PHNG PHP GII BI TON V ALUMINUM V HP CHT CA ALUMINUMI. MT S CH KHI GII BI TON V ALUMINUM V HP CHT CA ALUMINUM.Nh chng ta bit nhm l mt nguyn t c dng rt ph bin, nht l cc vt dng trong gia nhChnh v vy m nguyn t nhm v hp cht ca chng v cng phc tp, bi nhm v mt s hp cht ca chung tc dng c ng thi axit v baz (lng tnh).V d: khi cho mt kim loi kim M vo dung dch mui ca nhm th ty vo d kin ca bi ton, t l v s mol m sn phm c th xut hin kt ta, sau kt ta tan mt phn hoc kt ta tan ht.qua v d cc phng trnh phn ng c th xy ra.2 2333 2 22M 2H O 2MOH H (1)3OH Al Al(OH) (2)OH Al(OH) AlO 2H O (3) + + + + + + lm tt cc bi ton v nhm v hp cht ca nhm cn lu mt s im sau:1. Nm c c im cu to, cu hnh, v tr, , nhm, chu k.2. Nm c tnh cht vt l, tnh cht ha hc, phng php iu ch ca nhm.3. Nm c cch x l qung nhm (Bxit) thu c nhm.4. Nmctnhcht vt l,tnhchthahc, phngphpiuchcaxitnhm ( Al2O3), hiroxit nhm (Al(OH)3),mui nhm (Al3+)5. Nm c mt s phng php gii nhanh lin quan n nhm v hp cht ca chng.6. Khi cho dung dch kim vo mui nhm thu c kt ta th ta c phn ng c th xy ra:Al3+ +3OH- Al(OH)3(1)Al(OH)3 + OH- AlO2- + 2H2O (2)32 2Al 4OH AlO 2H O+ + + (3)7.Khi cho mui 2AlO tc dng vi dung dch axit thu c kt ta th ta c phn ng c th xy ra:2 2 3AlO H H O Al(OH) ++ + (1)33 2Al(OH) 3H Al 3H O+ ++ + (2)32 2AlO 4H Al 3H O + ++ + (3)II. BI TON P DNG.Bi ton 1: ( Trch thi tt nghip TH PT 2007). Ho tan 5.4 gam Al bng mt lng dung dch H2SO4 long d. Sau phn ng thu c dung dch X v V lt kh H2 ktc. Gi tr ca V lt:A. 2.24 lt B. 3.36 lt. C. 4.48 lt D. 6.72 ltBi gii:23Al H2Al 3e Al0, 2 0, 65, 4n 0, 2 n 0,3272H 2e H0, 6 0,3++ '+ molVH2 = 0,3.22,4 =6,72 lt => D ngBi ton 2: (Trch thi tuyn sinh i hc H CKhi A 2007)Cho m gam hn hp Mg, Al vo 250 ml dung dch X cha hn hp axit HCl 1 M v axit H2SO40,5 M thu c 5,32 lt H2( ktc) v dung dch Y (coi th tch dung dch khng i). Dung dch Y c pH l: A: 1 B: 2 C: 6 D: 7Bi gii:( X)2 4H (HCl)HH (H SO )n 0, 25.1 0, 25(mol)n 0, 25 0, 25 0,5moln 0, 25.0,5.2 0, 25(mol)+++ + ; 2H++ 2e H20,475mol.. 0,2375(mol)2H5,32n 0, 2375(mol)22, 4 1H (Y)0, 025n 0,5 0, 475 0, 025(mol) [H ] 0,1 10 (mol / lit)0, 250++ pH = 1A ngBi ton 3: (Trch thi tuyn sinh H- C - KB - 2007). Cho 200 ml dung dch AlCl3 1,5 M tc dng vi V lt dung dch NaOH 0,5M, lng kt ta thu c l 15,6 gam, gi tr ln nht ca V lt l:A: 1,2 gam B: 1,8 gam C: 2,0 gam D: 2,4 gamBi gii:Ta c: 33AlClAln n 0, 2.1, 5 0, 3mol+

