TRNG I HC S PHM HUKHOA HA HC

TIU LUN MN PHNG PHP NGHIN CU KHOA HCNGNH: S PHM HA HC

ti: Su tm v xy dng h thng bi tp nng cao chng kim loi kim- kim loi kim th- nhm bi dng hc sinh gii ha hc THPT

H v tn: V Th nh Hng Lp: Ha 2A M SV: 13S2011039

A.C S L THUYT

B. CC DNG BI TP Dng 1: Tm tn kim loi hoc cng thc ca hp chtPhng php tng qut Kim loi kim hoc hn hp kim loi kim tc dng vi nc hoc axit th n(H2) =1/2*n(kim loi) Kim loi kim th hoc hn hp kim loi kim th tc dng vi nc hoc axit th n(H2) =n(kim loi)1. Nu bi ton yu cu tm tn 2 kim loi A,B k tip nhau trong cng mt phn nhm chnh th tm khi lng nguyn t trung bnh ca 2 kim loi ri dng bng h thng tun hon suy ra A v B:

A = 1. Nu cho hn hp 2 kim loai A, B tan c trong nc:+ Nu A, B l kim loi kim th c hai u phn ng trc tip vi nc.+ Nu A l kim loi kim, B cha bit th c th1. B l kim loi kim th: Ca, Ba, th c A, B u tan1. B l nguyn t lng tnh (Be, Zn , Al, Cr).1. Phng trnh dng tng qut ca nguyn t lng tnh tc dng vi dung dch kim. Gi dd kim l AOH, nguyn t lng tnh l B:

B + nH2O B(OH)n + n/2 H2B(OH)n + (4 n) AOH A4 nBO2 + 2H2O

B + (n 2) H2O + (4 n) AOH A4 nBO2 +n/2 H2 Ty nhiu trng hp m ta s dng cc nh lut bo ton: khi lng, nguyn t, bo ton in tch, bo ton electronTa s xt cc trng hp c th sau:Bi tp

Bi 1: Ha tan hon ton 0,297g hn hp gm natri v mt kim loi thuc nhm IIA trong bng HTTH vo nc, ta c dung dich X v 56ml kh Y (dktc)Xc nh kim loi cha bit v tnh thnh phn mi kim loi trong hn hp ban u.Hng dn gii:t k hiu nguyn t khi ca kim loi nhm IIA cha bit l M v a,b ln lt l s mol ca Na v M trong hn hp.Theo bi ta c hpt ha hc:mhh =mNa + mM =23a +Mb =0,297 (1)nH2 = 0,5a +b =56:22400 =0,0025 mol (2)T (2) suy ra a = 0,005 2b, th vo (1) ri rt gn ta c:b(M-46) =0,182 hay b=0,182/(M-46)iu kin : 0mNa=0,002.137=0,274gV mNa=0,297-0,274=0,0023g.Bi 2: Oxi ha hon ton 7,83 gam hp kim gm mt kim loi ha tr II, mt kim loi ha tr III th to thnh 14,23g hn hp 2 oxit c t l s mol l 1:1.Cho hn hp oxit vo dung dch kim d th cn li 4,03g cht khng tan. Xc nh tn 2 kim loi ban u. Bit oxit ca kim loi III l oxit lng tnhHng dn gii:t k hiu v NTK ca hai kim loi ha t II v III ln lt l M v R.V ch R2O3 l oxit lng tnh tan trong dung dch kim nn:mMO= 4,03gVy: mR2O3=14,23-4,03=10,2gTheo nh lut BTKL, ta c:mO= 14,23-7,83=6,4gDo hai oxit c t l v s mol l 1:1 nn s gam oxi trong MO v trong R2O3 t l vi nhau l 1:3T ta tnh c mO trong MO =6,4/(3+1)=1,6gV mMO=4,03g, suy ra:mM trong 4,03g MO =4,03-1,6=2,43g=>M=2,43.16/1,6=24,3Vy M l MgTng t, mO trong 10,2g R2O3=1,6.3=4,8gmB=10,2-4,8=5,4g=>B=5,4.48/2,4.8=27

