ngày soạn

GV: Nguyn th Minh Huyn

Gio n t chn 11-

Ngy son: 20/08/2011 Tit1: S IN LY A. Muc tieu : 1/Kien thc : - Biet c cac khai niem ve s ien li . - Hieu nguyen nhan ve tnh dan ien cua dd chat ien li . - Biet c the nao la chat ien li manh . Chat ien li yeu 2/Ky nang : -Ren luyen k nang thc hanh quan sat so sanh . -Ren luyen k nang viet ptl va lam toan . - Dung thc nghiem e nhan biet chat ien li manh , yeu , khong ien li . 3.Thi : Rn luyn tnh cn thn nghim tc trong nghin cu khoa hc B: Chun b: 1. Phng tin Chuan b he thong cau hoi va cac bai tap . Neu van e va am thoai . 2. Thit b: SGK, SBT, Gio n C. Tin trnh bi hc: 1/On nh t chc : Lp 11A3 11A8 S s Ngy dy 2. Kim tra bi c: - Trnh by nh ngha Axit, baz theo thuyt Arniut . Cho v d - Trnh by nh ngha hiroxit lng tnh. Vit phng trnh chng minh Sn(OH)2 l hiroxit lng tnh. 3/ Ging bi mi Hot ng ca gio vin Hot ng ca hc sinh Hot ng 1: Bi 1: GV: Chp ln bng, yu cu HS Vit phng trnh in li ca cc cht trong chp vo v. dd sau: HBrO4, CuSO4, Ba(NO3)2, HClO, Bi 1: HCN. Cho bit cht no l cht in li mnh, Vit phng trnh in li ca cc cht no l cht in li yu. cht trong dd sau: HBrO4, CuSO4, Gii: Ba(NO3)2, HClO, HCN. Cho bit cht no l cht in li mnh, cht HBrO4 H+ + BrO4no l cht in li yu. CuSO4 Cu2+ + SO 2 4 HS: Chp Ba2+ + 2NO 3 Ba(NO3)2 GV: Yu cu 1 HS ln bng gii, cc HClO H+ + ClOHS cn li lm nhp v theo di bi HCN H+ + CNbn lm. HBrO4, CuSO4, Ba(NO3)2 l cht in li GV: Yu cu 1 HS nhn xt, GV mnh.1


Gio n t chn 11-

nhn xt ghi im. Hot ng 2: GV: Chp ln bng, yu cu HS chp vo v. Bi 2: Vit phng trnh in li ca hiroxit lng tnh Al(OH)3. GV: Yu cu HS suy ngh 3 pht, sau gi 1 HS ln bng gii. GV quan st cc HS lm bi. GV: Nhn xt, hng dn li Hot ng 3: Bi 3: Vit phng trnh phn ng xy ra khi cho Al2(SO4)3 tc dng vi NaOH d. GV: Yu cu HS suy ngh , sau gi 1 HS ln bng gii. Cc HS cn li ly nhp ra lm bi v theo di bi bn lm. HS: Ln bng trnh by GV: Nhn xt, hng dn li, lu cho HS phn hiroxit lng tnh. Hot ng 4: Bi 4: Da vo thuyt Arniut. Gii thch NH3 l mt baz. GV: Yu cu 1 HS ln bng gii, cc HS cn li lm nhp v theo di bi bn lm. HS: Ln bng trnh by GV: Nhn xt, hng dn li Hot ng 5: Bi 5: Trong mt dd c cha a mol Ca2+, b mol Mg2+, c mol Cl-, d mol NO 3 . a/ Lp biu thc lin h a, b, c, d. b/ Nu a = 0,01; c = 0,01; d = 0,03 th b bng bao nhiu. HS: Chp GV: Hng dn HS cch gii.HS: Ch nghe ging

HClO, HCN l cht in li yu. Bi 2: Vit phng trnh in li ca hiroxit lng tnh Al(OH)3. Gii: Al(OH)3 Al3+ + 3OHAl(OH)3 H3O+ + AlO 2 Bi 3: Vit phng trnh phn ng xy ra khi cho Al2(SO4)3 tc dng vi NaOH d. Gii: Al2(SO4)3 + 6NaOH 2Al(OH)3 + 3Na2SO4 Al(OH)3 + NaOH NaAlO2 + 2H2O Bi 4: Da vo thuyt Arniut. Gii thch NH3 l mt baz. Gii: NH3 + H2O NH + + OH4 Bi 5: Trong mt dd c cha a mol Ca2+, b mol Mg2+, c mol Cl-, d mol NO 3 . a/ Lp biu thc lin h a, b, c, d. b/ Nu a = 0,01; c = 0,01; d = 0,03 th b bng bao nhiu. Gii: a/ Trong mt dd, tng in tch ca cc cation bng tng in tch ca cc anion, v vy: 2a + 2b = c + d b/ b =c + d 2a 0,01 + 0,03 2.0,01 = = 0,01 2 2

4: Cng c h thng bi hc:- Theo thuyt Arniut, cht no di y l axit? A. Cr(NO3)3 B. HBrO3 C. CdSO4 D. CsOH2


Gio n t chn 11-

- Theo thuyt Arniut, cht no di y l baz? A. Cr(NO3)3 B. HBrO3 C. CdSO4 D. NH3 5. Hng dn v nh: Chun b bi s in li ca nc. pH. Cht ch th axit baz. Ngy son: 20/08/2011 Tit 2: Luyn tp I/Muc tiei bai hoc : 1/Ve kien thc : - Biet c tch so ion cua nc va y ngha cua ai lng nay .Biet c khai niem ve pH va chat ch th axit baz . - Biet cach van dung CT lam toan . 2/Ve k nang : - Van dung tch so ion cua nc e xac nh nong o ion H+ va OHtrong dung dch . - Biet anh gia o axit , baz cua dung dch da vao nong o H+ , OH-, pH, pOH .Giai toan ve pH . *)/Trong tam: nh ngha pH. Tnh pH cua cac dung dch. Giai cac bai toan ve axit baz va tnh pH .II/Chuan b :He thong cau hoi va bai tap *)/Phng phap : am thoai , Neu van e . III. Tin trnh ging dy: 1. n nh t chc: . Lp 11A8 S s Ngy dy 2/ Bi c: - Trnh by khi nim pH. - Tnh pH ca dd HCl 0,01 M v dd KOH 0,001 M 3/ Bi mi Hot ng ca gio vin v hc sinh Hot ng 1: GV: Chp ln bng, yu cu HS chp vo v. Bi 1: Mt dd axit sunfuric c pH = 2. a/ Tnh nng mol ca axit sunfuric trong dd . Bit rng nng ny, s phn li ca axit Kin thc trng tm Bi 1: Mt dd axit sunfuric c pH = 2. a/ Tnh nng mol ca axit sunfuric trong dd . Bit rng nng ny, s phn li ca axit sunfuric thnh ion c coi l hon ton. b/ Tnh nng mol ca ion OH- trong dd . Gii: a/ pH = 2 [H+] = 10-2 = 0,01M3


sunfuric thnh ion c coi l hon ton. b/ Tnh nng mol ca ion OHtrong dd . GV: Yu cu 1 HS ln bng gii, cc HS cn li lm nhp v theo di bi bn lm.GV: Yu cu 1 HS nhn xt, GV nhn xt ghi im. Hot ng 2: GV: Chp ln bng, yu cu HS chp vo v. Bi 2: Cho m gam natri vo nc, ta thu c 1,5 lt dd c pH = 13. Tnh m. GV: Hng dn HS cch gii. HS: Nghe ging v hiu Hot ng 3: GV: Chp ln bng, yu cu HS chp vo v. Bi 3: Tnh pH ca dd cha 1,46 g HCl trong 400,0 ml. HS: Chp GV: Yu cu HS suy ngh , sau gi 1 HS ln bng gii. Cc HS cn li ly nhp ra lm bi v theo di bi bn lm. HS: Ln bng trnh by GV: Yu cu 1 HS nhn xt, GV nhn xt ghi im Hot ng 4: Bi 4: Tnh pH ca dd to thnh sau khi trn 100,0 ml dd HCl 1,00M vi 400,0 ml dd NaOH 0,375M. HS: Chp GV:Hng dn HS cch gii tnh [OH-] GV: Yu cu HS tnh [H ] v pH HS: Tnh [H+] v pH+

H2SO4 2 H + SO 2 4+

Gio n t chn 111 2

[H2SO4] = b/ [OH-] =

1 2

[H+] =

.0,01 = 0,005M

10 14 = 10 12 M 2 10

Bi 2: Cho m gam natri vo nc, ta thu c 1,5 lt dd c pH = 13. Tnh m. Gii: pH = 13 [H+] = 10-13 [OH-] = 10-1 = 0,1M S mol OH- trong 1,5 lt dd bng: 0,1.1,5 = 0,15 (mol) 2Na + 2H2O 2Na+ + 2OH- + H2 S mol Na = s mol OH- = 0,15 ( mol) Khi lng Na = 0,15.23 = 3,45 gam Bi 3: Tnh pH ca dd cha 1,46 g HCl trong 400,0 ml. Gii: CM(HCl) =1,46 1000 . = 0,100 M =10 1 M 36 ,5 400 ,0

[H+] = [HCl] = 10-1M pH = 1,0

Bi 4: Tnh pH ca dd to thnh sau khi trn 100,0 ml dd HCl 1,00M vi 400,0 ml dd NaOH 0,375M. Gii: nNaOH = 0,4.0,375 = 0,15 (mol) nHCl = 0,1.1,000 = 0,10 ( mol) Sauk hi trn NaOH d nNaOH (d) = 0,15 0,10 = 0,05 (mol) S mol NaOH = s mol OH- = 0,05 (mol) [OH-] = [H ] =+

0,05 = 0,1M 0,4 + 0,1

1,0.10 14 =1,0.10 13 M 1,0.10 1

Vy pH = 13

IV. Cng c: pH ca dd CH3COOH 0,1M phi A. nh hn 1 C. bng 7 B. ln hn 1 nhng nh hn 7 D. ln hn 74


Gio n t chn 11-

V. Dn d: Chun b bi phn ng trao i ion trong dd cht in li

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Gio n t chn 11-

Tit 3: S In Ly Ngy son: I/Muc tiei bai hoc : 1/Ve kien thc : Cung co kien thc ve pH va phan ng trao oi xay ra trong dung dch cac chat ien li . 2/Ve k nang : Ren luyen k nang viet phng trnh phan ng di dang ion va ion rut gon . Cach tnh pH cua dd . II/Chuan b : Phieu hoc tap theo noi dung kiem tra bai cu va bai tap luyen tap . *)/Phng phap : Neu va giai quyet van e , am thoai . III. Tin trnh ging dy 1. n nh t chc: Lp 11A8 S s Ngy dy 2/ Bi c: - Trnh by iu kin phn ng trao i ion trong dung dch cc cht in li. - Vit phng trnh phn t v phng trnh ion rt gn ca phn ng sau: NaHCO3 + NaOH 3/ Bi mi Hot ng ca gio vin v hc sinh Hot ng 1: Bi 1: Vit phng trnh dng phn t ng vi phng trnh ion rt gn sau: 2 a/ Ba2+ + CO 3 BaCO3 b/ Fe3+ + 3OH- Fe(OH)3 c/ NH + + OH- NH3 + H2O 4 2d/ S + 2H+ H2S Hot ng 2: Bi 2: Vit phng trnh dng phn t ca cc phn ng theo s sau. a/ MgCO3 + ? MgCl2 + ?. b/ Fe2(SO4)3 + ? K2SO4 + ?. Hot ng 3: Bi 3: Ho tan 1,952 g mui BaCl2.xH2O Kin thc trng tm Gii: a/ Ba(NO3)2 + Na2CO3 BaCO3 + 2NaNO3 b/ Fe2(SO4)3 + 6NaOH 2Fe(OH)3 + 3Na2SO4 c/ NH4Cl + NaOH NH3 + H2O + NaCl d/ FeS + 2HCl FeCl2 + H2S Gii: a/ MgCO3 + 2HCl MgCl2 + H2O + CO2 b/ Fe2(SO4)3 + 6KOH 3K2SO4 + Fe(OH)3

. Gii: BaCl2.xH2O + H2SO4 BaSO4 + 2HCl + 2H2O (1)6

GV: Nguyn th Minh Huyn4

Gio n t chn 11-

trong nc. Thm H2SO4 long, d n Ba SO = 1,864 = 0,008 (mol ) 233 vo dung dch thu c. Kt ta to Theo phng trnh (1) s mol BaSO4 = s thnh c lm kh v cn c 1,864 gam. Xc nh cng thc ho mol BaCl2.xH2O 1,952 hc ca mui. M = 0,008 = 244 GV: Nhn xt, hng dn li 244 208 =2 x= 18 Hot ng 4: CTHH ca mui l : BaCl2.2H2O Bi 4: Bi 4: Trn 250 ml dung dch hn hp HCl Gii: 0,08M v H2SO4 0,01M vi 250 ml S mol HCl ban u = 0,25.0,08 = 0,02 ( mol) dung dch Ba(OH)2 c nng x (M) S mol H2SO4 ban u = 0,25.0,01= 0,0025 thu c m gam kt ta v 500 ml ( mol) dung dch c pH = 12. Hy tnh m Sau khi phn ng dung dch c pH =12 ngha v x. Coi Ba(OH)2 in li hon ton Ba(OH)2 cn d v cc axit phn ng ht. c 2 nc. 2HCl + Ba(OH)2 BaCl2 + 2H2O GV:Yu cu tnh s mol HCl ban 0,02 0,01 u , s mol H2SO4 ban u , vit cc H2SO4 + Ba(OH)2 BaSO4 + 2H2O phng trnh phn ng xy ra. 0,0025 0,0025 0,0025 Khi lng kt ta: m = 0,0025.233 = 0,5825 (gam) Sau khi phn ng dung dch c pH =12 ngha l: [H+] = 10-12M [OH-] = 10-2M S mol OH- trong dung dch = 0,01.0,5 = 0,005 (mol) Ba(OH)2 Ba2+ + 2OHGV: Hng dn HS tnh khi lng 1 S mol Ba(OH)2 cn d = 2 s mol OH- = kt ta, Tnh nng mol ca Ba(OH)2 . 0,0025 (mol) S mol Ba(OH)2 ban u = 0,01 + 0,0025 + 0,0025 = 0,015 (mol) Nng Ba(OH)2 : x = HS: Nghe ging v hiu IV:Cng c: Vit phng trnh phn t v phng trnh ion rt gn ca cc phn ng sau. a/ Pb(NO3)2 + Na2SO4 b/ Pb(OH)2 + H2SO4 V: Dn d: Chun b bi thc hnh s 10,015 = 0,06 ( M ) 0,25

7


Gio n t chn 11-

Tit 4:

Ni t- Phot pho

Ngy son: I/Muc tiei bai hoc : 1/Ve kien thc : Hieu c tnh chat hoa hoc cua Nit . Biet phng phap ieu che nit trong cong nghiep va trong phong th nghiem . Hieu c ng dung cua nit . 2/Ve k nang : Van dung ac iem cau tao phan t cua nit e giai thch tnh chat vat l ,hoa hoc cua nit . Ren luyen k nang suy luan logic va giai bai tap . .Trong tam: Bai tap : Tnh chat hoa hoc va ieu che nit Toan . II/Chuan b : Phieu hoc tap theo noi dung kiem tra bai cu va bai tap luyen tap . III. Tin trnh ging dy: 1/On nh t chc: Lp 11A8 S s Ngy dy 2/ Bi c: Trnh by tnh cht ha hc ca amoniac. 3/ Bi mi Hot ng ca gio vin v hc sinh Kin thc trng tm