3Al(OH)15, 6n 0, 2mol78 NaOHNaOH0, 6n 0, 6mol V 1, 2lit0, 51n 1mol V 2lit0, 5

gi tr ln nht l 2 lt C ng.Bi ton 4: (Trch tuyn sinh H- C Khi A 2008). Cho hn hp 2,7 gam nhm v 5,6 gam st vo 550 ml dung dch AgNO3 1M. Sau khi phn ng xy ra hon ton th c m gam cht rn( bit Fe3+/Fe2+ ng trc Ag+/Ag). Gi tr m gam l:A. 59,4 gam B. 64,8 gam C. 32,4 gam D. 54,0 gamBi gii:Phn ng xy ra hon ton, nn:AgNO3 Ag+ NO3- p dng LBT nguyn t bc: 0,550,55mol3Al(OH)n-OHn0,30,310,21,2Al3+0,60,93Ag AgNO AgAgn n n 0, 55mol; m 0, 55.108 59, 4g+ A ngCh : - Nu phn ng khng hon ton hoc AgNO3 phn ng ang cn d th khng p dng c LBT nguyn t- Nu Al FeAgn 3n 2n 0, 5mol + mAg = 0,5 . 108 = 54,0g D saiBi ton 5: (Trch thi tuyn sinh H C Khi A 2008) Cho hn hp bt gm 2,7 gam Al v 5,6 gam Fe vo 550 ml dung dch AgNO3 1M, sau khi cc phn ng xy ra hon ton thu c m gam cht rn, gi tr m l ( bit th t trong dy th in ha: Fe3+/ Fe2+ ng trc Ag+/Ag). A. 59,4 gam B. 64,8 gam C. 32,4 gam D. 54,0 gamBi gii:nAl = 0,1mol; nFe= 0,1mol;3AgNOAgn n 0, 55mol+ Phng trnh phn ng xy ra: Al + 3Ag+ Al3+ + 3AgmAg = 108. 0,30,1 0,3 0,1 0,3Fel + 2Ag+ Fe2+ + 2AgmAg = 108. 0,20,1 0,2 0,1 0,2Fe2++Ag+Fe3+ +Ag+mAg = 108. 0,050,05 0,05 0,05 0,05m = 108 . 0,55 = 59,4gA ngCh : + Phn ng xy ra hon ton nn ta p dng nh lut bo ton nguyn t bc: 3AgNOAgn n 0, 55mol+ , Agm m 108.0, 55 59, 4g+ A ng+ Nu mAg = 108 . 0,5 = 54gD saiBi ton 6: (Trch thi tuyn sinh H C- Khi A-2008). Nung nng m gam hn hp Al v Fe2O3 (trong mi trng khng c khng kh) n phn ng xy ra hon ton thu c hn hp rn Y, chia Y thnh hai phn bng nhn bng nhau:- Phn 1 tc dng vi H2SO4 long d sinh ra 3.08 lt kh hir (ktc).- Phn 2 tc dng NaOH d sinh ra 0.84 lt kh hir (ktc). Gi tr m gam l:A. 22.75 B. 21.40 C. 29.40 D. 29.43 .Bi gii:Phn tch bi ton: T P2 + NaOH d nn Al d cn Fe2O3 ht:Nh vy hn hp Y: Fe, Al2O3 v Al d Gi x, y, z ln lt l s mol Al2O3, Fe v Al d trong mi phn: P1: 322Al 3e Alz 3z zFe 2e Fey 2y y2H 2e H0, 275 0,1375+++ ;+ p dng LBT e: 3z + 2y = 0,275 (1)P2: 32Al 3e Alz 3z z2H 2e H0, 075 0, 0375++ ;+ p dng LBT e: 3z + 2y = 0,075z= 0,025MThay vo (1)y = 0,1mol:Fe2O3 + 2Al Al2O3 + 2Fe(3)T (3) 2 3Al O Fe1x n n 0, 05mol2 m = 2.(0,05 . 102 + 56. 0,1 + 27 . 