X ( 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: n_axit + 3n_triglixerit = n_NaOH → 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: m_muối = \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVGI8VfYFOqpC0xXdHaVhbbf9v8qqaqFr0xc9pk0xbb % a9q8WqFfea0-yr0RYxir-Jbba9q8aq0-yq-He9q8qqQ8frFve9Fve9 % Ff0dmeaabaqaaiaacaGaaeqabaWaaeaaeaaakeaacaWGTbWaaSbaaS % qaaiaadIfaaeqaaOGaey4kaSIaamyBamaaBaaaleaacaWGobGaamyy % aiaad+eacaWGibaabeaakiabgkHiTiaad2gadaWgaaWcbaGaam4zai % aadYgacaWGPbGaamiEaiaadwgacaWGYbGaam4BaiaadYgaaeqaaOGa % eyOeI0IaamyBamaaBaaaleaacaWGibWaaSbaaWqaaiaaikdaaeqaaS % Gaam4taaqabaGccqGH9aqpcaaI4aGaaGyoaaaa!4EDB! {m_X} + {m_{NaOH}} - {m_{glixerol}} - {m_{{H_2}O}} = 89\)