Thủy phân hoàn toàn a mol triglixerit X trong dung dịch NaOH vừa đủ, thu được glixerol và m gam hỗn hợp muối. Đốt cháy hoàn toàn a mol X thu được 1,375 mol CO2 và 1,275 mol H2O. Mặt khác, a mol X tác dụng tối đa với 0,05 mol Br2 trong dung dịch. Giá trị của m là
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Lời giải:
Báo saiĐộ bất bão hòa của X là \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaaciGaaiaabeqaamaabaabaaGcbaGaam4Aaiabg2 % da9maalaaabaGaaGimaiaacYcacaaIWaGaaGynaaqaaiaadggaaaGa % ey4kaSIaaG4maaaa!3D60! k = \frac{{0,05}}{a} + 3\)
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaaciGaaiaabeqaamaabaabaaGcbaGaamyyamaabm % aabaGaam4AaiabgkHiTiaaigdaaiaawIcacaGLPaaacqGH9aqpcaWG % UbWaaSbaaSqaaiaadoeacaWGpbWaaSbaaWqaaiaaikdaaeqaaaWcbe % aakiabgkHiTiaad6gadaWgaaWcbaGaamisamaaBaaameaacaaIYaaa % beaaliaad+eaaeqaaOGaeyOKH4Qaamyyaiabg2da9iaaicdacaGGSa % GaaGimaiaaikdacaaI1aGaeyOKH4QaamOBamaaBaaaleaacaWGpbaa % beaakiabg2da9iaaiAdacaWGHbGaeyypa0JaaGimaiaacYcacaaIXa % GaaGynaaaa!565B! a\left( {k - 1} \right) = {n_{C{O_2}}} - {n_{{H_2}O}} \to a = 0,025 \to {n_O} = 6a = 0,15\)
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaaciGaaiaabeqaamaabaabaaGcbaGaeyOKH4Qaam % yBamaaBaaaleaacaWGybaabeaakiabg2da9iaad2gadaWgaaWcbaGa % am4qaaqabaGccqGHRaWkcaWGTbWaaSbaaSqaaiaadIeaaeqaaOGaey % 4kaSIaamyBamaaBaaaleaacaWGpbaabeaakiabg2da9iaaikdacaaI % XaGaaiilaiaaisdacaaI1aaaaa!4739! \to {m_X} = {m_C} + {m_H} + {m_O} = 21,45\)
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaaciGaaiaabeqaamaabaabaaGcbaGaamOBamaaBa % aaleaacaWGobGaamyyaiaad+eacaWGibaabeaakiabg2da9iaaioda % caWGHbGaeyypa0JaaGimaiaacYcacaaIWaGaaG4naiaaiwdacaGG7a % Gaaeiiaiaad6gadaWgaaWcbaGaam4qamaaBaaameaacaaIZaaabeaa % liaadIeadaWgaaadbaGaaGynaaqabaWcdaqadaqaaiaad+eacaWGib % aacaGLOaGaayzkaaWaaSbaaWqaaiaaiodaaeqaaaWcbeaakiabg2da % 9iaadggaaaa!4DDF! {n_{NaOH}} = 3a = 0,075;{\text{ }}{n_{{C_3}{H_5}{{\left( {OH} \right)}_3}}} = a\)
BTKL → mmuối = 22,15
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