0,025)= 22,75A ngBi ton 7: (Trch thi tuyn sinh H - KA - 2008). Cho V lt dung dch NaOH 2 M vo dung dch cha 0,1 mol Al2(SO4)3 v 0,1 mol H2SO4 n khi phn ng hon ton thu c 7,8 gam kt ta. Gi tr ln nht ca V lt thu c khi lng kt ta trn l: A: 0,05 lt B: 0,25 lt C: 0,35 lt D: 0,45 ltBi gii:Cch 1.32 4 2 4 3H SO Al (SO )H Aln 2.n 2.0,1 0.2mol; n 2.n 0, 2mol+ + Th t cc phng trnh xy ra: 3Al(OH)7,8n 0,1mol78 H+ + OH- H2O(1)0,20,2mol Al3+ + 3OH-Al(OH)3 (2)0,20,6mol- -3 2 2Al(OH)+ OHAlO 2H O0,1 0,1mol +(3)T (1), (2), (3): OHn 0, 2 0, 6 0,1 0,9mol + + NaOH0, 9n 0, 452 lt D ng Ch :+ Nu OH0,1n 0,1mol V 0, 052 ltA sai + Nu OH0, 5n 0, 5mol V 0, 252 ltB sai + Nu OH0, 7n 0, 7mol V 0, 352 ltC sai+ Cch 2: p dng phng php th: 32 4 2 4 3 3H SO Al (SO ) Al(OH)Al7,8n 0,1mol; n 2n 0, 2mol; n 0,1mol78+ Trn th OHn 0, 7mol mt khc trung ho 0,1mol H2SO4 th cn 0,2mol OH-NaOH NaOHOH0, 9n n 0, 2 0, 7 0, 9mol V 0, 452 + ltD ng Bi ton 8: Nhng mt thanh nhm nng 45 gam vo 400 ml dung dch CuSO4 0,5 M. Sau mt thi gian ly thanh nhm ra cn nng 46,38 gam. Khi lng gam Cu thot ra l:A. 0,64 gam B. 12,80 gam C. 1.92 gamD. 1,38 gamBi gii:Kim loi mnh y kim loi yu ra khi dung dch mui (phn ng th bi k/loi)2Al + 3CuSO4 Al2(SO4)3 + 3Cu2Al + 3Cu2+ 2Al3+ + 3Cu (rt gn)p dng phng php tng gim khi lng ta c: C 2 mol Al to ra 3 mol Cu nh vy khi lng tng: ( 3. 64 - 2. 27) = 138gTheo bi ra th x mol Cu khi lng tng: 46,38 - 45 = 1,38gCu Cu1, 38x n 3. 0, 03mol m 0, 03.64 1, 92g138 C ngCh : + Nu mCu = 0,01 . 64 = 0,64gA sai+ Nu mCu = 0,5 . 0,4 . 64 = 12,8B sai+ Nu mCu = 46,38 - 45 = 1,38gD saip dng khi gp bi ton c PT rt gn: nA + mBn+ nAm+ + mB(Trong : n, m ln lt l in tchca kim loi B v A) th ta p dng phng php tng gim khi lng. Khi lng tng hay gim mt lngc tnh theo cng thc tr tuyt i B Am.M n.M sau da vo d kin ca ra tnh ton Bi ton 9: 2,7 gam mt thanh nhm ngoi khng kh, mt thi gian sau em cn thy thanh nhm nng 4,14 gam. Phn trmkhi lng thanh nhm b oxi ha bi oxi ca khng kh l:A. 65,21%B. 30%C. 67,5%D. 60%Bi gii:Khi lng ca oxit tham gia phn ng vi thanh nhm l: 4,14 - 2,7 = 1,44 gam Al(2 1, 44m . .27 0, 06.27 1, 62g3 16 p )1, 62%Al .100% 60%2, 7 D ng Ch :+ Nu 2, 7%Al .100% 65, 21%4,14 A sai + Nu Al(2 1, 44 1,8225m . .27 1.8225g %Al .100% 67, 5%3 32 2, 7 p )C saiAl(OH)3OHn0,20,30,6A0,80,1O0,7+ Nu Al(2 1, 44 0,81m . .27 0,81g %Al .100% 30%3 32 2, 7 p )B saiBi ton 10: Chia hn hp X gm Alv Al2O3 thnh 2 phn bng nhau. - Phn 1 tc dng NaOH d thu c 0.3 mol kh. - Phn 2 tan hon ton trong dung dch HNO3 thu c 0.075 mol kh Y duy