Vy B l Al. Bi 3: Hn hp X gm hai kim loi kim A,B thuc hai chu k lien tip.a) Nu cho X tc dng va vi V1 lt dung dch HCl ri c cn th thu c a gam hn hp mui clorua khan. Cn nu cho X tc dng vi V2 lt dung dch H2SO4 ri c cn th thu c b gam hn hp mui sunfat khan.Hy thit lp biu thc tnh tng s mol ca hai kim loi c trong X theo a v b.b) Nu cho X tc dng vi dung dch gm 0,5V1 lt dung dch HCl v 0,5V2 lt dung dch H2SO4 dng trn ri em c cn th thu c c gam hn hp mui clorua v sunfat khan ca A v B.Hy thit lp biu thc tnh c theo a v b.c) Cho b=1,1807a. Hi A,B l kim loi kim g?d) Cho c=45,25g. Tnh khi lng ca X v lng kt ta thu c sau khi ha tan c gam hn hp cc mui trn vo nc v cho tc dng vi BaCl2 d.Hng dn gii:K hiu chung 2 kim loi kim A,B l M v tng s mol ca A,B l x mol.a) Cc PTPU:M +HCl =MCl +1/2 H22M +H2SO4 =M2SO4 + H2Hiu khi lng mui sunfat v mui clorua:b-a =0,5x(2M+96)-x(M+35,5)=12,5xhay x=(b-a)/12,5 (I)b) V dng mi loi axit bng mt na lng ln mt, nn lng mui mi loi cng bng mt na lng ln mt, tc:c=(a+b)/2 (II)c) T l khi lng 2 mui l:Mui sunfat/Mui clorua =0,5x(2M+96)/(x(M+35,5)) =b/a =1,1807 M=33,8. V hai kim loi kim tuc hai chu k lien tip nn chng ch c th l : Na(23) v K(39)d) T (II) ta c: 45,25=(a+1,1807a)/2 => a=41,5Theo (I), tng s mol ca hai kim loi kim l:x= (b-a)/12,5 =(1,1807a-a)/12,5 =0,1807a.41,5/12,5 =0,6molVy, khi lng cc kim loi kim bng: 0,6.33,8 =20,28gTheo (2): nH2SO4 = nM2SO4 =0,3/2=0,15 mol (v ch dng mt na lng axit)Ta c: mBaSO4 =0,15.233= 34,95g.Bi 4: Ho tan 46g mt hn hp gm Ba v 2 kim loi kim A, B thuc 2 chu k lin tip vo nc th c dung dch D v 11,2 lt kh o ktc. Nu thm 0,18 mol Na2SO4 vo dung dch D th dung dch sau phn ng vn cha kt ta ht Ba. Nu thm 0,21 mol Na2SO4 vo dung dch D th dung dch sau phn ng cn d Na2SO4. Xc nh tn 2 kim loi kim? Hng dn gii:Gi R l kim loi i din cho hai kim loi kim A,B; x l s mol Ba, y l s mol ca R.Theo ta c:mhh = 137x + Ry=46 (1)nH2 =x +0,5y =11,2/22,4 =0,5 (2)nBa = nNa2SO4 suy ra 0,18 (137-2R)x =46-R (4)T (3) v (4) ta suy ra (137-2R).0,18 nNaOH=0,02*2+0,01=0,05 mol nCO2=0,02+0,01=0,03 molVy x=0,05/0,1=0,5(M) v H%=0,03/(3/84)*100% =84%Bi 6: Sc V lt kh CO2 vo dung dch Ca(OH)2 th thu c m1 gam kt ta. Nu sc tip 3,36 lt kh CO2 na th thu c m2 gam kt ta, cn nu sc tip V1 lt kh CO2 na th thu c kt ta cc i l m gam. Bit m1=3/7m, m1:m2=3:2. Tm V v m1,m2,m.Hng dn gii:V m2>m`1>m nn khi sc tip 3,36 lit CO2 na th kt ta t m1 tng n cc i (m) ri tan n m2.Ta c: m1=3/7m, m2=2/7m suy ra mi quan h gia 3,36lit v V l: 3,36=(1-3/7+5/7)V= 9/7V => V=2,61 ltm1=2,61/22,4*100=11,65 gamm=7/3m1= 27,183 gamm2=2/3m1= 7,77 gam. Dng 3: Hn hp nhm vi kim loi kim, kim th hoc vi kim loi khc

