1 / 22

oân thi ñaïi hoïc

oân thi ñaïi hoïc. Moân: Hoaù Hoïc. GV. NGUYEÃN TAÁN TRUNG (Trung Taâm Luyeän Thi Chaát Löôïng Cao VÓNH VIEÃN). Boå trôï kieán thöùc hoaù Voâ cô - Ñaïi cöông. Baøi 7. CO 2 phaûn öùng Vôùi dung dòch bazô. CaCO 3 . ?. TH1. TH2. Ca(HCO 3 ) 2. ?. Ca(HCO 3 ) 2.

ken
Download Presentation

oân thi ñaïi hoïc

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. oân thi ñaïi hoïc Moân: Hoaù Hoïc GV. NGUYEÃN TAÁN TRUNG (Trung Taâm Luyeän Thi Chaát Löôïng Cao VÓNH VIEÃN)

  2. Boå trôï kieán thöùc hoaù Voâ cô - Ñaïi cöông

  3. Baøi 7 CO2 phaûn öùng Vôùi dung dòch bazô

  4. CaCO3 ? TH1 TH2 Ca(HCO3)2 ? Ca(HCO3)2 CaCO3 CO2 phaûn öùng vôùi dd Ca(OH)2 • Giaùo khoa CO2+ Ca(OH)2 CaCO3+ H2O (1) Sau (1) coøn CO2 thì: CaCO3 + H2O + CO2 Ca(HCO)3 (2) Toùm laïi: CO2 + Ca(OH)2

  5. nCO2 nCa(OH)2 1 2 • Trong ñònh löôïng: • Phaûn öùng: (1), (2) neân vieát laïi CO2 + Ca(OH)2 CaCO3  + H2O (1’) 2CO2 + Ca(OH)2 Ca(HCO3)2 (2’) • Baûng toùm taét saûn phaåm: (1’), (2’)  baûng TTSP: Ca(HCO3)2 Saûn phaåm CaCO3 Ca(HCO3)2 Ca(OH)2 dö CaCO3 CaCO3 Ca(HCO3)2 CO2 dö

  6. CO2 phaûn öùng vôùi dd Ca(OH)2 • Giaùo khoa CO2+ Ca(OH)2 CaCO3+ H2O (1) Sau (1) coøn CO2 thì: CaCO3 + H2O + CO2 Ca(HCO)3 (2) Toùm laïi: CaCO3 CO2 + Ca(OH)2 ? TH1 TH2 Ca(HCO3)2 ? Ca(HCO3)2 CaCO3

  7. Trong ñònh löôïng: • Phaûn öùng: (1), (2) neân vieát laïi CO2 + Ca(OH)2 CaCO3  + H2O (1’) 2CO2 + Ca(OH)2 Ca(HCO3)2 (2’) • Baûng toùm taét saûn phaåm: (1’), (2’)  baûng TTSP: nCO2 nCa(OH)2 TH2 TH1 1 2 Ca(HCO3)2 Saûn phaåm CaCO3 Ca(HCO3)2 Ca(OH)2 dö CaCO3 CaCO3 Ca(HCO3)2 CO2 dö

  8. nmax = nCO2 hñ nCO2 nCa(OH)2 TH1 TH2 • Ñöôøng bieåu dieãn löôïng keát tuûa löôïng m=ñeà 2 1

  9. Khi cho CO2 vaøo dd Ca(OH)2 thaáy coù , suy ra baøi toaùn coù hai tröôøng hôïp: TH1: Baøi toaùn chæ coù phaûn öùng CO2 + Ca(OH)2 CaCO3 + H2O TH2:Baøi toaùn goàm caùc phaûn öùng CO2 + Ca(OH)2 CaCO3 + H2O (1) CaCO3+ CO2+ H2O Ca(HCO3)2 (2) Hoaëc: CO2 + Ca(OH)2 CaCO3 + H2O (1’) 2CO2 + Ca(OH)2 Ca(HCO3)2 (2’)

  10. 300ml ddCa(OH)2 0,1M 2 caùch giaûi CO2 1,12lit (ñkc) • Aùp duïng 1: Cho 1,12 lít CO2 (ÑKC) vaøo bình chöùa 300 ml dd Ca(OH)2 0,1M. Tính khoái löôïng keát tuûa thu ñöôïc 1,12 lít CO2 (ÑKC) 300 ml dd Ca(OH)2 0,1M Khoái löôïng Keát tuûa:?

