Hỗn hợp X gồm axit panmitic, axit stearic và triglixerit Y. Đốt cháy hoàn toàn m gam X cần dùng vừa đủ 7,675 mol O2, thu được H2O và 5,35 mol CO2. Mặt khác, m gam X tác dụng vừa đủ với 0,3 mol NaOH trong dung dịch, thu được glixerol và dung dịch chỉ chứa a gam hỗn hợp muối natri panmitat, natri stearat. Giá trị của a là
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Lời giải:
Báo saiX ( Axit (x mol); Triglixerit (y mol)) + NaOH → Muối của axit béo no + Glixerol + H2O
Hỗn hợp gồm axit béo có k1 = 1 và triglixerit no có k2 = 3.
Theo độ bất bão hoà: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVGI8VfYFOqpC0xXdHaVhbbf9v8qqaqFr0xc9pk0xbb % a9q8WqFfea0-yr0RYxir-Jbba9q8aq0-yq-He9q8qqQ8frFve9Fve9 % Ff0dmeaabaqaaiaacaGaaeqabaWaaeaaeaaakeaacaWGUbWaaSbaaS % qaaiaadoeacaWGpbWaaSbaaWqaaiaaikdaaeqaaaWcbeaakiabgkHi % Tiaad6gadaWgaaWcbaGaamisamaaBaaameaacaaIYaaabeaaliaad+ % eaaeqaaOGaeyypa0JaaiikaiaadUgadaWgaaWcbaGaaGymaaqabaGc % cqGHsislcaaIXaGaaiykaiaadIhacqGHRaWkcaGGOaGaam4AamaaBa % aaleaacaaIYaaabeaakiabgkHiTiaaigdacaGGPaGaamyEaiabgkDi % ElaaiwdacaGGSaGaaG4maiaaiwdacqGHsislcaWGUbWaaSbaaSqaai % aadIeadaWgaaadbaGaaGOmaaqabaWccaWGpbaabeaakiabg2da9iaa % ikdacaWG5bGaaeiiaiaacIcacaaIXaGaaiykaaaa!5C10! {n_{C{O_2}}} - {n_{{H_2}O}} = ({k_1} - 1)x + ({k_2} - 1)y \Rightarrow 5,35 - {n_{{H_2}O}} = 2y{\text{ }}(1)\)
Ta có: naxit + 3ntriglixerit = nNaOH → x + 3y = 0,3 (2)
BTNT O: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVGI8VfYFOqpC0xXdHaVhbbf9v8qqaqFr0xc9pk0xbb % a9q8WqFfea0-yr0RYxir-Jbba9q8aq0-yq-He9q8qqQ8frFve9Fve9 % Ff0dmeaabaqaaiaacaGaaeqabaWaaeaaeaaakeaaqaaaaaaaaaWdbi % aaikdacaGGUaGaaGynaiaacYcacaaIZaGaaGynaiabgUcaRiaad6ga % daWgaaWcbaGaamisamaaBaaameaacaaIYaaabeaaliaad+eaaeqaaO % Gaeyypa0ZdaiaaikdacaGGUaGaaG4naiaacYcacaaI2aGaaG4naiaa % iwdapeGaey4kaSIaaiikaiaaikdacaWG4bGaey4kaSIaaGOnaiaadM % hacaGGPaGaaeiiaiaabIcacaqGZaGaaeykaaaa!4F3B! 2.5,35 + {n_{{H_2}O}} = 2.7,675 + (2x + 6y){\text{ (3)}}\)
Từ (1), (2), (3) suy ra: x = 0,15 ; y = 0,05 ; \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVGI8VfYFOqpC0xXdHaVhbbf9v8qqaqFr0xc9pk0xbb % a9q8WqFfea0-yr0RYxir-Jbba9q8aq0-yq-He9q8qqQ8frFve9Fve9 % Ff0dmeaabaqaaiaacaGaaeqabaWaaeaaeaaakeaaqaaaaaaaaaWdbi % aad6gadaWgaaWcbaGaamisamaaBaaameaacaaIYaaabeaaliaad+ea % aeqaaOGaeyypa0JaaGynaiaacYcacaaIYaGaaGynaaaa!3E68! {n_{{H_2}O}} = 5,25\)
BTKL: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVGI8VfYFOqpC0xXdHaVhbbf9v8qqaqFr0xc9pk0xbb % a9q8WqFfea0-yr0RYxir-Jbba9q8aq0-yq-He9q8qqQ8frFve9Fve9 % Ff0dmeaabaqaaiaacaGaaeqabaWaaeaaeaaakeaacaWGTbWaaSbaaS % qaaiaadIfaaeqaaOGaeyypa0JaamyBamaaBaaaleaacaWGdbGaam4t % amaaBaaameaacaaIYaaabeaaaSqabaGccqGHRaWkcaWGTbWaaSbaaS % qaaiaadIeadaWgaaadbaGaaGOmaaqabaWccaWGpbaabeaakiabgkHi % Tiaad2gadaWgaaWcbaGaam4tamaaBaaameaacaaIYaaabeaaaSqaba % GccqGH9aqpcaaI4aGaaGinaiaacYcacaaIZaaaaa!49CC! {m_X} = {m_{C{O_2}}} + {m_{{H_2}O}} - {m_{{O_2}}} = 84,3\)
BTKL: mmuối = \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVGI8VfYFOqpC0xXdHaVhbbf9v8qqaqFr0xc9pk0xbb % a9q8WqFfea0-yr0RYxir-Jbba9q8aq0-yq-He9q8qqQ8frFve9Fve9 % Ff0dmeaabaqaaiaacaGaaeqabaWaaeaaeaaakeaacaWGTbWaaSbaaS % qaaiaadIfaaeqaaOGaey4kaSIaamyBamaaBaaaleaacaWGobGaamyy % aiaad+eacaWGibaabeaakiabgkHiTiaad2gadaWgaaWcbaGaam4zai % aadYgacaWGPbGaamiEaiaadwgacaWGYbGaam4BaiaadYgaaeqaaOGa % eyOeI0IaamyBamaaBaaaleaacaWGibWaaSbaaWqaaiaaikdaaeqaaS % Gaam4taaqabaGccqGH9aqpcaaI4aGaaGyoaaaa!4EDB! {m_X} + {m_{NaOH}} - {m_{glixerol}} - {m_{{H_2}O}} = 89\)
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