Phng php tng qut: Hn hp nhm vi kim loi kim hoc kim th:Thng thng ngi ta thng cho hn hp gm nhm v kim loi kim hoc kim th vo nc d hoc cho vo dung dch kim, xy ra cc phn ng sau: Hn hp vo nc d:2A + 2H2O 2AOH + H2 vi A l kim loi kimB + 2H2O B(OH)2 + H2 vi B l kim loi kim th2Al + 2OH- + 2H2O 2AlO2- + 3H2Thng thng trong dng ton ny th ngi ta thng hay cho Al d, v vy ta c th tnh s mol kim loi kim hoc kim th ban u nh sau: Gi x l s mol kim loi kim hoc kim th trong hn hp. Suy ra nOH-= x nu l kim loi kim hoc nOH-= 2x nu l kim loi kim th. T tnh c s mol ca Al p vi OH- l: nAlp= nOH- => s mol H2 do p gia nhm v bazo sinh ra l: nH2= 3/2 nOH-.Vy tng s mol H2 l: nH2=1/2 nOH- +3/2nOH-=2nOH- Cn nu hn hp vo bazo d th cc p tng t nh trn, nhng y Al tan ht, v vy:nH2=1/2nOH- + 3/2nAlbandauTy vo tng trng hp c th m ngi ta cho hn hp sau p (gm AlO2- v OH-) tc dng vi CO2 hoc axit Hn hp Al vi cc kim loi khcThng gp l cho hn hp nhm vi cc kim loi khc vo dung dch H2SO4 c, HNO3hoc vo dung dch mui ca kim loi khc.Phng php gii thng l gi 2 hay 3 n, s dng cc nh lut bo ton electron, bo ton nguyn t, phng php tng gim khi lngBi tp: Ta s xt cc bi ton c th sau:Bi 1: Ha tan 5,9 gam hn hp Al v kim loi R ha tr II vo dung dch H2SO4 c nng. Sau cho tc dng vi dung dch NaOH n d. Lc v ra sch kt ta cn li trong dung dch ri cho tacs dng tip vi dung dch HCl d, thu c dung dch A. em in phn dung dch A cho n khi c bt kh thot ra catot th dng li, khi lng ca catot tng ln 3,2 gam cn anot thot ra 1,12 lit mt kh.Xc nh R v th tch kh thot ra t dung dch H2SO4. (Cc kh o ktc). Hng dn gii:S phn ng:Al,RAl2(SO4)3, RSO4NaAlO2, R(OH)2Gi x l s mol ca R.R(OH)2RCl2R + Cl2x mol x mol x molSuy ra x=nCl2=1,12/22,4=0,05 molR=3,2/0,05=64 (Cu); nAl=(5,9-3,2)/27=0,1 molAl0 Al+3 + 3e0,1 0,3Cu0 Cu+2 + 2e0,05 0,1S+6 + 2e S+4 0,4 0,2VSO2= 0,2*22,4=4,48 lt.Bi 2: Ly 3,61 gam hn hp bt nhm, st cho tc dng vi 100 ml dung dch cha AgNO3 v Cu(NO3)2, khuy k ti phn ng hon ton ch thu c 8,12g cht rn gm 3 kim loi. Ha tan cht rn bng dung dch HCl d thy bay ra 0,672 lt kh H2 (ktc). Tnh nng mol/lit ca AgNO3 v Cu(NO3)2 trong dung dch ban u ca chng. Bit hiu sut cc phn ng l 100%.Hng dn gii:V Al hot ng hn Fe, v v cht rn gm 3 kim loi nn chng phi l Ag, Cu, Fe. Cc phn ng c th xy ra l:Al + 3AgNO3 Al(NO3)3 + 3Ag (1)2Al + 3Cu(NO3)2 2Al(NO3)3 + 3Cu (2)Fe + 2AgNO3 Fe(NO3)2 + 2Ag (3)Fe+ Cu(NO3)2 Fe(NO3)2 + Cu (4)Fe + 2HCl FeCl2 + H2 (5)Cc phn ng (1) v (5) chc chn xy ra, cn (2),(3),(4) xy ra hay khng cn ty thuc vo lng Al hay AgNO3 tha, thiu.Theo (5), nH2=0,672/22,4= 0,03mol = nFe cn liGi a, b l s mol AgNO3 v Cu(NO3)2 ta c phng trnh bo ton e:1*a + 2*b +2*0,03 =3*0,0 3 +2*0,05 (I)Mt khc, theo khi lng 3 kim loi, ta c:108a + 64b +0,03*56 =8,12 (II)T (I),(II) ta rt ra: a=0,03 v b=0,05Vy CM(AgNO3)=0,03/0,1=0,3M CM(Cu(NO3)2)=0,05/0,1=0,5M.Bi 3: Mt hn hp X gm K v Al c khi lng l 10,5g. Ho tan X trong nc th hn hp X tan ht cho ra dung dch A. a) Thm t t mt dung dch HCl 1M vo dung dch A. Khi u khng c kt ta. Khi th tch dung dch HCl 1M thm vo ti 100ml th dung dch A bt u c kt ta. Tnh % mi kim loi trong hn hp X. b) Mt hn hp Y cng gm K v Al. Trn 10,5g hn hp X trn vi 9,3g hn hp Y c hn hp Z. Hn hp Z tan ht trong nc cho ra dung dch B. Thm HCl vo dung dch B th ngay git u tin dung dch HCl thm vo c kt ta. Tnh khi lng K v Al trong hn hp Y.Hng dn gii:a) Khi thm t t dd HCl vo dd A m ban u khng c kt ta nn suy ra dd A cha KOH v KAlO2.nHCl=0,1*1=0,1 mol => nKOH=0,1 moldd A: KOH: 0,1 mol KAlO2: x molTa c pt: (0,1 +x)*39 + 27x=10,5 => x=0,1 mol%Al=25,7%, %K=74,3%b) Gi a,b ln lt l s mol ca K, Al trong hn hp YV khi nh dd HCl vo dd B th c kt ta ngay nn suy ra dd B ch cha KAlO2.Bo ton nguyn t K th s mol ca KAlO2 l: 0,2+a molBo ton nguyn t Al th s mol ca KAlO2 l: 0,1+b mol=> 0,2+a=0,1+b (1)Li c: 39a+27b=9,3 (2)T (1) v (2) suy ra a=0,1, b=0,2mK=0,1*39=3,9 gam, mAl=0,2*27=5,4 gam.Bi 4: Mt hn hp gm Na, Al, Fe. Tin hnh cc th nghim sau TN1: Cho hn hp vo nc, c V lt kh thot ra. TN2: Cho hn hp vo dung dch NaOH d, thy thot ra 7/4V