nht. Y l: A. NO2B. NO C. N2O D. N2 Bi gii:Trong X ch c Al c tnh kh:2H2O + 2e H2 + 2OH-0,6 0,3Khi tc dng vi HNO3, cht oxi ho l HNO3 N+5 + neY0,075n 0,075mol ta c: 0,075n = 0,6Vi n l s e m N+5 nhn to thnh Y n = 8. Vy Y l N2O C ngBi ton 11: Cho m gam Al tan hon ton dung dch HNO3 nng d thu c 11.2 lt (ktc) hh kh A gm: N2 , NO, N2O c t l v s mol tng ng l 2:1:2. Gi tr m gam l: A. 35.1B. 18.9C. 27.9 D. 26.1Bi gii:2 2A N NO N O11, 2n 0,5mol n 0, 2mol; n 0,1mol; n 0, 2mol22, 4 + Qu trnh oxi ho: Al- 3eAl3+ (1)a 3aa+ Qu trnh kh:2N+5+ 10eN2 (2)0,42 0,2N+5+ 3eN+2 (NO) (3)0,30,12N+5+ 8e N+1(N2O) (4) 1,6 0,2p dng LBT e: 3a = 2 + 0,3 + 1,6 = 3,9 a = 1,3mAl = 27 . 1,3 = 35,1g A ngPhn tch bi ton: + Nu (2, 3, 4) khng cn bng 3a = 1 + 0,3 + 0,8 a = 0,7 mAl = 18,9g B sai+ Nu (2, 3) cn bng cn (4) khng: 3a = 2 + 0,3 + 0,8 3,1a3mAl = 27,9 C sai+ Nu (2) khng cn bng, (3,4) cn bng 3a = 1 + 0,3 + 1,6 2,9a3mAl = 26,1 D saiB i ton 12. Mt dung dch cha 0,39 gam K+, 0,54gam Al3+ cng 2 loi anion 1,92 gam SO42- v ion NO3-. Nu c cn dung dch th s thu c khi lng mui khan l:A. 4,71 gam B. 3,47 gamC. 4,09 gamD. 5,95 gamBi gii3 24K Al SOn 0, 01mol, n 0, 02mol , n 0, 02mol+ + p dng LBTT : 0,01 + 0,02.3 = 0,02 + nNO3- => nNO3-=0,03 mol m3 24 3muoiK Al SO NOm m m m m0,39 0,54 1,92 62.0, 03 4, 71gam A dung+ + + + + + + + >Ch :- Nu 3muoiNOn 0, 01 0, 02 0, 02 0, 01mol m 3, 47gam B sai + > >- Nu 3muoiNOn 0, 02mol m 4, 09gam C sai > >- Nu 3muoiNOn 0, 01 0, 02 0, 02 0, 05mol m 5,95gam D sai + + > >.Bi ton 13. Cho tan hon ton 15,6 gam hn hp Al v Al2O3 trong 500 ml dung dch NaOH 1M thu c 6,72 lt H2 (ktc) v dung dch X . Th tch HCl 2 M cn cho vo X thu c kt ta ln nht l : A. 0,25 lt. B. 0,35 lt. C. 0,5 lt. D. 0,244 lt.Bi gii:Trong dung dch X cha AlO2- v OH- (nu d). Dung dch X trung ha v in tch nn. 2AlO OH Nan n n 0,5mol ++ khi cho axit HCl vo dung dch X ta c ptp sau: H+ + OH- H2OH+ + AlO2- + H2O Al(OH)3 thu c kt qu ln nht th2HClH AlO OH0,5n n n 0,5mol V 0, 25lit A dung2+ + > >Bi ton14:Mt hn hp 3 kim loi gm Al , Fe, Mg c khi lng 26.1 gam c chia lm 3 phn bng nhau.- Phn 1 cho tan ht trong dung dch HCl thy thot ra 13.44 lt kh.