giao an tu chon 11
TRANSCRIPT
Tit 1:
BI TP N TP
I. Mc tiu: Rn luyn k nng gii bi tp v nguyn t, cn bng phn ng oxi ho kh, tnh phn trm khi lng. II. Trng tm: Nguyn t, cn bng phn ng, % khi lng. III. Chun b: Gio n, hc sinh n li kin thc hc lp 10 IV.Tin trnh ln lp: 1/ n nh lp 2/ Bi c: 3/ Bi mi Hot ng ca thy v tr Hot ng 1: GV: Chp ln bng, GV yu cu HS tho lun theo bn, GV gi 1 HS ln trnh by. Ni dung Bi 1: Nguyn t ca nguyn t X c tng s ht p, n v e bng 40, tng s ht mang in nhiu hn tng s ht khng mang in l 12. Xc nh Z, A v vit cu hnh e ca nguyn t X, cho bit v tr nguyn t X trong BTH Gii: Ta c: p + n + e = 40 M p = e = Z 2p + n = 40 (1) Theo bi rat ta c 2p n = 12 (2) T (1) v (2) ta c: p = Z =13, n = 14 A = Z + n = 13 + 14 = 27 Cu hnh electron ca nguyn t X l: 1s22s22p63s23p1 - th 13 - Chu k 3 - Nhm chnh nhm IIIA Bi 2: Cn bng cc phng trnh sau y bng phng php cn bng phn ng oxi ho kh. Al + HNO3 Al(NO3)3 + NO + H2OFe + H2SO4 () t Fe2(SO4)3 + SO2 + H2O
HS: Ln bng trnh by
Hot ng 2: GV: Chp ln bng
Gii: GV: yu cu 2 HS ln trnh by, cc Al + HNO3 Al(NO3)3 + NO + H2O em cn li lm vo v nhp v quan +3 0 st 1x Al Al + 3e HS: Ln bng trnh by +2 +5 N 1x N + 3e Al + 4HNO3 Al(NO3)3 + NO + 2H2OH:01:25H:01:25H:01:250 +5 +3 +2
0 0
+6 +3
+3
Fe + H2SO4 () t Fe2(SO4)3 + SO2 + H2O
+4
GV: Nhc li 4 bc lp phng trnh phn ng oxi ho kh cng HS kim ta li bi lm ca cc bn trn bng
3 Fe Fe + 3e 2 S + 2e S 2Fe + 6H2SO4 () t Fe2(SO4)3 +3SO2 + 6H2O+6 +4
Bi 3: Cho 1,5 gam hn hp gm Nhm v Magi vo dd HCl c nng 1 mol/l ngi ta thu Hot ng 3: c 1,68 lt kh ( ktc) GV: Chp ln bng a/ Tnh % khi lng mi kim loi. b/ Th tch axit dung. Gii: HS: Hc sinh quan st v suy ngh 2AlCl3 + 3H2 2Al + 6HCl cch lm bi. x 3x 3/2x Mg + 2HCl MgCl2 + H2 y 2y y Gi x, y ln lt l s mol ca Al, Mg GV: Yu cu HS ln trnh by Ta c: 27x + 24y = 1,5 x = 1/30 3/2x + y = 0,075 y = 0,025 GV: Gi HS nhn xt % Mg =0,025.24 .100 = 40% 1,5 1 + 2.0,025 = 0,15(mol ) 30
% Al = 60 %n HCl = 3x + 2 y = 3. V = n 0,15 = = 0,15(l ) CM 1
Hot ng 4: Cng c - dn d - Cn bng phng trnh sau y bng phng php cn bng phn ng oxi ho kh. Al + HNO3 Al(NO3)3 + N2 + H2O FexOy + HNO3 Fe(NO3)3 + NO + H2O - BTVN: Ho tan hon ton 1,12 g kim loi ho tr II vo dd HCl thu c 0,448 lt kh ktc. Kim loi cho l: A. Mg B. Zn C.Cu D. Fe - Chun b bi in li sgk 11
Tit 2:
BI TP S IN LI AXIT, BAZ V MUI
H:01:25H:01:25H:01:25
I. Mc tiu: Vit phng trnh in li, phn bit c cht in li mnh, yu; gii thch c tnh axit, baz, theo thuyt Arniut, hiroxit lng tnh. II. Trng tm: S in li, axit, baz v hiroxit lng tnh. III. Chun b: GV:Gio n HS: n tp l thuyt cc bi trc IV.Tin trnh ln lp: 1/ n nh lp 2/ 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/ Bi mi Hot ng ca thy v tr Hot ng 1: GV: Chp ln bng, yu cu HS chp vo v. Bi 1: Vit phng trnh in li ca cc cht trong dd sau: HBrO4, CuSO4, Ba(NO3)2, HClO, HCN. Cho bit cht no l cht in li mnh, cht no l cht in li yu. HS: Chp 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. Ni dung Bi 1: Vit phng trnh in li ca cc cht trong dd sau: HBrO4, CuSO4, Ba(NO3)2, HClO, HCN. Cho bit cht no l cht in li mnh, cht no l cht in li yu.
Gii: HBrO4 H+ + BrO4CuSO4 Cu2+ + SO 2 4 Ba2+ + 2NO 3 Ba(NO3)2 HClO H+ + ClOHCN H+ + CNHBrO4, CuSO4, Ba(NO3)2 l cht in li mnh. HClO, HCN l cht in li yu. Bi 2: Vit phng trnh in li ca hiroxit lng tnh Al(OH)3.
Hot ng 2: GV: Chp ln bng, yu cu HS chp vo v. Bi 2: Vit phng trnh in li ca hiroxit Gii: lng tnh Al(OH)3. HS: Chp Al(OH)3 Al3+ + 3OHGV: Yu cu HS suy ngh 3 pht, Al(OH)3 H3O+ + AlO 2 sau gi 1 HS ln bng gii. GV quan st cc HS lm bi. GV: Nhn xt, hng dn li Bi 3: Vit phng trnh phn ng xy ra khi cho Hot ng 3:H:01:25H:01:25H:01:25
GV: Chp ln bng, yu cu HS chp vo v. Bi 3: Vit phng trnh phn ng xy ra khi cho Al2(SO4)3 tc dng vi NaOH d. 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: Nhn xt, hng dn li, lu cho HS phn hiroxit lng tnh. Hot ng 4: GV: Chp ln bng, yu cu HS chp vo v. Bi 4: Da vo thuyt Arniut. Gii thch NH3 l mt baz. HS: Chp 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
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 Hot ng 5: GV: Chp ln bng, yu cu HS Mg2+, c mol Cl-, d mol NO 3 . chp vo v. a/ Lp biu thc lin h a, b, c, d. Bi 5: b/ Nu a = 0,01; c = 0,01; d = 0,03 th b bng Trong mt dd c cha a mol Ca2+, bao nhiu. b mol Mg2+, c mol Cl-, d mol NO 3 . Gii: a/ Lp biu thc lin h a, b, c, d. b/ Nu a = 0,01; c = 0,01; d = 0,03 a/ Trong mt dd, tng in tch ca cc cation th b bng bao nhiu. bng tng in tch ca cc anion, v vy: HS: Chp 2a + 2b = c + d GV: Hng dn HS cch gii. b/ b = HS: Ch nghe ging Hot ng 6: Cng c - dn d * Cng c: - Theo thuyt Arniut, cht no di y l axit? A. Cr(NO3)3 B. HBrO3 C. CdSO4 - Theo thuyt Arniut, cht no di y l baz? A. Cr(NO3)3 B. HBrO3 C. CdSO4 * Dn d:H:01:25H:01:25H:01:25 c + d 2a 0,01 + 0,03 2.0,01 = = 0,01 2 2
D. CsOH D. NH3
Chun b bi s in li ca nc. pH. Cht ch th axit baz.
Tit 3:
BI TP. PH.
I. Mc tiu: Gii c cc bi ton lin quan n tnh pH. II. Trng tm: Cc bi tp tnh pH III. Chun b: GV:Gio n HS: n tp l thuyt cc bi trc IV.Tin trnh ln lp: 1/ n nh lp 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 thy v tr 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 sunfuric thnh ion c coi l hon ton. b/ Tnh nng mol ca ion OHtrong dd . HS: Chp GV: Yu cu 1 HS ln bng gii, cc HS cn li lm nhp v theo di bi bn lm. Ni dung 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 = 0,01M H2SO4 2 H+ + SO 2 41 + 1 [H ] = .0,01 = 0,005M 2 2 14 10 b/ [OH-] = 2 = 10 12 M 10
[H2SO4] =
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. HS: Chp GV: Hng dn HS cch gii. HS: Nghe ging v hiuH:01:25H:01:25H:01:25
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) Hot ng 3: Khi lng Na = 0,15.23 = 3,45 gam GV: Chp ln bng, yu cu HS Bi 3: chp vo v. Tnh pH ca dd cha 1,46 g HCl trong 400,0 Bi 3: ml. Tnh pH ca dd cha 1,46 g HCl trong 400,0 ml. HS: Chp Gii: GV: Yu cu HS suy ngh , sau gi 1 HS ln bng gii. Cc HS cn 1,46 1000 1 li ly nhp ra lm bi v theo di CM(HCl) = 36,5 . 400,0 = 0,100 M = 10 M bi bn lm. [H+] = [HCl] = 10-1M pH = 1,0 HS: Ln bng trnh by GV: Yu cu 1 HS nhn xt, GV nhn xt ghi im Hot ng 4: GV: Chp ln bng, yu cu HS Bi 4: chp vo v. Tnh pH ca dd to thnh sau khi trn 100,0 Bi 4: ml dd HCl 1,00M vi 400,0 ml dd NaOH Tnh pH ca dd to thnh sau khi 0,375M. 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 Gii: [OH-] 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) HS: Nghe ging v hiu GV: Yu cu HS tnh [H+] v pH HS: Tnh [H+] v pH0,05 = 0,1M 0,4 + 0,1 1,0.10 14 = 1,0.10 13 M [H+] = 1,0.10 1
[OH-] =
Vy pH = 13
Hot ng 5: Cng c - dn d * Cng c: pH ca dd CH3COOH 0,1M phi A. nh hn 1 B. ln hn 1 nhng nh hn 7 C. bng 7 D. ln hn 7 * Dn d: Chun b bi phn ng trao i ion trong dd cht in liH:01:25H:01:25H:01:25
Tit 4: BI TP PHN NG TRAO I ION TRONG DUNG DCH CC CHT IN LII. Mc tiu: HS vn dng c kin thc hc gii bi tp II. Trng tm: Cc bi tp phn ng trao i ion trong dung dch cc cht in li III. Chun b: GV:Gio n HS: n tp l thuyt cc bi trc IV.Tin trnh ln lp: 1/ n nh lp 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 thy v tr Hot ng 1: GV: Chp ln bng, yu cu HS chp vo v. 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 4 + OH- NH3 + H2O d/ S2- + 2H+ H2S HS: Chp 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: Vit phng trnh dng phn t ca cc phn ng theo s sau. a/ MgCO3 + ? MgCl2 + ?. b/ Fe2(SO4)3 + ? K2SO4 + ?. HS: Chp GV: Yu cu 1 HS ln bng gii, cc HS cn li lm nhp v theo di bi bn lm. Gi HS nhn xt , ghi imH:01:25H:01:25H:01:25
Ni dung 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 4 + OH- NH3 + H2O d/ S2- + 2H+ H2S
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:
Bi 2: a/ MgCO3 + ? MgCl2 + ?. b/ Fe2(SO4)3 + ? K2SO4 + ?