lt kh. TN3: Cho hn hp vo dung dch HCl d n phn ng xong, thy thot ra 9/4V lt kh. a) Vit phng trnh phn ng v gii thch b) Tnh thnh phn % mi kim loi trong hn hp. c) Nu vn gi nguyn lng Al, cn thay Na v Fe bng mt kim loi nhm 2 c khi lng bng 45% tng khi lng Na v Fe, sau cng cho hn hp vo dung dch HCl d cho n phn ng xong, cng thy thot ra 9/4V lt kh. Xc nh tn kim loi nhm 2 (khng c dng kt qu % ca cu b). Cc th tch u o cng iu kin.Hng dn gii:a) Hn hp gm 3 kim loi: Na, Al, Fe khi cho vo nc d th s mol H2 t hn so vi khi cho vo dd bazo d suy ra trong th nghim 1 Al vn cha tan ht.Cc pt p:TN1: 2Na + 2H2O 2NaOH + H2 x x 0,5x 2Al + 2OH- +2H2O 2AlO2- + 3H2 x x 1,5xTN2: 2Na + 2H2O 2NaOH + H2 x x x 2Al + 2OH- +2H2O 2AlO2- + 3H2 y 1,5yTN3: 2Na + 2HCl 2NaCl + H2 x x 2Al + 6HCl 2AlCl3 + 3H2 y 1,5y Fe +2HCl FeCl2 + H2 z zb) TrongTN1, nH2=0,5x +1,5x=2xTrong TN2, nH2=0,5x +1,5y=7/4*2x=3,5x =>y=2xTrong TN3, nH2=0,5x +1,5y +z=9/4*2x=4,5x =>z=x%Na=23x/(23x +2x*27 +56x)*100%=17,3%%Al= 2x*27/(23x +2x*27 +56x)*100%=40,6%%Fe= 42,1%C) Gi nguyn lng Al: 2x Thay Na v Fe bng kim loi M ha tr II c khoois lng l: (23x +56x)*45%=35,55xNH2=4,5x => nM=(4,5x-3x)= 1,5xVy MM=35,55x/(1,5x)=24. Vy kim loi M l Magie (Mg).Bi 5: A l mt loi hp kim ca Ba, Mg, Al c dng nhiu trong k thut chn khng. TN1: Ly mg A (dng bt) cho vo nc ti khi ht phn ng, thy thot ra 0,896 lt H2 TN2: Ly mg A (dng bt) cho vo dung dch NaOH d ti khi ht phn ng, thy thot ra 6,944 lt H2. TN3: Ly mg A ho tan bng mt lng va dung dch HCl, ta thu c dung dch B v 9,184 lt H2 a) Tnh m v % khi lng ca cc kim loi trong hn hp A. b) Thm 10g dung dch H2SO4 9,8% vo dung dch B, sau thm tip 210g dung dch NaOH 20%. Sau khi kt thc phn ng, ly kt ta thu c em nung nhit cao. Tnh khi lng cht rn thu c. Cc th tch kh c o ktc.Hng dn gii:a) Tng t nh cu trn ta suy ra c trong TN1, Al cha tan ht.Gi x l s mol Ba trong hn hp => nOH-= 2x mol2Al + 2OH- + 2H2O 2AlO2- + 3H22x 2x 3x=> nH2=4x = 0,04 => x=0,01 molTrong TN2, nH2=0,31 mol => nAl= (nH2- 0,01)*2/3= (0,31-0,01)*2/3= 0,2 molTrong TN3, nH2= 0,41 mol => nMg=(0,41-0,31) =0,1 molVy m= 0,01*137 + 0,2*27 + 0,1*24=9,17 gam%Ba=14,94%, %Al= 58,89%, %Mg=26,17%b) nH2SO4= 0,01 mol, nNaOH= 1,05 molTa c cc p:H+ + OH- H2O0,02 0,02Ba2+ + SO42- BaSO40,01 0,01Mg2+ + 2OH- Mg(OH)20,1 0,2Al3+ + 3OH- Al(OH)30,2 0,6Tng s mol OH- p l: 0,02 +0,2 +0,6=0,82 mol OH- d s ha tan tip Al(OH)3.nOH- d= 1,05-0,82=0,23 mol>0,2 mol nn Al(OH)3 s b ha tan t, v vy kt ta sau p ch c Mg(OH)2 v BaSO4em nung kt ta th thu c ct rn l: MgO (0,1 mol), BaSO4 (0,01 mol) c khoois lng l: 6,33 gam.Bi 6: Ho tan 2,16g hn hp 3 kim loi Na, Al v Fe vo nc d, Thu c 0,448 lt kh (ktc) v cn li mt lng cht rn. Cho lng cht rn ny tc dng ht vi 60ml dung dch CuSO4 1M thu c 3,2g Cu kim loi v dung dch A.Cho dung dch A tc dng vi mt lng va dung dch NH3 thu c kt ta. Nung kt ta thu c trong khng kh n khi lng khng i c cht rn B. a) Xc nh khi lng tng kim loi trong hn hp u.b) Tnh khi lng cht rn B.Hng dn giia) 3 kim loi Na, Al ,Fe tc dng vi nc d th:2Na + 2H2O 2NaOH + H2x x 0,5x2Al + 2OH- + 2H2O 2AlO2- + 3H2x x 1,5xTa c: 2x=0,02 => x= 0,01Cht rn sau p gm Fe v c th c Al d. Gi a,b ln lt l s mol ca Al d v Fe.nCuSO4=0,06 mol, nCu=0,05 molTheo bo ton electron, ta c: 3a +2b= 0,05*2=0,1 (1)Theo khi lng, ta c pt: 27a + 56b=2,16- 0,01*23- 0,01*27=1,66 (2)T (1),(2) ta suy ra: a=b= 0,02 molVy hn hp ban u gm: Na(0,01 mol), Al(0,03 mol), Fe(0,02mol).mNa=0,01*23=0,23g, mAl=0,03*27=0,81g, mFe=0,02*56=1,12gb) DD A gm: Al3+: 0,02 mol, Fe2+: 0,02 mol, Cu2+: 0,01 molKhi cho dd NH3 va vo dung dch A th kt ta gm Al(OH)3 v Fe(OH)2 v Cu2+ to phc vi NH3.Nung kt ta trong khng kh ti khi lng khng i th to 2 oxit Al2O3 (0,01 mol) v Fe2O3 (0,01 mol).Vy khi lng cht rn thu c l: 0,01*102+0,01*160=2,62 gam.Bi 7: Cho a gam bt Al vo dung dch cha 0,1 mol AgNO3 ;0,15 mol Cu(NO3)2 v 0,2 mol Fe(NO3)3 thu c dung dch X v kt ta Y. Hy la chn gi tr ca a kt ta Y thu c cha 3 kim loiHng dn gii:Khi cho nhm vo hn hp cc mui trn th th t b kh s l:Ag+ +1e Ag0,1 0,1

