Dẫn từ từ đến dư khí CO2 vào dung dịch chứa 0,2 mol Ba(OH)2. Sự phụ thuộc của số mol kết tủa (a mol) vào số mol khí CO2 tham gia phản ứng (b mol) được biểu diễn như đồ thị sau
Tỉ lệ y : x là
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Lời giải:
Báo saiĐoạn 1: CO2 + Ba(OH)2 → BaCO3 + H2O
→ x = 0,075
Đoạn 2: CO2 + H2O + BaCO3 → Ba(HCO3)2
+ Khi \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaaciGaaiaabeqaamaabaabaaGcbaGaamOBamaaBa % aaleaacaWGdbGaam4tamaaBaaameaacaaIYaaabeaaaSqabaGccqGH % 9aqpcaWG5bGaeyOKH4QaamOBamaaBaaaleaacaWGcbGaamyyaiaado % eacaWGpbWaaSbaaWqaaiaaiodaaeqaaaWcbeaakiabg2da9iaaigda % caGGSaGaaGynaiaadshacqGHRaWkcaaIWaGaaiilaiaaicdacaaIYa % GaaGynaaaa!4BAD! {n_{C{O_2}}} = y \to {n_{BaC{O_3}}} = 1,5t + 0,025\)
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaaciGaaiaabeqaamaabaabaaGceaqabeaacaWGcb % GaamivaiaabccacaWGcbGaamyyaiabgkziUkaad6gadaWgaaWcbaGa % amOqaiaadggadaqadaqaaiaadIeacaWGdbGaam4tamaaBaaameaaca % aIZaaabeaaaSGaayjkaiaawMcaamaaBaaameaacaaIYaaabeaaaSqa % baGccqGH9aqpcaaIWaGaaiilaiaaigdacaaI3aGaaGynaiabgkHiTi % aaigdacaGGSaGaaGynaiaadshaaeaacaWGcbGaamivaiaabccacaqG % dbGaeyOKH46aaeWaaeaacaaIXaGaaiOlaiaaiwdacaWG0bGaey4kaS % IaaGimaiaacYcacaaIWaGaaGOmaiaaiwdaaiaawIcacaGLPaaacqGH % RaWkcaaIYaWaaeWaaeaacaaIWaGaaiilaiaaigdacaaI3aGaaGynai % abgkHiTiaaigdacaGGSaGaaGynaiaadshaaiaawIcacaGLPaaacqGH % 9aqpcaWG5bWaaeWaaeaacaaIXaaacaGLOaGaayzkaaaaaaa!6A8B! \begin{gathered} BT{\text{ }}Ba \to {n_{Ba{{\left( {HC{O_3}} \right)}_2}}} = 0,175 - 1,5t \hfill \\ BT{\text{ C}} \to \left( {1.5t + 0,025} \right) + 2\left( {0,175 - 1,5t} \right) = y\left( 1 \right) \hfill \\ \end{gathered} \)
+ Khi \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaaciGaaiaabeqaamaabaabaaGcbaGaamOBamaaBa % aaleaacaWGdbGaam4tamaaBaaameaacaaIYaaabeaaaSqabaGccqGH % 9aqpcaaIZaGaamiDaiabgUcaRiaaicdacaGGSaGaaGimaiaaikdaca % aI1aGaeyOKH4QaamOBamaaBaaaleaacaWGcbGaamyyaiaadoeacaWG % pbWaaSbaaWqaaiaaiodaaeqaaaWcbeaakiabg2da9iaaicdacaGGSa % GaaGimaiaaiEdacaaI1aaaaa!4CE6! {n_{C{O_2}}} = 3t + 0,025 \to {n_{BaC{O_3}}} = 0,075\)
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaaciGaaiaabeqaamaabaabaaGceaqabeaacaWGcb % GaamivaiaabccacaWGcbGaamyyaiabgkziUkaad6gadaWgaaWcbaGa % amOqaiaadggadaqadaqaaiaadIeacaWGdbGaam4tamaaBaaameaaca % aIZaaabeaaaSGaayjkaiaawMcaamaaBaaameaacaaIYaaabeaaaSqa % baGccqGH9aqpcaaIWaGaaiilaiaaigdacaaIYaGaaGynaaqaaiaadk % eacaWGubGaaeiiaiaadoeacqGHsgIRcaaIWaGaaiilaiaaicdacaaI % 3aGaaGynaiabgUcaRiaaikdacaGGUaGaaGimaiaacYcacaaIXaGaaG % OmaiaaiwdacqGH9aqpcaaIZaGaamiDaiabgUcaRiaaicdacaGGSaGa % aGimaiaaikdacaaI1aWaaeWaaeaacaaIYaaacaGLOaGaayzkaaaaaa % a!613B! \begin{gathered} BT{\text{ }}Ba \to {n_{Ba{{\left( {HC{O_3}} \right)}_2}}} = 0,125 \hfill \\ BT{\text{ }}C \to 0,075 + 2.0,125 = 3t + 0,025\left( 2 \right) \hfill \\ \end{gathered} \)
(1)(2) → y = 0,225 và t = 0,1
→ y : x = 3
Đề thi thử tốt nghiệp THPT QG môn Hóa năm 2020
Sở GD&ĐT Bắc Ninh