a/ MgCO3 + 2HCl MgCl2 + H2O + CO2 b/ Fe2(SO4)3 + 6KOH 3K2SO4 + Fe(OH)3
Gii:
Hot ng 3: GV: Chp ln bng, yu cu HS chp vo v. Bi 3: Ho tan 1,952 g mui BaCl2.xH2O trong nc. Thm H2SO4 long, d vo dung dch thu c. Kt ta to thnh c lm kh v cn c 1,864 gam. Xc nh cng thc ho hc ca mui. HS: Chp GV: Yu cu HS suy ngh tho lun 5 pht, sau cho 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
Bi 3: Ho tan 1,952 g mui BaCl2.xH2O trong nc. Thm H2SO4 long, d vo dung dch thu c. Kt ta to thnh c lm kh v cn c 1,864 gam. Xc nh cng thc ho hc ca mui. Gii:BaCl2.xH2O + H2SO4
BaSO4 + 2HCl + 2H2O (1)
n BaSO 4 =
1,864 = 0,008(mol ) 233
Theo phng trnh (1) s mol BaSO4 = s mol BaCl2.xH2O1,952 = 244 0,008 244 208 =2 x= 18M=
Hot ng 4: GV: Chp ln bng, yu cu HS chp vo v. Bi 4: Trn 250 ml dung dch hn hp HCl 0,08M v H2SO4 0,01M vi 250 ml dung dch Ba(OH)2 c nng x (M) thu c m gam kt ta v 500 ml dung dch c pH = 12. Hy tnh m v x. Coi Ba(OH)2 in li hon ton c 2 nc. HS: Chp GV:Yu cu tnh s mol HCl ban u , s mol H2SO4 ban u , vit cc phng trnh phn ng xy ra. HS: Tr li
GV: Hng dn HS tnh khi lng 1 kt ta, Tnh nng mol ca S mol Ba(OH)2 cn d = s mol OH- = 2 Ba(OH)2 . 0,0025 (mol) S mol Ba(OH)2 ban u = 0,01 + 0,0025 +H:01:26H:01:26H:01:26
CTHH ca mui l : BaCl2.2H2O Bi 4: Trn 250 ml dung dch hn hp HCl 0,08M v H2SO4 0,01M vi 250 ml dung dch Ba(OH)2 c nng x (M) thu c m gam kt ta v 500 ml dung dch c pH = 12. Hy tnh m v x. Coi Ba(OH)2 in li hon ton c 2 nc. Gii: S mol HCl ban u = 0,25.0,08 = 0,02 ( mol) S mol H2SO4 ban u = 0,25.0,01= 0,0025 ( mol) Sau khi phn ng dung dch c pH =12 ngha Ba(OH)2 cn d v cc axit phn ng ht. 2HCl + Ba(OH)2 BaCl2 + 2H2O 0,02 0,01 H2SO4 + Ba(OH)2 BaSO4 + 2H2O 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+ + 2OH-
0,0025 = 0,015 (mol) HS: Nghe ging v hiu Nng Ba(OH)2 : x =0,015 = 0,06( M ) 0,25
Hot ng 5: Cng c - dn d * 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 * Dn d: Chun b bi thc hnh s 1
Tit 5: BI TP TNG KT CHNG S IN LII. Mc tiu: HS vn dng c kin thc hc gii bi tp II. Trng tm: Phn ng trao i ion trong dung dch cc cht in li, Ph ca dung dch. III. Chun b: GV:Gio n HS: n tp l thuyt cc bi trc IV.Tin trnh ln lp: 1/ n nh lp 2/ Bi c: Trn 100 ml dung dch HCl 0,1 M vi 200ml dung dch Ba(OH)2 0,1 M c dung dch A . Tnh nng mol ca cc ion trong dung dch A. 3/ Bi mi Hot ng ca thy v tr Hot ng 1: GV: Chp ln bng, yu cu HS chp vo v. Bi 1: Trong ba dung dch c cc loi ion sau: 2 Ba2+, Mg2+, Na+, SO 2 , CO 3 v NO 3 4 Mi dung dch ch cha mt loi cation v mt loi anion. a/ Cho bit l 3 dd mui g b/ Hy chn dung dch axit thch hp nhn bit 3 dung dch mui ny. HS: Chp GV: Yu cu 1 HS ln bng gii, cc HS cn li lm nhp v theo di bi bn lm.H:01:26H:01:26H:01:26
Ni dung Bi 1: Trong ba dung dch c cc loi ion sau: 2 Ba2+, Mg2+, Na+, SO 2 , CO 3 v NO 3 4 Mi dung dch ch cha mt loi cation v mt loi anion. a/ Cho bit l 3 dd mui g b/ Hy chn dung dch axit thch hp nhn bit 3 dung dch mui ny.
Gii: a/ V cc mui BaSO4, BaCO3, MgCO3 khng tan nn ba dung dch phi l dung dch Ba(NO3)2, dung dch MgSO4 v dung dch Na2CO3.
GV: Yu cu 1 HS nhn xt, GV b/ Cho dung dch H2SO4 vo c 3 dung dch . nhn xt ghi im. dung dch Na2CO3 c si bt: Na2CO3 + H2SO4 Na2SO4 + H2O + CO2 dung dch Ba(NO3)2, xut hin kt ta Hot ng 2: trng. GV: Chp ln bng, yu cu HS Ba(NO3)2 + H2SO4 BaSO4 + 2HNO3 chp vo v. Dung dch MgSO4 vn trong sut. Bi 2: Bi 2: 150 ml dung dch KOH vo 50 ml 150 ml dung dch KOH vo 50 ml dung dung dch H2SO4 1M, dung dch tr dch H2SO4 1M, dung dch tr thnh d baz. thnh d baz. C cn dung dch C cn dung dch thu c 11,5 gam cht thu c 11,5 gam cht rn. Tnh rn. Tnh nng mol/lt ca dung dch KOH. nng mol/lt ca dung dch KOH. Gii HS: Chp S mol H2SO4 = 0,05 (mol) GV: Yu cu HS tho lun , gi 1 V baz d nn axit phn ng ht. HS ln bng trnh by 2KOH + H2SO4 K2SO4 + 2H2O 0,1 0,05 0,05 (mol) HS: Ln bng trnh by C cn dung dch , thu c cht rn gm c K2SO4, KOH d GV: Yu cu 1 HS nhn xt, GV mKOH(d) = 11,5 8,7 = 2,8 (gam) nhn xt ghi im. nKOH(d) = 2,8:56 = 0,05 (mol) S mol KOH c trong 150 ml dung dch KOH l. Hot ng 3: 0,1 + 0,05 = 0,15 (mol) GV: Chp ln bng, yu cu HS Nng mol/l ca dung dch KOH: chp vo v. CM(KOH) = 0,15: 0,15 = 1M Bi 3: Bi 3: Thm t t 400 g dung dch H2SO4 Thm t t 400 g dung dch H2SO4 49% vo 49% vo nc v iu chnh lng nc v iu chnh lng nc thu c nc thu c ng 2 lt dung ng 2 lt dung dch A. Coi H2SO4 in li hon dch A. Coi H2SO4 in li hon ton ton c 2 nc. c 2 nc. a/ Tnh nng mol ca ion H+ trong dung a/ Tnh nng mol ca ion H+ trong dch A. dung dch A. b/ Tnh th tch dung dch NaOH 1,8M cn b/ Tnh th tch dung dch NaOH thm vo 0,5 lt dung dch A thu c 1,8M cn thm vo 0,5 lt dung dch dung dch . A thu c dung dch . + Dung dch c Ph = 1 + Dung dch c Ph = 1 + Dung dch c Ph = 13 + Dung dch c Ph = 13 Gii GV: Yu cu 1 HS ln bng gii cu 400.49 = 2(mol) a, cc HS cn li lm nhp v theo a/ S mol H2SO4: 100.98 di bi bn lm. H2SO4 2H+ + SO 2 4 HS: Ln bng trnh by 2 4 (mol) GV: Gi HS nhn xt GV: Hng dn HS lm cu bH:01:26H:01:26H:01:26 m K 2SO 4 = 0,05.174 = 8,7(gam)
Nng H+ trong dung dch A l :
4 = 2M 2
b/ S mol H+ trong 0,5 lt dung dch A l : 2.0,5 = 1 (mol) t th tch dung dch NaOH l x th s mol
HS: Nghe ging v hiu
NaOH trong l 1,8x. NaOH Na+ + OH1,8x 1,8x 1,8x Axit d + Ph = 1 H+ + OH- H2O Ban u : 1 1,8x Phn ng: 1,8x Cn d : 1 -1,8x Nng H+ sau phn ng:1 1,8 x = 0,1M x = 0,5(l ) 0,5 + x + Ph = 13 Baz d H+ + OH- H2O
Ban u : 1 1,8x Phn ng: 1 1 Cn d : 1,8x 1 Sau phn ng Ph = 13 [H+] = 10-13M [OH-] = 10-1M1,8 x 1 = 0,1M x = 0,62(l ) 0,5 + x
Hot ng 4: Cng c - dn d * Cng c: Trong dung dch A c cc ion K+, Mg2+, Fe3+ v Cl- . Nu c cn dung dch s thu c hn hp nhng mui no. Dn d: Chun b bi Amoniac v mui Amoni
Tit 6:
BI TP NIT V AMONIAC
I. Mc tiu: HS vn dng c kin thc hc gii bi tp II. Trng tm: Bi tp nit v Amoniac. III. Chun b: GV:Gio n HS: n tp l thuyt cc bi trc IV.Tin trnh ln lp: 1/ n nh lp 2/ Bi c: Trnh by tnh cht ha hc ca amoniac. 3/ Bi mi Hot ng ca thy v tr Ni dung
H:01:26H:01:26H:01:26
Hot ng 1: GV: Chp ln bng, yu cu HS chp vo v. 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. HS: Chp 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: 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: Chp GV: Yu cu HS tho lun. GV: Hng dn HS cch lm bi HS:Nghe ging v hiu HS: T tnh phn trm s mol nit phn ng, th tch (kt) kh ammoniac c to thnh. Hot ng 3: GV: Chp ln bng, yu cu HS chp vo v. 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.H:01:26H:01:26H:01:26
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. Gii: S mol kh N2: p=21 = 0,75(mol ) 28
p sut ca kh N2:
nRT 0,75.0,082(25 + 273) = = 1,83(atm) V 10
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. Gii N2 (k) + 3H2 (k) 2NH3(k)S mol kh ban u: 2 S mol kh phn ng: x S mol kh lc cn bng: 2-x 7 3x 7 3x 0 2x 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 ng0,4.100% = 20% 2
b/ Th tch (kt) kh ammoniac c to thnh: 2.0,4. 22,4 = 17,9 (lt) 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. Gii a/ pthh ca cc phn ng. t C 2NH3 + 3CuO N2 + 3Cu + 3H2O (1) Cht rn A thu c sau phn ng gm Cu
GV: Yu cu HS tho lun. GV: Hng dn HS cch vit pt.