Fe3+ +1e Fe2+0,2 0,2Cu2+ +2e Cu0,15 0,3Fe2+ + 2e Fe0,2 0,4Theo thu c 3 kim loi th 3 kim loi chc chn phi l Ag, Cu, Fe. Mun vy th lng nhm cn dng phi ln hn lng nhm phn ng vi cc mui to ra 2 kim loi l Ag v Cu v phi nh hn hoc bng lng nhm phn ng ht vi cc mui to ra 3 kim loi. phn ng ht vi cc mui m to ra 3 kim loi th: mAl= (0,1+ 0,2+ 0,3+ 0,4)/3*27= 9 g phn ng vi cc mui m to ra 2 kim loi l Ag v Cu th mAl= (0,1+ 0,2+ 0,3)/3*27= 5,4 g.V vy gi tr ca a nm trong khong: 5,4< a 9.Bi 8: Cho 0,87 gam hh gm Fe, Cu v Al vo bnh ng 300 ml dd H2SO4 0,1M. Sau khi cc phn ng xy ra hon ton, thu c 0,32 gam cht rn v c 448 ml kh (ktc) thot ra. Thm tip vo bnh 0,425 gam NaNO3, khi cc phn ng kt thc th th tch kh NO (ktc, sn phm kh duy nht) to thnh v khi lng mui trong dd l?Hng dn gii:n (H2SO4) = 0,03 nH+ = 0,06n( H2 )= 0,448/22,4 = 0,02 molnCu = 0,32/64 = 0,005 moln (NaNO3)= 0,005 Fe + 2H+ Fe2+ + H2 x-----2x--------x--------x Al + 3H+ Al3+ + 3/2H2 y-----3y---------y--------3/2yTa c : x + 3/2y = 0,02 (1) v 56x + 27y = 0,87 0,32 = 0,55 (2)1. V (2) x = 0,005 v y = 0,01Dung dch sau p c : nFe2+ = 0,005 nH+ cn li = 0,06 2x 3y = 0,06 2.0,005 3.0,01 = 0,02 3Fe2+ + 4H+ + NO3- 3Fe3+ + NO + 2H2O 0,005---1/150---0,005/3-------------0,005/3 nH+ cn = 0,02 1/150 = 1/75 ; nNO3- = 0,005 0,005/3 = 1/300 3Cu + 8H+ + 2NO3- 3Cu2+ + 2NO + 4H2O 0,005---1/75----1/300 ---------------1/300Sau phn ng H+ v NO3- ht nNO = 0,005/3 + 1/300 = 0,005 mol V(NO) = 0,005.22,4 = 0,112 ltm mui = mcc kim loi ban u + mSO42- + mNa+ = 0,87 + 0,03.96 + 0,005.23 = 3,865gam. Dng 4: phn ng nhit nhmPhng php tng qut:Ta c: FexOy, Cr2O3, CuO, MxOy + Al t Al2O3 + (Cu, Fe, Cr .)nh lut bo ton khi lng:mAl(ban u) + moxit = mcht rnnh lut tng gim khi lng:mrn gim = mrn trc - mrn sau Cht rn thu c sau phn ng gm 2 nhm:+ Nhm 1: Al2O3 v kim loi M+Nhm 2: Al d hoc oxit d hoc c 2 cng d- Thng gp l phn ng nhit nhm: Gi a l s mol ca Al; b l s mol ca oxit st.(Fe2O3)2Al + Fe2O3 Al2O3 + 2FeI. Nu phn ng xy ra hon ton: Thng do khng bit s mol Al v Fe2O3 l bao nhiu nn phi xt 3 trng hp ri tm nghim hp l:1. Trng hp 1: Al v Fe2O3 dng va : 2Al + Fe2O3 Al2O3 + 2Fe