8


Gio n t chn 11-

Hot ng 1: Bi 1: Trong mt bnh kn dung tch 10 lt cha 21 gam nit. Tnh p sut ca kh trong bnh, bit nhit ca kh bng 250C. Hot ng 2: Bi 2: Nn mt hn hp kh gm 2 mol nit v 7 mol hiro trong mt bnh phn ng c sn cht xc tc thch hp v nhit ca bnh c gi khng i 4500C. Sau phn ng thu c 8,2 mol hn hp kh. a/ Tnh phn trm s mol nit phn ng . b/ Tnh th tch (kt) kh ammoniac c to thnh. HS: T tnh phn trm s mol nit phn ng, th tch (kt) kh ammoniac c to thnh. Hot ng 3: Bi 3: Cho lng d kh ammoniac i t t qua ng s cha 3,2 g CuO nung nng n khi phn ng xy ra hon ton, thu c cht rn A v mt hn hp kh. Cht rn A phn ng va vi 20 ml dung dch HCl 1 M a/ Vit pthh ca cc phn ng. b/ Tnh th tch nit ( ktc) c to thnh sau phn ng. GV:Yu cu HS ln bng trnh by cu b

.Gii: 21 S mol kh N2: 28 = 0,75 (mol ) p sut ca kh N2: nRT 0,75 .0,082 ( 25 + 273 ) = 1,83 ( atm ) p= V = 10 Bi 2: Gii N2 (k) + 3H2 (k) 2NH3(k) S mol kh ban u: 2 7 0 S mol kh phn ng: x 3x 2x S mol kh lc cn bng: 2-x 7 3x 2x Tng s mol kh lc cn bng: 2 x + 7 3x + 2x = 9 2x Theo ra: 9 2x = 8,2 x = 0,4 a/ Phn trm s mol nit phn ng b/ Th tch (kt) kh ammoniac c to thnh: 2.0,4. 22,4 = 17,9 (lt) Bi 3: Gii a/ pthh ca cc phn ng. 2NH3 + 3CuO N2 + 3Cu + 3H2O (1) Cht rn A thu c sau phn ng gm Cu v CuO cn d . ch c CuO phn ng vi dung dch HCl. CuO + 2HCl CuCl2 + H2O b/ S mol HCl phn ng vi CuO: nHCl = 0,02( mol) Theo (2) s mol CuO d: nCuO = 1/2 s mol HCl = 0,02: 2 = 0,01 (mol) S mol CuO tham gia phn ng (1) = s mol CuO ban u s mol CuO d =tC

0,4.100% = 20% 2

3,2 0,01 = 0,03 (mol ) 80 1 Theo (1), s mol N2= 3

s mol CuO =

.0,03 = 0,01 (mol)9

1 3


Gio n t chn 11-

Th tch kh nit to thnh : 0,01. 22,4 = 0,224 (lt)

IV. Cng c: Amoniac phn ng c vi tt c cc cht trong nhm no sau y. A. HCl, O2, Cl2, CuO, dd AlCl3 B. H2SO4, PbO, FeO, NaOH C. HCl, KOH, FeCl3, Cl2 D. KOH, HNO3, CuO, CuCl2 V Dn d: Chun b tip phn cn li bi Amoniac v mui Amoni

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Gio n t chn 11-

Tit 5: Ni t- phot pho Ngy son: I/Muc tiei bai hoc : 1/Ve kien thc : -HS hieu : Tnh chat hoa hoc cua amoniac va muoi amoni. -HS biet : Phng phap ieu che amoniac trong phong th nghiem va trong cong nghiep . 2/Ve k nang : Da vao cau tao phan t e giai thch tnh chat hoa hoc cua amoniac va muoi amoni . Ren luyen kha nang viet cac phng trnh trao oi ion */Trong tam: Tnh chat hoa hoc cua NH3 , ieu che NH3 Lam cac BT ve amoniac II/Chuan b : Phieu hoc tap theo noi dung kiem tra bai cu va bai tap luyen tap . *)/Phng phap : Neu va giai quyet van e am thoai III: Tin trnh ging dy: Lp 11A8 S s Ngy dy 2/ KIm tra bi c: Trnh by tnh cht ha hc ca Amoniac 3/ Bi mi Hot ng ca gio vin v hc sinh Hot ng 1: GV: Chp ln bng, yu cu HS chp vo v. Bi 1: Khi cho oxit ca mt kim loi ha tr n t dng vi dung dch HNO3 d th to thnh 34,0 g mui nitrat v 3,6 g nc ( khng c sn phm khc ). Hi l oxit kim loi no v khi lng ca oxit kim loi phn ng l bao nhiu Kin thc trng tm Bi 1: Gii: PTHH. M2On + 2nHNO3 2M(NO3)n + nH2O (1) Theo phn ng (1), khi to thnh 1 mol ( tc (A + 62n) g ) mui nitrat th ng thi to thnh n/2 mol ( 9n gam ) nc (A + 62n) g mui nitrat 9n g nc 34,0 g mui nitrat 3,6 g nc A + 62 n 9n Ta c: 34 = 3,6 Gii pt: A = 23n. Ch c nghim n = 1, A = 2311


Gio n t chn 11-

GV: Yu cu HS vit pt v tnh khi lng ca oxit kim loi phn ng HS: Vit pt v tnh khi lng ca oxit kim loi phn ng Hot ng 2: Bi 2: Chia hn hp hai kim loi Cu v Al lm 2 phn bng nhau. + Phn th nht: Cho tc dng vi dung dch HNO3 c ngui thu c 8,96 lt kh NO2 ( ktc) + Phn th hai: Cho tc dng vi hon ton vi dung dch HCl, thu c 6,72 lt kh ( ktc) Xc nh thnh phn phn trm v khi lng ca mi kim loi trong hn hp. Hot ng 3: Bi 3: Cho 12,8 g Cu tc dng vi dung dch HNO3 c, sinh ra kh NO2. Tnh th tch NO2 ( ktc).

Vy kim loi M trong oxit l natri Na2O + 2HNO3 2NaNO3 + H2O (2) Theo phn ng (2) C to ra 18 g nc th c 62 g Na2O phn ng Vy to ra 3,6g nc th c x g Na2O phn ng x = (3,6.62) : 18 = 12,4 (g) Bi 2: Gii Phn th nht, ch c Cu phn ng vi HNO3 c. Cu + 4HNO3 c Cu(NO3)2 + 2NO2 + 2 H2O (1) Phn th 2, ch c Al phn ng vi 2Al + 3HCl AlCl3 + 3H2 (2) Da vo (1) ta tnh c khi lng Cu c trong hn hp l 12,8 g. Da vo (2) ta tnh c khi lng Al c trong hn hp l 5,4 g. % khi lng ca Cu = 70, 33% % khi lng ca Al = 29,67% Bi 3: Gii Cu + 4HNO3 c Cu(NO3)2 + 2NO2 + 2 H2O 0,2 0,4 (mol) 12 ,8 nCu = 64 = 0,2(mol )V NO 2 = 0,4.22 ,4 = 8,96 (l )

IV Cng c: Ha tan 12,8 g kim loi ha tr II trong mt lng va dung dch HNO 3 60% ( d = 1,365g/ml), thu c 8,96 lt ( ktc) mt kh duy nht mu nu . Tn ca kim loi v th tch dung dch HNO3 phn ng l A. Cu; 61,5 ml B. Cu; 61,1 ml C. Cu; 61,2 ml D. Cu; 61,0 ml *VDn d: Chun b tip phn cn li bi Axit v mui nitrat12


Gio n t chn 11-

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Gio n t chn 11-

Tit 6 Ni t- Phot pho Ngy son I/Muc tiei bai hoc : 1/Ve kien thc : -HS hieu : Tnh chat hoa hoc cua muoi amoni. Vai tro quan trong cua muoi amoni trong i song va trong k thuat . -HS biet : Phng phap ieu che muoi amoni . 2/Ve k nang : Da vao cau tao phan t e giai thch tnh chat vat l ,hoa hoc cua muoi amoni . Ren luyen kha nang lap luan logic va kha nang viet cac phng trnh trao oi ion /Trong tam: ieu che va tnh chat hoa hoc cua muoi amoni II/Chuan b : Phieu hoc tap theo noi dung kiem tra bai cu va bai tap luyen III. Tin trnh ging dy 1. n nh t chc Lp 11A8 S s Ngy dy 2/ Bi c: Trnh by tnh cht ha hc ca mui nitrat 3/ Bi mi: Hot ng ca gio vin v hc sinh Hot ng 1: Bi 1: Nhit phn hon ton 27,3 gam hn hp rn gm NaNO3 v Cu(NO3)2, thu c hn hp kh c th tch 6,72 lt ( ktc). Tnh thnh phn % v khi lng ca mi mui trong hn hp X. Hot ng 2: Bi 2: Nung nng 27,3 g hn hp NaNO3 v Cu(NO3)2 ; hn hp kh thot ra Kin thc trng tm Bi 1: Gii: 2NaNO3 2NaNO2 + O2 (1) x 0,5x ( mol) 2Cu(NO3)2 2CuO + 4NO2 + O2 (2) y y 2y 0,5y ( mol) Gi x v y l s mol ca NaNO 3 v Cu(NO3)2 trong hn hp X. Theo cc phn ng (1) v (2) v theo bi ra . Ta c. 85x + 188y = 27,3 0,5x + 2y + 0,5y = 0,3t0 t0

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Gio n t chn 11-

c dn vo 89,2 ml nc th cn d 1,12 l kh(ktc) khng b hp th. ( Lng O2 ha tan khng ng k) a/ Tnh khi lng ca mi mui trong hn hp u. b/ Tnh nng % ca dd axt. HS: Chp GV: Hng dn HS cch gii, yu cu HS ln bng trnh by Hot ng 3: GV: Chp ln bng, yu cu HS chp vo v. Bi 3: Nung mt lng mui Cu(NO3). Sau mt thi gian dng li, ngui v em cn th thy khi lng gim i 54g. + Khi lng Cu(NO3) b phn hy. + S mol cc cht kh thot ra l

x = y = 0,1 85 .0,1.100 % = 31,1% % m NaNO = 27 ,3 188 .0,1.100 % = 68,9% % mCu ( NO ) = 27 .3 Bi 2: Gii 2NaNO3 2NaNO2 + O2 (1) 2 1 ( mol) 2Cu(NO3)2 2CuO + 4NO2 + O2 (2) 2 4 1 ( mol) 4NO2 + O2 + 2H2O 4 HNO3 (3) 4 1 4 ( mol) a/ Theo pt (1), (2), (3) , nu cn d 1,12 l kh ( hay 0,05 mol ) th l kh O2, c th coi lng kh ny do mui NaNO3 phn hy to ra T (1) ta c: n NaNO = 2.0,05 = 0,1(mol )3

3 2

t0

t0

3

m NaNO 3 = 0,1.85 = 8,5( g ) mCu ( NO3 ) 2 = 27 ,3 8,5 = 18,8( g ) nCu ( NO3 ) 2 = 18,8 : 188 = 0,1(mol )

T (2) ta c:nO2

( Cc kh ny hp th vo nc) T (3) ta c : n HNO = n NO = 0,2(mol ) Khi lng HNO3 l: 0,2.63 = 12,6 (g) Khi lng ca dung dch = 0,2.46 + 0,05.32 + 89,2 = 100 (g) C% ( HNO3) = 12,6 % Bi 3: Gii 2Cu(NO3)2 2CuO + 4NO2 + O2 + C 188g mui b phn hu th khi lng gim : 188 80 = 108 (g) Vy x = 94 g mui b phn hu th khi lng gim 54 g Khi lng mui b phn hu3 2

0,1 .4 = 0,2( mol ) 2 0,1 = .1 = 0,05 (mol ) 2 n NO 2 =

t0

mCu ( NO3 ) 2 = 94 ( g )

+ nCu ( NO ) = 94 : 188 = 0,5(mol )0,5 n NO 2 = .4 = 1( mol ) 2 0,5 nO2 = . = 0,25 (mol ) 23 2

15


Gio n t chn 11-

IV Cng c: Nung nng 66,2 g Pb (NO3)2 thu c 55,4 g cht rn. Hiu sut ca phn ng phn hy l. A. 96% B. 50% C. 31,4% D. 87,1% V. Dn d: Chun b bi Axit photphoric v mui photphat Tit 7: Ni t- phot pho Ngy son I/Muc tiei bai hoc : 1/Ve kien thc : Hieu c tnh chat vat l , hoa hoc cua axit nitric va muoi nitrat . Biet phng phap ieu che axit nitric trong phong th nghiem va trong cong nghiep . 2/Ve k nang : Ren luyen k nang viet phng trnh phan ng oxi hoa kh va phan ng trao oi ion . Ren luyen k nang giai bai tap ve axit nitric . */Trong tam: Tnh chat hoa hoc va ieu che HNO3 trong cong nghiep va giai cac bai toan ve axit nitric . II/Chuan b : Phieu hoc tap theo noi dung kiem tra bai cu va bai tap luyen tap . *)/Phng phap : Neu va giai quyet van e am thoai III. Tin trnh ging dy 1. n nh t chc Lp 11A8 S s Ngy dy 2/ Bi c: Trnh by tnh cht ha hc ca axit photphoric v mui photphat 3/ Bi mi Hot ng ca gio vin v hc sinh Hot ng 1: Bi 1: Cho 11,76 g H3PO4 vo dung dch cha 16,8 g KOH. Tnh khi lng ca tng mui thu c sau khi cho dung dch bay hi n kh Hot ng 2: Bi 2: Bng phng php ha hc, hy Kin thc trng tm Bi 1: Gii: H3PO4 + KOH KH2PO4 + H2O (1) H3PO4 + 2KOH K2HPO4 + 2H2O (2) H3PO4 + 3KOH K3PO4 + 3H2O (3) S mol H3PO4 0,12 (mol) S mol KOH 0,3 (mol) Da vo t l s mol gia KOH v H3PO4 12,72 g K3PO4 v 10,44g K2HPO416


Gio n t chn 11-

phn bit dung dch HNO3 v dung dch H3PO4 Hot ng 3: Bi 3: Bng phng php ha hc phn bit cc mui: Na3PO4, NaCl, NaBr, Na2S, NaNO3. Nu r hin tng dng phn bit v vit phng trnh ha hc ca cc phn ng Hot ng 4: Bi 4: Cho 62 g canxi photphat tc dng vi 49 g dung dch H2SO4 64%. Lm bay hi dung dch thu c n cn kh th c mt hn hp rn, bit rng cc phn ng u xy ra vi hiu sut 100%

Bi 2: Gii Cho mnh kim loi Cu vo dung dch ca tng axit Cu + HNO3 () Cu(NO3)2 + 2NO2 + 2H2O Cu khng t dng vi H3PO4 Bi 3: Gii Dng dung dch AgNO3 phn bit cc mui: Na3PO4, NaCl, NaBr, Na2S, NaNO3. Ly mi mui mt t vo tng ng nghim, thm nc vo mi ng v lc cn thn ha tan ht mui. Nh dung dch AgNO3 vo tng ng nghim - dung dch no c kt ta mu trng khng tan trong axit mnh, th l dung dch NaCl NaCl + AgNO3 AgCl + NaNO3 - dung dch no c kt ta mu vng nht khng tan trong axit mnh, th l dung dch NaBr. NaBr + AgNO3 AgBr + NaNO3 - dung dch no c kt ta mu en, th l dung dch Na2S Na2S + 2AgNO3 Ag2S + 2NaNO3 - dung dch no c kt ta mu vng tan trong axit mnh, th l dung dch Na3PO4 Na3PO4 + 3AgNO3 Ag3PO4 + 3NaNO3 Bi 4: Gii Ca3(PO4)2 + H2SO4 2CaHPO4 + CaSO4 (1) Ca3(PO4)2 + 2H2SO4 Ca(H2PO4)2 + 2CaSO4 (2) Ca3(PO4)2 + 3H2SO4 H3PO4 + 3CaSO4 (3) S mol Ca3(PO4)2 = S mol H2SO4 =62 = 0,2(mol ) 310 49 .64 = 0,32 (mol ) 100 .98