v CuO cn d . ch c CuO phn ng vi dung dch HCl. CuO + 2HCl CuCl2 + H2O HS:Nghe ging v hiu 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 GV:Yu cu HS ln bng trnh by cu CuO ban u s mol CuO d =b
HS: Ln bng trnh by GV: Gi HS nhn xt
3,2 0,01 = 0,03(mol ) 80
Theo (1), s mol N2=
1 1 s mol CuO = .0,03 3 3
= 0,01 (mol) Th tch kh nit to thnh : 0,01. 22,4 = 0,224 (lt)
Hot ng 4: Cng c - dn d * 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 * Dn d: Chun b tip phn cn li bi Amoniac v mui Amoni
Tit 7:
BI TP AXIT NITRIC
I. Mc tiu: HS vn dng c kin thc hc gii bi tp II. Trng tm: Bi tp axit nitric. III. Chun b: GV:Gio n HS: n tp l thuyt cc bi trc IV.Tin trnh ln lp: 1/ n nh lp 2/ Bi c: Trnh by tnh cht ha hc ca Axit nitric 3/ Bi miH:01:26H:01:26H:01:26
Hot ng ca thy v tr 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 HS: Chp GV: Hng dn HS cch vit pt, gi cch gii, yu cu HS lm HS: Tho lun lm bi
Ni dung 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 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 Ta c:A + 62n 9n = 34 3,6
Gii pt: A = 23n. Ch c nghim n = 1, A = 23 Vy kim loi M trong oxit l natri GV: Yu cu HS cho bit kt qu Na2O + 2HNO3 2NaNO3 + H2O (2) Theo phn ng (2) C to ra 18 g nc th c 62 g Na2O GV: Yu cu HS vit pt v tnh khi phn ng lng ca oxit kim loi phn ng Vy to ra 3,6g nc th c x g Na2O phn ng HS: Vit pt v tnh khi lng ca x = (3,6.62) : 18 = 12,4 (g) oxit kim loi phn ng Hot ng 2: Bi 2: GV: Chp ln bng, yu cu HS Chia hn hp hai kim loi Cu v Al lm 2 chp vo v. phn bng nhau. Bi 2: + Phn th nht: Cho tc dng vi dung dch Chia hn hp hai kim loi Cu v Al HNO3 c ngui thu c 8,96 lt kh NO2 lm 2 phn bng nhau. ( ktc) + Phn th nht: Cho tc dng vi + Phn th hai: Cho tc dng vi hon ton dung dch HNO3 c ngui thu c vi dung dch HCl, thu c 6,72 lt kh 8,96 lt kh NO2 ( ktc) ( ktc) + Phn th hai: Cho tc dng vi Xc nh thnh phn phn trm v khi hon ton vi dung dch HCl, thu lng ca mi kim loi trong hn hp c 6,72 lt kh ( ktc) Gii Xc nh thnh phn phn trm v Phn th nht, ch c Cu phn ng vi HNO3 khi lng ca mi kim loi trong c. hn hp. Cu + 4HNO3 c Cu(NO3)2 + 2NO2 + 2 H2O (1) HS: Chp Phn th 2, ch c Al phn ng vi GV: Yu cu 1 HS ln bng trnh 2Al + 3HCl AlCl3 + 3H2 (2) by. Cc HS cn li lm v theo di Da vo (1) ta tnh c khi lng Cu c bi ca bn trong hn hp l 12,8 g. Da vo (2) ta tnh c khi lng Al c HS:Ln bng trnh by trong hn hp l 5,4 g.H:01:26H:01:26H:01:26
GV: Gi HS nhn xt, ghi im
% khi lng ca Cu = 70, 33% % khi lng ca Al = 29,67% Hot ng 3: Bi 3: GV: Chp ln bng, yu cu HS Cho 12,8 g Cu tc dng vi dung dch HNO3 chp vo v. c, sinh ra kh NO2. Tnh th tch NO2 Bi 3: ( ktc). Cho 12,8 g Cu tc dng vi dung Gii dch HNO3 c, sinh ra kh NO2. Tnh Cu + 4HNO3 c Cu(NO3)2 + 2NO2 + 2 H2O th tch NO2 ( ktc). 0,2 0,4 (mol) HS: Chp 12,8 GV: Yu cu 1 HS ln bng trnh n = = 0,2(mol ) Cu 64 by. Cc HS cn li lm v theo di V NO = 0,4.22,4 = 8,96(l ) bi ca bn HS:Ln bng trnh by GV: Gi HS nhn xt, ghi im2
Hot ng 4: Cng c - dn d * Cng c: Ha tan 12,8 g kim loi ha tr II trong mt lng va dung dch HNO3 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 * Dn d: Chun b tip phn cn li bi Axit v mui nitrat
H:01:26H:01:26H:01:26
Tit 8: BI TP MUI NITRATI. Mc tiu: HS vn dng c kin thc hc gii bi tp II. Trng tm: Bi tp mui nitrat III. Chun b: GV:Gio n HS: n tp l thuyt bi axit nitric v mui nitrat. IV.Tin trnh ln lp: 1/ n nh lp 2/ Bi c: Trnh by tnh cht ha hc ca mui nitrat 3/ Bi mi: - n nh lp 11B2:Ngy ging.........................................S s........................................................... - n nh lp 11B3:Ngy ging.........................................S s........................................................... - n nh lp 11B4:Ngy ging.........................................S s........................................................... - n nh lp 11B5:Ngy ging.........................................S s........................................................... - n nh lp 11B6:Ngy ging.........................................S s........................................................... Hot ng ca thy v tr Hot ng 1: GV: Chp ln bng, yu cu HS chp vo v. 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. HS: Chp GV: Hng dn HS cch vit pt, gi cch gii, yu cu HS lm HS: Tho lun lm bi Ni dung 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.t0
Gii: 2NaNO3 2NaNO2 + O2 (1) x 0,5x ( mol) t0 2Cu(NO3)2 2CuO + 4NO2 + O2 (2) y y 2y 0,5y ( mol) Gi x v y l s mol ca NaNO3 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 GV: Yu cu HS ln bng gii 0,5x + 2y + 0,5y = 0,3 x = y = 0,1 85.0,1.100% HS: Ln bng trnh by = 31,1% % m NaNO3 = 27,3 GV: Nhn xt ghi im 188.0,1.100% = 68,9% % mCu ( NO3 ) 2 = 27.3 Hot ng 2: Bi 2: GV: Chp ln bng, yu cu HS Nung nng 27,3 g hn hp NaNO3 v Cu(NO3)2 ; chp vo v. hn hp kh thot ra c dn vo 89,2 ml Bi 2: H:01:26H:01:26H:01:26
Nung nng 27,3 g hn hp NaNO3 v Cu(NO3)2 ; hn hp kh thot ra 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
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 Gii 2NaNO3 2NaNO2 + O2 (1) 2 1 ( mol) t0 2Cu(NO3)2 2CuO + 4NO2 + O2 (2) 2 4 1 ( mol) 4 HNO3 (3) 4NO2 + O2 + 2H2O 4 1 4 ( mol) a/ Theo pt (1), (2), (3) , nu cn d 1,12 l kh ( hay 0,05 mol ) th l kh O 2, c th coi lng kh ny do mui NaNO3 phn hy to ra T (1) ta c: n NaNO3 = 2.0,05 = 0,1(mol )t0
HS:Ln bng trnh by
m NaNO3 = 0,1.85 = 8,5( g ) mCu ( NO3 ) 2 = 27,3 8,5 = 18,8( g ) GV: Gi HS nhn xt, ghi im nCu ( NO3 ) 2 = 18,8 : 188 = 0,1(mol ) T (2) ta c: n NO2 = nO2 = 0,1 .4 = 0,2( mol ) 2
0,1 .1 = 0,05(mol ) 2 ( Cc kh ny hp th vo nc) T (3) ta c : n HNO3 = n NO2 = 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 % Hot ng 3: GV: Chp ln bng, yu cu HS Bi 3: Nung mt lng mui Cu(NO3). Sau mt thi chp vo v. gian dng li, ngui v em cn th thy khi Bi 3: lng gim i 54g. Nung mt lng mui Cu(NO3). Sau + Khi lng Cu(NO3) b phn hy. mt thi gian dng li, ngui v + S mol cc cht kh thot ra l em cn th thy khi lng gim i 54g. + Khi lng Cu(NO3) b phn hy. Gii + S mol cc cht kh thot ra l t0 HS: Chp 2Cu(NO3)2 2CuO + 4NO2 + O2 GV: Yu cu 1 HS ln bng trnh by. + C 188g mui b phn hu th khi lng gim Cc HS cn li lm v theo di bi ca : 188 80 = 108 (g) bn Vy x = 94 g mui b phn hu th khi lng gim 54 g HS:Ln bng trnh by Khi lng mui b phn hu mCu ( NO3 ) 2 = 94( g ) GV: Gi HS nhn xt, ghi im + nCu ( NO3 ) 2 = 94 : 188 = 0,5(mol ) 0,5 n NO2 = .4 = 1(mol ) 2 H:01:26H:01:26H:01:26
nO2 =
0,5 . = 0,25(mol ) 2
