BAM SAT 11 MOI

Trng THPT Quang Trung

Bm st chng trnh chun

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 +3 0 em cn li lm vo v nhp v quan Al + 3e 1x Al st +2 +5 HS: Ln bng trnh by 1x N + 3e NGio Vin : V Quc Sanh 10 +5 +3 +2


Al + 4HNO3 Al(NO3)3 + NO + 2H2O0 0 +6 +3 +3

Bm st chng trnh chun+4

Fe + H2SO4 () t Fe2(SO4)3 + SO2 + H2O

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 S2Fe + 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 Hot ng 3: ngi ta thu 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 2Al + 6HCl 2AlCl3 + 3H2 cch lm bi. x 3x 3/2x MgCl2 + H2 Mg + 2HCl 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 3/2x + y = 0,075 % Mg = x = 1/30 y = 0,025

GV: Gi HS nhn xt

0,025 .24 .100 = 40 % 1,5

% Al = 60 %n HCl = 3 x + 2 y = 3. V = 1 + 2.0,025 = 0,15 ( mol ) 30 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 MUI2

Gio Vin : V Quc Sanh



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:

H+ + BrO4HBrO4 Cu2+ + SO 2 CuSO4 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: Hot ng 2: Vit phng trnh in li ca hiroxit lng GV: Chp ln bng, yu cu HS tnh Al(OH) . 3 chp vo v. Bi 2: Vit phng trnh in li ca hiroxit lng tnh Al(OH)3. Gii: HS: Chp GV: Yu cu HS suy ngh 3 pht, Al(OH) Al3+ + 3OH3 sau gi 1 HS ln bng gii. GV Al(OH)3 H3O+ + AlO 2 quan st cc HS lm bi. GV: Nhn xt, hng dn liGio Vin : V Quc Sanh 3



Hot ng 3: 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 Hot ng 5: GV: Chp ln bng, yu cu HS chp vo v. 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

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

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. CdSO4Gio Vin : V Quc Sanh

D. CsOH4



- Theo thuyt Arniut, cht no di y l baz? A. Cr(NO3)3 B. HBrO3 C. CdSO4 D. NH3 * Dn d: 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 .

a/ pH = 2 [H+] = 10-2 = 0,01M H2SO4 2 H+ + SO 2 4 [H2SO4] = b/ [OH-] =1 1 [H+] = .0,01 = 0,005M 2 2

Gii:

GV: Yu cu 1 HS nhn xt, GV nhn xt ghi im. Bi 2: Hot ng 2: Cho m gam natri vo nc, ta thu c 1,5 GV: Chp ln bng, yu cu HS lt dd c pH = 13. Tnh m. chp vo v. Bi 2:Gio Vin : V Quc Sanh 5

10 14 = 10 12 M 10 2



Cho m gam natri vo nc, ta thu c 1,5 lt dd c pH = 13. Tnh m. Gii: HS: Chp GV: Hng dn HS cch gii. pH = 13 [H+] = 10-13 [OH-] = 10-1 = 0,1M HS: Nghe ging v hiu 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 Hot ng 3: Bi 3: GV: Chp ln bng, yu cu HS Tnh pH ca dd cha 1,46 g HCl trong 400,0 chp vo v. ml. Bi 3: Tnh pH ca dd cha 1,46 g HCl trong 400,0 ml. Gii: HS: Chp 1,46 1000 GV: Yu cu HS suy ngh , sau 1 CM(HCl) = 36 ,5 . 400 ,0 = 0,100 M =10 M gi 1 HS ln bng gii. Cc HS cn li ly nhp ra lm bi v theo di [H+] = [HCl] = 10-1M pH = 1,0 bi bn lm. HS: Ln bng trnh by GV: Yu cu 1 HS nhn xt, GV nhn xt ghi im Bi 4: Hot ng 4: GV: Chp ln bng, yu cu HS Tnh pH ca dd to thnh sau khi trn 100,0 ml dd HCl 1,00M vi 400,0 ml dd NaOH chp vo v. 0,375M. 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: HS: Chp nNaOH = 0,4.0,375 = 0,15 (mol) GV:Hng dn HS cch gii tnh nHCl = 0,1.1,000 = 0,10 ( mol) [OH-] Sauk hi trn NaOH d nNaOH (d) = 0,15 0,10 = 0,05 (mol) S mol NaOH = s mol OH- = 0,05 (mol) [OH-] = HS: Nghe ging v hiu GV: Yu cu HS tnh [H+] v pH HS: Tnh [H+] v pH [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

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 7Gio Vin : V Quc Sanh 6



C. bng 7 D. ln hn 7 * Dn d: Chun b bi phn ng trao i ion trong dd cht in li

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 + + OH- NH3 + H2O 4 2d/ S + 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.Gio Vin : V Quc Sanh

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 + + OH- NH3 + H2O 4 2d/ S + 2H+ H2S

a/ Ba(NO3)2 + Na2CO3b/ Fe2(SO4)3 + 6NaOH

c/ NH4Cl + NaOH NH3 + H2O + NaCl d/ FeS + 2HCl FeCl2 + H2S

BaCO3 + 2NaNO3 2Fe(OH)3 + 3Na2SO4

Gii:

Bi 2: a/ MgCO3 + ? MgCl2 + ?. b/ Fe2(SO4)3 + ? K2SO4 + ?

