Cho 0,075 mol hỗn hợp hai este đơn chức X và Y tác dụng vừa đủ với dung dịch NaOH thu được hỗn hợp các chất hữu cơ Z. Đốt cháy hoàn toàn Z thu được 0,18 mol CO2, 0,045 mol Na2CO3. Làm bay hơi hỗn hợp Z thu được m gam chất rắn. Giá trị gần nhất của m là
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Lời giải:
Báo saiBTNT Na: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY-Hhbbf9v8qqaqFr0xc9pk0xbb % a9q8WqFfea0-yr0RYxir-Jbba9q8aq0-yq-He9q8qqQ8frFve9Fve9 % Ff0dmeaabaqaciaacaGaaeqabaWaaeaaeaaakeaacaWGUbWaaSbaaS % qaaiaad6eacaWGHbGaam4taiaadIeaaeqaaOGaeyypa0JaaGOmaiaa % d6gadaWgaaWcbaGaamOtaiaadggadaWgaaadbaGaaGOmaaqabaWcca % WGdbGaam4tamaaBaaameaacaaIZaaabeaaaSqabaGccqGH9aqpcaaI % WaGaaiilaiaaicdacaaI5aGaaGPaVlaad2gacaWGVbGaamiBaiaayk % W7cqGHshI3caaIXaGaeyipaWZaaSaaaeaacaWGUbWaaSbaaSqaaiaa % d6eacaWGHbGaam4taiaadIeaaeqaaaGcbaGaamOBamaaBaaaleaaca % WGLbGaaGPaVlaadohacaWG0bGaamyzaaqabaaaaOGaeyypa0ZaaSaa % aeaacaaIWaGaaiilaiaaicdacaaI5aaabaGaaGimaiaacYcacaaIWa % GaaG4naiaaiwdaaaGaeyypa0JaaGymaiaacYcacaaIYaGaeyipaWJa % aGOmaiaaykW7cqGHshI3caaMc8+aaiqaaqaabeqaaiaabgdacaaMc8 % UaaeyzaiaabohacaqG0bGaaeyzaiaaykW7caqGWbGaaeiAaiaabwga % caqGUbGaae4BaiaabYgacaqG6aGaaGPaVlaabIhacaaMc8UaaeyBai % aab+gacaqGSbaabaGaaeymaiaaykW7caqGLbGaae4CaiaabshacaqG % LbGaaGPaVlaadggacaqGUbGaae4yaiaab+gacaqGSbGaaeOoaiaayk % W7caqG5bGaaGPaVlaab2gacaqGVbGaaeiBaaaacaGL7baaaaa!96D1! {n_{NaOH}} = 2{n_{N{a_2}C{O_3}}} = 0,09\,mol\, \Rightarrow 1 < \frac{{{n_{NaOH}}}}{{{n_{e\,ste}}}} = \frac{{0,09}}{{0,075}} = 1,2 < 2\, \Rightarrow \,\left\{ \begin{gathered} {\text{1}}\,{\text{este}}\,{\text{phenol:}}\,{\text{x}}\,{\text{mol}} \hfill \\ {\text{1}}\,{\text{este}}\,a{\text{ncol:}}\,{\text{y}}\,{\text{mol}} \hfill \\ \end{gathered} \right.\)
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY-Hhbbf9v8qqaqFr0xc9pk0xbb % a9q8WqFfea0-yr0RYxir-Jbba9q8aq0-yq-He9q8qqQ8frFve9Fve9 % Ff0dmeaabaqaciaacaGaaeqabaWaaeaaeaaakeaadaGabaabaeqaba % GaamOBamaaBaaaleaacaWGLbGaaGPaVlaadohacaWG0bGaamyzaaqa % baGccqGH9aqpcaWG4bGaey4kaSIaamyEaiabg2da9iaaicdacaGGSa % GaaGimaiaaiEdacaaI1aaabaGaamOBamaaBaaaleaacaWGobGaamyy % aiaad+eacaWGibaabeaakiabg2da9iaaikdacaWG4bGaey4kaSIaam % yEaiabg2da9iaaicdacaGGSaGaaGimaiaaiMdaaaGaay5EaaGaaGPa % VlabgkDiEpaaceaaeaqabeaacaWG4bGaeyypa0JaaGimaiaacYcaca % aIWaGaaGymaiaaiwdacaaMc8UaamyBaiaad+gacaWGSbaabaGaamyE % aiabg2da9iaaicdacaGGSaGaaGimaiaaiAdacaaMc8UaamyBaiaad+ % gacaWGSbaaaiaawUhaaaaa!6B0A! \left\{ \begin{gathered} {n_{e\,ste}} = x + y = 0,075 \hfill \\ {n_{NaOH}} = 2x + y = 0,09 \hfill \\ \end{gathered} \right.\, \Rightarrow \left\{ \begin{gathered} x = 0,015\,mol \hfill \\ y = 0,06\,mol \hfill \\ \end{gathered} \right.\)
BTNT C:
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY-Hhbbf9v8qqaqFr0xc9pk0xbb % a9q8WqFfea0-yr0RYxir-Jbba9q8aq0-yq-He9q8qqQ8frFve9Fve9 % Ff0dmeaabaqaciaacaGaaeqabaWaaeaaeaaakeaacaWGUbWaaSbaaS % qaaiaadoeacaGGOaGaamiwaiaacYcacaWGzbGaaiykaaqabaGccqGH % 9aqpcaWGUbWaaSbaaSqaaiaadoeacaGGOaGaamOwaiaacMcaaeqaaO % Gaeyypa0JaamOBamaaBaaaleaacaWGdbGaam4tamaaBaaameaacaaI % YaaabeaaaSqabaGccqGHRaWkcaWGUbWaaSbaaSqaaiaad6eacaWGHb % WaaSbaaWqaaiaaikdaaeqaaSGaam4qaiaad+eadaWgaaadbaGaaG4m % aaqabaaaleqaaOGaeyypa0JaaGimaiaacYcacaaIYaGaaGOmaiaaiw % dacaaMc8UaeyO0H4TaaGPaVpaanaaabaGaam4qamaaBaaaleaacaWG % ybGaaiilaiaadMfaaeqaaaaakiabg2da9maalaaabaGaamOBamaaBa % aaleaacaWGdbaabeaaaOqaaiaad6gadaWgaaWcbaGaamyzaiaaykW7 % caWGZbGaamiDaiaadwgaaeqaaaaakiabg2da9maalaaabaGaaGimai % aacYcacaaIYaGaaGOmaiaaiwdaaeaacaaIWaGaaiilaiaaicdacaaI % 3aGaaGynaaaacqGH9aqpcaaIZaGaaGPaVlabgkDiEdaa!7184! {n_{C(X,Y)}} = {n_{C(Z)}} = {n_{C{O_2}}} + {n_{N{a_2}C{O_3}}} = 0,225\, \Rightarrow \,\overline {{C_{X,Y}}} = \frac{{{n_C}}}{{{n_{e\,ste}}}} = \frac{{0,225}}{{0,075}} = 3\, \)
Vậy có 1 este có C < 3 : C2H4O2
Theo đường chéo ta có:
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aIWaGaaiilaiaaiAdaaeaacaaIWaGaaiilaiaaigdacaaI1aaaaiab % g2da9maalaaabaGaam4qamaaBaaaleaacaWGzbaabeaakiabgkHiTi % aaiodaaeaacaaIXaaaaiabgkDiElaadoeadaWgaaWcbaGaamywaaqa % baGccqGH9aqpcaaI3aGaaeiiaiaaboeadaWgaaWcbaGaaG4naaqaba % GccaqGibWaaSbaaSqaaiaabIdaaeqaaOGaae4tamaaBaaaleaacaqG % Yaaabeaaaaa!4C37! \frac{{0,6}}{{0,15}} = \frac{{{C_Y} - 3}}{1} \Rightarrow {C_Y} = 7{\text{ }}{{\text{C}}_7}{{\text{H}}_{\text{8}}}{{\text{O}}_{\text{2}}}\)
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY-Hhbbf9v8qqaqFr0xc9pk0xbb % a9q8WqFfea0-yr0RYxir-Jbba9q8aq0-yq-He9q8qqQ8frFve9Fve9 % Ff0dmeaabaqaciaacaGaaeqabaWaaeaaeaaakeaadaGabaabaeqaba % GaamisaiaadoeacaWGpbGaam4taiaadoeacaWGibWaaSbaaSqaaiaa % iodaaeqaaOGaaiOoaiaaicdacaGGSaGaaGOnaiaaykW7caWGTbGaam % 4BaiaadYgaaeaacaWGibGaam4qaiaad+eacaWGpbGaam4qamaaBaaa % leaacaaI2aaabeaakiaadIeadaWgaaWcbaGaaGynaaqabaGccaGG6a % GaaGPaVlaaicdacaGGSaGaaGymaiaaiwdacaaMc8UaamyBaiaad+ga % caWGSbaaaiaawUhaaiabgUcaRiaad6eacaWGHbGaam4taiaadIeada % GdKaWcbaaabeGccaGLsgcadaGabaabaeqabaGaamisaiaadoeacaWG % pbGaam4taiaad6eacaWGHbGaaiOoaiaaykW7caaIWaGaaiilaiaaic % dacaaI3aGaaGynaiaaykW7caWGTbGaam4BaiaadYgaaeaacaWGdbWa % aSbaaSqaaiaaiAdaaeqaaOGaamisamaaBaaaleaacaaI1aaabeaaki % aad+eacaWGobGaamyyaiaacQdacaaIWaGaaiilaiaaicdacaaIXaGa % aGynaiaaykW7caWGTbGaam4BaiaadYgaaaGaay5EaaGaaGPaVlabgk % DiElaad2gacqGH9aqpcaaI2aGaaiilaiaaiIdacaaI0aGaaGPaVlaa % dEgacaWGHbGaamyBaiaac6caaaa!86D0! \left\{ \begin{gathered} HCOOC{H_3}:0,6\,mol \hfill \\ HCOO{C_6}{H_5}:\,0,15\,mol \hfill \\ \end{gathered} \right. + NaOH\xrightarrow{{}}\left\{ \begin{gathered} HCOONa:\,0,075\,mol \hfill \\ {C_6}{H_5}ONa:0,015\,mol \hfill \\ \end{gathered} \right.\, \Rightarrow m = 6,84\,gam.\)
Đề thi thử tốt nghiệp THPT QG môn Hóa năm 2020
Sở GD&ĐT Thái Nguyên lần 1