V t l s mol H2SO4 v Ca3(PO4)2 l 1,6 Nn xy ra phn ng (1) v (2). Gi a v b l s mol Ca3(PO4)2 tham gia cc phn ng (1) v (2) Ta c h pt: a + 2b =0,32 a + b = 0,217


Gio n t chn 11-

a = 0,08; b = 0,12 IV. Cng c: Dung dch H3PO4 c cha cc ion ( khng k ion H+v OH- ca nc) A. H+, PO 3 B. H+, PO 3, H2PO 4 4 4 3 3 + + B. H , PO 4 , HPO 4 D. H , PO 4 , H2PO , HPO 4 4 V Dn d: Chun b bi Luyn tp trang 5

Tit 8

Ni t- Photpho

Ngy son: I/Muc tieu can at: 1/Ve kien thc : Biet cau tao phan t cua axitphotphoric . Tnh chat hoa hoc cua axitphotphoric . Biet tnh chat va nhan biet moi photphat. Biet ng dung va ieu che axitphotphoric . 2/K nang :Van dung kien thc e giai bai tap va lam BT ve axit phot phoric . */Trong tam: BT ve axit va muoi - Tnh chat hoa hoc , ieu che axit photphoric va nhan biet ion PO43II. Chuan b : Phieu hoc tap theo noi dung kiem tra bai cu va bai tap luyen tap . */Phng phap : Neu va giai quyet van e am thoai III. Tin trnh ging dy: 1. n nh t chc: Lp 11A8 S s Ngy dy 2/ Bi c Cn bng phng trnh phn ng sau : R + HNO3 R(NO3)n + NO2 + H2O18


Gio n t chn 11-

3/ Bi mi

19

Hot ng ca gio vin v hc sinh GV: Nguyn th Minh Huyn Hot ng 1: Bi 1: Mt lng 8,32 g Cu tc dng va vi 240 ml dd HNO3 , cho 4,928 l ( o ktc) hn hp gm hai kh NO v NO2 bay ra. + Tnh s mol ca NO v NO2 to ra l + Tnh nng mol/l ca dd axt ban u l Bi 1: Gii:

Kin thc trng tmGio n t chn 11-

Cu + 4HNO3 Cu(NO3)2 + 2NO2 + 2H2O x 4x 2x 3Cu + 8HNO3 3Cu(NO3)2 + 2NO + 4H2O y 8/3y 2y Theo bi ra ta c: ( x + y ).64 = 8,32 (1) 2x +2 y 3

=

4,928 = 0,22 22 .4

(2)

Gii (1) v (2) c x = 0,1; y = 0,03 a/ S mol ca NO2 l 2.0,1 = 0,2 (mol) S mol ca NO l8 .0,03 32 3

Hot ng 2: Bi 2: Mt lng 13,5 g Al tc dng va vi 2,2 l dd HNO3 cho bay ra mt hn hp gm hai kh NO v N2O. Bit t khi ca hn hp kh so vi hir bng 19,2. + Tnh s mol ca NO v N2O to ra l + Tnh nng mol/l ca dd axt u.

.0,03 = 0,02 (mol)

b/ Tng s mol HNO3 phn ng = 4.0,1 + = 0,48 (mol)0,48 = 2( M ) 0,24

Nng mol/l ca dung dch axitC M ( HNO 3 ) =

Bi 2:

Gii Al(NO3)3 + NO + 2H2O Al + 4HNO3 x 4x 2x (mol) 8Al(NO3)3 + 3N2O + 8Al + 30HNO3 Hot ng 3: 15H2O Bi 3: y 30/8y (= 3,75 y) 3/8y (= 0,375 Nung 9,4 gam mt mui nitrat trong y) mt bnh kn. Sau khi phn ng xy Theo bi ra ta c: ( x + y ).27 = 13,5 (1) ra hon ton cn li 4 gam oxit. Tm 30 .x + 44 .0,375 y cng thc ca mui nitrat d hh / H = =19 ,2 (2) ( x + 0,375 y )2 HS: Chp GV: Yu cu 1 HS ln bng trnh by. Cc HS cn li lm v theo di Gii (1) v (2) c x = 0,1; y = 0,4 a/ S mol ca NO l = 0,1 (mol) bi ca bn S mol ca N2O l 0,375.0,4 = 0,15 (mol) b/ Tng s mol HNO3 phn ng = 4.0,1 + HS:Ln bng trnh by 3,75.0,4 = 1,9 (mol) GV: Gi HS nhn xt, ghi im Nng mol/l ca dung dch axit2

C M ( HNO 3 ) =

1,9 = 0,86 ( M ) 2,2

Bi 3: Gii 2R(NO3)2 R2On + 2nNO2 + n/2O2 a a/2 na na/4 20 Ta c: a.( MR + 62n) = 9,4 (1) 0,5.a( 2MR + 16n) = 4 (2) Ly (1) : (2) ta c M = 32n. Khi n = 2 tht0


Gio n t chn 11-

IV. Cng c: + Nhit phn hn hp gm 2 mui KNO3 v Cu(NO3)2 c khi lng l 95,4 gam. Khi phn ng xy ra hon ton thu c hn hp kh c M = 37,82. Vy khi lng mi mui trong hn hp ban u l A. 20 v 75,4 B. 20,2 v 75,2 C. 15,4 v 80 D. 30 v 65,4 + Dung dch HNO3 long tc dng vi hn hp Zn v ZnO to ra dd c cha 8 g NH4NO3 v 113,4 g Zn(NO3)2. Khi lng ca Zn v ZnO trong hn hp l A. 26; 16,2 B. 27; 23,2 C. 28; 22,2 D. 23; 24,2 V Dn d: Chun b bi Thc hnh s 2

21


Gio n t chn 11-

Tit 9: Ni t- photpho Ngy son: I/Muc tieu : 1/Ve kien thc : -Cung co cac kien thc ve tnh chat vat l, tnh chat hoa hoc , ieu che va ng dung cua nit , photpho va mot so hp chat cua chung . -Bai tap on tap chng III 2/K nang : Van dung kien thc c ban giai bai tap *) Trng tm: -Cung co cac kien thc ve tnh chat vat l, tnh chat hoa hoc , ieu che va ng dung cua nit , photpho va mot so hp chat cua chung . II/Chun b: : bai tap SGK, bai tap on tap. III/Tin trnh ging dy: 1. n nh t chc Lp 11A8 S s Ngy dy 2/ Bi mi: Hot ng ca gio vin v hc sinh Hot ng 1: Bi 1: Cho 3 mol N2 v 8 mol H2 vo mt bnh kn c th tch khng i cha sn cht xc tc ( th tch khng ng k ). Bt tia la in cho phn ng xy ra, sau a v nhit ban u th thy p sut gim 10% so vi p sut ban u. Tm % v th tch ca N2 sau phn ng. Hot ng 2: Bi 2: Khi ha tan hon ton 1,5875 gam mt kim loi ha tr III trong dung dch HNO3 long thu c 604,8 ml hn hp kh N2 v NO (ktc) c t khi hi so vi H2 l 14,5. Tm tn M Hot ng 3: GV: Chp ln bng, yu cu HS Kin thc trng tm Bi 1: Gii: N2 + 3H2 2NH3 Trc phn ng 3 8 0 ( mol) Phn ng x 3x Sau phn ng 3 x 8 - 3x 2x S mol kh trc phn ng n1= 11 (mol) S mol kh sau phn ng n2= 11 2x (mol) Do bnh kn nn p sut t l vi s mol, ta cn1 P1 11 P 1 = = = x = 0,55 n 2 P2 11 2 x 0,9 P 0,9%N2 = 3 0,55 .100 % = 24,75% 11 2.0,55

Bi 2: Gii M + 4HNO3 M(NO3)3 + NO + 2H2O x 4x 2x (mol) 10M+ 36HNO3 10M(NO3)3 + 3N2 + 18H2O y 3/10y22


Gio n t chn 11-

chp vo v. Bi 3: NH4Cl (1) NH3 (2) N2 (3) NO ( ( ( 4) NO2 5) HNO3 6) NaNO3 ( 7) NaNO3 Hot ng 4: Bi 4: Cho 500ml dung dch KOH 2M vo 500ml dung dch H3PO4 1,5M. Sau phn ng trong dung dch thu c cc sn phm no GV: Yu cu 1 HS ln bng trnh by. Cc HS cn li lm v theo di bi ca bn HS:Ln bng trnh by GV: Gi HS nhn xt, ghi im

Theo bi ra ta c: x + (1)d hh / H 2 =

3 y 10

= 0,27

30 .x + 28 .

3 y 10 = 14 ,5.2 3 x+ y 10

(2) Gii (1) v (2) c x = 0,0135; y = 0,045 S mol ca M l 0,045 + 0,0135 = 0,0585 (mol)M = 1,5875 = 27 0,0585

Vy M l Al Bi 3: NH4Cl (1) NH3 (2) N2 (3) NO ( ( ( 4) NO2 5) HNO3 6) NaNO3 ( 7) NaNO3 Gii 1/ NH4Cl + NaOH NH3 + H2O + NaCl 2/ NH3 + 3O2 2 N2 + 6H2O 3/ N2 + O2 2NO 4/ 2NO+ O2 2NO2 5/ 4NO2 + 2H2O + O2 4 HNO3 6/ HNO3 + NaOH NaNO3 + H2O 7/ 2NaNO3 2NaNO2 + O2 Bi 4: Cho 500ml dung dch KOH 2M vo 500ml dung dch H3PO4 1,5M. Sau phn ng trong dung dch thu c cc sn phm no Giit t t

S mol ca NaOH = 0,5.2 =1 (mol) S mol H3PO4 = 0,5.1,5 = 0,75 (mol) T l 1/0,75 = 1,333 nn to hai mui NaH2PO4 , Na2HPO4

IV:Cng c:

23


Gio n t chn 11V N 2O V NO 1 4

Ha tan 4,59 g Al bng dung dch HNO3 long thu c hn hp kh NO v N2O c t khi i vi H2 bng 16,75. T l th tch kh A.1 3

trong hn hp l D.3 4

B.

2 3

C.

V: Dn d: Chun b bi Cacbon

Tit 10: Cacbon- Silic( lay sang phan co ban) Ngy son: I/Muc tieu : 1/Ve kien thc : Hoc sinh biet -Cau tao cua phan t CO va CO2 -Tnh chat vat l va hoa hoc cua CO va CO2 -Cac phng phap ieu che CO va CO2 2/K nang : -Cung co kien thc ve lien ket hoa hoc -Van dung kien thc e giai thch tnh chat. Va lam BT . */Trong tam: Nam c tnh chat hoa hoc va phng phap ieu che , biet cach van dung lam cac BT ve cac hp chat cua cacbon II/Chuan b : Phieu hoc tap theo noi dung kiem tra bai cu va bai tap luyen tap . */Phng phap : Neu va giai quyet van e am thoai Hoat ong nhom III. Tin trnh ging dy 1. n nh t chc Lp 11A8 S s Ngy dy 2/ Bi c: Trnh tnh cht ca mui cacbonat 3/ Bi mi Hot ng ca gio vin v hc sinh Hot ng 1 Bi 1: Kin thc trng tm Bi 1: Gii:24


Gio n t chn 110

Nung 52,65 g CaCO3 1000 C v cho ton b lng kh thot ra hp th ht vo 500 ml dung dch NaOH 1,8 M. Khi lng mui to thnh l ( Hiu sut ca phn ng nhit phn CaCO3 l 95% ) Hot ng 2 Bi 2: xc nh hm lng cacbon trong mt mu thp khng cha lu hunh, ngi ta phi t mu thp trong oxi d v xc nh CO2 to thnh. Hy xc nh hm lng cacbon trong mu thp X, bit rng khi t 10g X trong oxi d ri dn ton b sn phm qua nc vi trong d th thu c 0,5 g kt ta Hot ng 3 Bi 3: C a gam hn hp bt X gm CuO, Al2O3 . Ngi ta thc hin cc th nghim sau: Th nghim 1: Cho X phn ng hon ton vi dung dch HCl, c cn dung dch thu c 4,02 g cht rn khan. Th nghim 2: Cho X phn ng va vi bt cacbon nhit cao th thu c 0,112 lt kh (kt) Tnh a ?

CaCO3

t 0C

CaO + CO252 ,65 = 0,5265 ( mol ) 100

nCO 2 = nCaCO 3 =

V phn ng trn c h = 95% nn s mol CO2 thc t thu cnCO 2 = 0,5265 .95 = 0,5002 ( mol ) 100

nNaOH = 0,5.1,8 = 0,9 (mol) T l s mol NaOH v CO2= Trnh by khi nim ng ng, ng phn. V d minh ha Hot ng ca gio vin v hc sinh Hot ng 2: Bi 1: Trong cc cht di y, cht no l ng ng ca nhau? cht no l ng phn ca nhau? 1. CH3CH2CH3 2. CH3CH2CH2Cl 3. CH3CH2CH2CH3 4. CH3CHClCH3 5. (CH3)2CHCH3 6. CH3CH2CH=CH2 7. CH3CH=CH2 8. CH2-CH2 CH2-CH2 9. CH3 C=CH2 CH3 Hot ng 3: Bi 2: Khi t chy 1,5 g ca mi cht A hoc B hoc D u thu c sn phm gm 0,9 g nc v 2,2 g kh CO2. Ba cht trn c phi l ng phn ca nhau khng? Cho v d. Hot ng 4: Bi 3: Hn hp kh A cha hai hirocacbon k tip nhau trong mt dy ng ng. Ly 1,12 lt A (ktc) em t chy hon ton. Sn phm chy c Kin thc trng tm Bi 1: Gii: + Cc cht ng ng: (1) v (3); (1) v (5); (6) v (7); (7) v (9) + Cc cht ng phn: (2) v (4); (3) v (5); (6) v (7); (6), (8) v (9) Bi 2: Gii V cc cht c cng s mol C ( cng khi lng CO2), cng s mol H ( cng khi lng nc) v cng s mol oxi trong cng mt lng mi cht c ngha l 3 cht c cng thc n gin ging nhau. Nu 3 cht c cng phn t khi na th chng mi l ng phn ca nhau. V d: Ba cht l axit axetic C2H4O2, glucoz C6H12O6 v anehitfomic khng phi l ng phn ca nhau mc d u c cng thc n gin l CH2O; khi t 30 g mi cht u sinh ra 1 mol CO2 v 1 mol nc. Bi 3: Gii Hai hirocacbon k tip nhau trong dy ng ng c CTPT l CxHy v Cx+1Hy + 2 Gi a l s mol CxHy Gi b l s mol Cx+1Hy + 2 Ta c: a + b = 0,05 (1) CxHy + aC x +1 H y +2

y y (x + )O 2 xCO 2 + H 2 O 4 2

axy + (x + + 1,5)O 42

y/2a (x + 1)CO2

+

y +2 H 2O 2

34


Gio n t chn 11-

dn qua bnh (1) ng H2SO4 (c), sau qua bnh (2) ng dung dch NaOH ( c d). Sau th nghim, khi lng bnh (1) tng 2,16 g v bnh (2) tng 7,48g. Hy xc nh CTPT v % v th tch ca tng cht trong hn hp A.

b

y +2 b 2

(x + 1)b

S mol CO2: ax + b(x + 1) = 0,17 (2) S mol H2O:ay + b(y + 2) = 0,12 2

(3)

T (2) ta c (a + b)x + b =0,17 b = 0,17 - 0,05x b l s mol a mt trong hai ht nn 0 < b < 0,05 Do 0 < 0,17 0,05x < 0,05

b =0,17 (0,05.3)=0,02 a =0,05 0,02 = 0,03 Thay gi tr ca a v b vo (3) ta c: 0,03y + 0,02( y + 2) = 0 y = 4 CTPT ca 2 cht l C3H4, C4H6 % v th tch ( cng l % v s mol) caC3H4 trong hn hp A.0,03 .100% = 60% 0,05

2,4 < x < 3,4 x = 3

% v th tch ca C4H6 trong hn hp l 40% IV: Cng c: Cht no di y l ng phn ca CH3COOCH3? A. CH3CH2OCH3 B. CH3CH2COOH C. CH3COCH3 D. CH3CH2CH2OH V:Dn d: Chun b bi Luyn tp BTVN: Hn hp M th lng,cha 2 hp cht hu c k tip nhau trong mt dy ng ng. Nu lm bay hi 2,58g M th th tch hi thu c ng bng th tch ca 1,4 g kh N2 cng iu kin. t chy hon ton 6,45 g M th thu c 7,65 g H2O v 6,72 lt CO2(ktc). Xc nh CTPT v % khi lng ca tng cht trong hn hp M.