7

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 im 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


Bm st chng trnh chun a/ MgCO3 + 2HCl MgCl2 + H2O + CO2 b/ Fe2(SO4)3 + 6KOH 3K2SO4 + Fe(OH)3

Gii:

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 Ba SO 4 =

1,864 = 0,008 ( mol ) 233

Theo phng trnh (1) s mol BaSO4 = s mol BaCl2.xH2OM=

1,952 = 244 0,008

x=

244 208 =2 18

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

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 =128


ngha l: [H ] = 10 M [OH-] = 10-2M S mol OH- trong dung dch = 0,01.0,5 = 0,005 (mol) GV: Hng dn HS tnh khi lng Ba(OH)2 Ba2+ + 2OH1 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 + 0,0025 = 0,015 (mol)+ -12



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 4Gio Vin : V Quc Sanh

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.9



NO 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. 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: 150 ml dung dch KOH vo 50 ml dung dch H2SO4 1M, dung dch tr thnh d baz. C cn dung dch thu c 11,5 gam cht rn. Tnh nng mol/lt ca dung dch KOH. HS: Chp GV: Yu cu HS tho lun , gi 1 HS ln bng trnh by HS: Ln bng trnh by GV: Yu cu 1 HS nhn xt, GV nhn xt ghi im. Hot ng 3: GV: Chp ln bng, yu cu HS chp vo v. Bi 3: Thm t t 400 g dung dch H2SO4 49% vo nc v iu chnh lng nc thu c ng 2 lt dung dch A. Coi H2SO4 in li hon ton c 2 nc. a/ Tnh nng mol ca ion H+ trong dung dch A. b/ Tnh th tch dung dch NaOH 1,8M cn thm vo 0,5 lt dung dchGio Vin : V Quc Sanh

3

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. b/ Cho dung dch H2SO4 vo c 3 dung dch . dung dch Na2CO3 c si bt: Na2CO3 + H2SO4 Na2SO4 + H2O + CO2 dung dch Ba(NO3)2, xut hin kt ta trng. Ba(NO3)2 + H2SO4 BaSO4 + 2HNO3 Dung dch MgSO4 vn trong sut. Bi 2: 150 ml dung dch KOH vo 50 ml dung dch H2SO4 1M, dung dch tr thnh d baz. C cn dung dch thu c 11,5 gam cht rn. Tnh nng mol/lt ca dung dch KOH. Gii S mol H2SO4 = 0,05 (mol) V baz d nn axit phn ng ht. 2KOH + H2SO4 K2SO4 + 2H2O 0,1 0,05 0,05 (mol) C cn dung dch , thu c cht rn gm c K2SO4, KOH dm K 2SO 4 = 0,05.174 = 8,7(gam)

mKOH(d) = 11,5 8,7 = 2,8 (gam) nKOH(d) = 2,8:56 = 0,05 (mol) S mol KOH c trong 150 ml dung dch KOH l. 0,1 + 0,05 = 0,15 (mol) Nng mol/l ca dung dch KOH: CM(KOH) = 0,15: 0,15 = 1M Bi 3: Thm t t 400 g dung dch H2SO4 49% vo nc v iu chnh lng nc thu c ng 2 lt dung dch A. Coi H2SO4 in li hon ton c 2 nc. a/ Tnh nng mol ca ion H+ trong dung dch A. b/ Tnh th tch dung dch NaOH 1,8M cn thm vo 0,5 lt dung dch A thu c dung dch . + Dung dch c Ph = 1 + Dung dch c Ph = 1310



A thu c dung dch . Gii 400.49 + Dung dch c Ph = 1 = 2(mol) a/ S mol H2SO4: 100.98 + Dung dch c Ph = 13 GV: Yu cu 1 HS ln bng gii cu H2SO4 2H+ + SO 2 4 a, cc HS cn li lm nhp v theo 2 4 (mol) 4 di bi bn lm. Nng H+ trong dung dch A l : = 2M 2 HS: Ln bng trnh by 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 GV: Gi HS nhn xt NaOH trong l 1,8x. NaOH Na+ + OHGV: Hng dn HS lm cu b 1,8x 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: HS: Nghe ging v hiu + 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 1 1,8 x = 0,1M x = 0,5(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 nGio Vin : V Quc Sanh 11



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 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:Gio Vin : V Quc Sanh

Ni dung 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 cht12



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 tho lun. GV: Hng dn HS cch vit pt.

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. 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 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 =tC

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 = 3 3

.0,03 = 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 lpGio Vin : V Quc Sanh 13



2/ Bi c: Trnh by tnh cht ha hc ca Axit nitric 3/ Bi mi 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 + 62 n 9n = 34 3,6

Gii pt: A = 23n. Ch c nghim n = 1, A = 23 Vy kim loi M trong oxit l natri Na2O + 2HNO3 2NaNO3 + H2O (2) GV: Yu cu HS cho bit kt qu 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 Bi 2: Hot ng 2: GV: Chp ln bng, yu cu HS Chia hn hp hai kim loi Cu v Al lm 2 phn bng nhau. chp vo v. + Phn th nht: Cho tc dng vi dung dch Bi 2: Chia hn hp hai kim loi Cu v Al HNO3 c ngui thu c 8,96 lt kh NO2 ( ktc) lm 2 phn bng nhau. + 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 ( ktc) 8,96 lt kh NO2 ( 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 Gii c 6,72 lt kh ( ktc) Xc nh thnh phn phn trm v Phn th nht, ch c Cu phn ng vi khi lng ca mi kim loi trong HNO3 c. Cu + 4HNO3 c Cu(NO3)2 + 2NO2 + 2 H2O (1) hn hp. Phn th 2, ch c Al phn ng vi HS: Chp Gio Vin : V Quc Sanh 14

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. 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. Cu + 4HNO3 c Cu(NO3)2 + 2NO2 + 2 H2O Tnh th tch NO2 ( ktc). 0,2 0,4 (mol) HS: Chp GV: Yu cu 1 HS ln bng trnh nCu = 12 ,8 = 0,2(mol ) 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

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: Hot ng ca thy v tr Ni dung Hot ng 1: Bi 1: GV: Chp ln bng, yu cu HS Nhit phn hon ton 27,3 gam hn hp rn Gio Vin : V Quc Sanh 15

Trng THPT Quang Trung 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 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: 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

Bm st chng trnh chun 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. Gii: 2NaNO3 2NaNO2 + O2 (1) x 0,5x ( mol) t 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 = 0,3 x = y = 0,1t00

% m NaNO =3

85 .0,1.100 % = 31,1% 27 ,3

% mCu ( NO ) =3 2

188 .0,1.100 % = 68,9% 27 .3

Bi 2: 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 Gii 2NaNO3 t 2NaNO2 + O2 (1) 2 1 ( mol) 2Cu(NO3)2 t 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 O 2, c th coi lng kh ny do mui NaNO3 phn hy to ra T (1) ta c: n NaNO 3 = 2.0,05 = 0,1(mol ) m NaNO = 0,1.85 = 8,5( g )00

