Dung dịch X chứa x mol HCl, dung dịch Y chứa y mol Na2CO3 và 2y mol NaHCO3. Nhỏ từ từ đến hết X vào Y, thu được V lít khí CO2 (đktc). Nhỏ từ từ đến hết Y vào X, thu được dung dịch Z và 2V lít khí CO2 (đktc). Cho dung dịch Ca(OH)2 dư vào Z thu được m gam kết tủa. Giá trị của m là
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Lời giải:
Báo saiLượng CO2 thoát ra khác nhau nên HCl không dư trong cả 2 thí nghiệm.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaacaGaaiaabeqaamaabaabaaGcbaGaamOBamaaBa % aaleaacaWGdbGaam4tamaaBaaameaacaaIYaaabeaaaSqabaGccqGH % 9aqpdaWcaaqaaiaadAfaaeaacaaIYaGaaGOmaiaacYcacaaI0aaaai % abg2da9iaadQhaaaa!4085! {n_{C{O_2}}} = \frac{V}{{22,4}} = z\)
TN1: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaacaGaaiaabeqaamaabaabaaGcbaGaamOBamaaBa % aaleaacaWGibWaaWbaaWqabeaacqGHRaWkaaaaleqaaOGaeyypa0Ja % amOBamaaBaaaleaacaWGdbGaam4tamaaDaaameaacaaIZaaabaGaaG % OmaiabgkHiTaaaaSqabaGccqGHRaWkcaWGUbWaaSbaaSqaaiaadoea % caWGpbWaaSbaaWqaaiaaikdaaeqaaaWcbeaakiabgkziUkaadIhacq % GH9aqpcaWG5bGaey4kaSIaamOEamaabmaabaGaaGymaaGaayjkaiaa % wMcaaaaa!4D19! {n_{{H^ + }}} = {n_{CO_3^{2 - }}} + {n_{C{O_2}}} \to x = y + z\left( 1 \right)\)
TN2: Số mol phản ứng: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaacaGaaiaabeqaamaabaabaaGcbaGaamOBamaaBa % aaleaacaWGdbGaam4tamaaDaaameaacaaIZaaabaGaaGOmaiabgkHi % TaaaaSqabaGccqGH9aqpcaWGRbGaamyEaiaacUdacaWGUbWaaSbaaS % qaaiaadIeacaWGdbGaam4tamaaDaaameaacaaIZaaabaGaeyOeI0ca % aaWcbeaakiabg2da9iaaikdacaWGRbGaamyEaaaa!482C! {n_{CO_3^{2 - }}} = ky;{n_{HCO_3^ - }} = 2ky\)
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaacaGaaiaabeqaamaabaabaaGceaqabeaacaWGUb % WaaSbaaSqaaiaadIeadaahaaadbeqaaiabgUcaRaaaaSqabaGccqGH % 9aqpcaaIYaGaam4AaiaadMhacqGHRaWkcaaIYaGaam4AaiaadMhacq % GH9aqpcaWG4bGaeyOKH4QaamiEaiabg2da9iaaisdacaWGRbGaamyE % aaqaaiaad6gadaWgaaWcbaGaam4qaiaad+eadaWgaaadbaGaaGOmaa % qabaaaleqaaOGaeyypa0Jaam4AaiaadMhacqGHRaWkcaaIYaGaam4A % aiaadMhacqGH9aqpcaaIYaGaamOEaiabgkziUkaadUgacaWG5bGaey % ypa0ZaaSaaaeaacaaIYaGaamOEaaqaaiaaiodaaaaaaaa!5D46! \begin{gathered} {n_{{H^ + }}} = 2ky + 2ky = x \to x = 4ky \hfill \\ {n_{C{O_2}}} = ky + 2ky = 2z \to ky = \frac{{2z}}{3} \hfill \\ \end{gathered} \)
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaacaGaaiaabeqaamaabaabaaGcbaGaeyOKH4Qaam % iEaiabg2da9maalaaabaGaaGioaiaadQhaaeaacaaIZaaaaaaa!3C6E! \to x = \frac{{8z}}{3}\)
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaacaGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca % aIXaaacaGLOaGaayzkaaGaeyOKH4QaamyEaiabg2da9maalaaabaGa % aGynaiaadQhaaeaacaaIZaaaaiabgkziUkaadUgacqGH9aqpcaaIWa % Gaaiilaiaaisdaaaa!44BB! \left( 1 \right) \to y = \frac{{5z}}{3} \to k = 0,4\)
Z chứa \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaacaGaaiaabeqaamaabaabaaGcbaGaamOBamaaBa % aaleaacaWGdbGaam4tamaaDaaameaacaaIZaaabaGaaGOmaiabgkHi % TaaaaSqabaGccqGH9aqpcaWG5bGaeyOeI0Iaam4AaiaadMhacqGH9a % qpcaaIWaGaaiilaiaaiAdacaWG5bGaeyypa0JaamOEaiaacUdacaWG % UbWaaSbaaSqaaiaadIeacaWGdbGaam4tamaaDaaameaacaaIZaaaba % GaeyOeI0caaaWcbeaakiabg2da9iaaikdacaWG5bGaeyOeI0IaaGOm % aiaadUgacaWG5bGaeyypa0JaaGOmaiaadQhaaaa!55B2! {n_{CO_3^{2 - }}} = y - ky = 0,6y = z;{n_{HCO_3^ - }} = 2y - 2ky = 2z\)
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaacaGaaiaabeqaamaabaabaaGceaqabeaacqGHsg % IRcaWGUbWaaSbaaSqaaiaadoeacaWGHbGaam4qaiaad+eadaqhaaad % baGaaG4maaqaaaaaaSqabaGccqGH9aqpcaWG6bGaey4kaSIaaGOmai % aadQhacqGHRaWkcaaIZaGaamOEaiabg2da9maalaaabaGaaGyoaiaa % dIhaaeaacaaI4aaaaaqaaiabgkziUkaad2gadaWgaaWcbaGaam4qai % aadggacaWGdbGaam4tamaaBaaameaacaaIZaaabeaaaSqabaGccqGH % 9aqpdaWcaaqaaiaaigdacaaIWaGaaGimaiaac6cacaaI5aGaamiEaa % qaaiaaiIdaaaGaeyypa0JaaGymaiaaigdacaaIYaGaaiilaiaaiwda % caWG4baaaaa!5B96! \begin{gathered} \to {n_{CaCO_3^{}}} = z + 2z + 3z = \frac{{9x}}{8} \hfill \\ \to {m_{CaC{O_3}}} = \frac{{100.9x}}{8} = 112,5x \hfill \\ \end{gathered} \)
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