35


Gio n t chn 11-

Tit 15 i cng v ha hu c Ngy son: I.Mc tiu: HS vn dng c kin thc hc gii bi tp .Trng tm: n tp cc bi tp tnh pH, lin quan ti HNO3, nhn bit, s phn ng II. Chun b: GV:Gio n HS: n tp l thuyt, lm bi tp tnh pH, lin quan ti HNO 3, nhn bit, s phn ng III:Tin trnh ging dy: 1/ n nh lp Lp 11A8 S s Ngy dy 2:Bi c: Trnh by tnh oxi ha ca HNO3. Cho v d minh ha 3. Bi mi: Hot ng ca gio vin v hc Kin thc trng tm sinh Hot ng 2: Bi 1: Bi 1: Gii: Trn 200 ml dung dch HCl 0,1 M Nng cc cht sau khi pha trn vi 800ml dung dch HNO3 0,01M. C = 200 .0,1 = 0,02 M HCl 200 + 800 Tnh pH ca dung dch thu c 800 .0,01 Hot ng 3: C HNO = = 0,008 M 200 + 800 GV: Chp ln bng, yu cu HS Phng trnh in li chp vo v. HCl H+ + ClBi 2: 0,02 0,02 (M) Ha tan hon ton 5,6 g Fe vo dung H+ + NO3HNO3 dch HNO3 c nng, thu c V lt 0,008 0,008 (M) kh (ktc). Tm V Tng nng ion H+ = 0,028M Hot ng 4 pH = -lg0,028 =1,55 Bi 3: Bi 2: Nhit phn 66,2 gam Pb(NO3)2 thu Gii c 55,4 gam thu c 55,4 gam Fe + 6HNO3 Fe(NO3)3 + 3NO2 + 3H2O cht rn. Tnh hiu sut ca phn ng 0,1 0,3 phn hy. (mol) Hot ng 5: 5,6 n Fe = = 0,1(mol) n NO = n Fe = 0,3(mol) Bi 4: 56 Ch dng mt ha cht duy nht VNO = 0,3.22,4 = 6,72(l) phn bit cc l mt nhn ng cc Bi 3: dung dch sau: NaCl, Na2SO4, Gii NH4Cl, (NH4)2SO4 Pb(NO3)2 PbO+ 2NO2 + 1/2O2 Hot ng 6: x 2x x/23 2 2

36


Gio n t chn 11-

Bi 5: Hon thnh s phn ng Gi x l s mol Pb(NO3)2 nhit phn sau. Khi lng kh thot ra = 2x.46 + 0,5x.32 = NaOH A B C D NaNO3 66,2 55,4 = 10,8 x = 0,1 (mol) . Hiu sut ca phn ng l:30000 C

H = 0,2 .100% = 50% Bi 4:Gii - Trch mi l ra mt t lm mu th - Cho Ba(OH)2 ln lt vo cc mu th + Mu th khng c hin tng: dung dch NaCl + Mu th c kt ta trng : dung dch Na2SO4 Na2SO4 + Ba(OH)2 BaSO4 + 2NaOH + Mu th c kh mi khai : dung dch NH4Cl 2NH4Cl + Ba(OH)2 BaCl2 + 2NH3 + 2H2O + Mu th c kt ta trng, c kh mi khai : dung dch (NH4)2SO4 (NH4)2SO4 + Ba(OH)2 BaSO4 + 2NH3 + 2H2O Bi 5: N2 + O2 2NO 2NO + O2 2NO2 4NO2 + O2 + 2H2O 4HNO3 HNO3 + NaOH NaNO3 + H2O Vy A l N2, B l NO, C l NO2, D l HNO30 30 00 C

0,1

IV: Cng c: - Cc bi tp tnh pH, lin quan ti HNO3, nhn bit, s phn ng - Cho 4,8 gam Cu kim loi vo dung dch HNO3 long d n khi phn ng xy ra hon ton. Lng kh thot ra iu kin chun l A. 2,24 lt B. 6,72 lt C. 1,12 lt D. 3,36 lt + 3+ - nhn bit s c mt ca 3 ion Fe , NH 4 , NO 3 c trong dung dch ta c th dng cht no sau y A. NaOH B. H2SO4 C. Qu tm D. CaO t0 - Cho phn ng:2NH4Cl + Ca(OH)2 CaCl2 + A +2H2O. A l cht kh no di y A. N2 B. NH3 C. H2 D. N2O V:Dn d: n tp chun b kim tra hc k I

37


Gio n t chn 11-

Tit 16: Hirocacbon no Ngy son: I/Muc tieu : 1/Ve kien thc : Hoc sinh biet : -Goi ten cac ankan mach chnh khong qua 10 C Hoc sinh hieu : -Tnh chat vat l, tnh chat hoa hoc, phng phap ieu che va ng dung cua ankan 2/K nang : Viet CTPT, cong thc cau tao , phng trnh cua ankan va lam toan . */Trong tam: Cac dang BT ly thuyet + Toan lien quan en ankan . II/Chuan b : Phieu hoc tap theo noi dung kiem tra bai cu va bai tap luyen tap . */Phng phap : Neu va giai quyet van e am thoai Hoat ong nhom III: Tin trnh ging dy: 1. n nh t chc Lp 11A8 S s Ngy dy 2.KtBi c: Trnh by cch gi tn mch cacbon phn nhnh. Gi tn CTCT sau CH3 CH - CH2 CH CH2 CH2 CH3 CH3 C2H5 3. Bi mi Hot ng ca gio vin v hc Kin thc trng tm sinh Hot ng 2: Bi 1:38


Gio n t chn 11-

Bi 1: Gi tn cc CTCT sau CH3 CH2 CH CH2 CH3 CH CH3 CH3 CH3 CH3 CH2 CH CH2 CH CH3 CH CH3 CH3 CH3 Hot ng 3: Bi 2: Vit CTCT thu gn ca a/ 4-etyl-2,3,3-trimetylheptan b/ 3,5-ietyl-2,2,3-trimetyloctan

Gii: + 3-etyl -2-metylpentan. + 4-etyl-2,2,5-trimetylhexan Bi 2: Gii a/ CH3 CH3 CH C CH CH2 CH2 CH3 CH3 CH3 CH2 CH3 b/

CH3 CH3 CH3 C C - CH2 CH CH2 CH2 CH3 Hot ng 4: CH3 CH2 CH2 Bi 3: CH3 CH3 Cht A l mt ankan th kh. Bi 3: t chy hon ton 1,2 lt A cn Gii 3n +1 dng va ht 6 lt oxi cng iu C H O2 t nCO2 + (n+1)H2O n 2n + 2 + 2 kin. 1,2lt 6 lt a/ Xc nh CTPT ca A. b/ Cho cht A tc dng vi kh clo 6 = 5 n = 3 250C v c nh sng. Hi c th 3n +1 = 1,2 2 thu c my dn xut monoclo ca CTPT ca A l C3H8 A.Cho bit tn ca mi dn xut . CH3 CH2 Dn xut no thu c nhiu hn. CH2 - Cl Hot ng 5: CH3 CH2 CH3 + Cl2 a s 1- clopropan (43%) Bi 4: + HCl t chy hon ton 1,45 gam mt CH3 CHCl ankan phi dng va ht 3,64 lt O2( CH3 ktc) 2- clopropan (57%) a/ Xc nh CTPT ca ankan Bi 4: Gii b/ Vit CTCT v gi tn tt c cc 3n +1 CnH2n + 2 + 2 O2 t nCO2 + (n+1)H2O ng phn ng vi cng thc . HS: Chp 3n +1 (mol) GV: Gi hng dn HS cch gii (14n + 2)g 2 HS: Ln bng trnh by 1,45 g 0,1625 (mol)14 n + 2 3n +1 = n = 4 1,45 2.0,1625

CTPT ca A l C4H10 CH3 CH2 CH2 CH3 Butan CH3 CH CH3 CH3 Isobutan (2-metylpropan) IV: Cng c:39


Gio n t chn 11-

Nhc li cch gi tn mch cacbon phn nhnh. Cho tn gi vit CTCT V: Dn d: Chun b bi Xicloankan BTVN: t chy hon ton 2,86 g hn hp gm hexan v octan ngi ta thu c 4,48 lt CO2 ( ktc). Xc nh % v khi lng ca tng cht trong hn hp.

Tit 17: Hirocacbon no Ngy son: I/Muc tieu can at: 1/Ve kien thc :HS biet -Cau truc ong phan danh phap cua mot so monoxicloankan. -Tnh chat vat l, tnh chat hoa hoc va ng dung cua xicloankan. 2/K nang : Viet phng trnh phan ng chng minh tnh chat hoa hoc cua xicloankan. Lam cac BT lien quan ve xiclo ankan . */Trong tam: Tnh chat hoa hoc va phng phap ieu che xicloankan Cac dang toan ve xicloankan40


Gio n t chn 11-

II/Chuan b : Phieu hoc tap theo noi dung kiem tra bai cu va bai tap luyen tap . *)/Phng phap : Neu va giai quyet van e am thoai Hoat ong nhom III: Tin trnh ging dy: 1. n nh t chc: Lp 11A8 S s Ngy dy 2.ktBi c: Trnh by tnh cht ha hc ca xicloankan. Ly v d minh ha 3.Bi mi: Hot ng ca gio vin v hc sinh Hot ng 2: Bi 1: Gi tn cc CTCT sauC2H5 H3C

Kin thc trng tm Gii: 4-etyl-1,2-imetylxiclohexan Bi 2: Gii a/

CH3

Hot ng 3: CH3 Bi 2: Vit CTCT thu gn ca CH3 a/ 1,1-imetylxiclopropan b/ b/ 1-etyl-1-metylxiclohexan Hot ng 4: CH3 CH2CH3 Bi 3: Mt monoxicloankan c t khi hi so vi nit bng 3. a/ Xc nh CTPT ca A. b/Vit CTCT v tn tt c cc Bi 3: xicloankan ng vi CTPT tm c Gii a/ CnH2n = 28.3 = 84 14n = 84 n = 6 Hot ng 5: Bi 4: Hn hp kh A cha mt CTPT: C6H12 ankan v mt xicloankan. T khi b/ Cc CTCT ca A i vi H2 l 25,8. t chy hon ton 2,58gam A ri hp th ht sn phm chy vo dung dch Ba(OH)2 d, thu c 35,46 gam kt ta. Xc nh CTPT ca ankan v xicloankan GV: Gi hng dn HS cch gii Tm MA Vit pthh Gi x, yln lt l s mol ca ankan, xicloankan41


Gio n t chn 11CH3 CH3 CH3

Lp phng trnh Gii phng trnh v bin lun tm n, m

xiclohexan metylpentan 1,1-dimetylxiclobutan CH3 CH3

CH2-CH3

HS: Lm bi theo cc bc GV hng dn

CH3 H3C 1,2-dimetylxiclobutan 1,3-dimetylxiclobutan etylxiclobutan CH3 H3C CH3 CH3 CH3 CH3 CH3

CH2CH3

1,2,3-trimetylxiclopropan 1,1,2-trimetylxiclopropan 1-etyl-2-metylxiclopropan

CH3 CH2CH3 1-etyl-1-metylxiclopropan CH2CH2CH3 propylxiclopropan

CHCH3 CH3 isopropylxiclopropan

Bi 4: Gii Gi s trong 2,58g hn hp A c x mol CnH2n + 2 (n1) v y mol CmH2m (m3) . MA = 25,8.2 = 51,6(g/mol) x+y=2,58 = 0,05 (1) 51 ,6

CnH2n + 2 + x (mol) CmH2m +

3n +1 O2 t nCO2 2

+ (n+1)H2O nx

3m 2

O2 t mCO2 + mH2O

y my (mol) BaCO3 + H2O CO2 + Ba(OH)2 S mol CO2 = s mol BaCO3 =35 ,46 = 0,18 (mol ) 197

nx + my = 0,18 (2) Khi lng hn hp A: (14n + 2)x + 14my = 2,58 (3) 14(nx + my) + 2x = 2,58 2x = 2,58 14.0,18 x = 0,03; y = 0,02 (2) ta c : 0,03n + 0,02m = 0,18 3n + 2m = 18 Nghim thch hp m = 3; n = 4 CTPT l C4H10; C3H6 IV: Cng c Nhc li cch gi tn ca xicloankan.Cch gii bi ton tm CTPT ca ankan v xicloankan

42


Gio n t chn 11-

Tit 18:

Hirocacbon no

Ngy son: I. MUC TIEU : 1. Kien thc : Cung co :cac kien thc ve ankan va xicloankan 2. Ky nang : - Ren luyen k nang viet CTCT va goi ten cac ankan - Ren luyen k nang lap CTPT cua hp chat hu c , viet ptp co chu y van dung quy luat the vao phan t ankan . 3. Trong tam : Giai cac bai tap van dung . II/Chuan b : Phieu hoc tap theo noi dung kiem tra bai cu va bai tap luyen tap . *). PHNG PHAP : am thoai gi m neu va giai quyet van e hoat ong nhom III. Tin trnh ging da: 1/On nh to chc : Lp 11A8 S s Ngy dy 2/Kiem tra : Va on tap va k tra 3. Va bi:

43


Gio n t chn 11-

Hot ng ca gio vin v hc sinh Gv cho hs thao luan theo nhom . Sau o trnh bay . Nhom 01 ai dien nhom len trnh bay GV tong ket Nhom 02 ai dien nhom len trnh bay GV tong ket Nhom 03 ai dien nhom len trnh bay GV tong ket