HS:Ln bng trnh by

3

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

T (2) ta c: n NO =2

0,1 .4 = 0,2( mol ) 2

GV: Gi HS nhn xt, ghi im

nO2 =

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 + Gio Vin : V Quc Sanh 16



89,2 = 100 (g) C% ( HNO3) = 12,6 % Bi 3: Nung mt lng mui Cu(NO3). Sau mt thi Hot ng 3: gian dng li, ngui v em cn th thy khi GV: Chp ln bng, yu cu HS lng gim i 54g. chp vo v. + Khi lng Cu(NO3) b phn hy. Bi 3: + S mol cc cht kh thot ra l Nung mt lng mui Cu(NO3). Sau mt thi gian dng li, ngui v em cn th thy khi lng gim i 54g. Gii + Khi lng Cu(NO3) b phn 2Cu(NO3)2 t 2CuO + 4NO2 + O2 hy. + C 188g mui b phn hu th khi lng gim + S mol cc cht kh thot ra l : 188 80 = 108 (g) HS: Chp Vy x = 94 g mui b phn hu th khi lng GV: Yu cu 1 HS ln bng trnh by. gim 54 g Cc HS cn li lm v theo di bi ca Khi lng mui b phn hu bn mCu ( NO ) = 94 ( g )0

3 2

HS:Ln bng trnh by GV: Gi HS nhn xt, ghi im

+ nCu ( NO3 ) 2 = 94 : 188 = 0,5(mol )n NO 2 = 0,5 .4 = 1( mol ) 2 0,5 = . = 0,25 (mol ) 2

nO2

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

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 Hot ng ca thy v tr Ni dung Hot ng 1: Bi 1: GV: Chp ln bng, yu cu HS Cho 11,76 g H3PO4 vo dung dch cha 16,8 g Gio Vin : V Quc Sanh 17

Trng THPT Quang Trung 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 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

Bm st chng trnh chun KOH. Tnh khi lng ca tng mui thu c sau khi cho dung dch bay hi n kh H3PO4 + KOH KH2PO4 + H2O (1) H3PO4 + 2KOH K2HPO4 + 2H2O (2) H3PO4 + 3KOH K3PO4 + 3H2O (3) Gii:

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 axit Cu + HNO3 () Cu(NO3)2 + 2NO2 + 2H2O Cu khng t dng vi H3PO4

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. GV: Gi HS nhn xt, ghi im 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: Hot ng 4: Cho 62 g canxi photphat tc dng vi 49 g dung GV: Chp ln bng, yu cu HS dch H2SO4 64%. Lm bay hi dung dch thu chp vo v c n cn kh th c mt hn hp rn, Gio Vin : V Quc Sanh 18

Trng THPT Quang Trung 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

Bm st chng trnh chun bit rng cc phn ng u xy ra vi hiu sut 100% 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 =

62 = 0,2(mol ) 310 49 .64 = 0,32 (mol ) GV: Hng dn HS cch vit pt. Yu S mol H2SO4 = 100 .98

cu HS gii

HS:Ln bng trnh by GV: Gi HS nhn xt, ghi im

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 mCaHPO 4 = 2.0,08 .136 = 21,76 ( g )mCa ( H 2 PO 4 ) 2 = 0,12 .234 = 28 .08 ( g )

mCaSO 4 = (a + 2b).136 = (0,08 + 0,24 ).136 = 45,52 ( g )

Hot ng 5: Cng c - dn d * 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 B. H+, PO 3, HPO D. H+, PO 3, H2PO , HPO 4 4 4 4 4 * Dn d: Chun b bi Luyn tp trang 59

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 3/ Bi mi

R(NO3)n + NO2 + H2O

Hot ng ca thy v tr Ni dung Hot ng 1: Bi 1: GV: Chp ln bng, yu cu HS Mt lng 8,32 g Cu tc dng va vi 240 chp vo v. ml dd HNO3 , cho 4,928 l ( o ktc) hn hp Gio Vin : V Quc Sanh 19

Trng THPT Quang Trung 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 GV: Hng dn HS cch vit pt, gi cch gii, yu cu HS lm HS: Tho lun lm bi GV: Yu cu 1 HS 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: 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. HS: Chp GV: Gi hng dn HS cch gii, yu cu 1 HS ln bng trnh by HS: Ln bng trnh by, cc HS cn li ly nhp lm bi

Bm st chng trnh chun 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) 2x +2 4,928 y = = 0,22 (2) 3 22 .4

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 l2 .0,03 = 0,02 (mol) 3

b/ Tng s mol HNO3 phn ng = 4.0,1 +8 .0,03 = 0,48 (mol) 3

Nng mol/l ca dung dch axitC M ( HNO 3 ) = 0,48 = 2( M ) 0,24

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. Al + 4HNO3 Al(NO3)3 + NO + 2H2O Gii

x 4x 2x (mol) 8Al + 30HNO3 8Al(NO3)3 + 3N2O + 15H2O y 30/8y (= 3,75 y) 3/8y (= 0,375 y) Theo bi ra ta c: ( x + y ).27 = 13,5 (1)d hh / H 2 = 30 .x + 44 .0,375 y =19 ,2 ( x + 0,375 y )2

(2)

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 axitC M ( HNO 3 ) =

GV: Gi HS nhn xt, ghi im Hot ng 3: GV: Chp ln bng, yu cu HS Gio Vin : V Quc Sanh

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

Bi 3: Nung 9,4 gam mt mui nitrat trong mt bnh kn. Sau khi phn ng xy ra hon ton cn li 4 gam oxit. Tm cng thc ca mui nitrat 20

Trng THPT Quang Trung chp vo v. Bi 3: Nung 9,4 gam mt mui nitrat trong mt bnh kn. Sau khi phn ng xy ra hon ton cn li 4 gam oxit. Tm cng thc ca mui nitrat HS: Chp 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

Bm st chng trnh chun Gii 2R(NO3)2 R2On + 2nNO2 + n/2O2 a a/2 na na/4 Ta c: a.( MR + 62n) = 9,4 (1) 0,5.a( 2MR + 16n) = 4 (2) Ly (1) : (2) ta c MR = 32n. Khi n = 2 th MR = 64 Vy cng thc mui nitrat Cu(NO3)2t0