Hot ng 4: Cng c - dn d * 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% * Dn d: Chun b bi Axit photphoric v mui photphat
H:01:26H:01:26H:01:26
Tit 9:
BI TP. AXIT PHOTPHORIC V MUI PHOTPHAT
I. Mc tiu: HS vn dng c kin thc hc gii bi tp II. Trng tm: Bi tp axit photphori v mui photphat III. Chun b: GV:Gio n HS: n tp l thuyt bi axit photphoric v mui photphat. IV.Tin trnh ln lp: 1/ n nh lp 2/ Bi c: Trnh by tnh cht ha hc ca axit photphoric v mui photphat 3/ Bi mi - n nh lp 11B2:Ngy ging.........................................S s........................................................... - n nh lp 11B3:Ngy ging.........................................S s........................................................... - n nh lp 11B4:Ngy ging.........................................S s........................................................... - n nh lp 11B5:Ngy ging.........................................S s........................................................... - n nh lp 11B6:Ngy ging.........................................S s........................................................... Hot ng ca thy v tr Hot ng 1: GV: Chp ln bng, yu cu HS chp vo v. 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 HS: Chp GV: Yu cu HS cch vit pt, gi cch gii, yu cu HS lm HS: Tho lun lm bi GV: Yu cu HS ln bng gii HS: Ln bng trnh by GV: Nhn xt ghi im Hot ng 2: GV: Chp ln bng, yu cu HS chp vo v. Bi 2: Bng phng php ha hc, hy phn H:01:26H:01:26H:01:26 Ni dung 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 Gii: KH2PO4 + H2O (1) H3PO4 + KOH 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 K2HPO4 Bi 2: Bng phng php ha hc, hy phn bit dung dch HNO3 v dung dch H3PO4 Gii Cho mnh kim loi Cu vo dung dch ca tng
bit dung dch HNO3 v dung dch H3PO4 HS: Chp GV: Yu cu HS ln bng trnh by HS:Ln bng trnh by GV: Gi HS nhn xt, ghi im Hot ng 3: GV: Chp ln bng, yu cu HS chp vo v. 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 HS: Chp GV: Yu cu HS chia nhm tho lun. Gi i din mt nhm ln trnh by HS:Ln bng trnh by
axit
Cu + HNO3 () Cu(NO3)2 + 2NO2 + 2H2O Cu khng t dng vi H3PO4
GV: Gi HS nhn xt, ghi im
Hot ng 4: GV: Chp ln bng, yu cu HS chp vo v 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% HS: Chp GV: Hng dn HS cch vit pt. Yu cu HS gii HS:Ln bng trnh by GV: Gi HS nhn xt, ghi im H:01:26H:01:26H:01:26
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 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: 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% Gii 2CaHPO4 + CaSO4 (1) Ca3(PO4)2 + H2SO4 Ca3(PO4)2 + 2H2SO4 Ca(H2PO4)2 + 2CaSO4 (2) Ca3(PO4)2 + 3H2SO4 H3PO4 + 3CaSO4 (3) 62 = 0,2(mol ) S mol Ca3(PO4)2 = 310 49.64 = 0,32(mol ) S mol H2SO4 = 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 = 0,2
a = 0,08; b = 0,12 mCaHPO4 = 2.0,08.136 = 21,76( g ) mCa ( H 2 PO4 ) 2 = 0,12.234 = 28.08( g ) mCaSO4 = (a + 2b).136 = (0,08 + 0,24).136 = 45,52( g ) Hot ng 5: Cng c - dn d Dung dch H3PO4 c cha cc ion ( khng k ion H+v OH- ca nc) A. H+, PO 3 B. H+, PO 3 , H2PO C. H+, PO 3 , HPO D. H+, PO 3 , H2PO , HPO 4 4 4 4 4 4 4 4
Tit 10:
BI TP AXIT NITRIC - MUI NITRAT
I. Mc tiu: HS vn dng c kin thc hc gii bi tp II. Trng tm: Bi tp axit nitric - mui nitrat III. Chun b: GV:Gio n HS: n tp l thuyt cc bi trc IV.Tin trnh ln lp: 1/ n nh lp 2/ Bi c Cn bng phng trnh phn ng sau : R + HNO3 R(NO3)n + NO2 + H2O 3/ Bi mi - n nh lp 11B2:Ngy ging.........................................S s........................................................... - n nh lp 11B3:Ngy ging.........................................S s........................................................... - n nh lp 11B4:Ngy ging.........................................S s........................................................... - n nh lp 11B5:Ngy ging.........................................S s........................................................... - n nh lp 11B6:Ngy ging.........................................S s........................................................... Hot ng ca thy v tr Hot ng 1: GV: Chp ln bng, yu cu HS chp vo v. 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 HS: Chp H:01:26H:01:26H:01:26 Ni dung 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 Gii: 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)
GV: Hng dn HS cch vit pt, gi cch gii, yu cu HS lm
2 4,928 y = = 0,22 (2) 3 22.4 Gii (1) v (2) c x = 0,1; y = 0,03 HS: Tho lun lm bi a/ S mol ca NO2 l 2.0,1 = 0,2 (mol) 2 S mol ca NO l .0,03 = 0,02 (mol) GV: Yu cu 1 HS ln bng gii 3 b/ Tng s mol HNO3 phn ng = 4.0,1 + HS: Ln bng trnh by, cc HS cn li 8 ly nhp lm bi .0,03 = 0,48 (mol) 3 Nng mol/l ca dung dch axit GV: Gi HS nhn xt ghi im 0,48 C M ( HNO3 ) = = 2( M ) 0,24 Hot ng 2: Bi 2: GV: Chp ln bng, yu cu HS Mt lng 13,5 g Al tc dng va vi 2,2 l dd chp vo v. HNO3 cho bay ra mt hn hp gm hai kh NO Bi 2: v N2O. Bit t khi ca hn hp kh so vi hir Mt lng 13,5 g Al tc dng va bng 19,2. vi 2,2 l dd HNO3 cho bay ra mt hn + Tnh s mol ca NO v N2O to ra l hp gm hai kh NO v N2O. Bit t + Tnh nng mol/l ca dd axt u. khi ca hn hp kh so vi hir bng 19,2. Gii + Tnh s mol ca NO v N2O to ra l Al + 4HNO3 Al(NO3)3 + NO + 2H2O + Tnh nng mol/l ca dd axt u. x 4x 2x (mol) 8Al(NO3)3 + 3N2O + 15H2O HS: Chp 8Al + 30HNO3 GV: Gi hng dn HS cch gii, y 30/8y (= 3,75 y) 3/8y (= 0,375 y) yu cu 1 HS ln bng trnh by Theo bi ra ta c: ( x + y ).27 = 13,5 (1) 30.x + 44.0,375 y d hh / H 2 = = 19,2 HS: Ln bng trnh by, cc HS cn li ( x + 0,375 y )2 ly nhp lm bi (2) 2x + Gii (1) v (2) c x = 0,1; y = 0,4 a/ S mol ca NO l = 0,1 (mol) S mol ca N2O l 0,375.0,4 = 0,15 (mol) b/ Tng s mol HNO3 phn ng = 4.0,1 + 3,75.0,4 = 1,9 (mol) Nng mol/l ca dung dch axit 1,9 GV: Gi HS nhn xt, ghi im C M ( HNO3 ) = = 0,86( M ) 2,2 Bi 3: Nung 9,4 gam mt mui nitrat trong mt bnh Hot ng 3: GV: Chp ln bng, yu cu HS kn. Sau khi phn ng xy ra hon ton cn li 4 gam oxit. Tm cng thc ca mui nitrat chp vo v. Bi 3: Nung 9,4 gam mt mui nitrat trong Gii mt bnh kn. Sau khi phn ng xy ra t0 hon ton cn li 4 gam oxit. Tm cng 2R(NO3)2 R2On + 2nNO2 + n/2O2 thc ca mui nitrat a a/2 na na/4 HS: Chp Ta c: a.( MR + 62n) = 9,4 (1) GV: Yu cu 1 HS ln bng trnh by. 0,5.a( 2MR + 16n) = 4 (2) Cc HS cn li lm v theo di bi ca Ly (1) : (2) ta c MR = 32n. Khi n = 2 th MR = bn 64 Vy cng thc mui nitrat Cu(NO3)2 H:01:26H:01:26H:01:26