Kin thc trng tm

Hng dan lam cac bai tap sgk va sbt 01. Viet PTPU cua n butan a. Tac dung vi clo theo ty le 1:1 b. e hidro hoa c. Crackinh 02. Khi ot ankan trong kh clo sinh ra muoi en va mot chat kh lam o giay qu t , nhng san pham o la g ? Tnh the tch clo can e ot chay hoan toan hh kh gom 2 l CH4 va 1 lt C3H8 ? 03. Mot chat ankan kh A . Khi ot chay 2 lit tao san pham co 8 lt CO2 ( Cac the tch o cung k nhiet o va apxuat ) a. Xac nh CTPT , CTCT cua A co the co ? b. Xac nh CTCT ung cua A ? Biet rang khi Nhom 04 clo hoa ( as) theo ty le mol 1:1 tao ra 2 dan ai dien nhom len trnh bay xuat mono clo ? GV tong ket c. ot chay h toan 5,8 g A , san pham chay hap thu het vao 200 ml dd NaOH 3M hoi tao muoi g ? khoi lng bao nhieu ? Nhom 05 04. Ankan X , khoi lng n to C la 83,33 % ai dien nhom len trnh bay a. Xac nh CTPT cua X biet khi tac dung GV tong ket vi clo ( as , 1:1 ) cho mot dan xuat monoclo ? Nhom 06 b. Khi ot chay 14,4 g X , san pham chay hap ai dien nhom len trnh bay thu het vao 200 ml dd NaOH tao 97,2 g hh 2 GV tong ket muoi cacbonat . Tnh CM dd NaOH 05 . Viet CTCT thu gon va ten cua cac chat co CT C3H6Br2 ; C2H4Cl2 ? 06. Ankan Y mach khong phan nhanh co CTG nhat la C2H5 . a) Viet cac CTCT co the co va goi ten Y ? b) Viet ptp cua Y vi brom khi chieu sang , ch ro san pham chnh cua phan ng ? IV. Cung co : V. Dan do :

44


Gio n t chn 11-

Tit 19: Hirocacbon khng no Ngy son: I/Muc tieu : 1/Ve kien thc :HS biet -Cau truc electron va cau truc khong gian cua anken -Viet ong phan cau tao va ong phan hnh hoc va goi ten anken -Phng phap ieu che va ng dung cua anken HS hieu: Tnh chat hoa hoc cua anken 2/K nang : -Viet phng trnh phan ng chng minh tnh chat hoa hoc cua anken -Viet c cac ong phan anken va goi ten chung -Lam c cac bai tap anken *)/Trong tam: Tnh chat hoa hoc , cac bai tap van dung cua anken. II/Chuan b : Phieu hoc tap theo noi dung kiem tra bai cu va bai tap luyen tap45


Gio n t chn 11-

*. PHNG PHAP : am thoai gi m neu va giai quyet van e hoat ong nhom III. Tin trnh ging dy: 1. n nh t chc Lp 11A8 S s Ngy dy 2.ktBi c: Vit cc CTCT ca C5H10. Gi tn cc CTCT Trnh by tnh cht ha hc ca anken 3.Bi mi: Hot ng ca gio vin v hc Kin thc trng tm sinh Hot ng 2: Bi 1: Bi 1: Gi tn cc CTCT sau Gii: CH3 4,4 imetylpent 1- en 2-etylbut-3-en CH3 - C - CH2 - CH = CH2 Bi 2:CH3CH3 - CH2 - C - CH2 -CH3 CH2

Gii Bi 3: Gii Gi s hn hp A c x mol CnH2n + 2 v y mol CmH2m.

Hot ng 3: Bi 2: Vit CTCT thu gn ca 2,4 8,96 x +y = = 0,4(1) 22 ,4 imetylhex-1-en (14 n + 2) x +14 m = 9( 2) y Hot ng 4: Bi 3: Hn hp kh A cha mt ankan v CnH2n + 2 + 3n +1 O2 t nCO2 + (n+1)H2O 2 mt anken. Khi lng hn hp A l x nx 9 gam v th tch l 8,96 lt. t chy hon ton A, thu c 13,44 lt (mol) 3m CO2. Cc th tch c o ktc. CmH2m + O2 t mCO2 + mH2O 2 Xc nh CTPT v % th tch tng y my (mol) cht trong A. 13 ,44 nx + my = 22 ,4 = 0,6 (3) Hot ng 5: T (1), (2), (3) ta c x = 0,3; y = 0,1 Bi 4: Thay x, y vo (3) ta c: 3n + m = 6 Dn 3,584 lt hn hp X gm 2 Chn m = 3, n =1 anken A v B lin tip nhau trong CH4 chim 60% th tch A v C3H6 chim dy ng ng vo nc brom (d), 40% thy khi lng bnh ng nc Bi 4: Gii brom tng 10,5 g a/ t ng thc ca 2 anken l CnH2n v a/ Tm CTPTca A, B ( bit th tch46


Gio n t chn 11-

kh o 0 C v 1,25 atm ) v tnh % th tch ca mi anken b/ Tnh t khi c hn hp so vi H2 t cng thc 2 anken, cng thc trung bnh Vit pthh Tm gi tr x Tm CTPT ca 2 anken Tnh % th tch ca mi anken Tnh t khi c hn hp so vi H2

0

Cn+1H2n+2 Cng thc chung ca 2 anken CxH2x vi n < x < n + 1 CxH2x + Br2 CxH2xBr2 tng khi lng ca bnh ng dd chnh l khi lng ca 2 anken.n= 1,25 .3,584 = 0,2(mol ) 22 ,4 10 ,5 = 52 ,5 =14 x x = 3,75 n = 3 M = 0,2

Hai anken l C3H6 v C4H8 Gi a v b l s mol ca C3H6 v C4H8 trong hn hp. Ta c: a + b = 0,2 a = 0,05 HS: Lm bi theo cc bc GV 42a + 56b = 10,5 b = 0,15 hng dn %V = 25%C3 H 6

%VC4 H 7 = 75%

b/ d X / H = 26,252

*IV Cng c: Nhc li cch gi tn ca anken. Tnh cht ha hc ca anken. Cch gii bi ton tm CTPT ca 2 anken ng ng lin tip nhau. *V:Dn d: Chun b phn cn li ca bi anken.

Tit 20: Hirocacbon khng no Ngy son: I. MUC TIEU : 1. Kien thc : Cho hoc sinh biet -ac iem cau truc cua he lien ket oi lien hp47


Gio n t chn 11-

-Phng phap ieu che , ng dung va tnh chat hoa hoc cua butaien va isopren 2. Ky nang : Viet phan ng cong , phan ng trung hp cua butaien va isopren . Lam cac BT co lien quan . *). Trong tam : Tnh chat va ng dung cua butaien va isopren . Cac BT lien quan . II/Chuan b : Phieu hoc tap theo noi dung kiem tra bai cu va bai tap luyen tap . *). PHNG PHAP : am thoai gi m neu va giai quyet van e hoat ong nhom III. Tin trnh ging dy: 1. n nh t chc Lp 11A8 S s Ngy dy 2: Bi c: Trnh by tnh cht ha hc ca ankaien 3 Bi mi: Hot ng ca gio vin v hc sinh Kin thc trng tm Hot ng 2: Bi 1 Bi 1: Gii: 12 n 83,33 Ankan X c cacbon chim 83,33% v a/ %C = 14 n + 2 .100 % = 100 n = 5 khi lng phn t CTPT: C5H12 a/ Tm CTPT, vit cc CTCT c th b/ c ca X. CH3 - CH2 - CH2 - CH2 - CH3 (1) b/ Khi X tc dng vi brom un nng c chiu sng c th to ra 4 dn xut CH3 - CH - CH2 - CH3 (2) ng phn cha mt nguyn t brom trong phn t. Vit CTCT v gi tn CH3

CH3

GV: Hng dn HS chn CTCT ng CH3 - C - CH3 (3) khi cho X tc dng vi brom to ra 4 dn xut. CH3 Hot ng 3: Bi 2: Cht A l mt ankaien lin hp c mch cacbon phn nhnh. t chy hon ton 3,4 g A cn dng va ht Bi 2: 7,84 lt oxi (ktc). Xc nh CTPT , Gii CTCT, gi tn 3n 1 Hot ng 4: CnH2n - 2 + 2 O2 t nCO2 + (n-1)H2O48


Gio n t chn 110,7 (14n -2). 3n 1 = 3,4 n = 5

. Bi 3: Hn hp kh A cha mt ankan v mt ankaien . t chy hon ton 6,72 lt A phi dng va ht 28 lt O2 ( cc th tch kh ly ktc). Dn sn phm chy qua bnh 1 ng H2SO4 c, sau qua bnh 2 ng dung dch NaOH d th khi lng bnh 1 tng p gam, bnh 2 tng 35,2 gam. Xc dnh CTPT, tnh p. t cng thc ankan, cng thc ankaien Vit pthh Da vo d kin ra tm CTPT, tnh p

0,7 3n 1

0,35 (mol) C5H8

CTPT: CTCT:

CH2 = C - CH = CH2 CH3 2-metylbuta-1,3-dien

Bi 3: Gii Gi s hn hp A c x mol CnH2n + 2 v y mol CmH2m - 2.x+y = 6,72 = 0,3(1) 22 ,4

CnH2n + 2 + x +1)x CmH2m - 2 + y 1).y

3n +1 O2 t nCO2 + (n+1)H2O 2 3n +1 .x nx (n 2 3m 1 O2 t mCO2 + (m-1)H2O 2 3m 1 .y my (m2 3n +1 3m 1

S mol oxi: 2 .x + 2 .y = 1,25 (3n + 1)x + (3m -1)y =2,5 (2) S mol CO2: nx + my =35 ,2 = 0,8 (3) 44

T (1), (2), (3) ta c x = 0,2; y = 0,1 Thay x, y vo (3) ta c: 2n + m = 8 Chn m = 4, n =2 CTPT: C2H6 v C4H6 S mol H2O = (n + 1)x + (m -1)y = 0,9(mol) p = 0,9.18 = 16,2 (g) *IV Cng c: Nhc li tnh cht ha hc ca ankan v ankaien. Cch gii bi ton tm CTPT ca ankan, ankaien . *V Dn d: Chun b bi luyn tp

49


Gio n t chn 11-

Tit 21: Hirocacbon khng no Ngy son; I Mc tiu: HS vn dng c kin thc hc gii bi tp *)Trng tm: Bi tp ankin II. Chun b: GV:Gio n HS: n tp l thuyt, lm bi tp ankin III. Tin trnh ging dy: 1. n nh t chc: Lp S s Ngy dy 11A8

2. Bi c: Trnh by tnh cht ha hc ca ankin 3. Bi mi: Hot ng ca thy v tr Hot ng 2: Bi 1: Trnh by phng php ha hc phn bit cc cht sau: but -2 en, propin, butan. Vit cc phng trnh ha hc minh ha. Hot ng 3: Bi 2: Mt bnh kn ng hn hp kh H2 vi axetilen v mt t bt niken. Nung nng bnh mt thi gian sau a v nhit ban u. Nu cho mt na kh trong bnh sau khi nung nng i qua dung dch AgNO3 trong NH3 th c 1,2 gam kt ta mu vng nht. Nu cho na cn li qua bnh ng nc brom d thy khi lng bnh tng 0,41 g. Tnh khi lng axetilen cha phn ng, khi Kin thc trng tm Bi 1: Gii: - Dn tng kh qua dung dch bc nitrat trong amoniac: bit c cht to kt ta l propin, do c phn ng: CH3 C = CH + AgNO3 + H2O CH3 C = CAg + NH4NO3 - Dn hai kh cn li vo dung dch brom: bit cht lm nht mu dung dch brom l but 2 en, do c phn ng: CH3CH=CHCH3 + Br2 CH3CHBrCHBrCH3 Kh cn li l butan. Bi 2: Gii C2H2 + H2 C2H4 (1) C2H2 + 2H2 C2H6 (2) C2H4 + H2 C2H6 (3) CH = CH + 2AgNO3 + 2H2O CAg = CAg50


Gio n t chn 11-

lng etilen to ra sau phn ng.

Hot ng 4: Bi 3: t 3,4 gam mt hirocacbon A to ra 11 gam CO2. Mt khc, khi cho 3,4 gam tc dng vi lng d dung dch AgNO3 trong NH3 thy to ra a gam kt ta. a/ Xc nh CTPT ca A. b/ Vit CTCT ca A v tnh khi lng kt ta to thnh, bit khi A tc dng vi hiro d, c xc tc Ni Gii to thnh isopentan. a/ Gi CTPT ca A l CxHy. y y GV: Gi hng dn HS cch gii CxHy + (x + 4 )O2 xCO2 + 2 H2O Vit pthh 11 Tm CTPT mC = .12 = 3( g ) 44 Da vo d kin ra bin lun tm m = 3,4 3 = 0,4( g ) H CTCT ng 3 0,4 x:y = 12 : 1 = 5 : 8 CTGN: C5H8 CTPT (C5H8)n b/ V A tc dng c vi dung dch AgNO3 trong NH3, A c dng R - C = CH V A tc dng vi H2 to thnh isopentan nn A phi c mch nhnh. CTCT: CH = C CH(CH3) CH3 CH = C CH(CH3) CH3 + AgNO3 + H2O CAg = C CH(CH3) CH3 + NH4NO3 S mol A = s mol kt ta = 3,4 : 68 = 0,05(mol) Khi lng kt ta = 0,05 . 175 =8,75 (gam) * IV. Cng c: + Hon thnh s phn ng sau: CaCO3 CaO CaC2 C2H2 vinylclorua PVC + Trnh by phng php ha hc nhn bit but 1-in, but-2-in, metan. + Cht no khng tc dng vi dung dch AgNO3 trong amoniac? A. but 1-in B. but 2-in C. Propin D. Etin *V. Dn d: Chun b bi luyn tp

+ 2NH4NO3 (4) C2H2 + 2Br2 C2H2Br4 (5) C2H4 + Br2 C2H4Br2 (6) S mol C2Ag2 = 0,005 (mol) T (4) ta c s mol axetilen trong hn hp cn li l: 2.0,005 =0,01 (mol) Theo (5), khi lng bnh ng brom tng 0,005.26 = 0,13 gam Vy khi lng etilen phn ng (6) l: 0,410,13 = 0,28(g) Khi lng etilen to ra: 2.0,28 = 0,56 gam Bi 3:

51


Gio n t chn 11-

Tit 22 Ngy son:

Hiro cacbon thm- H thng ha v Hirocabon

I: Mc tiu: HS vn dng c kin thc hc gii bi tp Trng tm: Bi tp benzen v ng ng. Mt s hirocacbon thm khc II. Chun b: GV:Gio n HS: n tp l thuyt, lm bi tp benzen v ng ng. Mt s hirocacbon thm khc III.Tin trnh ln lp: 1. n nh t chc: Lp 11A8 S s Ngy dy 2. Bi c: Trnh by tnh cht ha hc ca benzen 3. Bi mi:

52

Hot ng ca thy v tr Hot ng 2: Bi 1: GV: Nguyn th Minh Huyn A Cng c:Nhc li cch gi tn t IV:l mt ng ng ca benzen ccc khi hi so vi metan bng 5,75. A tham gia cc qu trnh chuyn ha theo s sau: A+ Cl2 B (1mol) C + HNO 3 , ( 3 mol ), H 2 SO 4 D + KMnO4 ,t E + H 2 , Ni,t

Kin thc trng tm Bi 1: Gio n t chn 11Gii ng ng benzen. Cc cch gii bi tp tm MA = 5,75.16 = 92 (g/mol) 14n 6 = 92 n =7 A l C7H8 hay C6H5 CH3 ( Toluen) C6H5 CH3 + Cl2 t C6H5 CH2Cl + HCl B: benzyl clorua N i,t C6H5 CH3 + 3H2 C6H11CH3 C: Metylxiclohexan O C6H5-CH3 + 3HNO3 H S C6H2(NO2)3CH3 + 3H2O D: TNT (trinitrotoluen) C6H5 CH3 + KmnO4 t C6H5-COOK + KOH + 2MnO2 + H2O E: kali benzoat Bi 2:2 4

Trn s ch ghi cc cht sn phm hu c ( phn ng cn c th to ra cc cht v c) Hy vit phng trnh ha hc ca cc qu trnh chuyn ha. Cc cht hu c vit di dng CTCT, km theo tn gi. Hot ng 3: Bi 2: Cht A l mt ng ng ca benzen. Khi t chy hon ton 1,5 g cht A, ngi ta thu c 2,52 lt kh CO2 ( ktc). a/ Xc nh CTPT. b/ Vit cc CTCT ca A. Gi tn. c/ Khi A tc dng vi Br2 c cht xc tc Fe v nhit th mt nguyn t H nh vi vng benzen b thay th bi Br, to ra dn xut monobrom duy nht. Xc nh CTCT ca A.