a a

Hn hp sau phn ng: Fe: a mol; Al2O3: mol2. Trng hp 2: Al dng d: 2Al + Fe2O3 Al2O3 + 2Fe 2b b b 2b Hn hp sau phn ng: Fe: 2b mol; Al2O3: b mol; Ald: (a-2b) mol. iu kin: (a-2b>0)3. Trng hp 3: Fe2O3 dng d: 2Al + Fe2O3 Al2O3 + 2Fe

a a

Hn hp sau phn ng: Fe: a mol; Al2O3: ; Fe2O3: (b-)mol.

iu kin: (b-)>0)II. Nu phn ng xy ra khng hon ton: Gi x l s mol Fe2O3 tham gia phn ng2Al + Fe2O3 Al2O3 + 2Fe2x x x 2x Hn hp sau phn ng: Fe: 2x mol; Al2O3: x mol; Fe2O3 d: (b-x)mol; Al d: (a-2x)molCh : Nu yu cu tnh hiu sut phn ng ta gii trng hp phn ng xy ra khng hon ton.Ta thng gp cc vn sau:1. VN 1: bi cho sn phm sau phn ng tc dng vi NaOH v cho kh thot ra th sn phm sau phn ng c Al d1. VN 2: bi cho sn phm sau phn ng nhit nhm ch tc dng vi dung dch H+ cho kh H2 bay ra1. VN 3: Thc hin phn ng nhit nhm gm hn hp cc oxit kim loi km hot ng vi nhm. Sn phm sau phn ng tc dng vi dd H+ cho kh H2 v cho tc dng vi dung dch OH- v cho khi lng cht rn khng tc dng.1. VN 4: bi cho sn phm sau phn ng nhit nhm c chia thnh hai phn, mt phn cho tc dng vi dung dch OH- (th nghim 1) v mt phn cho tc dng vi dung dch H+ (th nghim 2). Tnh khi lng mi cht trong hn hp.Bi tp:i vi tng vn th ta s c nhng cch gii quyt ring, by gi ta s i vo c th tng bi:Bi 1: C mt hn hp Al v FexOy. Sau phn ng nhit nhm thu c 92,35 gam cht rn. Ha tan cht rn trong NaOH d thy c 8,4 lit (ktc) bay ra v cn li phn khng tan D. Ha tan lng D bng H2SO4 c nng phi dng 60 gam dung dchH2SO4 98%. Ga s ch to mui st(III). Tnh khi lng Al2O3 to thnh v cng thc FexOy?Hng dn gii:Ta c: 2yAl + 3FexOy t y Al2O3 + 3xFe1. Sn phm sau phn ng tc dng vi NaOH d, c kh bay ra chng t sn phm c nhm d, cht rn cn li l Al2O3, Al d, Fe.D ch c FeAl + NaOH + H2O 3/2 H2 0,250,375mAl d = 6,75 gam ; n(H2SO4) = 0,6 molLBT electron: 2nH2SO4 = 6nFe nFe = 0,3 molKhi lng Fe to thnh do phn ng nhit nhm l:mFe = 0,3.56.4 = 44,8(gam)mAl2O3 = 92,35 44,8 = 40,8 gam1. Xc nh cng thc ca FexOy: 2yAl + 3FexOy t y Al2O3 + 3xFe2yy3xT l: = Fe2O3Bi 2: Ly 26,8 gam hn hp gm Al v Fe2O3 thc hin hon ton phn ng nhit nhm, thu c cht rn A, cho cht rn ny ha tan hon ton trong dung dch HCl d thy kh that ra 11,2 lit kh H2(ktc). Hy xc nh thnh phn phn trm cc cht trong hn hp?Hng dn gii:Ta c: 2Al + Fe2O3 t Al2O3 + 2FeC 3 trng hp:1. Sn phm Al2O3 v Fe (phn ng va ) 2Al + Fe2O3 t Al2O3 + 2Fe2bb mol2bAl2O3 + 6HCl 2AlCl3 + 3H2OFe + 2HCl H2nAl= nFe = nH2 = 0,5 mol = 2b b= 0,25 molTng khi lng ban u:mAl + mFe2O3 = 0,25.27 + 0,5.160 = 53,5> 26,8 (loi)1. Sn phm Al2O3, Fe, Al (Al d)nAl ban u = a mol; nFe2O3 = b mol

Fe + 2HCl H22b2bAl + 2HCl 3/2H2(a 2b)3/2(a 2b)27a +160b = 26,8a = 0,4 mol2b +3/2(a 2b) = 0,5b = 0,1 mol%Al = = 40,3%%Fe2O3 = 100 40,3 = 59,7%1. Sn phm Al2O3, Fe, Fe2O3 d nAl = nFe = nH2 = 0,5 nFe2O3 = 0,25 molNgoi khi lng Al, Fe2O3 = 53,5 gam cn khi lng d Fe2O3 nn khi lng ln hn 26,8 gam ban u (loi)Bi 3: : Mt hn hp M gm Fe3O4, CuO v Al c khi lng 5,55g. Sau khi thc hin phn ng nhit nhm xong (hiu sut 100%) th thu c cht rn A1. Nu ha tan A trong HCl d th lng H2 sinh ra ti a 1,344 lit khi ( ktc).1. Nu ha tan A trong NaOH d th sau khi phn ng xong cn 2,96 g cht rn. Tnh thnh phn phn trm ca cc cht trong hn hp A?Hng dn gii:8Al + 3Fe3O4 t 4 Al2O3 + 9Fe8/3yy mol3y2Al + 3CuO t Al2O3 + 3Cu2/3xx molxCht rn A c Fe hoc Al d( trong dung dch HCl d H2)1. Ga s Al phn ng ht, A : ch c FenFe = nH2 = 0,06 mol mFe = 3,36 > 2,96 gam vy Al d: cht rn A c Fe v Al d, CuM(Cu + Fe) = 2,96 (gam)Gi Al: z mol (trong hn hp)Ta c:80x +233y + 27z = 5,55x = 0,0264x + 3y.5b = 2,96y = 0,013y + (z 2/3x 8/3y).3/2 = 0,06z = 0,06

%Cu = 28,8%; %Al = 29,2%; %Fe3O4 = 42%Bi 4: Cho hn hp A dng bt gm bt Al v oxit st t. Nung A nhit cao phn ng xy ra hon ton thu c hn hp B. nghin nh hn hp B chia thnh hai phn:Phn t (phn I) tc dng vi dung dch NaOH d thu c 1,176 lt kh H2 (o ktc). Tch ring cht khng tan em ha tan trong dung dch HCl d thu c 1,008 lt kh (o ktc).1. Phn nhiu (Phn II) cho tc dng vi dung dch HCl thu c 6,552 lt kh (o ktc).1. Vit cc phng trnh phn ng xy ra.1. Tnh khi lng hn hp B v thnh phn % khi lng cc cht trong hn hp A.Hng dn gii:1. Phng trnh phn ng:8Al+3Fe3O49Fe +4Al2O3(1)1. Phn I:Al2O3+2NaOH 2NaAlO2+H2O(2)