- Phn 2 cho tc dng vi dung dch NaOH d thu c 3.36 lt kh- Phn 3 cho tc dung dch CuSO4 d , lc ly ton b cht rn thu c sau phn ng em ho tan trong dung dch HNO3 d th thu c V lt kh NO2 ( cc kh u o ktc). Gi tr V lt thu c l: A. 26.88.B. 53.70.C. 13.44 D. 44.8.Bi gii:2Al + 6HCl AlCl3 + 3H2Mg + 2HCl MgCl2 + H2Fe + 2HCl FeCl2 + H2Khi lng mi phn 26,1m 8, 7g3 Gi x, y, z l s molAl, Mg, Fe trong 7,4 gam hn hp27x 24y 56z 8, 7 x 0,11,5x y z 0,3 y 0, 0751,5 0,15 x 0, 075+ + + + ' ' trong 34,7 gam hn hp nAl = 0,4; nMg = 0,3; nFe = 0,32Al + 3CuSO4 Al2(SO4)3 + 3CuMg + CuSO4 MgSO4 + CuFe + CuSO4 FeSO4 + CuCu + 4HNO3 Cu(NO3)2 + 2NO2 + 2H2O- P3 khi cc kim loi tc dng vi dung dch CuSO4 to thnh Cu, lng Cu ny tc dng vi HNO3 to ra Cu2+.Do : Al, Mg, Fe l cht kh, nhng e.ne nhng = 3. 0,1 + 2. 0,075 + 2. 0,075 = 0,6mol- HNO3 l cht oxi ho, nhn e: N+5 + 1e N+4 (NO2) a = 0,6 aa2 2NO NOn 0, 6mol V 0, 6.22, 4 13, 44l it C ng Bi ton 15: Chia hn hp X gm Al, Al2O3, ZnO thnh 2 phn bng nhau. Phn 1 cho tc dng dung dch NaOH d thu c 0.3 mol kh. Phn 2 tan hon ton trong dung dch HNO3 thu c 0.075 molkh Y duy nht. Kh Y l:A. NO2B. NOC. N2O C. N2 .Bi gii:Trong X ch c Al c tnh kh nc b nhm kh theo phng trnh 2H2O + 2e H 2 + 2OH- 0,60,3molKhi tc dng vi HNO3, cht oxi ho l HNO3N+5 + ne Y0,075n 0,0750,075n = 0,6, n l s e m N+5 nhn to thnh Y.n = 8.Vy Y l N2 OC ng Bi ton 16:Ho tan hon ton 17.4 gam hn hp 3 kim loi Al, Fe, Mg trong dung dch HCl thy thot ra 13.44 lt kh,nu cho 34.8 gam hn hp trn tc dng vi dung dch CuSO4d, lc ly ton b cht rn thu c sau phn ng tc dng vi dung dch HNO3 nng d, thu c V lt kh NO2 ktc. Gi tr V l:A. 11.2 lt B. 22.4 ltC. 53.76 ltD. 26.88 lt.Bi gii:Al, Fe, Mg nhng e, s mol e ny chnh bng s mol e Cu nhng khi tham gia phn ng vi HNO3 s mol e m H+ nhn cng chnh l s mol e m HNO3 nhn. 2H+ + 2e H21,2mol 13, 440, 6mol22, 417,4 gam hn hp H+ nhn 1,2 mol e. Vy 34,8 gam s molm H+ nhn l: 2,4 mol17,4g hn hp Hn 1, 2+ 34,8ghn hp Hn 2, 4mol+ N+5 + 1e NO2 2,4 2,4mol2NOV 2, 4.22, 4 53, 76 lt C ngCh : Nu Hn 1, 2+ 2NOV 1, 2.22, 4 26,88 ltD saiB i ton 17: Ho tan 11.2 gam hn hp X gm Al v Fe trong HCl d th thu c hn hp dung dch mui Y1v kh Y2. Cho dung dch Y1tc dng vi NaOH d, lc kt ta ri nung trong khng kh n khi lng khng i th thu c 8 gam cht rn Z. Thnh phn % Fe trong hn hp u l:A. 58,03 % B. 26.75 % C. 75.25 % D. 50.00 %.Bi gii:Sn phm ca qu trnh nung l: {33 2 3FeCl Fe(OH)HCl NaOH Fe2 2Al AlCl Al(OH)3tanFe(OH) Fe O+ + ' ' 2 3Fe O8n 0, 05mol160 .p dng nh lut bo ton nguyn t Fe ta c: 2 3Fe Fe On 2n 0, 05.2 0,1mol mFe = 0,1.56 = 5,6gam, %Fe =50,00% D