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: Hot ng ca thy v tr Hot ng 1: GV: Chp ln bng, yu cu HS chp vo v. 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 Gio Vin : V Quc Sanh Ni dung 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. Gii: 21

Trng THPT Quang Trung 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. 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 2: GV: Chp ln bng, yu cu HS chp vo v. 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 HS: Chp GV: Gi hng dn HS cch gii, yu cu 1 HS ln bng trnh by HS: Ln bng trnh by, cc HS cn li ly nhp lm bi

Bm st chng trnh chun 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 c n1 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: 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 Gii M + 4HNO3 M(NO3)3 + NO + 2H2O x 4x 2x (mol) 10M+ 36HNO3 10M(NO3)3 + 3N2 + 18H2O y 3/10y Theo bi ra ta c: x + (1)d hh / H 2 = 30 .x + 28 . 3 y 10 = 14 ,5.2 3 x+ y 103 y = 0,27 10

GV: Gi HS nhn xt, ghi im

(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

Hot ng 3: GV: Chp ln bng, yu cu HS chp vo v. Bi 3: ( NH4Cl (1) NH3 2) N2 (3) NO ( 4) (5) ( NO2 HNO3 6) NaNO3 ( 7) NaNO3 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 Hot ng 4: Gio Vin : V Quc Sanh

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 dungt t t

22

Trng THPT Quang Trung GV: Chp ln bng, yu cu HS chp vo v. 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

Bm st chng trnh chun dch H3PO4 1,5M. Sau phn ng trong dung dch thu c cc sn phm no 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 VN O i vi H2 bng 16,75. T l th tch kh trong hn hp l V NO2

A.

1 3

B.

2 3

C.

1 4

D.

3 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 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 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% ) Gio Vin : V Quc Sanh 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 95% ) Gii: CaCO3 tC CaO + CO20

nCO 2 = nCaCO 3 =

52 ,65 = 0,5265 ( mol ) 100

23

Trng THPT Quang Trung HS: Chp GV: Yu cu HS tho lun lm bi. HS: Tho lun lm bi

Bm st chng trnh chun 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 n NaOH 0,9 = s mol CO2 X, Y thuc dyng ng ca ankan.3n +1

12 ,6 = 0,7(mol ) 18

O2 t nCO2 + (n+1)H2O HS: Ln bng trnh by, cc HS cn li CnH2n + 2 + 2 ly nhp lm bi 0,5 0,7 GV: Gi HS nhn xt ghi im Ta c : 0,5(n + 1 ) = 0,7n n = 2,5 CTPT ca X, Y l: C2H6, C3H8 Bi 2: Hot ng 3: ton vi dung dch KMnO4 long d, thu GV: Chp ln bng, yu cu HS c 5,2 gam sn phm hu c. Tm CTPT chp vo v. ca X. Bi 2:

Cho 3,5 gam mt anken X tc dng hon

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 14n 14n + 34 GV: Yu cu HS tho lun lm bi. 3,5 5,2 HS: Tho lun lm bi Ta c: 3,5( 14n + 34 ) = 5,2.14n GV: Cho HS xung phong ln bng gii n = 5 HS: Ln bng trnh by, cc HS cn li CTPT ca X l C5H10 ly nhp lm bi GV: Gi HS nhn xt ghi im

3CnH2n + 2KMnO4 + 4H2O

CHn

Gii2n

(OH)2 + 2MnO2 + 2KOH

Bi 3: Hot ng 4: GV: Chp ln bng, yu cu HS t chy hon ton hon hai hirocacbon mch h M, N lin tip trong dy ng ng thu c chp vo v. 22,4 lt CO2 ( ktc) v 12,6 gam nc. Tm CTPT ca M, N. Gii Bi 3:=1(mol ) t chy hon ton hon hai nCO = 22 ,4 hirocacbon mch h M, N lin tip trong dy ng ng thu c 22,4 lt n H O = 12 ,6 = 0,7(mol ) 18 CO2 ( ktc) v 12,6 gam nc. Tm S mol nc < s mol CO2 M, N thuc CTPT ca M, N. dy ng ng ca ankin. HS: Chp GV: Gi hng dn HS cch gii, CnH2n - 2 + 3n 1 O2 t nCO2 + (n -1)H2O 2 yu cu HS ln bng trnh by 1 0,7 n = 3,3 Ta c : (n - 1 ) = 0,7n GV: Gi HS nhn xt ghi im CTPT ca M, N l: C3H4, C4H6 Bi 4: t chy hon ton a lt (ktc) mt ankin X th2

22 ,4

2


48

Trng THPT Quang Trung Hot ng 5: 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.

Bm st chng trnh chun 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. GiinCaCO 3 =

mCO2 = 44 .0,225 = 9,9( gam )12 ,6 9,9 = 0,15 (mol ) 18 3n 1 CnH2n - 2 + O2 t nCO2 + (n -1)H2O 2 n H 2O =

22 ,5 = 0,225 (mol ) 100

HS: Chp GV: Yu cu HS tho lun lm bi. HS: Tho lun lm bi 0,225 0,15 GV: Cho HS xung phong ln bng gii Ta c : 0,225(n - 1 ) = 0,15n n = 3 HS: Ln bng trnh by, cc HS cn li CTPT ca X l: C3H4 ly nhp lm bi GV: Gi HS nhn xt ghi im 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 XIX. XX.