Al+NaOH NaAlO2 +3/2H20,035 mol0,0525 mol

Cht khng tan l Fe cho tc dng vi dung dch HCl dFe+2HClFeCl3+H20,045mol0,045molTheo phng trnh (1):nAl2O3 = 4/9nFe = 0,045 = 0,02 molphn IIAl2O3 +HClAl2O3+3H2OAl+3HClAlCl3+ 3/2H2OFe+2HClFeCl2+H2Gi s phn II nhiu gp m phn I nn trong phn IInA= 0,035mnFe = 0,045mnH2 gii phng: x0,035m + 0,045m = = 0,2925 molgii ra ta c: m = 3Vy khi lng c hn hp B gp 4 ln phn Ip dng nh lut bo ton khi lngmB = 0,035.27.4 + 0,045.4.56 + 0,02.4.102 = 22,02g%Al = 36,8%; %Fe3O4 = 63,2%Bi 5: Tin hnh phn ng nhit nhm vi hn hp A gm bt Al v Fe2O3 trong iu kin khng c khng kh sau phn ng thu c hn hp cht rn B, chia B thnh hai phn bng nhau.Phn 1: Cho tc dng vi dung dch NaOH d, thu c 1,68 lit kh (hidro 27,3% v 2,2 atm)Phn 2: Cho tc dng vi dd HCl 1M va , thu c 12,32 lt H2 (dktc) v dd D, cho dung dch D tc dng vi dung dch NaOH d trong khng kh lc kt ta, ri em nung cho n khi khi lng khng i th thu c cht rn E. Cc phn ng u c hiu sut 100%1. Vit phng trnh phn ng1. Tnh % khi lng trong hn hp B 1. Tnh th tch dd HCl1. Tnh khi lng EHng dn gii:1. 2Al + Fe2O3 t Al2O3 + 2FePhn 1: nH2 = = 0,15 molAl2O3 + 2NaOH 2NaAlO2 + H2O

2Al + 2NaOH +2 H2O 2 NaAlO2 +3 H20,10,15Phn 2: Al2O3 +HClAl2O3+3H2OFe+2HClFeCl2+H2xxAl+3HClAlCl3+ 3/2H2a1,5a(H2) = x + 1,5a = 0,55 x = 0,55 0,15 = 0,4 mol hn hp B c: 0,1 mol Al; 0,4 mol Fe1. 2Al + Fe2O3 t Al2O3 + 2Fe0,20,4 hn hp B c: 0,1 mol Al; 0,4 mol Fe; 0,2 mol Al2O3Tnh phn trm khi lng trong hn hp B:mhh = 0,1.27 + 0,2.102 + 0,4.27 = 45,5 gam%Al = 5,93%; %Fe = 49,23%; %Al2O3 = 44,84%1. Al2O3 + 6HCl 2AlCl3 + 3H2O0,21,2nHCl = 1,2 + 0,8 + 0,3 = 2,3 molVHCl = = 2,3 lt

1. Ta c:0,4 mol FeFe FeCl2 Fe(OH)2 Fe(OH)3 1/2 Fe2O30,4mol0,2 molmE = 0,2.160 = 32 gamBi 6: Ly 93,9 gam Fe3O4 trn vi Al c hn hp X. Nung hn hp trong mi trng khng c khng kh. Sauk hi phn ng xy ra hon ton ta thu c hn hp Y. Chia Y thnh 2 phn c khi lng khc nhauPhn 1: Tc dng vi dd NaOH d cho 0,673 lit kh H2(ktc).Phn 2: Tc dng vi dung dch HCl d cho 18,816 lit kh H2(ktc)Tnh khi lng ca cc cht trong hn hp u, bit H = 100%Hng dn gii:8Al+3Fe3O49Fe +4Al2O38/3aamol3a4/3anAl d = b - 8/3aPhn 1: Al + NaOH 3/2 H20,020,03molPhn 2: Al+3HClAlCl3+ 3/2H20,02k mol0,03k molFe+2HClFeCl2+H23a mol3a molGi s s mol mi cht trong phn 2 bng k phn s mol trong phn 1nH2 = 3a. + = = 0,84