HS:Ln bng trnh by GV: Gi HS nhn xt, ghi im Hot ng 4: Cng c - dn d * 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 * Dn d: Chun b bi Thc hnh s 2
Tit 11: BI TP TNG KT CHNG NIT - PHOTPHOI. Mc tiu: HS vn dng c kin thc hc gii bi tp II. Trng tm: Bi tp tng kt chng nit - photpho III. Chun b: GV:Gio n HS: n tp l thuyt cc bi trc IV.Tin trnh ln lp: 1/ n nh lp 2/ Bi c: (khng kim tra) 3/ Bi mi: - n nh lp 11B2:Ngy ging.........................................S s........................................................... - n nh lp 11B3:Ngy ging.........................................S s........................................................... - n nh lp 11B4:Ngy ging.........................................S s........................................................... - n nh lp 11B5:Ngy ging.........................................S s........................................................... - n nh lp 11B6:Ngy ging.........................................S s........................................................... Hot ng ca thy v tr Ni dung Hot ng 1: Bi 1: GV: Chp ln bng, yu cu HS Cho 3 mol N2 v 8 mol H2 vo mt bnh kn c chp vo v. th tch khng i cha sn cht xc tc ( th Bi 1: tch khng ng k ). Bt tia la in cho phn Cho 3 mol N2 v 8 mol H2 vo mt bnh ng xy ra, sau a v nhit ban u th kn c th tch khng i cha sn cht thy p sut gim 10% so vi p sut ban u. xc tc ( th tch khng ng k ). Bt Tm % v th tch ca N2 sau phn ng. tia la in cho phn ng xy ra, sau Gii: a v nhit ban u th thy p N2 + 3H2 2NH3 sut gim 10% so vi p sut ban u. Trc phn ng 3 8 0 ( mol) H:01:26H:01:26H:01:26
Tm % v th tch ca N2 sau phn ng. HS: Chp GV: Yu cu HS tho lun lm bi. HS: Tho lun lm bi GV: Cho HS xung phong ln bng gii
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 c n1 P1 11 P 1 = = = x = 0,55 n 2 P2 11 2 x 0,9 P 0,9 HS: Ln bng trnh by, cc HS cn li 3 0,55 ly nhp lm bi %N 2 = .100% = 24,75% 11 2.0,55 GV: Gi HS nhn xt ghi im Bi 2: Hot ng 2: GV: Chp ln bng, yu cu HS Khi ha tan hon ton 1,5875 gam mt kim loi ha tr III trong dung dch HNO3 long thu c chp vo v. 604,8 ml hn hp kh N2 v NO (ktc) c t Bi 2: Khi ha tan hon ton 1,5875 gam mt khi hi so vi H2 l 14,5. Tm tn M kim loi ha tr III trong dung dch HNO3 Gii long thu c 604,8 ml hn hp kh M + 4HNO3 M(NO3)3 + NO + 2H2O N2 v NO (ktc) c t khi hi so vi x 4x 2x (mol) H2 l 14,5. Tm tn M 10M+ 36HNO3 10M(NO3)3 + 3N2 + 18H2O HS: Chp 3/10y GV: Gi hng dn HS cch gii, y 3 yu cu 1 HS ln bng trnh by y = 0,27 Theo bi ra ta c: x + (1) HS: Ln bng trnh by, cc HS cn li 10 ly nhp lm bi 3 30.x + 28. y 10 = 14,5.2 d hh / H 2 = (2) 3 x+ y 10 Gii (1) v (2) c x = 0,0135; y = 0,045 S mol ca M l 0,045 + 0,0135 = 0,0585 (mol) GV: Gi HS nhn xt, ghi im 1,5875 M = = 27 0,0585 Vy M l Al Hot ng 3: Bi 3: ( 2) GV: Chp ln bng, yu cu HS NH4Cl (1) NH3 N2 ( 3) NO ( 4) ( 5) ( 6) chp vo v. NO2 HNO3 NaNO3 Bi 3: (7) NaNO3 (1) ( 2) ( 3) NH4Cl NH3 N2 NO ( 4) ( 5) ( 6) NO2 HNO3 NaNO3 (7) NaNO3 Gii NH3 + H2O + NaCl GV: Yu cu 1 HS ln bng trnh by. 1/ NH4Cl + NaOH Cc HS cn li lm v theo di bi ca 2/ NH3 + 3O2 t 2 N2 + 6H2O bn 3/ N2 + O2 t 2NO 4/ 2NO+ O2 2NO2 HS:Ln bng trnh by 5/ 4NO2 + 2H2O + O2 4 HNO3 6/ HNO3 + NaOH NaNO3 + H2O GV: Gi HS nhn xt, ghi im 7/ 2NaNO3 t 2NaNO2 + O2 Hot ng 4: Bi 4: GV: Chp ln bng, yu cu HS Cho 500ml dung dch KOH 2M vo 500ml dung chp vo v. dch H3PO4 1,5M. Sau phn ng trong dung dch Bi 4: thu c cc sn phm no Cho 500ml dung dch KOH 2M vo H:01:26H:01:26H:01:26
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
Gii 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
Hot ng 5: Cng c - dn d * Cng c: Ha tan 4,59 g Al bng dung dch HNO3 long thu c hn hp kh NO v N2O c t khi V N 2O i vi H2 bng 16,75. T l th tch kh trong hn hp l V NO 1 2 1 3 A. B. C. D. 3 3 4 4 Dn d: Chun b bi Cacbon
Tit 12:
BI TP CACBON V CC HP CHT CA CACBON
I. Mc tiu: HS vn dng c kin thc hc gii bi tp II. Trng tm: Bi tp cacbon v cc hp cht ca cacbon III. Chun b: GV:Gio n HS: n tp l thuyt, lm bi tp cc bi cacbon v cc hp cht ca cacbon IV.Tin trnh ln lp: 1/ n nh lp 2/ Bi c: Trnh tnh cht ca mui cacbonat 3/ Bi mi - n nh lp 11B2:Ngy ging.........................................S s........................................................... - n nh lp 11B3:Ngy ging.........................................S s........................................................... - n nh lp 11B4:Ngy ging.........................................S s........................................................... - n nh lp 11B5:Ngy ging.........................................S s........................................................... - n nh lp 11B6:Ngy ging.........................................S s........................................................... Hot ng ca thy v tr Hot ng 1 GV: Chp ln bng, yu cu HS chp vo v. Bi 1: Nung 52,65 g CaCO3 10000C v cho H:01:26H:01:26H:01:26 Ni dung Bi 1: Nung 52,65 g CaCO3 10000C 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
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% ) HS: Chp GV: Yu cu HS tho lun lm bi.
95% )t 0C
Gii: CaCO3 CaO + CO2 52,65 nCO2 = nCaCO3 = = 0,5265(mol ) 100 V phn ng trn c h = 95% nn s mol CO2 thc t thu c 0,5265 HS: Tho lun lm bi nCO2 = .95 = 0,5002( mol ) 100 nNaOH = 0,5.1,8 = 0,9 (mol) T l s mol NaOH v CO2 n NaOH 0,9 GV: Cho HS xung phong ln bng gii = s mol CO2 X, Y thuc dy ng ng ca ankan. 3n + 1 CnH2n + 2 + O2 t nCO2 + (n+1)H2O 2 HS: Ln bng trnh by, cc HS cn li 0,5 0,7 ly nhp lm bi Ta c : 0,5(n + 1 ) = 0,7n n = 2,5 GV: Gi HS nhn xt ghi im CTPT ca X, Y l: C2H6, C3H8 Bi 2: Hot ng 2 GV: Chp ln bng, yu cu HS Cho 3,5 gam mt anken X tc dng hon chp vo v. ton vi dung dch KMnO4 long d, thu Bi 2: c 5,2 gam sn phm hu c. Tm CTPT
Cho 3,5 gam mt anken X tc dng ca X. hon ton vi dung dch KMnO4 long d, thu c 5,2 gam sn phm hu c. Tm CTPT ca X.
HS: Chp Gii 3CnH2n + 2KMnO4 + 4H2O CnH2n(OH)2 + 2MnO2 + 2KOH GV: Yu cu HS tho lun lm bi. 14n 14n + 34 HS: Tho lun lm bi 5,2 GV: Cho HS xung phong ln bng gii 3,5 Ta c: 3,5( 14n + 34 ) = 5,2.14n HS: Ln bng trnh by, cc HS cn li n = 5 ly nhp lm bi CTPT ca X l C5H10 GV: Gi HS nhn xt ghi im Hot ng 3 GV: Chp ln bng, yu cu HS Bi 3: chp vo v. 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 Bi 3: CTPT ca M, N. t chy hon ton hon hai Gii hirocacbon mch h M, N lin tip trong dy ng ng thu c 22,4 lt H:01:27H:01:27H:01:27
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
Hot ng 4 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.