CHN

Ch : BI TP TNG KT CHNG HIROCACBON NO V HIROCACBON KHNG NO

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


49

Trng THPT Quang Trung Bm st chng trnh chun Hot ng 2: GV: Chp ln bng, yu cu HS Bi 1: chp vo v. Hon thnh s phn ng sau Cao su buna C4H4 C4H4 Bi 1: Hon thnh s phn ng sauC4H4 CaCO3 CaO CaC2 C2H2 C2H4 P E Vinylclorua PVC Cao su buna

C4H4

Gii: 1/CaCO3 t CaO + CO2 2/ CaO + 3C t CaC2 + CO GV: Yu cu HS tho lun lm bi. 3/ CaC2 + 2H2O C2H2 + Ca(OH)2 HS: Tho lun lm bi 4/ C2H2 + H2 Pd,PbCO CH2 = CH2 GV: Cho HS xung phong ln bng gii 5/ nCH2 = CH2 xt,p (-CH2 CH2 - )n t , 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 im3

CaCO3 CaO CaC2 C2H2 C2H4 PE Vinylclorua PVC

Cl

Cl Pd,PbCO3 CH2 = CH CH = CH2

8/ 2C2H2 CH2 = CH- C = CHx t,t

9/ CH2 = CH- C = CH + H2

10/ nCH2 = CH CH = CH2 xt, (-CH2 - CH = CH - CH2-)n t , p

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. Hot ng 3: Nu cho mt lng X nh trn tc dng vi GV: Chp ln bng, yu cu HS HBr, thu c cht Z, thy khi lng Y, Z thu chp vo v. 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 ng nc Gii ban u tng 4,2 gam. Nu cho mt CxH2x + H2O CxH2x +1 OH (Y) , (1) lng X nh trn tc dng vi HBr, thu CxH2x + HBr CxH2x +1 Br (Z) , (2) c cht Z, thy khi lng Y, Z thu tng khi lng bnh = khi lng anken c khc nhau 9,45gam. 4,2 ( mol ) phn ng n X = HS: Chp 14 x GV: Yu cu HS tho lun lm bi. 4,2 mY = (14 x +18 ) HS: Tho lun lm bi 14 x GV: Cho HS xung phong ln bng gii4,2 (14 x + 81) 14 x 4,2 4,2 (14 x + 81 ) (14 x +18 ) =9,45 mZ - mY = 14 x 14 x mZ =

x=2 HS: Ln bng trnh by, cc HS cn li CTPT ca X: C2H4 ly nhp lm bi Bi 3: GV: Gi HS nhn xt ghi im Khi t mt th tch hirocacbon A mch h cn 30 th tch khng kh, sinh ra 4 th tch kh CO2. A tc dng vi H2 ( xt Ni ), to thnh mt hirocacbon no mch nhnh. Xc nh CTPT, Hot ng 4: Gio Vin : V Quc Sanh 50

Trng THPT Quang Trung GV: Chp ln bng, yu cu HS chp vo v. Bi 3: Khi t mt th tch hirocacbon A mch h cn 30 th tch khng kh, sinh ra 4 th tch kh CO2. A tc dng vi H2 ( xt Ni ), to thnh mt hirocacbon no mch nhnh. Xc nh CTPT, CTCT ca A.

Bm st chng trnh chun CTCT ca A. Gii CxHy + (x +VO2 =

y )O2 4

xCO2 +

y H2O 2

Th tch oxi phn ng:20 20 Vkk = .30 = 6 (lt) 100 100

Ta c phng trnh x = 4, x + y/4 = 6 y = 8 A c CTPT C4H8 mch h nn A thuc loi anken. V A tc dng vi H2 to thnh mt hirocacbon no mch nhnh. CTCT ca A. HS: Chp CH2 = C CH3 GV: Gi hng dn HS cch gii, CH3 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 Tit 26: T CHNCh : BI TP BENZEN V NG NG. MT S HIROCACBON THM KHC

XXI. XXII.

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 III. Chun b: GV:Gio n HS: n tp l thuyt, lm bi tp benzen v ng ng. Mt s hirocacbon thm khc IV.Tin trnh ln lp: Hot ng 1: n nh lp + Bi c Gio Vin : V Quc Sanh 51

Trng THPT Quang Trung Bi c: Trnh by tnh cht ha hc ca benzen Bi mi:



52

Hot ng ca thy v tr Hot ng 2: GV: Chp ln bng, yu cu HS Trng THPT Quang Trung 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 , Ni,t C + HNO 3 ,(4 D 3mol ), H 2 SO + KMnO ,t 4 E

Ni dung Bi 1: Bm st chng trnh chun 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 + HNO 3 , ( 3 mol ), H 2 SO 4 D + KMnO ,t 4 E

A

A

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

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 C6H5 CH3 + 3H2 N i,t C6H11CH3 C: Metylxiclohexan2 4

HS: Ln bng trnh by, cc HS cn li C6H5-CH3 + 3HNO3 H S C6H2(NO2)3CH3 + 3H2O O ly nhp lm bi D: TNT GV: Gi HS nhn xt ghi im (trinitrotoluen) C6H5 CH3 + KmnO4 t C6H5-COOK + KOH + 2MnO2 + H2O E: kali benzoat Hot ng 3: Bi 2: GV: Chp ln bng, yu cu HS Cht A l mt ng ng ca benzen. Khi t chp vo v. chy hon ton 1,5 g cht A, ngi ta thu Bi 2: c 2,52 lt kh CO2 ( ktc). Cht A l mt ng ng ca benzen. a/ Xc nh CTPT. Khi t chy hon ton 1,5 g cht A, b/ Vit cc CTCT ca A. Gi tn. ngi ta thu c 2,52 lt kh CO2 c/ Khi A tc dng vi Br2 c cht xc tc Fe v ( ktc). nhit th mt nguyn t H nh vi vng a/ Xc nh CTPT. benzen b thay th bi Br, to ra dn xut b/ Vit cc CTCT ca A. Gi tn. monobrom duy nht. Xc nh CTCT ca A. c/ Khi A tc dng vi Br2 c cht xc Gii tc Fe v nhit th mt nguyn t H 3n 3 O2 nCO2 + (n-3)H2O nh vi vng benzen b thay th bi CnH2n 6 + 2 Br, to ra dn xut monobrom duy nht. C ( 14n -6)g A to ra n mol CO2 Xc nh CTCT ca A. 2,52 C 1,5 g A to ra 22 ,4 = 0,1125 molCO 2 HS: Chp GV: Yu cu HS tho lun lm bi. 14 n 6 n = n = 9 HS: Tho lun lm bi 1,5 0,1125 GV: Cho HS xung phong ln bng gii CTPT: C9H12 Cc CTCT:CH3 CH3 CH3 CH3 H3C CH3CH3

