Dẫn 26,88 lít (đktc) hỗn hợp X gồm hơi nước và khí cacbonic qua than nung đỏ thu được a mol hỗn hợp khí Y gồm CO, H2, CO2; trong đó có V1 lít (đktc) CO2. Hấp thụ hoàn toàn khí CO2 vào dung dịch có chứa 0,06b mol Ca(OH)2, khối lượng kết tủa tạo ra phụ thuộc vào thể tích khí CO2 được ghi ở bảng sau:
| Thể tích khí CO2 (đktc - lít) | V | V+8,96 | V1 |
| Khối lượng kết tủa (gam) | 5b | 3b | 2b |
Giá trị của a gần nhất giá trị nào sau đây:
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Lời giải:
Báo saiĐặc \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaacaGaaiaabeqaamaabaabaaGcbaGaamiEaiabg2 % da9maalaaabaGaamOvaaqaaiaaikdacaaIYaGaaiilaiaaisdaaaaa % aa!3BC4! x = \frac{V}{{22,4}}\)
Giả sử khi \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaacaGaaiaabeqaamaabaabaaGcbaGaamOBamaaBa % aaleaacaWGdbGaam4tamaaBaaameaacaaIYaaabeaaaSqabaGccqGH % 9aqpcaWG4baaaa!3BAC! {n_{C{O_2}}} = x\) thì kết tủa đã bị hòa tan.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaacaGaaiaabeqaamaabaabaaGcbaGaeyOKH4Qaam % OBamaaBaaaleaacaWGdbGaamyyaiaadoeacaWGpbWaaSbaaWqaaiaa % iodaaeqaaaWcbeaakiabg2da9iaaicdacaGGSaGaaGimaiaaiwdaca % WGIbGaae4oaiaabccacaWGUbWaaSbaaSqaaiaadoeacaWGHbWaaeWa % aeaacaWGibGaam4qaiaad+eadaWgaaadbaGaaG4maaqabaaaliaawI % cacaGLPaaadaWgaaadbaGaaGOmaaqabaaaleqaaOGaeyypa0JaaGim % aiaacYcacaaIWaGaaGOnaiaadkgacqGHsislcaaIWaGaaiilaiaaic % dacaaI1aGaamOyaiabg2da9iaaicdacaGGSaGaaGimaiaaigdacaWG % Ibaaaa!5A7C! \to {n_{CaC{O_3}}} = 0,05b{\text{; }}{n_{Ca{{\left( {HC{O_3}} \right)}_2}}} = 0,06b - 0,05b = 0,01b\)
Bảo toàn C → x = 0,05b + 2.0,01b
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaacaGaaiaabeqaamaabaabaaGcbaGaamOBamaaBa % aaleaacaWGdbGaam4tamaaBaaameaacaaIYaaabeaaaSqabaaaaa!399F! {n_{C{O_2}}}\) tăng = 0,4 thì kết tủa giảm 0,02b mol → 0,02b = 0,4 → b = 0,2 → x = 0,14
Khi \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaacaGaaiaabeqaamaabaabaaGcbaGaamOBamaaBa % aaleaacaWGdbGaam4tamaaBaaameaacaaIYaaabeaaaSqabaGccqGH % 9aqpcaWG4bGaey4kaSIaaGimaiaacYcacaaI0aGaeyypa0JaaGimai % aacYcacaaI1aGaaGinaaaa!42A3! {n_{C{O_2}}} = x + 0,4 = 0,54\) thì các sản phẩm gồm CaCO3 (0,03b = 0,006) và Ca(HCO3)2 (0,06b – 0,03b = 0,006)
Bảo toàn C → 0,54 = 0,006 + 0,006.2 : Vô lý
Vậy khi \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaacaGaaiaabeqaamaabaabaaGcbaGaamOBamaaBa % aaleaacaWGdbGaam4tamaaBaaameaacaaIYaaabeaaaSqabaGccqGH % 9aqpcaWG4baaaa!3BAC! {n_{C{O_2}}} = x\) thì kết tủa chưa bị hòa tan.