CnH2n 6 +

3n 3 O2 2

Gii nCO2 + (n-3)H2O

C ( 14n -6)g A to ra n mol CO2 C 1,5 g A to ra2,52 = 0,1125 molCO 22 ,42

14 n 6 n = n = 9 1,5 0,1125

CTPT: C9H12 Cc CTCT:CH3 CH3 CH33 1,2,3-trimetlybenzen

CH3 CH3 H3C CH3 CH3 1,2,5-trimetlybenzen

CH3

CH3CH3

CH

1,3,5-trimetlybenzen

C2H5 C2H5 1- etyl - 2 - metylbenzen 1- etyl - 3 - metylbenzen

C2H5 1- etyl - 4 - metylbenzen

Hot ng 4: Bi 3: Hn hp M cha bnzen v xiclohexen. Hn hp M c th lm mt mu ti a 75 g dung dch brom 3,2%. Nu t chy hon ton hn

CH3

Br + Br2 CH3Fe,t

53H3C

Br + HBr CH3

H3C


Gio n t chn 11-

V Dn d: Chun b bi ngun hirocacbon thin nhin Tit 23: Hiro cacbon thm- H thng ha v Hirocabon Ngy son: I.Mc tiu: HS vn dng c kin thc hc gii bi tp *).Trng tm: Bi tp h thng ha v hirocacbon II. Chun b: GV:Gio n HS: n tp l thuyt, lm bi tp h thng ha v hidrocacbon III: Tin trnh ging dy: 1/ n nh lp Lp 11A8 S s Ngy dy 2.ktBi c: Vit cc CTCT ca C5H10. Gi tn cc CTCT Bi c: Trnh by thnh phn ca du m 3. Bi mi:

54

Hot ng ca thy v tr Kin thc trng tm Hot ng 2: Bi 1: Bi 1: Gii GV: Nguyn th Minh Huyn Gio n t chn 1120 ,72 Hn hp M cha hai hidrocacbon k nCO = = 0,925 ( mol ) 22 ,4 tip nhau trong mt dy ng ng. Khi t chy hon ton 13,2 gam mC =12 .0.925 =11,1( gam ) hn hp M thu c 20,72t CO2 m H = 13 ,2 11,1 = 2,1( gam ) ( ktc). Hy xc nh CTPT v % n H O = 2,1 = 1,05 (mol ) 2 khi lng tng cht trong hn hp V s mol H2O > s mol CO2 nn hai cht M. trong hn hp M u l ankan. 3n +1 CnH2n +2 O2 nCO2 + ( n +1 ) H2O2

2

2

n 0,925 = n = 7, 4 n +1 1,05

Hot ng 3: Bi 2: Khi cho mt hidrocacbon mch h X tc dng vi nc brom ( d) sinh ra mt hp cht Y cha 4 nguyn t brom trong phn t. Trong Y, phn trm khi lng ca ccbon bng 10% khi lng ca Y. a/ Tm CTPT v CTCT ca X. b/ Trn 2,24 lt X vi 3,36 lt H2 ( ktc) sau un nng hn hp vi mt t bt Ni n khi phn ng xy ra hon ton. Tnh % khi lng ca cc cht sau phn ng. Hot ng 4: Bi 3: Cho hn hp kh X gm H2 v C2H2. Dn X qua ng ng bt kh Ni nung nng, sau khi dng phn ng thu c hn hp kh Y. Dn kh Y qua dung dch AgNO3 trong ammoniac d thy c kt ta, kh cn li lm nht mu nc brom v lm khi lng dung dch tng ln. Kh ra khi nc brom c t chy hon ton. Gii thch qu trnh th nghim trn v vit cc phng trnh minh ha

CTPT ca hai cht C7H16( x mol) v C8H18( y mol) Khi lng hai cht: 100x + 114y =13,2 S mol CO2: 7x + 8y = 0,925 x = 0,075; y = 0,05%C 7 H 16 = %C8 H 18 0,075 .100 .100 % = 56,8% 13,2 0,05 .114 = .100 % = 43,2% 13,2

Bi 2: Gii a/ X c CTPT CnH2n 2, tc dng vi brom: CnH2n 2 + 2Br2 CnH2n 2Br4 %C = Vy CTPT ca X l C3H4, CTCT ca X l CH3- C = CH b/ Phn ng hiro ha C3H4 + 2H2 C3H8 x 2x C3H4 + H2 C3H6 y y Ta c h phng trnh x + y = 0,1 2x + y = 0,15 x = 0,05; y = 0,05%mC3 H 6 = 48,84%12 n 10 .100 % = n=3 14n 2 100

%mC3 H 8 = 51,16%

Bi 3: Gii Cc phn ng khi un nng hn hp vi bt niken. C2H2 + H2 C2H4 C2H2 + 2H2 C2H6 Y gm C2H2, C2H4, H2, C2H6 tc dng vi AgNO3 trong ammoniac. 55 C2H2 + 2AgNO3 + 2NH3 C2Ag2 + 2NH4NO3


Gio n t chn 11-

* IV: Cng c: Nhc li cc kin thc hidrocacbon hc *V: Dn d: Chun b bi: Dn xut halogen ca hirocacbon

Tit 24: Dn xut halogen- ancol- phenol Ngy son: I.Mc tiu: HS vn dng c kin thc hc gii bi tp *).Trng tm: Bi tp dn xut halogen ca hirocacbon v ancol II. Chun b: GV:Gio n HS: n tp l thuyt, lm bi dn xut halogen ca hidrocacbon v ancol III: Tin trnh ging dy: 1/ n nh lp Lp 11A8 S s Ngy dy 2Bi c: Vit cc CTCT ca C5H10. Gi tn cc CTCT Bi c: Trnh by s chuyn ha gia cc hirocacbon 3. Bi mi:

56

Hot ng ca thy v tr Hot ng 2: Bi 1: GV: Nguyn th Minh Huyn Hon thnh s chuyn ha sau bng cc phng trnh ha hc. a/ Etan cloetan etyl magie clorua b/ Butan 2 brombutan but -2- en CH3CH(OH)CH2CH3 Hot ng 3: GV: Chp ln bng, yu cu HS chp vo v. Bi 2: t chy hon ton 3,96 g cht hu c A, thu c 1,792 lt CO2 ( ktc) v 1,44 g H2O. Nu chuyn ht lng clo c trong 2,475 g cht A thnh AgCl th thu c 7,175 g AgCl. a/ Xc nh cng thc n gin nht ca A. b/ Xc nh CTPT ca A bit rng t khi hi ca A so vi etan l 3,3. c/ Vit cc CTCT m A c th c v gi tn

Kin thc trng tm Bi 1:as

Gio n t chn 11-

Gii a/ C2H6 + Cl2 C2H5Cl + HCl C2H5Cl + Mg C2H5MgCl b/ CH3CH2CH2CH3 + Br2 CH3CHBrCH2CH3 + HBr O CH3CHBrCH2CH3 + KOH C HH CH3- CH = CH CH3 + KBr + H2O CH3- CH = CH CH3 + H2O CH3- CH(OH) CH CH3 Bi 2: Gii a/ Khi t chy A ta thu c CO2 v H2O, vy A phi cha C v H. Khi lng C trong 1,792 lt CO2:2 5

H+

12 .1,792 = 0,96 ( g ) 22 ,4

Khi lng H trong 1,44 g H2O:2.1,44 = 0,16 ( g ) 18

cng l khi lng C v H trong 3,96 g A Theo bi ra, A phi cha Cl. Khi lng Cl trong 7,175 g AgCl35 ,5.7,175 =1,775 143 ,5

(g)

cng l khi lng Cl trong 2,475 g A Vy, khi lng Cl trong 3,96 g A l:1,775 .3,96 = 2,84 ( gam 2,475

)

Vy cht A c dng CxHyClz x: y: z =0,96 0,16 2,84 : : =1 : 2 : 1 12 1 35 ,5

Hot ng 4: Bi 3: t chy hon ton mt lng hn hp hai ancol A, B no n chc k tip nhau trong dy ng ng thu c 4,48 lt kh CO2 (ktc) v 4,95 gam nc.

CTGN ca A l CH2Cl b/ MA = 3,3.30 = 99 (g/mol) (CH2Cl)n = 99 49 ,5n = 99 n = 2 CTPT ca A l C2 H4Cl2 c/ Cc CTCT CH3CHCl2 ; 1,1 icloetan CH2Cl CH2Cl; 1,2 - icloetan Bi 3: Gii Gi cng thc ca ancol A l : CnH2n + 1OH ( n 1) Gi cng thc ca ancol B l : CmH2m + 1OH ( m = n +1) Cng thc trung bnh ca 2 ancol l: : C57 2n + 1OH ( n < n < n + 1) nHnCO 2 = 4,48 = 0,2( mol ) 22 ,4

4,95


Gio n t chn 11-

* IV:Cng c: NV: Dn d: Chun b bi: Phn cn li bi Ancol

Tit 9: ancol- phenol Ngy son: I.Mc tiu: HS vn dng c kin thc hc gii bi tp *).Trng tm: Bi tp Ancol v phenol II. Chun b: GV:Gio n HS: n tp l thuyt, lm bi Ancol v phenol III:Tin trnh ging dy: 1. n nh t chc: Lp S s Ngy dy 2.Bi c: Vit cc CTCT ca C5H10. Gi tn cc CTCT : Trnh by nh ngha, phn loi, danh php Ancol. Ly VD minh ha. 3. Bi mi:

58

Hot ng ca thy Hot ng 2: Bi 1: GV: Nguyn th Minh Huyn Hn hp A cha gixerol v mt ancol n chc. Cho 20,30 gam A tc dng vi natri d thu c 5,04 lt H2 ( ktc). Mt khc 8,12 gam A ha tan va ht 1,96 g Cu(OH)2. Xc nh CTPT, Tnh % v khi lng ca ancol n chc trong hn hp A.

Hot ng ca tr Bi 1: Gii Gio n t chn 112C3H5 (OH)3 + Cu(OH)2 [C3H5(OH)2O]2Cu + 2H2O S mol gixerol trong 8,12 g A = 2 s mol Cu(OH)2 = 2.1,96 = 0,04 ( mol ) 98

S mol gixerol trong 20,3 g A:0,04 .20 ,3 = 0,1( m ) ol 8,12

Khi lng gixerol trong 20,3 g A l : 0,1.92 = 9,2 (g) Khi lng ROH trong 20,3 g A l: 20,3 9,2 =11,1(g) 2C3H5 (OH)3 + Na 2C3H5 (ONa)3 + 3H2 Hot ng 3: 0,1 Bi 2: 0,15 un nng hn hp 2 ancol no, n 2ROH + 2Na RONa + H2 chc, mch h vi H2SO4 1400C, x 0,5x thu c 72 gam hn hp 3 ete vi S mol H2 = 0,15 + 0,5x = s mol bng nhau. Khi lng nc 5, 04 = 0, 225 x = 0,1 tch ra trong qu trnh to thnh cc 22, 4 ete l 21,6 gam. Xc nh CTCT ca 11,1 Khi lng 1 mol ROH: 0,15 = 74 2 ancol. HS: Chp R = 29; R l C4H9 CTPT: C4H10O Phn trm khi lng C4H9OH = Hot ng 4: Bi 3: Hn hp X gm phenol v ancol etylic. Cho 14g hn hp tc dng vi natri d thy c 2,24 lt kh thot ra ( ktc). a/ Tnh % khi lng ca cc cht trong hn hp. b/ nu cho 14 g X tc dng vi dung dch brom th c bao nhiu gam kt ta.11,1 .100 % = 54,68 % 20,3

Bi 2: Gii 2CnH2n+1OH (2CnH2n +1)2O + H2O 2Cm H2m +1OH (2CmH2m+1)2O + H2O CnH2n+1OH + Cm H2m +1OH CnH2n+1OCmH2m +1 + H2O S mol 3 ete = s mol nc =21,6 =1,2( mol ) 18

S mol mi ete = 0,4 (mol) Khi lng 3 ete: (28n + 18).0,4 + ( 28m +18).0,4 + (14n + 14m + 18).0,4=72n+m=3

Hai CTCT ca ancol l: CH3OH, CH3CH2OH Bi 3: Gii a/C6H5OH + Na C6H5ONa + 1/2H2 59 x x/2 C2H5ONa + 1/2H2 C2H5OH + Na


Gio n t chn 11-

IV: Cng c: Nhc li tnh cht ca ancol v phenol V: Dn d: Chun b bi: Luyn tp

Tit 10: Luyn tp ancol- phenol Ngy son: I.Mc tiu: HS vn dng c kin thc hc gii bi tp .Trng tm: Bi tp dn xut Halogen + Ancol + Phenol II. Chun b: GV:Gio n HS: n tp l thuyt, lm bi dn xut Halogen + Ancol + Phenol III.Tin trnh ln lp: 1: n nh t chc: Lp S s Ngy dy 2. Bi mi:

60

Hot ng ca thy Hot ng 2: Bi 1: GV: Nguyn th Minh Huyn T etilen vit phng trnh iu ch cc cht sau: 1,2 ibrometan; 1,1 ibrometan; vinylclorua Hot ng 3: Bi 2: Cho 13,8 g hn hp X gm glixerol v mt ancol n chc A tc dng vi natri d thu c 4,48 lt kh hiro (ktc). Lng hidro do A sinh ra bng 1/3 lng hidro do glixerol sinh ra. Tm CTPT v gi tn A. Hot ng 4: Bi 3: Hn hp M gm ancol metylic, ancol etylic v phenol. Cho 14,45 g M tc dng vi natrib d, thu c 2,787 lt H2 ( 270C v 750 mm Hg ). Mt khc 11,56 g M tc dng va ht vi 80ml dung dch NaOH 1M. Tnh % khi lng tng cht trong hn hp M.