HS: Chp GV: Yu cu HS tho lun lm bi. HS: Tho lun lm bi 12,6 9,9 GV: Cho HS xung phong ln bng gii n H 2O = = 0,15(mol ) 18 HS: Ln bng trnh by, cc HS cn li 3n 1 ly nhp lm bi CnH2n - 2 + O2 t nCO2 + (n -1)H2O 2 GV: Gi HS nhn xt ghi im 0,225 0,15 Ta c : 0,225(n - 1 ) = 0,15n n = 3 CTPT ca X l: C3H4 IV 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 V.Dn d: Chun b bi tit sau n tp H-C
22,4 = 1(mol ) 22,4 12,6 n H 2O = = 0,7(mol ) 18 S mol nc < s mol CO2 M, N thuc dy ng ng ca ankin. 3n 1 CnH2n - 2 + O2 t nCO2 + (n -1)H2O 2 1 0,7 n = 3,3 Ta c : (n - 1 ) = 0,7n 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 22,5 nCaCO3 = = 0,225(mol ) 100 mCO2 = 44.0,225 = 9,9( gam) nCO2 =
Ngy son:09/01/2010
Tit 25: BI TP TNG KT CHNG
HIROCACBON NO V HIROCACBON KHNG NOI. Mc tiu: HS vn dng c kin thc hc gii bi tp Bi tp tng kt chng hirocacbon no v hirocacbon khng no. II. Chun b: GV:Gio n HS: n tp l thuyt, lm bi tp hirocacbon no v hirocacbon khng no. III.Tin trnh ln lp H:01:27H:01:27H:01:27
1. n nh t chc Ngy dy Lp 11B1 11B2 11B5 11B6
S s
Ghi ch
2. Bi c: (Lng vo bi mi) 3. Bi mi Hot ng ca thy v tr Hot ng 1 GV: Chp ln bng, yu cu HS chp vo v. Bi 1: Hon thnh s phn ng sau Cao su buna C4H4 C4H4 CaCO3 CaO CaC2 C2H2 C2H4 PE Vinylclorua PVC 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: Hon thnh s phn ng sau Cao su buna C4H4 C4H4 CaO CaC2 C2H2 C2H4 PE CaCO3 Vinylclorua PVC Gii: t 1/CaCO3 CaO + CO2 2/ CaO + 3C t CaC2 + CO 3/ CaC2 + 2H2O C2H2 + Ca(OH)2 4/ C2H2 + H2 Pd ,PbCO3 CH2 = CH2 t, p 5/ nCH2 = CH2 xt , (-CH2 CH2 - )n 6/ C2H2 + HCl t CH2 = CH Cl HS: Ln bng trnh by, cc HS cn li 7/ xt,t, p ly nhp lm bi nCH2 = CH (- CH2 - CH - )n GV: Gi HS nhn xt ghi imClxt ,t
Cl
Hot ng 2 GV: Chp ln bng, yu cu HS chp vo v. 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 gam. Nu cho mt lng X nh trn tc dng vi HBr, thu c cht Z, thy khi lng Y, Z thu c khc nhau 9,45gam. HS: Chp GV: Yu cu HS tho lun lm bi. HS: Tho lun lm bi GV: Cho HS xung phong ln bng gii
8/ 2C2H2 CH2 = CH- C = CH 9/ CH2 = CH- C = CH + H2 Pd ,PbCO3 CH2 = CH CH = CH2 t, p 10/ nCH2 = CH CH = CH2 xt , (-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 gam. Nu cho mt lng X nh trn tc dng vi HBr, thu c cht Z, thy khi lng Y, Z thu c khc nhau 9,45gam. Gii CxH2x + H2O CxH2x +1 OH (Y) , (1) CxH2x + HBr CxH2x +1 Br (Z) , (2) tng khi lng bnh = khi lng anken 4,2 (mol ) phn ng n X = 14 x 4,2 mY = (14 x + 18) 14 x mZ = 4,2 (14 x + 81) 14 x
HS: Ln bng trnh by, cc HS cn li H:01:27H:01:27H:01:27
ly nhp lm bi GV: Gi HS nhn xt ghi im
mZ - mY =
x=2 CTPT ca X: C2H4 Bi 3: Hot ng 3 Khi t mt th tch hirocacbon A mch h cn GV: Chp ln bng, yu cu HS 30 th tch khng kh, sinh ra 4 th tch kh CO2. chp vo v. A tc dng vi H2 ( xt Ni ), to thnh mt Bi 3: hirocacbon no mch nhnh. Xc nh CTPT, Khi t mt th tch hirocacbon A CTCT ca A. mch h cn 30 th tch khng kh, Gii sinh ra 4 th tch kh CO2. A tc dng vi H2 ( xt Ni ), to thnh mt y y hirocacbon no mch nhnh. Xc nh CxHy + (x + )O2 xCO2 + H2O 4 2 CTPT, CTCT ca A. 20 20 Vkk = .30 = 6 Th tch oxi phn ng: VO2 = 100 100 (lt) Ta c phng trnh x = 4, x + y/4 = 6 y = 8 HS: Chp GV: Gi hng dn HS cch gii, A c CTPT C4H8 mch h nn A thuc loi anken. V A tc dng vi H2 to thnh mt yu cu HS ln bng trnh by hirocacbon no mch nhnh. CTCT ca A. CH2 = C CH3 GV: Gi HS nhn xt ghi im CH3
4,2 4,2 (14 x + 81) (14 x + 18) =9,45 14 x 14 x
IV 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 V. Dn d: Chun b bi Benzen v ng ng ca benzen
Ngy son:09/01/2010 Tit 26: BI TP
BENZEN V NG NG. MT S HIROCACBON THM KHCI. Mc tiu H:01:27H:01:27H:01:27
HS vn dng c kin thc hc gii bi tp 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 Ngy dy Lp S s Ghi ch 11B2 11B3 11B4 11B5 11B6 2. Bi c: Trnh by tnh cht ha hc ca benzen
3. Bi mi:
H:01:27H:01:27H:01:27
Hot ng ca thy v tr Hot ng 1 GV: Chp ln bng, yu cu HS chp vo v. Bi 1: A l mt ng ng ca benzen c t khi hi so vi metan bng 5,75. A tham gia cc qu trnh chuyn ha theo s sau: + Cl2 B (1mol ) + H 2,t C , Ni + HNO3 ,( 3 mol ), H 2 SO4 A D + KMnO4 ,t E 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. 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: A l mt ng ng ca benzen c t khi hi so vi metan bng 5,75. A tham gia cc qu trnh chuyn ha theo s sau: + Cl2 B (1mol ) + H 2,t C , Ni + HNO3 ,( 3 mol ), H 2 SO4 A D + KMnO4 ,t E 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. Gii 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 Ni,t C6H5 CH3 + 3H2 C6H11CH3 C: Metylxiclohexan C6H5-CH3 + 3HNO3 H 2 SO4 C6H2(NO2)3CH3 +
HS: Ln bng trnh by, cc HS cn li 3H O 2 ly nhp lm bi D: TNT GV: Gi HS nhn xt ghi im (trinitrotoluen) C6H5 CH3 + KmnO4 t C6H5-COOK Hot ng 2 + KOH + 2MnO2 + H2O GV: Chp ln bng, yu cu HS E: kali benzoat chp vo v. Bi 2: Bi 2: Cht A l mt ng ng ca benzen. Khi t Cht A l mt ng ng ca benzen. chy hon ton 1,5 g cht A, ngi ta thu Khi t chy hon ton 1,5 g cht A, c 2,52 lt kh CO2 ( ktc). ngi ta thu c 2,52 lt kh CO2 a/ Xc nh CTPT. ( ktc). b/ Vit cc CTCT ca A. Gi tn. a/ Xc nh CTPT. c/ Khi A tc dng vi Br2 c cht xc tc Fe v b/ Vit cc CTCT ca A. Gi tn. nhit th mt nguyn t H nh vi vng c/ Khi A tc dng vi Br2 c cht xc benzen b thay th bi Br, to ra dn xut tc Fe v nhit th mt nguyn t H monobrom duy nht. Xc nh CTCT ca A. nh vi vng benzen b thay th bi Gii Br, to ra dn xut monobrom duy nht. 3n 3 Xc nh CTCT ca A. CnH2n 6 + O2 nCO2 + (n-3)H2O 2 HS: Chp C ( 14n -6)g A to ra n mol CO2 GV: Yu cu HS tho lun lm bi. 2,52 HS: Tho lun lm bi = 0,1125molCO2 C 1,5 g A to ra GV: Cho HS xung phong ln bng gii 22,4 14n 6 n = n=9 1,5 0,1125 CTPT: C9H12 Cc CTCT: HS: Ln bng trnh by, cc HS cn li ly nhp lm bi GV: Gi HS nhn xt ghi im H:01:27H:01:27H:01:27CH3 1,2,3-trimetlybenzen C2H5
CH3 CH3 CH3
CH3 CH3 H3C CH3 CH3 1,2,5-trimetlybenzen
CH3
CH3 1,3,5-trimetlybenzenCH3
C2H5 1- etyl - 2 - metylbenzen 1- etyl - 3 - metylbenzen
C2H5 1- etyl - 4 - metylbenzen
IV Cng c: Nhc li cch gi tn cc ng ng benzen. Cc cch gii bi tp tm CTPT, vit CTCT V.Dn d: Chun b bi ngun hirocacbon thin nhin Ngy son:09/01/2010
Tit 27: BI TP H THNG HA V HIROCACBONI. Mc tiu HS vn dng c kin thc hc gii bi tp 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 ln lp 1. n nh t chc Ngy dy Lp 11B1 11B2 11B5 11B6 S s Ghi ch
2. Bi c: Trnh by thnh phn ca du m 3. Bi mi:
H:01:27H:01:27H:01:27
Hot ng ca thy v tr Hot ng 1 GV: Chp ln bng, yu cu HS chp vo v. Bi 1: Hn hp M cha hai hidrocacbon k tip nhau trong mt dy ng ng. Khi t chy hon ton 13,2 gam hn hp M thu c 20,72t CO2 ( ktc). Hy xc nh CTPT v % khi lng tng cht trong hn hp M. 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
Ni dung Bi 1: Hn hp M cha hai hidrocacbon k tip nhau trong mt dy ng ng. Khi t chy hon ton 13,2 gam hn hp M thu c 20,72t CO2 ( ktc). Hy xc nh CTPT v % khi lng tng cht trong hn hp M. Gii 20,72 nCO2 = = 0,925( mol ) 22,4 mC = 12.0.925 = 11,1( gam) m H = 13,2 11,1 = 2,1( gam) 2,1 n H 2O = = 1,05(mol ) 2 V s mol H2O > s mol CO2 nn hai cht trong hn hp M u l ankan. 3n + 1 CnH2n +2 O2 nCO2 + ( n +1 ) H2O 2 n 0,925 = n = 7,4 n + 1 1,05 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 0,075.100 %C 7 H 16 = .100% = 56,8% 13,2 0,05.114 %C8 H 18 = .100% = 43,2% 13,2 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 2 GV: Chp ln bng, yu cu HS chp vo v. 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 Gii sau phn ng. a/ X c CTPT CnH2n 2, tc dng vi brom: HS: Chp CnH2n 2 + 2Br2 CnH2n 2Br4 12n 10 GV: Yu cu HS tho lun lm bi. .100% = n=3 %C = 14n 2 100 HS: Tho lun lm bi Vy CTPT ca X l C3H4, CTCT ca X l GV: Cho HS xung phong ln bng gii CH3- C = CH b/ Phn ng hiro ha C3H4 + 2H2 C3H8 x 2x C3H4 + H2 C3H6 y y H:01:27H:01:27H:01:27 Ta c h phng trnh HS: Ln bng trnh by, cc HS cn li x + y = 0,1 ly nhp lm bi 2x + y = 0,15 GV: Gi HS nhn xt ghi im
IV. Cng c Nhc li cc kin thc hidrocacbon hc V. Dn d: Chun b bi: Dn xut halogen ca hirocacbon
Ngy son:09/01/2010
Tit 28: BI TP DN XUT HALOGEN CA HIROCACBON V ANCOL I. Mc tiuHS vn dng c kin thc hc gii bi tp 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 ln lp 1. n nh t chc Ngy dy Lp S s 11B1 11B2 11B5 11B6 2. Bi c: Trnh by s chuyn ha gia cc hirocacbon 3. Bi mi: Ghi ch
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Hot ng ca thy v tr Hot ng 1 IV. Cng c GV: Chp ln bng, yu cu HS chp vo v. Bi 1: 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 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 Hot ng 2 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 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: 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 Gii a/ C2H6 + Cl2 as C2H5Cl + HCl C2H5Cl + Mg C2H5MgCl b/ CH3CH2CH2CH3 + Br2 CH3CHBrCH2CH3 + HBr CH3CHBrCH2CH3 + KOH C2 H 5OH CH3- CH = CH CH3 + KBr + H2O H+ CH3- CH = CH CH3 + H2O CH3- CH(OH) CH CH3 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
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: 12.1,792 = 0,96( g ) 22,4 Khi lng H trong 1,44 g H2O: 2.1,44 = 0,16( g ) HS: Ln bng trnh by, cc HS cn li ly 18 nhp lm bi cng l khi lng C v H trong 3,96 g A GV: Gi HS nhn xt ghi im Theo bi ra, A phi cha Cl. Khi lng Cl trong 7,175 g AgCl 35,5.7,175 = 1,775 (g) 143,5 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 0,96 0,16 2,84 : : = 1: 2 :1 x: y: z = 12 1 35,5 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 Hot ng 3 GV: Chp ln bng, yu cu HS chp CH3CHCl2 ; 1,1 icloetan H:01:28H:01:28H:01:28 CH2Cl CH2Cl; 1,2 - icloetan vo v. Bi 3: Bi 3: t chy hon ton mt lng hn hp hai t chy hon ton mt lng hn hp hai
Nhc li tnh cht ca dn xut halogen, ancol V. Dn d: Chun b bi: Phn cn li bi Ancol
Ngy son:09/01/2010
Tit 29: BI TP ANCOL V PHENOLI. Mc tiu HS vn dng c kin thc hc gii bi tp Bi tp Ancol v phenol II. Chun b GV:Gio n HS: n tp l thuyt, lm bi Ancol v phenol III.Tin trnh ln lp 1. n nh t chc Ngy dy Lp S s Ghi ch 11B1 11B2 11B5 11B6 2. Bi c: Trnh by nh ngha, phn loi, danh php Ancol. Ly VD minh ha. 3. Bi mi
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Hot ng ca thy v tr Hot ng GV: Chp ln bng, yu cu HS chp vo v. Bi 1: 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. 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: 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.