  11. 300. 0,1 = 0,03 (mol) 1000 1,12 nCO2 bñ = = 0,05 mol 22,4 • Giaûi: nCa(OH)2 bñ = Caùch 1:Giaûi baèng phöông phaùp 3 doøng

  12. (mol) Bñ: Bñ: 0,03 0,05 0 Pöù: Pöù: O,02 O,03 Sau: Sau: 0,02 0,03 m = 1 g CaCO3 - Theo ñeà ta coù pöù: Ca(OH)2 + CO2  CaCO3+ H2O (1) (mol) 0,03 0,03 (mol) 0 0,02 0,03 CO2 +CaCO3 +H2O  Ca(HCO3)2 (2) (mol) 0,02 0,01 (mol) 0

  13. Ca(HCO3)2 Saûn phaåm CaCO3 Ca(HCO3)2 nCO2 nCO2 Ca(OH)2 dö CaCO3 CaCO3 Ca(HCO3)2 CO2 dö nCa(OH)2 nCa(OH)2 1 2 nCa(OH)2=0,03 0,05  = 1,67 = nCO2 = 0,05 0,03 Caùch 2:Döïa vaøo baûng toùm taét saûn phaåm • Ta coù baûng toùm taét saûn phaåm: • Theo ñeà ta coù: Neân baøi toaùn coù 2 phaûn öùng sau:

  14. CO2 +Ca(OH)2  CaCO3+ H2O (1’) 2CO2 + Ca(OH)2 Ca(HCO3)2 (2’) m = 1 g CaCO3 Deã daøng tính ñöôïc:

  15. NaHCO3 ? Na2CO3 ? Na2CO3 NaHCO3 CO2 phaûn öùng vôùi dd NaOH (hay KOH) • Giaùo khoa CO2+ NaOH NaHCO3(1) Sau (1) coøn NaOHthì: NaHCO3 + NaOH  Na2CO3 + H2O(2) Toùm laïi: CO2 + NaOH

  16. nNaOH nCO2 1 2 • Trong ñònh löôïng: • Phaûn öùng: (1), (2) neân vieát laïi CO2 + NaOH NaHCO3(1’) CO2 + 2NaOH  Na2CO3 + H2O (2’) • Baûng toùm taét saûn phaåm: (1’), (2’)  baûng TTSP: NaHCO3 Saûn phaåm NaHCO3 Na2CO3 CO2 dö NaHCO3 Na2CO3 Na2CO3 NaOH (dö)

  17. 300ml ddNaOH 0,2M 2 caùch giaûi CO2 1,12lit (ñkc) • Aùp duïng 2: Cho 1,12 lít CO2 (ÑKC) vaøo bình chöùa 300 ml dd NaOH 0,2M. Tính khoái löôïng muoái thu ñöôïc 1,12 lít CO2 (ÑKC) 300 ml dd NaOH 0,2M Khoái löôïng muoái:?

  18. 300. 0,2 = 0,06 (mol) 1000 1,12 nCO2 bñ = = 0,05 mol 22,4 • Giaûi: nNaOH bñ = Caùch 1:Giaûi baèng phöông phaùp 3 doøng

  19. (mol) Bñ: Bñ: 0,05 0,06 0 Pöù: Pöù: O,01 O,01 O,05 Sau: Sau: 0,01 0,05 - Theo ñeà ta coù pöù: CO2 + NaOH NaHCO3 (1) (mol) 0,05 0,05 (mol) 0 0,01 0,05 NaOH + NaHCO3 Na2CO3 + H2O (2) 0,01 (mol) 0,04 0,01 0 (mol) (1),(2)  m Muoái =0,04.84 +0,01.106 =4,42 gam

  20. NaHCO3 Saûn phaåm NaHCO3 Na2CO3 nNaOH nNaOH CO2 dö NaHCO3 Na2CO3 Na2CO3 NaOHdö nCO2 nCO2 1 2 nNaOH=0,06 0,06  = 1,2 = nCO2 = 0,05 0,05 Caùch 2:Döïa vaøo baûng toùm taét saûn phaåm • Ta coù baûng toùm taét saûn phaåm: • Theo ñeà ta coù: Neân baøi toaùn coù 2 phaûn öùng sau:

  21. nNaOH=x+2y =0,06   x = 0,04; y = 0,01 m Muoái= 0,04.84 +0,01.106 =4,42 gam nCO2 =x +y = 0,05 Caùc phaûn öùng: CO2 + NaOH NaHCO3(1’) x x x CO2 + 2NaOH  Na2CO3 + H2O (2’) y 2y y Theo (1’), (2’) ,ñeà coù:

  22. Aùp duïngï 3: (Trích ñeà ÑH Sö phaïm TP HCM-2001) Cho V lít khí CO2 ño ôû 54,6oc vaø 2,4 atm haáp thuï hoaøn toaøn vaøo 200 ml dd hh KOH 1M vaø Ba(OH)2 0,75M thu ñöïôc 23,64 g keát tuûa. Tìm V lít? CO2 ño ôû 54,6oc vaø 2,4 atm 200 ml dd hh KOH 1M vaø Ba(OH)2 0,75M V = 1,344 (l) ; 4,256 (l) GV. NGUYEÃN TAÁN TRUNG (Trung Taâm Luyeän Thi Chaát Löôïng Cao VÓNH VIEÃN)

More Related