Hot ng ca tr Bi 1: Gio n t chn 11Gii iu ch 1,2 ibrom CH2 = CH2 + Br2 CH2Br CH2Br iu ch 1,1 ibrom CH2Br CH2Br + 2KOH CH = CH + 2KBr + H2O CH = CH + 2HBr CH3CHBr2 iu ch vinylclorua CH2 =CH2 + Cl2 CH2Cl CH2Cl CH2Cl CH2Cl + KOH CH2= CHCl + KCl + H2O Bi 2: Gii 2C3H5 (OH )3+ 6Na 2C3H5 (ONa )3 + 3H2 a 3a/2 2ROH + 2Na 2RONa + H2 b b/2 Ta c phng trnh: 92a + MA.b = 13,8 3a + b = 0,4 a=b a = b =0,1 (mol); MA = 46(g/mol) CTPT ca A C2H5OH; etanol Bi 3: Gii Khi cho 11,56 g M tc dng vi dung dch NaOH C6H5OH + NaOH C6H5ONa + H2O S mol C6H5OH trong 11,56 g M = s mol NaOH = 0,08 (mol) S mol C6H5OH trong 14,45 g M = C6H5OH + Na C6H5ONa + 1/2H2 0,1 0,05 CH3OH + Na CH3ONa + 1/2H2 x x/2 C2H5OH + Na C2H5ONa + 1/2H2 y y/2 S mol H2 = 0,05 + x/2 + y/2 = 0,112 x + y = 0,124 (1) 94.0,1 + 32x + 46y = 14,45 32x + 46y =5,05 (2) 61 T (1) v (2) ta c: x =0,05; y =0,075%C 6 H 5 OH = 94.0,1 .100 % = 65% 14,45 2,787 .750 = 0,112 ( m ) ol 0,082 .300 .760 0,08 .14 ,45 = 0,1( m ) ol 11 ,56


Gio n t chn 11-

* IV: Cng c: Nhc li tnh cht ca dn xut halogen ancol v phenol V Dn d: Chun b bi: Anehit - Xeton

Tit 26: Dn xut halogen- ancol- phenol Ngy son: I.Mc tiu: HS vn dng c kin thc hc gii bi tp .Trng tm: Bi tp Anehit - Xeton II. Chun b: GV:Gio n HS: n tp l thuyt, lm bi Anehit - Xeton III.Tin trnh ln lp: 1: n nh t chc: Lp 11A8 S s Ngy dy62


Gio n t chn 11-

2.: Bi c: Trnh by tnh cht ha hc ca Anehit - Xeton 3.Bi mi:

63

Hot ng ca thy v tr Hot ng 2: Bi 1: GV: Nguyn th Minh Huyn *Cht A l mt anehit n chc. Cho IV: Cng c: 10,5 gam A tham gia ht vo phn ng trng bc. Lng to thnh c ha tan ht vo axit nitric long lm thot ra 3,85 lt kh NO ( o 27,30C v 0,8 atm ). Xc nh CTPT, CTCT v tn cht A.

Kin thc trng tm Bi 1:Gio n t chn 11-

Gii RCHO + 2AgNO3 + 3NH3 + H2O RCOONH4 + 2NH4NO3 + 2Ag 3Ag + 4HNO3 3AgNO3 + NO + 2H2On NO = 3,85 .0,8 = 0,125 (mol ) 0,082 .300 ,3

S mol Ag = 3 s mol NO = 0,375 (mol) S mol RCHO = s mol Ag = 0,1875(mol) Khi lng 1 mol RCHO = 0,1875 = 56 R = 56 -29 = 27 R l C2H3 CTPT l C3H4O CTCT l CH2 = CH CHO ( propenal ) Bi 2: t chy hon ton mt lng cht hu c A phi dng va ht 3,08 lt O2. Sn phm thu c ch gm 1,8 gam H2O v 2,24 lt CO2. Cc th tch o ktc. a/ Xc nh CTGN ca A. b/ Xc nh CTPT ca A. Bit rng t khi ca A i vi oxi l 2,25. c/ Xc nh CTCT ca A, gi tn, bit rng A l hp cht cacbonyl. Gii a/ Theo nh lut bo ton khi lngm A = mCO 2 + mH 2O mO 2 = +1,8 3,08 .32 = 1,8( gam ) 22 ,4 2,24 .44 22 ,4

10 ,5

Hot ng 3: Bi 2: t chy hon ton mt lng cht hu c A phi dng va ht 3,08 lt O2. Sn phm thu c ch gm 1,8 gam H2O v 2,24 lt CO2. Cc th tch o ktc. a/ Xc nh CTGN ca A. b/ Xc nh CTPT ca A. Bit rng t khi ca A i vi oxi l 2,25. c/ Xc nh CTCT ca A, gi tn, bit rng A l hp cht cacbonyl.

Khi lng C trong 1,8 gam A :12 .2,24 =1,2( gam ) 22 ,4

Khi lng H trong 1,8 gam A :2.1,8 = 0,2( gam ) 18

Hot ng 4: Bi 3: Cho 10,2 gam hn hp hai anehit k tip nhau trong dy ng ng ca anehit fomic tc dng vi dung dch AgNO3 trong ammoniac d thu c 43,2 gam bc kt ta. Tm cng thc hai anehit v tnh % khi lng ca mi cht trong hn hp. HS: Chp

Khi lng O trong 1,8 gam A: 1,8 1,2 0,2 = 0,4 (gam) Cng thc cht A c dng: CxHyOz x:y:z =1,2 0,2 0,4 : : = 4 : 8 :1 12 1 16

CTGN l: C4H8O b/ MA = 2,25.32 = 72g/mol CTPT trng CTGN: C4H8O c/ Cc hp cht cacbonyl CH3 CH2 CH2 CHO butanal CH3 CH CHO 2 metylpropanal CH3 64 CH CO CH CH3 butan 2 on 2 3 Bi 3:


Gio n t chn 11-

Trong phn t anehit no, n chc, mch h X c phn trm khi lng oxi bng 27,586 %. X c CTPT l A. CH2O B. C2H4O C. C3H6O D. C4H8O *V Dn d: Chun b bi: axit cacboxylic

Tit 11: Ch : AXIT CACBOXYLIC I.Mc tiu: HS vn dng c kin thc hc gii bi tp II.Trng tm: Bi tp Axit cacboxylic III. Chun b: GV:Gio n HS: n tp l thuyt, lm bi Axit cacboxylic IV.Tin trnh ln lp: Hot ng 1: n nh lp + Bi c Bi c: Trnh by tnh cht ha hc ca Axit cacboxylic Bi mi:

65

Hot ng ca thy Hot ng 2: GV: Chp ln bng, GV: Nguyn th Minh Huyn yu cu HS chp vo v. Bi 1: trung ha 50 ml dung dch ca mt axit cacboxylic n chc phi dng va ht 30 ml dung dch KOH 2M. Mt khc, khi trung ha 125 ml dung dch axit ni trn bng mt lng KOH va ri c cn, thu c 16,8 gam mui khan. Xc nh CTPT, CTCT, tn v nng mol ca axit trong dung dch . Hot ng 3: Bi 2: Cht A l mt axit no, n chc, mch h. t chy hon ton 2,225 gam A phi dng va ht 3,64 lt O2 ( ktc). Xc nh CTPT, CTCT v tn gi. HS: Chp GV: Yu cu HS tho lun lm bi. HS: Tho lun lm bi

Hot ng ca tr Gii RCOOK + H RCOOH + KOHGio n t chn 11- 2O S mol RCOOH trong 50 ml dung dch axit l:2.30 = 0,06 (mol ) 1000

Nng mol ca dung dch axit l:0,06 .1000 = 1,2(mol / l ) 50

S mol RCOOH trong 125 ml dung dch axit l:1,2.125 = 0,15 ( mol ) 1000

cng l s mol mui thu c sau khi c cn dung dch . Khi lng 1 mol mui l: 0,15 =112 RCOOK = 112 R = 29 R l C2H5 CTPT ca axit l: C3H6O2 CTCT: CH3 CH2 COOH axit propanoic Bi 2: Gii CnH2nO2 + dng vi3n 2 O2 nCO2 216 ,8

+ nH2O

Theo phng trnh ( 14n + 32)g axit tc3n 2 2

mol O2

Theo bi ra 2,25 gam axit tc dng vi 0,1625 mol O214 n + 32 3n 2 = n = 5 2,55 0,1625 .2

Hot ng 4: GV: Chp ln bng, yu cu HS chp vo v. Bi 3: Dung dch X c cha ng thi hai axit cacboxylic no, n chc, mch h, k tip nhau trong dy ng ng. Ly 80 ml dung dch X em chia lm 2 phn nh nhau.Trung ha phn (1) bng dung dch NaOH ri c cn thu c 4,26 gam hn hp mui khan. Trung ha phn (2) bng dung dch Ba(OH)2 ri c cn, thu c 6,08 g hn hp mui khan. Xc nh CTPT v nng mol ca

CTPT C5H10O2 CH3 CH2 CH2 CH2 COOH axit pentanoic CH3 CH CH2 COOH axit -3metylbutanoic CH3 CH3 CH2 CH COOH axit -2metylbutanoic CH3 CH3 CH3 C COOH axit -2,2 -dimetylpropanoic CH3 Bi 3: Gii t cng thc chung ca 2 axit l CnH2n + 1 COOH CnH2n + 1 COOH + NaOH CnH2n + 1 66 COONa + H2O (14n + 68 ).x = 4,26 (1) 2CnH2n + 1 COOH + Ba(OH)2 (CnH2n + 1


Gio n t chn 11-

Hot ng 5: Cng c - dn d * Cng c: Trung ha 10g dung dch axit hu c n chc X nng 3,7% cn dng 50 ml dung dch KOH 0,1 M. CTCT ca X l A. CH3CH2COOH B. CH3COOH C. HCOOH D. CH3CH2CH2COOH * Dn d: Chun b bi: Luyn tp

Tit 33: T CHN Ch : BI TP ANEHIT XETON AXIT CACBOXYLIC I.Mc tiu: HS vn dng c kin thc hc gii bi tp67


Gio n t chn 11-

II.Trng tm: Bi tp anehit xeton axit cacboxylic III. Chun b: GV:Gio n HS: n tp l thuyt, lm bi anehit xeton axit cacboxylic IV.Tin trnh ln lp: Hot ng 1: n nh lp + Bi c Bi c: Trnh by tnh cht ha hc ca anehit v axit cacboxylic Bi mi:

68

Hot ng ca thy v tr Hot ng 2: GV: Chp ln bng, GV: Nguyn th Minh Huyn yu cu HS chp vo v. Bi 1: Hn hp M cha ba cht hu c A, B v C l 3 ng phn ca nhau. A l anehit n chc v C l ancol. t chy hon ton 1,45g hn hp M, thu c 1,68 lt ( ktc) kh CO2 v 1,35 gam H2O. Xc nh CTPT, CTCT v tn A, B, C. GV: Yu cu HS tho lun lm bi. HS: Tho lun lm bi GV: Cho HS xung phong ln bng gii

Ni dung Bi 1: Gio n t chn 11Hn hp M cha ba cht hu c A, B v C l 3 ng phn ca nhau. A l anehit n chc v C l ancol. t chy hon ton 1,45g hn hp M, thu c 1,68 lt ( ktc) kh CO2 v 1,35 gam H2O. Xc nh CTPT, CTCT v tn A, B, C. Gii Ba cht A, B, C l ng phn nn c CTPT ging nhau. A l anehit n chc nn phn t A ch c 1 nguyn t oxi. Vy A, B v C c CTPT CxHyO. Khi t chy hon ton hn hp M y 1 CxHyO + ( x + )O2 xCO2 + y/2H2O

4 2 HS: Ln bng trnh by, cc HS cn Theo phng trnh: (12x + y +16 ) g M to li ly nhp lm bi ra x mol CO2 v y/2 mol H2O. GV: Gi HS nhn xt ghi im 1,45g M to ra 0,075 mol CO2 v 0,075 mol H2O12 x + y +16 x y = = x = 3, y = 6 1,45 0,075 0,15

CTPT ca A, B v C l C3H6O A l CH3CH2CHO propanal B l CH3COCH3 axeton C l CH2= CH CH2 OH propenol Bi 2: Trung ha 250g dung dch 3,7% ca mt Hot ng 3: axit n chc X cn 100ml dung dch GV: Chp ln bng, yu cu HS NaOH 1,25M ( hiu sut 100%) chp vo v. a/ Tm CTPT, vit CTCT v tn gi ca X. Bi 2: b/ C cn dung dch sau khi trung ha th Trung ha 250g dung dch 3,7% ca thu c bao nhiu gam mui khan. mt axit n chc X cn 100ml dung dch NaOH 1,25M ( hiu sut 100%) a/ Tm CTPT, vit CTCT v tn gi ca X. Gii b/ C cn dung dch sau khi trung a/ Axit n chc, cng thc CxHyCOOH ha th thu c bao nhiu gam CxHyCOOH + NaOH CxHyCOONa + mui khan. H2O HS: Chp GV: Yu cu HS tho lun lm bi. S mol NaOH = 0,125 (mol); khi lng HS: Tho lun lm bi axit X = 9,25 gam. Theo phng trnh s mol axit = s mol GV: Cho HS xung phong ln bng NaOH gii 9,25 Maxit = 0,125 = 74 ( g / mol ) 69 12x + y = 29 suy ra x = 2; y = 5 CTPT l C3H6O2


Gio n t chn 11-

Hot ng 5: Cng c - dn d * Cng c: Trnh by phng php ha hc phn bit cc dung dch trong nc ca cc cht sau: fomanehit, axit fomic, axit axetic, ancol etylic. * Dn d: Chun b bi: Bi thc hnh 6

70


Gio n t chn 11-

Tit 12: Luyn tp Ngy son I.Mc tiu: HS vn dng c kin thc hc gii bi tp II.Trng tm: Bi tp anehit xeton axit cacboxylic III. Chun b: GV:Gio n HS: n tp l thuyt, lm bi anehit xeton axit cacboxylic IV.Tin trnh ln lp: Hot ng 1: n nh lp + Bi c Bi c: khng kim tra Bi mi:

71

Hot ng ca thy Hot ng 2: GV: Chp ln bng, GV: Nguyn th Minh Huyn yu cu HS chp vo v. Bi 1: Mt hp cht hu c Y gm cc nguyn t C, H, O ch cha mt loi nhm chc c kh nng tham gia phn ng trng bc. Khi cho 0,01 mol Y tc dng vi dung dch AgNO3 trong ammoniac th thu c 4,32 g Ag. Xc nh CTPT v vit CTCT ca Y, bit Y c cu to mch cacbon khng phn nhnh v cha 37,21% oxi v khi lng. Hot ng 3: GV: Chp ln bng, yu cu HS chp vo v. Bi 2: Cho 10,2 g hn hp X gm anehit axetic v anehit propioic tc dng vi dung dch AgNO3 trong ammoniac d, thy c 43,2 g bc kt ta. a/ Vit phng trnh ha hc ca phn ng xy ra. b/ Tnh % khi lng ca mi cht trong hn hp ban u. HS: Chp GV: Yu cu HS tho lun lm bi. HS: Tho lun lm bi

Hot ng ca tr Bi 1:Gio n t chn 11-

nY = 0,01(mol ) n 4,32 0,01 1 n Ag = = 0,04(mol ) Y = = 108 n Ag 0,04 4

Gii

C 2 trng hp + Nu Y l HCHO %mO = %mO =16.100% = 53% 37,21% (loi) 3032.100% = 37,21% M Y = 86 MY

+ Nu Y l R(CHO)2 = CxHyO2 12x + y = 86 suy ra x = 4, y = 6 CTCT: CHO CH2 CH2 CHO Bi 2: Gii a/ CH3CHO + 2AgNO3 + 4NH3 + H2O CH3COONH4 + 2NH4NO3 + 2Ag C2H5CHO + 2AgNO3 + 4NH3 + H2O C2H5COONH4 + 2NH4NO3 + 2Ag b/ Gi x, y ln lt l s mol anehit axetic, anehit propioic. 44x + 58y = 10,2 2x + 2y = 0,4 Gii h x = y = 0,1 %CH3CHO =0,1.44 .100 % = 43,14 % 10 ,2