ng Bi ton 18: Cho 4.04 gam hn hp X gm 3 kim loi Fe, Cu, Al dng bt tc dng hon ton vi oxi thu chn hp Y gm cc oxit c khi lng 5.96 gam. Th tch dung dch HCl 2 M va phn nght vi hn hp Y l:A. 60 mlB. 120 mlC. 224 ml D. 30 ml.Bi gii:p dng nh lut bo ton khi lng: mO (oxit) = 5,96 - 4,04 = 1,92 gam 22O2H O H O 1, 92n 0,12mol :160, 24 0,12+ + 'VHCl = 0,24 = 0,12 2lt = 120 ml B ng Ch : - Nu 2O1, 92n 0, 06mol32 VHCl = 60 mlA sai- Nu 2O1, 92n 0, 06mol32 2HClH On n n+ V = 30ml D saiBi ton 19:Ho tan 10.14 gam hp kim Cu, Mg, Al bng mt lng va dung dch HCl thu c 7.84 lt kh X (ktc) v 1.54 gam cht rn Y v dung dchZ. C cn dung dch Z thu c m gam mui. Gi tr mgam l: A. 21.025 gamB.33.45 gamC.14.8125 gamD.18.6 gamBi gii: p dng inh lut bo ton khi lng:(Al Mg)Clm m m (10,14 1,54) 0, 7.35,5+ + += 8,6 + 24,85 = 33,45(g) B ngPhn tch bi ton: +Cu khng tc dng vi HCl, cn Mg v Al tc dng vi HCl, to ra khH2 c s mol = 0,35mol+ Nu 2HCl HHn n n 0, 35+ m = (10,14 - 1,54) + 0,35.35,5 = 21,025gA sai+ Nu 2HCl H1H n n 0,1752+ m = (10,14 - 1,54) + 0,175. 35,5 = 14,81255gC saiBi ton 20:Thc hin phn ng nhit nhm vi 9.66 gam hn hp X gm FexOyv Al thu c hn hp rn Y. Cho Y tc dng vi dung dch NaOH d thu c dung dch D, 0.672 lt kh ktc v cht rn khng tan Z. Sc CO2 n d vo dung dch D lc ly kt ta v nung n khi lng khng i c 5.1 gam cht rn.1. Khi lng gam ca FexOy v Al trong hn hp X l:A. 6.96 gam v 2.7 gam. B. 5.04 gam v 4.62 gamC. 2.52 gam v 7.14 gam D. 4.26 gam v 5.4 gam.2. Cng thc ca oxit st l:A. FeO B. Fe2O3C. Fe3O4D. cng thc khc.Bi gii:Phn tch bi ton: Bi c nhiu phng trnh phn ng, i hi nm vng kin thc v kim loi, tnh ton phc tp nn cn lm tng bc v p dng nh lut bo ton khi lng, bo ton nguyn t, lm.1)x y 2 32yAl 3Fe O 3xFe yAl O (1) + + 02 2 22 2 2 3 3t3 2 3 23Al NaOH H O NaAlO H (2)20, 02 0, 02 0, 03NaAlO CO 2H O Al(OH) NaHCO (3)2Al(OH) Al O 3H O (4)+ + ++ + + + Nhn xt:Tt c lng Al ban u u chuyn ht v Al2O3(4). Do p dng nh lut bo ton nguyn t Al: 2 3Al(bandau) Al O5,1n 2n 2. 0,1mol102 mAl = 0,1 . 27 = 2,7 (g) x yF Om 9, 66 2, 7 6, 96g A ng2) mAl = 2,7 gam . Theo nh lut bo ton nguyn t oxi ta c: no(trong FexOy) = no(trong Al2O3)=3.0,05=0,15 mol=> nFe=(6,96- 0,15.16):56=0.08x : y = 8 : 15 khng xc nh c B i ton 21: Cho 3,42 gam Al2(SO4)3 vo 50ml dung dch NaOH, thu c 1,56 gam kt ta v dung dch X. Nng M ca dung dch NaOH l: A: 0,6 B: 1,2 C: 2,4 D: 3,6Bi