CH3

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

CH3 1,2,3-trimetlybenzen 2 CH CH3

CH3

C2H5

- CH2 - CH3 1,2,5-trimetlybenzen+ Br2C2H5Fe,t

CH CH33

CH -CH3 1,3,5-trimetlybenzenBr

CH3

Br

53

H etyl CH 1- 3C - 2 - metylbenzen 3

propylbenzen

C2H5 + HBr

H3C 1- etyl - 3 - metylbenzen

isopropylbenzen CH3 1- etyl - 4 - metylbenzen

Trng THPT Quang Trung Bm st chng trnh chun Hot ng 5: Cng c - dn d * Cng c: Nhc li cch gi tn cc ng ng benzen. Cc cch gii bi tp tm CTPT, vit CTCT * Dn d: Chun b bi ngun hirocacbon thin nhin Tit 27: T CHNCh : BI TP H THNG HA V HIROCACBON

XXIII. XXIV.

Mc tiu: HS vn dng c kin thc hc gii bi tp Trng tm: Bi tp h thng ha v hirocacbon III. Chun b: GV:Gio n HS: n tp l thuyt, lm bi tp h thng ha v hidrocacbon IV.Tin trnh ln lp: Hot ng 1: n nh lp + Bi c Bi c: Trnh by thnh phn ca du m Bi mi:


54

Hot ng ca thy v tr Hot ng 2: GV: Chp ln bng, yu cu HS Trng THPT Quang Trung 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

Ni dung Bi 1: Bm st chng trnh chun 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. Gii20 ,72 nCO 2 = = 0,925 ( mol ) 22 ,4 mC =12 .0.925 =11,1( gam )

HS: Ln bng trnh by, cc HS cn li ly nhp lm bi V s mol H2O > s mol CO2 nn hai cht trong GV: Gi HS nhn xt ghi im hn hp M u l ankan. CnH2n +23n +1 O2 2

m H = 13 ,2 11,1 = 2,1( gam ) 2,1 n H 2O = = 1,05 ( mol ) 2

nCO2 + ( n +1 ) H2O

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%C 7 H 16 = %C8 H 18 0,075 .100 .100 % = 56 ,8% 13,2 0,05.114 = .100 % = 43,2% 13,2

Vy CTPT ca X l C3H4, CTCT ca X l CH3- C = CH GV: Yu cu HS tho lun lm bi. b/ Phn ng hiro ha HS: Tho lun lm bi C3H4 + 2H2 C3H8 2x GV: Cho HS xung phong ln bng gii x C3H4 + H2 C3H6 y y Ta c h phng trnh x + y = 0,1 Gio Vin : V Quc Sanh 2x + y = 0,15 HS: Ln bng trnh by, cc HS cn li x = 0,05; y = 0,05 %mC H = 48,84% ly nhp lm bi3 6

Hot ng 3: 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 sau phn ng. HS: Chp

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. Gii a/ X c CTPT CnH2n 2, tc dng vi brom: CnH2n 2 + 2Br2 CnH2n 2Br4 %C =12 n 10 .100 % = n=3 14 n 2 100

55

Trng THPT Quang Trung Hot ng 5: Cng c - dn d * Cng c: Nhc li cc kin thc hidrocacbon hc * Dn d: Chun b bi: Dn xut halogen ca hirocacbon Tit 28: T CHN XXV. XXVI.


Ch : BI TP DN XUT HALOGEN CA HIROCACBON V ANCOL

Mc tiu: HS vn dng c kin thc hc gii bi tp Trng tm: Bi tp dn xut halogen ca hirocacbon v ancol III. Chun b: GV:Gio n HS: n tp l thuyt, lm bi dn xut halogen ca hidrocacbon v ancol IV.Tin trnh ln lp: Hot ng 1: n nh lp + Bi c Bi c: Trnh by s chuyn ha gia cc hirocacbon Bi mi:


56

Hot ng ca thy v tr Hot ng 2: GV: Chp ln bng, yu cu HS Trng THPT Quang Trung 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 -2en 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 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 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: Bm st chng trnh chun 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 a s C2H5Cl + HCl C2H5Cl + Mg C2H5MgCl b/ CH3CH2CH2CH3 + Br2 CH3CHBrCH2CH3 + HBr CH3CHBrCH2CH3 + KOH C H CH3 OH CH = CH CH3 + KBr + H2O H CH3- CH = CH CH3 + H2O CH3CH(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 tn2 5+

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 ) 18

cng l khi lng C v H trong 3,96 g A Theo bi ra, A phi cha Cl. Khi lng Cl HS: Ln bng trnh by, cc HS cn li trong 7,175 g AgCl ly nhp lm bi 35 ,5.7,175 =1,775 (g) GV: Gi HS nhn xt ghi im 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 x: y: z = 12 : 1 : 35 ,5 =1 : 2 : 1 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: t chy hon ton mt lng hn hp hai 57 ancol A, B no n chc k tip nhau trong dy ng ng thu c 4,48 lt kh CO2 (ktc) v0,96 0,16 2,84


Trng THPT Quang Trung Hot ng 5: Cng c - dn d * Cng c: Nhc li tnh cht ca dn xut halogen, ancol * Dn d: Chun b bi: Phn cn li bi Ancol Tit 29: T XXVII. XXVIII.


CHN

Ch : BI TP ANCOL V PHENOL

Mc tiu: HS vn dng c kin thc hc gii bi tp Trng tm: Bi tp Ancol v phenol III. Chun b: GV:Gio n HS: n tp l thuyt, lm bi Ancol v phenol IV.Tin trnh ln lp: Hot ng 1: n nh lp + Bi c Bi c: Trnh by nh ngha, phn loi, danh php Ancol. Ly VD minh ha. Bi mi:


58

Hot ng ca thy v tr Hot ng 2: GV: Chp ln bng, yu cu HS Trng THPT Quang Trung 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: Bm st chng trnh chun 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 Cu(OH)2 = 2.1,96 = 0,04 ( mol ) 98

HS: Ln bng trnh by, cc HS cn li S mol gixerol trong 20,3 g A: 0,04 .20 ,3 ly nhp lm bi = 0,1( m ) ol 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

Khi lng 1 mol ROH: 0,15 = 74 R = 29; R l C4H9 CTPT: C4H10OPhn trm khi lng C4H9OH =