%C2H5CHO = 56,86%

GV: Cho HS xung phong ln bng Bi 3: gii Gii + Hn hp hai axit c phn ng trng bc, vy trong hn hp c axit fomic HCOOH + 2AgNO3 + 4NH3 + H2O (NH4)2CO3 + 2NH4NO3 + 2Ag Trong mt na A ( khi lng 6,7 g ) c s mol HCOOH = s mol Ag = 0,05 mol. Khi lng HCOOH = 2,3 gam; RCOOH = 4,4 gam. Phn trm khi lng HCOOH = 34,33%; RCOOH = 65,67% + Trung ha phn 2 RCOOH + NaOH RCOONa + H2O HCOOH + NaOH HCOONa + H2O 72 S mol hai axit = s mol NaOH = 0,1 (mol) S mol RCOOH = 0,1 0,05 = 0,05 (mol)

HS: Ln bng trnh by, cc HS cn li ly nhp lm bi GV: Gi HS nhn xt ghi im Hot ng 4: GV: Chp ln bng, yu cu HS chp vo v. Bi 3: Ha tan 13,4 g hn hp hai axit cacboxylic no, n chc, mch h vo nc c 50 g dung dch A. Chia A thnh 2 phn bng nhau. Cho phn th nht phn ng hon ton


Gio n t chn 11-

Hot ng 5: Cng c - dn d * Cng c: Tnh cht ha hc ca anehit, xeton, axit cacboxylic. * Dn d: Chun b bi: n tp cc kin thc hc chun b n tp hc k II

Tit 35: T CHN Ch : N TP HC K II I.Mc tiu: HS vn dng c kin thc hc gii bi tp II.Trng tm: Kin thc chng 5, 6, 7, 8 III. Chun b: GV:Gio n HS: n tp l thuyt, lm bi chng 5, 6, 7, 8 IV.Tin trnh ln lp: Hot ng 1: n nh lp + Bi c Bi c: Khng kim tra Bi mi:

73

Hot ng ca thy v tr Hot ng 2: GV: Chp ln bng, GV: Nguyn th Minh Huyn yu cu HS chp vo v. Bi 1: Bng phng php ha hc hy phn bit cc ha cht sau: Ancol etylic, phenol, glixerol. Vit phng trnh minh ha nu c GV: Yu cu HS tho lun lm bi. HS: Tho lun lm bi GV: Cho HS xung phong ln bng gii

Ni dung Bi 1: Gio n t chn 11Bng phng php ha hc hy phn bit cc ha cht sau: Ancol etylic, phenol, glixerol. Vit phng trnh minh ha nu c Gii

Trch mi l ra mt t lm mu th Cho dung dch Br2 ln lt vo cc mu th + Mu th no xut hin kt ta trng Phenol HS: Ln bng trnh by, cc HS cn C6H5OH + 3Br2 C6H2Br3OH + 3HBr li ly nhp lm bi + Mu th khng c hin tng l: Ancol etylic v glixerol. Cho dung dch CuSO4/ NaOH vo 2 mu th cn li + Mu th lm cho dung dch c mu xanh lam glixerol GV: Gi HS nhn xt ghi im CuSO4 + 2NaOH Cu(OH)2 + Na2SO4 2C3H5(OH)3 + Cu(OH)2 [C3H5(OH)2O]2Cu + 2H2O + Mu th khng c hin tng Ancol Hot ng 3: etylic GV: Chp ln bng, yu cu HS chp vo v. Bi 2: Bi 2: T CaC2 v cht v c cn thit c y T CaC2 v cht v c cn thit c vit phng trnh iu ch caosu buna, y vit phng trnh iu ch nha PE, PVC, CH3CHO caosu buna, nha PE, PVC, CH3CHO Gii C2H2 + Ca(OH)2 HS: Chp CaC2 + 2H2O GV: Yu cu HS tho lun lm bi. 2C2H2 x t CH2 = CH C = CH HS: Tho lun lm bi CH2 = CH C = CH + H2 P d CH2 = CH CH = CH2 p GV: Cho HS xung phong ln bng nCH2 = CH CH = CH2 xt,,t (- CH2 CH = gii CH CH2 - )n C2H2 + H2 P d CH2 = CH2 p HS: Ln bng trnh by, cc HS cn nCH2 = CH2 xt,,t ( - CH2 CH2 - )n li ly nhp lm bi C2H2 + HCl x t CH2 = CH Cl p CH2 = CH Cl xt,,t ( - CH2 CH - )n Cl H gSO C2H2 + H2O CH3CHO GV: Gi HS nhn xt ghi im4

Hot ng 4: GV: Chp ln bng, yu cu HS chp vo v. Bi 3: Cho 21,4 gam hn hp kh A gm

Bi 3: Cho 21,4 gam hn hp kh A gm metan, etilen, axetilen qua dung dch brom, thy c 74 112 gam brom tham gia phn ng. Mt khc, nu cho 21,4 gam kh A trn qua


Gio n t chn 11-

Hot ng 5: Cng c - dn d * Cng c: Nhc li cch nhn bit, iu ch, hon thnh s phn ng, gii cc bi ton hn hp * Dn d: Chun b bi: n tp bi chun b thi hc k II

Tit 22: T CHN Ch : BI TP ANKAN +ANKEN + ANKIN I. II. Mc tiu: HS vn dng c kin thc hc gii bi tp Trng tm: Bi tp ankan + anken + ankin III. Chun b: GV:Gio n HS: n tp l thuyt, lm bi tp ankan + anken + ankin IV.Tin trnh ln lp: Hot ng 1: n nh lp + Bi c Bi c: Bi mi: Hot ng ca thy v tr Hot ng 2: GV: Chp ln bng, yu cu HS chp vo v. Bi 1: t chy hon ton hn hp hai hirocacbon mch h X, Y lin tip trong dy ng ng thu c 11,2 lt CO2 (ktc) v 12,6 gam nc. Tm CTPT ca X, Y HS: Chp GV: Yu cu HS tho lun lm bi. HS: Tho lun lm bi GV: Cho HS xung phong ln bng gii Ni dung Bi 1: t chy hon ton hn hp hai hirocacbon mch h X, Y lin tip trong dy ng ng thu c 11,2 lt CO2 (ktc) v 12,6 gam nc. Tm CTPT ca X, Y Gii:nCO 2 = 11,2 = 0,5( mol ) 22 ,4

n H 2O =

S mol nc > s mol CO2 X, Y thuc dy ng ng ca ankan.CnH2n + 2 +3n +1 O2 t nCO2 2

12 ,6 = 0,7(mol ) 18

+ (n+1)H2O

0,5 0,7 HS: Ln bng trnh by, cc HS cn Ta c : 0,5(n + 1 ) = 0,7n n = 2,5 li ly nhp lm bi75


Gio n t chn 11-

GV: Gi HS nhn xt ghi im

Hot ng 3: GV: Chp ln bng, yu cu HS chp vo v. Bi 2: Cho 3,5 gam mt anken X tc dng hon ton vi dung dch KMnO4 long d, thu c 5,2 gam sn phm hu c. Tm CTPT ca X. HS: Chp GV: Yu cu HS tho lun lm bi. HS: Tho lun lm bi GV: Cho HS xung phong ln bng gii HS: Ln bng trnh by, cc HS cn li ly nhp lm bi GV: Gi HS nhn xt ghi im

CTPT ca X, Y l: C2H6, C3H8 Bi 2: Cho 3,5 gam mt anken X tc dng hon ton vi dung dch KMnO4 long d, thu c 5,2 gam sn phm hu c. Tm CTPT ca X.

Gii 3CnH2n + 2KMnO4 + 4H2O CnH2n(OH)2 + 2MnO2 + 2KOH 14n 14n + 34 3,5 5,2 Ta c: 3,5( 14n + 34 ) = 5,2.14n n = 5 CTPT ca X l C5H10

Bi 3: t chy hon ton hon hai hirocacbon mch h M, N lin tip trong dy ng ng Hot ng 4: thu c 22,4 lt CO2 ( ktc) v 12,6 gam GV: Chp ln bng, yu cu HS nc. Tm CTPT ca M, N. chp vo v. GiinCO 2 = 22 ,4 =1(mol ) 22 ,4

Bi 3: t chy hon ton hon hai hirocacbon mch h M, N lin tip trong dy ng ng thu c 22,4 lt CO2 ( ktc) v 12,6 gam nc. Tm CTPT ca M, N. HS: Chp GV: Gi hng dn HS cch gii, yu cu HS ln bng trnh by GV: Gi HS nhn xt ghi im

n H 2O =

S mol nc < s mol CO2 M, N thuc dy ng ng ca ankin.CnH2n - 2 +3n 1 O2 t nCO2 2

12 ,6 = 0,7(mol ) 18

+ (n -1)H2O

Hot ng 5: Bi 4: 22 ,5 n = = 0,225 (mol ) t chy hon ton a lt (ktc) mt CaCO 100 ankin X th kh thu c CO2 v mCO = 44 .0,225 = 9,9( gam ) H2O c tng khi lng 12,6 gam. n = 12 ,6 9,9 = 0,15 (mol ) 18 Nu cho sn phm chy qua dung H O3

1 0,7 Ta c : (n - 1 ) = 0,7n n = 3,3 CTPT ca M, N l: C3H4, C4H6 Bi 4: t chy hon ton a lt (ktc) mt ankin X th kh thu c CO2 v H2O c tng khi lng 12,6 gam. Nu cho sn phm chy qua dung dch nc vi trong d, thu c 22,5g kt ta. Tm CTPT ca X. Gii

2

2

76


dch nc vi trong d, thu c C H n 2n - 2 + 22,5g kt ta. Tm CTPT ca X. HS: Chp GV: Yu cu HS tho lun lm bi. HS: Tho lun lm bi GV: Cho HS xung phong ln bng gii HS: Ln bng trnh by, cc HS cn li ly nhp lm bi GV: Gi HS nhn xt ghi im

3n 1 O2 t nCO2 2

Gio n t chn 11-

+ (n -1)H2O

0,225 0,15 Ta c : 0,225(n - 1 ) = 0,15n n = 3 CTPT ca X l: C3H4

Hot ng 6: Cng c - dn d * Cng c: Khi t chy hirocacbon thu c - S mol H2O > s mol CO2 hirocacbon thuc dy ng ng ankan - S mol H2O = s mol CO2 hirocacbon thuc dy ng ng anken - S mol H2O < s mol CO2 hirocacbon thuc dy ng ng ankin * Dn d: Chun b bi tit sau kim tra vit. Tit 25: T CHN Ch : BI TP TNG KT CHNG HIROCACBON NO V HIROCACBON KHNG NO III. IV. Mc tiu: HS vn dng c kin thc hc gii bi tp Trng tm: Bi tp tng kt chng hirocacbon no v hirocacbon khng no. III. Chun b: GV:Gio n HS: n tp l thuyt, lm bi tp hirocacbon no v hirocacbon khng no. IV.Tin trnh ln lp: Hot ng 1: n nh lp + Bi c Bi c: Bi mi: Hot ng ca thy v tr Ni dung

77


Gio n t chn 11-

Hot ng 2: GV: Chp ln bng, yu cu HS Bi 1: chp vo v. Hon thnh s phn ng sau Bi 1: Cao su buna C4H4 C4H4 Hon thnh s phn ng sau CaO CaC2 C2H2 C2H4 PE CaCO3 Cao su buna C4H4 C4H4 Vinylclorua PVC Gii: t 1/CaCO3 CaO + CO2 Vinylclorua 2/ CaO + 3C t CaC2 + CO PVC 3/ CaC2 + 2H2O C2H2 + Ca(OH)2 GV: Yu cu HS tho lun lm bi. 4/ C2H2 + H2 Pd,PbCO CH2 = CH2 t , HS: Tho lun lm bi 5/ nCH2 = CH2 xt,p (-CH2 CH2 - )n GV: Cho HS xung phong ln bng 6/ C2H2 + HCl t CH2 = CH Cl xt,t, p gii 7/ CaCO3 CaO CaC2 C2H2 C2 H4 PE3

nCH2 = CH Cl

(- CH2 - CH - )n Cl

HS: Ln bng trnh by, cc HS cn li ly nhp lm bi GV: Gi HS nhn xt ghi im

x 8/ 2C2H2 t,t CH2 = CH- C = CH 9/ CH2 = CH- C = CH + H2 Pd,PbCO CH2 = CH CH = CH2 t , 10/ nCH2 = CH CH = CH2 xt,p (-CH2 - CH = CH - CH2-)n Bi 2: Cho mt lng anken X tc dng vi H2O (xc tc H2SO4) c cht hu c Y, thy khi lng bnh ng nc ban u tng 4,2 Hot ng 3: gam. Nu cho mt lng X nh trn tc dng GV: Chp ln bng, yu cu HS vi HBr, thu c cht Z, thy khi lng Y, chp vo v. Z thu c khc nhau 9,45gam. Bi 2: Cho mt lng anken X tc dng vi H2O (xc tc H2SO4) c cht hu c Y, thy khi lng bnh Gii CxH2x +1 OH (Y) , (1) ng nc ban u tng 4,2 gam. CxH2x + H2O Nu cho mt lng X nh trn tc CxH2x + HBr CxH2x +1 Br (Z) , (2) dng vi HBr, thu c cht Z, thy tng khi lng bnh = khi lng anken khi lng Y, Z thu c khc nhau phn ng n X = 4,2 (mol ) 14 x 9,45gam. 4,2 HS: Chp mY = (14 x +18 ) 14 x GV: Yu cu HS tho lun lm bi. HS: Tho lun lm bi 4,2 GV: Cho HS xung phong ln bng m Z = 14 x (14 x + 81) gii3

78


Gio n t chn 11-

mZ

-

mY =

4,2 4,2 (14 x + 81) (14 x +18 ) 14 x 14 x

=9,45 x=2 CTPT ca X: C2H4 HS: Ln bng trnh by, cc HS cn Bi 3: Khi t mt th tch hirocacbon A mch h li ly nhp lm bi cn 30 th tch khng kh, sinh ra 4 th tch GV: Gi HS nhn xt ghi im kh CO2. A tc dng vi H2 ( xt Ni ), to thnh mt hirocacbon no mch nhnh. Xc nh CTPT, CTCT ca A. Gii Hot ng 4: GV: Chp ln bng, yu cu HS y y chp vo v. CxHy + (x + 4 )O2 xCO2 + 2 H2O Bi 3: Th tch oxi phn ng: Khi t mt th tch hirocacbon A 20 20 mch h cn 30 th tch khng kh, VO = 100 Vkk = 100 .30 = 6 (lt) sinh ra 4 th tch kh CO2. A tc Ta c phng trnh x = 4, x + y/4 = 6 y = 8 dng vi H2 ( xt Ni ), to thnh mt A c CTPT C4H8 mch h nn A thuc loi hirocacbon no mch nhnh. Xc anken. V A tc dng vi H2 to thnh mt nh CTPT, CTCT ca A. hirocacbon no mch nhnh. CTCT ca A. CH2 = C CH3 CH32

HS: Chp GV: Gi hng dn HS cch gii, yu cu HS ln bng trnh by GV: Gi HS nhn xt ghi im Hot ng 5: Cng c - dn d * Cng c: 1/ t chy hon ton 0,15 mol 2 ankan thu c 9 gam nc.Cho hn hp sn phm sau phn ng vo dung dch nc vi trong d th khi lng kt ta thu c l bao nhiu gam. A. 38g B. 36 gam C. 37 gam D. 35 gam 2/ t chy hon ton m gam, mt hirocacbon thu c 33gam CO2 v 27 gam H2O. Gi tr ca m l A. 11g B. 12g C. 13g D. 14g * Dn d: Chun b bi Benzen v ng ng ca benzen

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Gio n t chn 11-

0

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