Gii 2C3H5 (OH)3 + Cu(OH)2 [C3H5(OH)2O]2Cu + 2H2O S mol gixerol trong 8,12 g A = 2 s mol 1,96 = 0,04(mol ) Cu(OH)2 = 2. 98 S mol gixerol trong 20,3 g A: HS: Ln bng trnh by, cc HS cn li 0,04.20,3 = 0,1(mol ) ly nhp lm bi 8,12 GV: Gi HS nhn xt ghi im 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 0,1 0,15 2ROH + 2Na RONa + H2 x 0,5x S mol H2 = 0,15 + 0,5x = 5,04 = 0,225 x = 0,15 22,4 11,1 = 74 Khi lng 1 mol ROH: 0,15 R = 29; R l C4H9 CTPT: C4H10OPhn trm khi lng C4H9OH =
Gii 2CnH2n+1OH (2CnH2n +1)2O + H2O 2Cm H2m +1OH (2CmH2m+1)2O + H2O CnH2n+1OH + Cm H2m +1OH CnH2n+1OCmH2m +1 + H2O HS: Ln bng trnh by, cc HS cn li S mol 3 ete = s mol nc = 21,6 = 1,2(mol ) 18 ly nhp lm bi S mol mi ete = 0,4 (mol) GV: Gi HS nhn xt ghi im Khi lng 3 ete: (28n + 18).0,4 + ( 28m +18).0,4 + (14n + 14m + 18).0,4=72 H:01:28H:01:28H:01:28 n+m=3 Hot ng 3 GV: Chp ln bng, yu cu HS Hai CTCT ca ancol l: CH3OH, CH3CH2OH chp vo v.
Hot ng 2 GV: Chp ln bng, yu cu HS chp vo v. Bi 2: un nng hn hp 2 ancol no, n chc, mch h vi H2SO4 1400C, thu c 72 gam hn hp 3 ete vi s mol bng nhau. Khi lng nc tch ra trong qu trnh to thnh cc ete l 21,6 gam. Xc nh CTCT ca 2 ancol. HS: Chp GV: Yu cu HS tho lun lm bi. HS: Tho lun lm bi GV: Cho HS xung phong ln bng gii
11,1 .100% = 54,68% 20,3 Bi 2: un nng hn hp 2 ancol no, n chc, mch h vi H2SO4 1400C, thu c 72 gam hn hp 3 ete vi s mol bng nhau. Khi lng nc tch ra trong qu trnh to thnh cc ete l 21,6 gam. Xc nh CTCT ca 2 ancol.
IV. Cng c: Nhc li tnh cht ca ancol v phenol V. Dn d: Chun b bi: Luyn tp Ngy son:09/01/2010
Tit 30: BI TP TNG KT CHNG
DN XUT HALOGEN ANCOL V PHENOLI. Mc tiu HS vn dng c kin thc hc gii bi tp 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 Ngy dy Lp S s 11B1 11B2 11B5 11B6 2. Bi c: ( Lng vo bi mi) 3. Bi mi Ghi ch
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Hot ng ca thy v tr Hot ng 1 GV: Chp ln bng, yu cu HS chp vo v. Bi 1: T etilen vit phng trnh iu ch cc cht sau: 1,2 ibrometan; 1,1 ibrometan; vinylclorua 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 Hot ng 2 GV: Chp ln bng, yu cu HS chp vo v. 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. 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 Hot ng 3 GV: Chp ln bng, yu cu HS chp vo v. 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. HS: Chp
Ni dung Bi 1: T etilen vit phng trnh iu ch cc cht sau: 1,2 ibrometan; 1,1 ibrometan; vinylclorua Gii 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: 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. 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: 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.
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 = 0,08.14,45 = 0,1(mol ) GV: Gi hng dn HS cch gii, yu 11,56 cu HS ln bng trnh by C6H5OH + Na C6H5ONa + 1/2H2 0,1 0,05 CH3OH + Na CH3ONa + 1/2H2 x x/2 C2H5OH + Na C2H5ONa + 1/2H2 H:01:28H:01:28H:01:28 y y/2 GV: Gi HS nhn xt ghi im 2,787.750 = 0,112(mol ) S mol H2 = 0,082.300.760
IV. Cng c: Nhc li tnh cht ca dn xut halogen ancol v phenol V. Dn d: Chun b bi: Anehit Xeton
Tit 31: BI TP ANEHIT XETONI. Mc tiu: HS vn dng c kin thc hc gii bi tp 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 Ngy dy Lp S s 11B1 11B2 11B5 11B6 2. Bi c: Trnh by tnh cht ha hc ca Anehit - Xeton 3. Bi mi:
Ghi ch
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Hot ng ca thy v tr Hot ng 1 GV: Chp ln bng, yu cu HS chp vo v. Bi 1: Cht A l mt anehit n chc. Cho 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. 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
Ni dung Bi 1: Cht A l mt anehit n chc. Cho 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. Gii RCHO + 2AgNO3 + 3NH3 + H2O RCOONH4 + 2NH4NO3 + 2Ag 3Ag + 4HNO3 3AgNO3 + NO + 2H2O 3,85.0,8 n NO = = 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) 10,5 = 56 Khi lng 1 mol RCHO = 0,1875 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 lng
Hot ng 2 GV: Chp ln bng, yu cu HS chp vo v. 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. HS: Chp GV: Yu cu HS tho lun lm bi. HS: Tho lun lm bi GV: Cho HS xung phong ln bng gii
m A = mCO2 + mH 2O mO 2 = + 1,8 3,08 .32 = 1,8( gam) 22,4
2,24 .44 22,4
Khi lng C trong 1,8 gam A : 12.2,24 = 1,2( gam) HS: Ln bng trnh by, cc HS cn li 22,4 ly nhp lm bi Khi lng H trong 1,8 gam A : 2.1,8 = 0,2( gam) 18 Khi lng O trong 1,8 gam A: 1,8 1,2 0,2 = 0,4 (gam) Cng thc cht A c dng: CxHyOz 1,2 0,2 0,4 : : = 4 : 8 :1 x:y:z = 12 1 16 CTGN l: C4H8O GV: Gi HS nhn xt ghi im b/ MA = 2,25.32 = 72g/mol H:01:28H:01:28H:01:28 CTPT trng CTGN: C4H8O c/ Cc hp cht cacbonyl CH3 CH2 CH2 CHO butanal
IV. Cng c: 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 32: BI TP AXIT CACBOXYLICI. Mc tiu HS vn dng c kin thc hc gii bi tp Bi tp Axit cacboxylic II. Chun b GV:Gio n HS: n tp l thuyt, lm bi Axit cacboxylic III.Tin trnh ln lp 1. n nh t chc Ngy dy Lp S s 11B1 11B2 11B5 11B6 2. Bi c: Trnh by tnh cht ha hc ca Axit cacboxylic 3. Bi mi
Ghi ch
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Hot ng ca thy v tr Hot ng 1 GV: Chp ln bng, 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 . 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: 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 .