gii:+ Cch 1: p dng phng php th: 2 4 3Al (SO )3, 42n 0, 01mol;342 3Al(OH)1,56n 0, 02mol78 OHn 0, 06mol M.NaOH0, 06C 1, 2M0, 05 B ng+ Cch 2: - TH1: OH- thiu nn xy ra phn ng: Al3+ + 3OH- Al(OH)33 3Al(OH) NaOH Al(OH)OH1, 56n 0, 02mol; n n 3n 0, 06mol78 NaOHM0, 06C 1, 2M0, 05 B ng - TH2: OH- d ho tan mt phn kt ta nn xy ra phn ng: Al3+ + 3OH- Al(OH)3(1)0,02 0,06 0,02Al(OH)3 + OH- AlO-2 + 2H2O(2) 32 4 3Al (SO )Aln 2n 0, 02mol+ 3Al(OH)n-OHn0,020,060,083Al(OH)n b ho tan = 0,02 - 0,02 = 0 loi trng hp nyB i ton 22:Cho 200ml dung dch KOH vo 200 ml dung dch AlCl3 1M thu c 7,8 gam kt ta keo. Nng M ca dung dch KOH l: A: 1,5 v 3,5B: 1,5 v 2,5 C: 2,5 v 3,5 D: 2,5 v 4,5Bi gii:+ Cch 1: p dng phng php th3 3AlCl Al(OH)7,8n 0, 2mol; n 0,1mol78 KOHKOHKOH MKOH M0, 3n 0, 3mol C 1, 5M0, 20, 7n 0, 7mol C 3, 5M0, 2

Chn A ng Bi ton 23: Cho 200ml dung dch HCl vo 200ml dung dch NaAlO2 2M thu c 15,6 gam kt ta keo. Nng M ca dung dch HCl l:A: 1 hoc 2 B: 2 hoc 5 C: 1 hoc 5 D: 2 hoc 4Bi gii:3Al(OH)15, 6n 0, 2mol78 HClHClMHMH0, 2n 0, 2mol C 1M0, 21n 1mol C 5M0, 2++

C ngBi ton 24: Cho 200ml dung dch H2SO4 vo 400ml dung dch NaAlO2 1M thu c 7,8 gam kt ta. Nng M ca dung dch H2SO4 l:3Al(OH)n-OHn0,20,60,30,7 0,8+Hn0,40,2 11,60,4Al(OH)30,2A: 0,125 v 1,625 B: 0,5 v 6,5 C: 0,25 v 0,5D: 0,25 v 3,25Bi gii:p dng phng php th: 2 3NaAlO Al(OH)7,8n 0, 4mol; n 0,1mol78 2 42 4H SO MH SO M1 0,1 0, 05n H 0, 05mol C 0, 25M2 2 0, 21 1, 3 0, 65n H 0, 65mol C 3, 25M2 2 0, 2++

D ngBi ton 25: Ho tan 4,59 gam Al bng dung dch HNO3 thu c hn hp kh NO v N2O c t khi hi i vi hir bng 16,75.Th tch lt NO v N2O (ktc) thu c ln lt l: A: 6,72 lt v 2,24 ltB: 67,2 lt v 22,4 ltC: 2,016 lt v 0,672 lt D: 0,672 lt v 2,016 ltBi gii:- p dng phng php bo ton electron: +Al -3eAl3+Al4,59n 0,17mol27 0,17.--> 3.0,17mol+ 5 25 12N 3e N (NO)3x..........x2N 2.4e 2N (N O)8y.......2y (y)+ ++ + + '+ 3x + 8y = 0,51 (1)p dng phng php ng chox m o l N O . . . . . . . MN O= 3 04 4 = 3 3 , 5M= 2 . 1 6 , 7 5 = 3 3 , 5y m o l N2O . . . . . . . MN 2 O= 4 43 3 , 5 - 3 0x 10,5 3y 3.5 1 x = 3yT (1) v (2):x = 0,09, y = 0,03. Vy VM = 0,09 22,4 = 2,016 lt,2N OV 0, 03 22, 4 0, 672 lt C ng +Hn0,40,11,3 1,60,4Al(OH)3Ch : + Nu VNO = 0,03 22,4 = 0,672 lt;2N OV= 0,09 22,4 = 2,016D sai+ T (2) x = 3y VNO = 0,3 22,4 = 6,72 lt;2N OV= 22,4 A sai+ T (2) x = 3y VNO = 67,2 lt;2N OV= 22,4B sai