11,1 .100 % = 54,68 % 20 ,3

Bi 2: un nng hn hp 2 ancol no, n chc, Hot ng 3: mch h vi H2SO4 1400C, thu c 72 gam GV: Chp ln bng, yu cu HS hn hp 3 ete vi s mol bng nhau. Khi chp vo v. lng nc tch ra trong qu trnh to thnh Bi 2: cc ete l 21,6 gam. Xc nh CTCT ca 2 un nng hn hp 2 ancol no, n ancol. chc, mch h vi H2SO4 1400C, thu c 72 gam hn hp 3 ete vi s mol Gii bng nhau. Khi lng nc tch ra 2CnH2n+1OH (2CnH2n +1)2O + H2O trong qu trnh to thnh cc ete l 2Cm H2m +1OH (2CmH2m+1)2O + H2O 21,6 gam. Xc nh CTCT ca 2 ancol. CnH2n+1OH + Cm H2m +1OH CnH2n+1OCmH2m HS: Chp +1 GV: Yu cu HS tho lun lm bi. + H2O HS: Tho lun lm bi S mol 3 ete = s mol nc = GV: Cho HS xung phong ln bng gii 21,6 HS: Ln bng trnh by, cc HS cn li S mol mi ete = 0,4 (mol) ly nhp lm bi Khi lng 3 ete: GV: Gi HS nhn xt ghi im (28n + 18).0,4 + ( 28m +18).0,4 + (14n + 14m + 18).0,4=72 n+m=3 Gio Vin : V Quc Sanh 59 Hai CTCT ca ancol l: CH3OH, CH3CH2OH Hot ng 4: GV: Chp ln bng, yu cu HS18 =1,2( mol )

Trng THPT Quang Trung Hot ng 5: Cng c - dn d * Cng c: Nhc li tnh cht ca ancol v phenol * Dn d: Chun b bi: Luyn tp Tit 30: T XXIX. XXX.


CHN

Ch : BI TP TNG KT CHNG DN XUT HALOGEN ANCOL V PHENOL

Mc tiu: HS vn dng c kin thc hc gii bi tp Trng tm: Bi tp dn xut Halogen + Ancol + Phenol III. Chun b: GV:Gio n HS: n tp l thuyt, lm bi dn xut Halogen + Ancol + Phenol IV.Tin trnh ln lp: Hot ng 1: n nh lp + Bi c Bi c: Khng kim tra Bi mi:


60

Hot ng ca thy v tr Hot ng 2: GV: Chp ln bng, yu cu HS Trng THPT Quang Trung 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

Ni dung Bi 1: Bm st chng trnh chun 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 = C6H5OH + Na C6H5ONa + 1/2H2 0,1 0,05 CH3OH + Na CH3ONa + 1/2H2 x x/2 C2H5OH + Na C2H5ONa + 1/2H2 y y/20,08 .14 ,45 = 0,1( m ) ol 11 ,56

Hot ng 3: 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 4: 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

GV: Gi hng dn HS cch gii, yu cu HS ln bng trnh by

Gio Vin : V Quc Sanh GV: Gi HS nhn xt ghi im

61

Trng THPT Quang Trung Hot ng 5: Cng c - dn d * Cng c: Nhc li tnh cht ca dn xut halogen ancol v phenol * Dn d: Chun b bi: Anehit - Xeton Tit 31: T CHNCh : BI TP ANEHIT - XETON


XXXI. XXXII.

Mc tiu: HS vn dng c kin thc hc gii bi tp Trng tm: Bi tp Anehit - Xeton III. Chun b: GV:Gio n HS: n tp l thuyt, lm bi Anehit - Xeton IV.Tin trnh ln lp: Hot ng 1: n nh lp + Bi c Bi c: Trnh by tnh cht ha hc ca Anehit - Xeton Bi mi:


62

Hot ng ca thy v tr Hot ng 2: GV: Chp ln bng, yu cu HS Trng THPT Quang Trung 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

Ni dung Bi 1: Bm st chng trnh chun 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 + 2H2On NO = 3,85 .0,8 = 0,125 (mol ) 0,082 .300 ,3

S mol Ag = 3 s mol NO = 0,375 (mol) HS: Ln bng trnh by, cc HS cn li S mol RCHO = s mol Ag = 0,1875(mol) 10 ,5 ly nhp lm bi Khi lng 1 mol RCHO = 0,1875 = 56 GV: Gi HS nhn xt ghi im R = 56 -29 = 27 R l C2H3 CTPT l C3H4O CTCT l CH2 = CH CHO ( propenal ) Hot ng 3: Bi 2: GV: Chp ln bng, yu cu HS t chy hon ton mt lng cht hu c chp vo v. A phi dng va ht 3,08 lt O2. Sn phm thu Bi 2: c ch gm 1,8 gam H2O v 2,24 lt CO2. t chy hon ton mt lng cht Cc th tch o ktc. hu c A phi dng va ht 3,08 lt O2. a/ Xc nh CTGN ca A. Sn phm thu c ch gm 1,8 gam b/ Xc nh CTPT ca A. Bit rng t khi ca H2O v 2,24 lt CO2. Cc th tch o A i vi oxi l 2,25. ktc. c/ Xc nh CTCT ca A, gi tn, bit rng A l a/ Xc nh CTGN ca A. hp cht cacbonyl. b/ Xc nh CTPT ca A. Bit rng t Gii khi ca A i vi oxi l 2,25. a/ Theo nh lut bo ton khi lng c/ Xc nh CTCT ca A, gi tn, bit 2,24 m A = mCO + mH O mO 2 = .44 rng A l hp cht cacbonyl. 22 ,4 HS: Chp 3,08 +1,8 .32 = 1,8( gam ) GV: Yu cu HS tho lun lm bi. 22 ,4 HS: Tho lun lm bi GV: Cho HS xung phong ln bng gii2 2

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

HS: Ln bng trnh by, cc HS cn li 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 x:y:z =1,2 0,2 0,4 : : = 4 : 8 :1 12 1 16

GV: Gi HS nhn xt ghi im Gio Vin : V Quc Sanh

CTGN l: C4H8O b/ MA = 2,25.32 = 72g/mol CTPT trng CTGN: C4H8O c/ Cc hp cht cacbonyl CH3 CH2 CH2 CHO butanal 63 CH3 CH CHO 2 metylpropanal CH3 CH3 CH2 CO CH3 butan 2 on

