaeqaaOGaeyypa0JaaGOmaiaad6gadaWgaaWcbaGaam % iwaaqabaGccqGHRaWkcaaIYaGaamOBamaaBaaaleaacaWGzbaabeaa % kiabgUcaRiaaikdacaWGUbWaaSbaaSqaaiaadQfaaeqaaOGaeyypa0 % JaaGimaiaacYcacaaIXaGaaGimaiaaisdaaeaacqGHsgIRcaWGUbWa % aSbaaSqaaiaadIfaaeqaaOGaeyypa0JaaGimaiaacYcacaaIWaGaaG % ymaiaaikdacaGG7aGaamOBamaaBaaaleaacaWGzbaabeaakiabgUca % Riaad6gadaWgaaWcbaGaamOwaaqabaGccqGH9aqpcaaIWaGaaiilai % aaicdacaaI0aaabaGaeyOKH4QaamOBamaaBaaaleaacaWGibWaaSba % aWqaaiaaikdaaeqaaSGaam4taaqabaGccqGH9aqpcaWGUbWaaSbaaS % qaaiaadIfaaeqaaOGaeyypa0JaaGimaiaacYcacaaIWaGaaGymaiaa % ikdaaaaa!7DC4! \begin{gathered} {n_{O\left( A \right)}} = 2{n_X} + 4{n_Y} + 4{n_Z} = 0,184 \hfill \\ {n_{NaOH}} = 2{n_X} + 2{n_Y} + 2{n_Z} = 0,104 \hfill \\ \to {n_X} = 0,012;{n_Y} + {n_Z} = 0,04 \hfill \\ \to {n_{{H_2}O}} = {n_X} = 0,012 \hfill \\ \end{gathered} \)

Bảo toàn khối lượng:

m_A + m_NaOH = m_muối + m_ancol + m_nước

→ m muối = 7,792 gam.