Gii RCOOH + KOH RCOOK + H2O 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 ) HS: Ln bng trnh by, cc HS cn li ly 50 nhp lm bi S mol RCOOH trong 125 ml dung dch axit l: GV: Gi HS nhn xt ghi im 1,2.125 = 0,15( mol ) 1000 cng l s mol mui thu c sau khi c cn dung dch . 16,8 = 112 Khi lng 1 mol mui l: 0,15 RCOOK = 112 R = 29 R l C2H5 CTPT ca axit l: C3H6O2 CTCT: CH3 CH2 COOH axit propanoic Bi 2: Hot ng 2 Cht A l mt axit no, n chc, mch h. t GV: Chp ln bng, yu cu HS chp chy hon ton 2,225 gam A phi dng va ht vo v. 3,64 lt O2 ( ktc). Bi 2: Xc nh CTPT, CTCT v tn gi. 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 Gii GV: Yu cu HS tho lun lm bi. 3n 2 O2 nCO2 + nH2O CnH2nO2 + HS: Tho lun lm bi 2 Theo phng trnh ( 14n + 32)g axit tc dng vi GV: Cho HS xung phong ln bng gii 3n 2 mol O2 2 Theo bi ra 2,25 gam axit tc dng vi 0,1625 mol O2 14n + 32 3n 2 = n=5 2,55 0,1625.2 HS: Ln bng trnh by, cc HS cn li ly nhp lm bi CTPT C5H10O2 CH3 CH2 CH2 CH2 COOH axit pentanoic CH3 CH CH2 COOH axit -3-metylbutanoic CH3 CH3 CH2 CH COOH axit -2-metylbutanoic CH3 CH3 GV: Gi HS nhn xt ghi im CH3 C COOH axit -2,2 -dimetylpropanoic H:01:28H:01:28H:01:28 CH3 Hot ng 3 Bi 3: GV: Chp ln bng, yu cu HS chp Dung dch X c cha ng thi hai axit cacboxylic
IV. 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 V. Dn d: Chun b bi: Luyn tp
Tit 33: BI TP ANEHIT XETON AXIT CACBOXYLICI. Mc tiu: HS vn dng c kin thc hc gii bi tp Bi tp anehit xeton axit cacboxylic II. Chun b: GV:Gio n HS: n tp l thuyt, lm bi anehit xeton axit cacboxylic III.Tin trnh ln lp: 1. n nh t chc Ngy dy Lp S s Ghi ch 11B1 11B2 11B5 11B6 2. Bi c: Trnh by tnh cht ha hc ca anehit v axit cacboxylic 3. Bi mi
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Hot ng ca thy v tr Hot ng 1 GV: Chp ln bng, 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 HS: Ln bng trnh by, cc HS cn li ly nhp lm bi GV: Gi HS nhn xt ghi im
Ni dung 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. 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 Theo phng trnh: (12x + y +16 ) g M to ra x mol CO2 v y/2 mol H2O. 1,45g M to ra 0,075 mol CO2 v 0,075 mol H2O 12 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 axit n chc X cn 100ml dung dch NaOH 1,25M ( hiu sut 100%) a/ Tm CTPT, vit CTCT v tn gi ca X. b/ C cn dung dch sau khi trung ha th thu c bao nhiu gam mui khan.
Hot ng 2 GV: Chp ln bng, yu cu HS chp vo v. Bi 2: Trung ha 250g dung dch 3,7% ca 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. b/ C cn dung dch sau khi trung ha Gii th thu c bao nhiu gam mui khan. a/ Axit n chc, cng thc CxHyCOOH HS: Chp CxHyCOOH + NaOH CxHyCOONa + H2O GV: Yu cu HS tho lun lm bi. HS: Tho lun lm bi S mol NaOH = 0,125 (mol); khi lng axit X GV: Cho HS xung phong ln bng gii = 9,25 gam. Theo phng trnh s mol axit = s mol NaOH 9,25 = 74( g / mol ) Maxit = 0,125 12x + y = 29 suy ra x = 2; y = 5 CTPT l C3H6O2 HS: Ln bng trnh by, cc HS cn li CTCT l CH3CH2COOH axit propionic. ly nhp lm bi b/ CH3CH2COOH + NaOH C2H5COONa + H2O C cn dung dch sau trung ha thu c mui khan C2H5COONa c s mol bng s mol GV: Gi HS nhn xt ghi im NaOH l 0,125 mol Khi lng mui khan l 0,125 .96 = 12gam Hot ng 3 Bi 3: H:01:28H:01:28H:01:28 GV: Chp ln bng, yu cu HS Trnh by phng php ha hc phn bit cc chp vo v. cht lng: HCOOH, CH3COOH, CH3CH2OH, Bi 3: CH2 = CHCOOH. Vit phng trnh minh ha.
IV. 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. V. Dn d Chun b bi: Bi thc hnh 6
Tit 34: BI TP TNG KT CHNG ANEHIT XETON AXIT CACBOXYLICI. Mc tiu: HS vn dng c kin thc hc gii bi tp Bi tp anehit xeton axit cacboxylic III. Chun b: GV:Gio n HS: n tp l thuyt, lm bi anehit xeton axit cacboxylic III.Tin trnh ln lp 1. n nh t chc Ngy dy Lp S s Ghi ch 11B1 11B2 11B5 11B6 2. Bi c: (Lng vo bi mi) 3. Bi mi
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Hot ng ca thy v tr Hot ng 1 GV: Chp ln bng, 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. 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
Ni dung 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. nY = 0,01(mol ) n 4,32 0,01 1 n Ag = = 0,04(mol ) Y = = 108 n Ag 0,04 4 C 2 trng hp + Nu Y l HCHO 16.100% = 53% 37,21% (loi) %mO = 30 + Nu Y l R(CHO)2 = CxHyO2 32.100% = 37,21% M Y = 86 %mO = MY 12x + y = 86 suy ra x = 4, y = 6 CTCT: CHO CH2 CH2 CHO 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. Gii
Hot ng 2 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. Gii HS: Chp a/ GV: Yu cu HS tho lun lm bi. CH3CHO + 2AgNO3 + 4NH3 + H2O HS: Tho lun lm bi CH3COONH4 + 2NH4NO3 + 2Ag C2H5CHO + 2AgNO3 + 4NH3 + H2O GV: Cho HS xung phong ln bng gii C2H5COONH4 + 2NH4NO3 + 2Ag b/ Gi x, y ln lt l s mol anehit axetic, anehit propioic. 44x + 58y = 10,2 HS: Ln bng trnh by, cc HS cn li 2x + 2y = 0,4 ly nhp lm bi Gii h x = y = 0,1 0,1.44.100% GV: Gi HS nhn xt ghi im = 43,14% %CH3CHO = 10,2 %C2H5CHO = 56,86% Hot ng 3 GV: Chp ln bng, yu cu HS Bi 3: chp vo v. Ha tan 13,4 g hn hp hai axit cacboxylic no, Bi 3: Ha tan 13,4 g hn hp hai axit n chc, mch h vo nc c 50 g dung cacboxylic no, n chc, mch h vo dch A. Chia A thnh 2 phn bng nhau. Cho H:01:28H:01:28H:01:28 nc c 50 g dung dch A. Chia A phn th nht phn ng hon ton vi lng thnh 2 phn bng nhau. Cho phn th d bc nitrat trong dung dch ammoniac, thu nht phn ng hon ton vi lng d c 10,8 g bc. Phn th 2 c trung ha
IV Cng c Tnh cht ha hc ca anehit, xeton, axit cacboxylic. V. Dn d Chun b bi: n tp cc kin thc hc chun b n tp hc k II
Tit 35: N TP HC K II I. Mc tiu: HS vn dng c kin thc hc gii bi tp Kin thc chng 5, 6, 7, 8 II. Chun b: GV:Gio n HS: n tp l thuyt, lm bi chng 5, 6, 7, 8 III.Tin trnh ln lp: Hot ng 1: n nh lp + Bi c Bi c: Khng kim tra Bi mi:
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Hot ng ca thy v tr Hot ng 2: GV: Chp ln bng, 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: Bng 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 li C6H5OH + 3Br2 C6H2Br3OH + 3HBr 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 vit T CaC2 v cht v c cn thit c y phng trnh iu ch caosu buna, nha PE, vit phng trnh iu ch caosu PVC, CH3CHO buna, nha PE, PVC, CH3CHO HS: Chp Gii GV: Yu cu HS tho lun lm bi. C2H2 + Ca(OH)2 CaC2 + 2H2O HS: Tho lun lm bi 2C2H2 xt CH2 = CH C = CH CH2 = CH C = CH + H2 Pd CH2 = CH CH = CH2 GV: Cho HS xung phong ln bng gii xt , p ,t nCH2 = CH CH = CH2 (- CH2 CH = CH CH2 - )n Pd CH2 = CH2 HS: Ln bng trnh by, cc HS cn li C2H2 + H2 p nCH2 = CH2 xt , ( - CH2 CH2 - )n ,t ly nhp lm bi C2H2 + HCl xt CH2 = CH Cl p CH2 = CH Cl xt , ( - CH2 CH - )n ,t Cl GV: Gi HS nhn xt ghi im HgSO4 C2H2 + H2O CH3CHO Hot ng 4: GV: Chp ln bng, yu cu HS chp vo v. Bi 3: Cho 21,4 gam hn hp kh A gm metan, etilen, axetilen qua dung dch brom, thy c 112 gam brom tham gia phn ng. Mt khc, nu cho 21,4 gam kh A trn qua dung dch bc nitrat trong amoniac thy c 24 gam kt ta. a/ Vit cc phng trnh ha hc xy ra. b/ Tnh thnh phn % theo khi lng mi cht trong hn hp H:01:28H:01:28H:01:28 A. HS: Chp Bi 3: Cho 21,4 gam hn hp kh A gm metan, etilen, axetilen qua dung dch brom, thy c 112 gam brom tham gia phn ng. Mt khc, nu cho 21,4 gam kh A trn qua dung dch bc nitrat trong amoniac thy c 24 gam kt ta. a/ Vit cc phng trnh ha hc xy ra. b/ Tnh thnh phn % theo khi lng mi cht trong hn hp A. Gii C2H4 + Br2 C2H4Br2 y y C2H2 + 2Br2 C2H2Br4 z 2z
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
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