→ x = 0,05b (1)
Khi \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaacaGaaiaabeqaamaabaabaaGcbaGaamOBamaaBa % aaleaacaWGdbGaam4tamaaBaaameaacaaIYaaabeaaaSqabaGccqGH % 9aqpcaWG4bGaey4kaSIaaGimaiaacYcacaaI0aGaeyOKH4QaamOBam % aaBaaaleaacaWGdbGaamyyaiaadoeacaWGpbWaaSbaaWqaaiaaioda % aeqaaaWcbeaakiabg2da9iaaicdacaGGSaGaaGimaiaaiodacaWGIb % Gaai4oaiaabccacaWGUbWaaSbaaSqaaiaadoeacaWGHbWaaeWaaeaa % caWGibGaam4qaiaad+eadaWgaaadbaGaaG4maaqabaaaliaawIcaca % GLPaaadaWgaaadbaGaaGOmaaqabaaaleqaaOGaeyypa0JaaGimaiaa % cYcacaaIWaGaaG4maiaadkgaaaa!59BB! {n_{C{O_2}}} = x + 0,4 \to {n_{CaC{O_3}}} = 0,03b;{\text{ }}{n_{Ca{{\left( {HC{O_3}} \right)}_2}}} = 0,03b\)
→ x + 0,4 = 0,03b + 0,03b.2 (2)
(1) (2) → a = 0,6; b = 10
Khi \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaacaGaaiaabeqaamaabaabaaGcbaGaamOBamaaBa % aaleaacaWGdbGaam4tamaaBaaameaacaaIYaaabeaaaSqabaGccqGH % 9aqpdaWcaaqaaiaadAfadaWgaaWcbaGaaGymaaqabaaakeaacaaIYa % GaaGOmaiaacYcacaaI0aaaaaaa!3F71! {n_{C{O_2}}} = \frac{{{V_1}}}{{22,4}}\) lít thì \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaacaGaaiaabeqaamaabaabaaGcbaGaamOBamaaBa % aaleaacaWGdbGaamyyaiaadoeacaWGpbWaaSbaaWqaaiaaiodaaeqa % aaWcbeaakiabg2da9iaaicdacaGGSaGaaGOmaiaacUdacaqGGaGaam % OBamaaBaaaleaacaWGdbGaamyyamaabmaabaGaamisaiaadoeacaWG % pbWaaSbaaWqaaiaaiodaaeqaaaWccaGLOaGaayzkaaWaaSbaaWqaai % aaikdaaeqaaaWcbeaakiabg2da9iaaicdacaGGSaGaaGinaaaa!4BC6! {n_{CaC{O_3}}} = 0,2;{\text{ }}{n_{Ca{{\left( {HC{O_3}} \right)}_2}}} = 0,4\)
Bảo toàn \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaacaGaaiaabeqaamaabaabaaGcbaGaam4qaiabgk % ziUkaad6gadaWgaaWcbaGaam4qaiaad+eadaWgaaadbaGaaGOmaaqa % baaaleqaaOGaeyypa0JaaGymaaaa!3E1F! C \to {n_{C{O_2}}} = 1\)
nC = nY – nX = a – 1,2
Bảo toàn electron: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaacaGaaiaabeqaamaabaabaaGcbaGaaGinaiaad6 % gadaWgaaWcbaGaam4qaaqabaGccqGH9aqpcaaIYaGaamOBamaaBaaa % leaacaWGdbGaam4taaqabaGccqGHRaWkcaaIYaGaamOBamaaBaaale % aacaWGibWaaSbaaWqaaiaaikdaaeqaaaWcbeaaaaa!41A4! 4{n_C} = 2{n_{CO}} + 2{n_{{H_2}}}\)
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaacaGaaiaabeqaamaabaabaaGcbaGaeyOKH4Qaam % OBamaaBaaaleaacaWGdbGaam4taaqabaGccqGHRaWkcaWGUbWaaSba % aSqaaiaadIeadaWgaaadbaGaaGOmaaqabaaaleqaaOGaeyypa0JaaG % OmaiaadggacqGHsislcaaIYaGaaiilaiaaisdaaaa!442D! \to {n_{CO}} + {n_{{H_2}}} = 2a - 2,4\)
→ nY = 2a – 2,4 + 1 = a
→ a = 1,4
Đề thi thử tốt nghiệp THPT QG môn Hóa năm 2020
Trường THPT Chuyên Bến Tre lần 2