Trng THPT Quang Trung Bm st chng trnh chun Hot ng 5: Cng c - dn d * 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 * Dn d: Chun b bi: axit cacboxylic Tit 32: T CHNCh : BI TP AXIT CACBOXYLIC

XXXIII. XXXIV.

Mc tiu: HS vn dng c kin thc hc gii bi tp 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:


64

Hot ng ca thy v tr Hot ng 2: GV: Chp ln bng, yu cu HS Trng THPT Quang Trung 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: Bm st chng trnh chun 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 ) 50

HS: Ln bng trnh by, cc HS cn li ly nhp lm bi S mol RCOOH trong 125 ml dung dch axit l: GV: Gi HS nhn xt ghi im 1,2.1251000 = 0,15 ( mol )

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: Cht A l mt axit no, n chc, mch h. Hot ng 3: t chy hon ton 2,225 gam A phi dng GV: Chp ln bng, yu cu HS va ht 3,64 lt O2 ( ktc). chp vo v. Xc nh CTPT, CTCT v tn gi. 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. Gii HS: Chp 3n 2 O2 nCO2 + nH2O CnH2nO2 + GV: Yu cu HS tho lun lm bi. 2 HS: Tho lun lm bi Theo phng trnh ( 14n + 32)g axit tc dng mol O2 GV: Cho HS xung phong ln bng gii vi 2 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 16 ,8

3n 2

CTPT C5H10O2 HS: Ln bng trnh by, cc HS cn li CH3 CH2 CH2 CH2 COOH axit ly nhp lm bi pentanoic CH3 CH CH2 COOH axit -3metylbutanoic CH3 CH3 CH2 CH COOH axit -2metylbutanoic CH3 Gio Vin : V Quc Sanh CH3 CH3 C COOH axit -2,2 GV: Gi HS nhn xt ghi im -dimetylpropanoic

65

Trng THPT Quang Trung Bm st chng trnh chun 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 CHNCh : BI TP ANEHIT XETON AXIT CACBOXYLIC

XXXV. XXXVI.

Mc tiu: HS vn dng c kin thc hc gii bi tp 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:


66

Hot ng ca thy v tr Hot ng 2: GV: Chp ln bng, yu cu HS Trng THPT Quang Trung 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: Bm st chng trnh chun 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 HS: Ln bng trnh by, cc HS cn li CxHyO + ( x + )O2 xCO2 + y/2H2O ly nhp lm bi 4 2 GV: Gi HS nhn xt ghi im 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 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 Hot ng 3: 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 th thu c bao nhiu gam mui khan. HS: Chp GV: Yu cu HS tho lun lm bi. HS: Tho lun lm bi 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.

Gii a/ Axit n chc, cng thc CxHyCOOH CxHyCOOH + NaOH CxHyCOONa + H2O

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 Maxit = 0,125 = 74 ( g / mol ) 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 4: Bi 3: Gio Vin : V Quc Sanh 67 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 3:9,25

Trng THPT Quang Trung Bm st chng trnh chun 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 Tit 34: T CHNCh : BI TP TNG KT CHNG ANEHIT XETON AXIT CACBOXYLIC

XXXVII.

Mc tiu: HS vn dng c kin thc hc gii bi tp XXXVIII. 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:


68

Hot ng ca thy v tr Hot ng 2: GV: Chp ln bng, yu cu HS Trng THPT Quang Trung 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

Ni dung Bi 1: Bm st chng trnh chun 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

Gii

C 2 trng hp + Nu Y l HCHO16.100%

HS: Ln bng trnh by, cc HS cn li %mO = = 53% 37,21% (loi) 30 ly nhp lm bi + Nu Y l R(CHO)2 = CxHyO2 GV: Gi HS nhn xt ghi im 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.

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

Gii a/ CH3CHO + 2AgNO3 + 4NH3 + H2O CH3COONH4 + 2NH4NO3 + 2Ag C2H5CHO + 2AgNO3 + 4NH3 + H2O C2H5COONH4 + 2NH4NO3 + 2Ag GV: Cho HS xung phong ln bng gii 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 HS: Ln bng trnh by, cc HS cn li 0,1.44 .100 % ly nhp lm bi = 43,14 % %CH CHO =3

10 ,2

GV: Gi HS nhn xt ghi im Hot ng 4: GV: Chp ln bng, yu cu HS chp vo v. Bi 3: Gio Vin : V Quc Sanh 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

%C2H5CHO = 56,86% 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 69 phn th nht phn ng hon ton vi lng d bc nitrat trong dung dch ammoniac, thu c 10,8 g bc. Phn th 2 c trung ha

Trng THPT Quang Trung Bm st chng trnh chun 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 CHNCh : N TP HC K II

XXXIX. XL.

Mc tiu: HS vn dng c kin thc hc gii bi tp 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:


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Hot ng ca thy v tr Hot ng 2: GV: Chp ln bng, yu cu HS Trng THPT Quang Trung 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: Bm st chng trnh chun 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. CaC2 + 2H2O C2H2 + Ca(OH)2 HS: Tho lun lm bi 2C2H2 x t CH2 = CH C = CH GV: Cho HS xung phong ln bng gii HS: Ln bng trnh by, cc HS cn li ly nhp lm biP CH2 = CH C = CH + H2 d CH2 = CH CH = CH2 xt, p ,t nCH2 = CH CH = CH2 (- CH2 CH = CH CH2 - )n P d CH2 = CH2 C2H2 + H2 p , nCH2 = CH2 xt,t ( - CH2 CH2 - )n xt C2H2 + HCl CH2 = CH Cl p , CH2 = CH Cl xt,t ( - CH2 CH - )n

Cl GV: Gi HS nhn xt ghi im 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 V Quc hp Gio Vin : trong hnSanh A. HS: Chp C2H2 + H2O CH3CHOH gSO 4

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 71 z 2z CH = CH + 2AgNO3 + NH3 Ag C = C Ag + 2NH4NO3

Trng THPT Quang Trung Bm st chng trnh chun 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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