gii h: 232a + 27b = 93,9 = 0,02 + 0,01.3k = 0,84k = 4; mFe3O4 = 69,6 gam; mAl = 24,3 gam Bi 7: Mt hn hp bt nhm v bt oxit st, trn tht u ri chia lm 3 phn bng nhau:Phn 1: cho tc dng vi HCl to ra 3,36 lit kh H2Dng tia la in t chy phn 2 v phn 3 (gi s hiu sut qu trnh t chy l 100% v b qua tc dng ca khng kh vi hn hp)Sn phm thu c phn th hai cho tc dng vi NaOH d thu c 0,672 lit kh H2.Sn phm thu c phn th ba em ha tan trong dd H2SO4 1,2M thu c 2,688 lit kh H2. Hi:a) Thnh phn % ca hn hp ban u v hn hp thu c sau p nht nhm l bao nhiu?b) Tnh th tch dd H2SO4 ti thiu phi dng. (th tch cc kh u o ktc)Hng dn gii:PTPU xy ra: phn 1: 2Al + 6HCl 2AlCl3 +3H2 (1) phn 2 v 3: 2y Al+ 3FexOy y Al2O3 +3x Fe (2)Sn phm thu c phn 2 cho tc dng vi dd NaOH d thu c kh H2 => sau chy cn d Al.2Al + 2NaOH + 2H2O 2NaAlO2 + 3H2 (3)Sn phm thu c phn 3(c Al d v Fe) tan trong H2SO42Al + 3H2SO4 Al2(SO4)3 + 3H2(4)Fe + H2SO4 FeSO4 + H2(5)Theo (1): nAl (trong 1/3 hn hp A)=3,36/22,4*2/3=0,1 molTheo (3): nAl (cn d)= 0,672/22,4*2/3=0,02 molTheo (5): nFe (to ra)=nAl (tham gia phn ng nhit nhm) = 0,1-0,02=0,08 molTheo (2) ta c t l hay x:y=3:4Cng thc oxit st l Fe3O4a) V gi s hiu sut t chy l 100% nn:nFe3O4 (trong 1/3 hn hp A)= 1/3 nFe= 1/3*0,09= 0,03 molTrong 1/3 hn hp ban u c:mAl= 0,1*27=2,7 gamv mFe3O4=0,03*232= 6,96 gam. Vy, trong hn hp A c:%Al=27,95%, %Fe3O4=72,05%Trong hn hp thu c sau phn ng c: (v hiu sut 100% nn Al d, Fe3O4 phi ht)%Al=%Fe=%Al2O3=42,24%b) Khi cho phn 3 tan trong dung dch H2SO4, ngoi p 4,5 cn c cc p sau:Al2O3 + 3H2SO4 Al2(SO4)3 + 3H2O(6)T (4,5,6) ta d dng tinh c s mol H2SO4 phan ng:3/2*0,02 + 0,09 + 3*2*0,02=0,24 molVy th tch ti thiu dung dch H2SO4 1,2M cn cho qu trnh ha tan l: 0,24/1,2=0,2 lit.Bi 8: Trn u 83 g hn hp Al, Fe2O3, CuO ri tin hnh phn ng nhit nhm.Gi s lc ch xy ra 2 p kh oxit thnh kim loi. Chia hn hp sau p thnh 2 phn c khi lng chnh lch nhau 66,4 g. Ly phn c khi lng ln ha tan bng dung dch H2SO4 d, thu c 23,3856 lt kh H2 (ktc), dd X v cht rn. Ly 1/10 dd X cho tc dng va vi 200ml dd KMnO4 0,018 M (bit rng trong mi trng axit Mn+7 b kh thnh Mn+2).Ha tan phn c khi lng nh bng dd NaOH d thy cn li 4,736 g cht rn khng tan.3. Vit cc pt p xy ra.3. Cho bit trong hn hp ban u s mol ca CuO gp n ln s mol ca Fe2O3, tnh % mi oxit kim loi b kh. p dng vi n=3/2.Hng dn gii: 1. Theo gi thit ta c phn ng nhit nhm:2Al + Fe2O3 Al2O3 + 2Fe(1)2Al + 3CuO Al2O3 + 3Cu(2)Cc pt p ca phn c khi lng ln ha tan bng dd H2SO4 d l:Fe + H2SO4 FeSO4 + H2(3)Al2O3 + 3H2SO4 Al2(SO4)3 + 3H2O(4)C th xy ra cc phn ng sau:2Al + 3H2SO4 Al2(SO4)3 + 3H2(5)Fe2O3 + 3 H2SO4 Fe2(SO4)3 + 3H2O(6)CuO + H2SO4 CuSO4 + H2O(7)2KMnO4 + 10FeSO4 +8 H2SO4 K2SO4 +2MnSO4 +5 Fe2(SO4)3+ 8H2O (8)Cc pt p ha tan phn c khi lng nh vi dd NaOH d:Al + NaOH +H2O NaAlO2 +3/2 H2(9)Al2O3 + 2NaOH 2NaAlO2 + H2O(10)1. Phn ng nhit nhm khng lm thay i ton b khi lng hn hp rn.Gi m1 l khi lng phn ln, m2 l khi lng phn nh. Theo gi thit ta c hpt:m1 + m2 =83 v m1-m2=66,4 => m1=74,7g v m2=8,3gGi x,y l s mol Al, Fe c trong phn ln th ton b Fe ca phn ln bin thnh Fe2+ (3) ri chuyn thnh Fe3+ trong (8)Nu tnh cho c dd X th theo (8) ta c:y= s mol Fe= s mol FeSO4= 0,2*0,018*10/2*10=0,18 mol FeTheo (3) v (5) ta c: 3/2x + y=23,3856/22,4=1,044 =>x=0,576 mol, trrong c hn hp c: 0,576*10/9=0,64 mol Al d vy (5) xy raNu ta ha tan ht c 83g hh trong NaOH d th phn cn li s l 4,736*10/1=47,36g Lng Al2O3 c trong c hn hp l 83-47,36-0,64*27= 18,36g=> n=0,18 molTheo(1): nAl2O3=1/2*0,18/9*10=0,1 molS mol ca Al2O3 trong (2)=0,18-0,1=0,08 mol => s mol CuO b kh l 0,08*3=0,24 molTng s mol Al ban u l: 0,36+0,64=1 molGi p,q ln lt l s mol Fe2O3 v CuO trong hh banu, q=npTa c pt theo khi lng: 160p+80q+27=83 =>2p+np=0,7 p= v q=np= % Fe2O3 b kh =0,1/p*100%%CuO b kh = 0,24/q*100%Vi n=3/2, ta c: %Fe2O3= 50%, %CuO= 80%.Bi 9: Mt hn hp A gm bt nhm v bt oxit st, trn tht u ri chia lm 3 phn bng nhau:Phn 1: cho tc dng vi HCl to ra 3,36 lt kh H2Dng tia la in t chy phn 2 v phn 3 (gi s hiu sut qu trnh t chy l 100% v b qua tc dng ca khng kh i vi hn hp)Sn phm thu c phn th 2 cho tc dng vi dung dch NaOH d thu c 0,672 lt kh H2Sn phm thu c phn th 3 em ha tan trong dung dch H2SO4 1,2M thu c 2,688 lt kh H2. Hi:a) Thnh phn % ca hn hp ban u v hn hp thu c sau p nhit nhm l bao nhiu?b).Tnh th tch dung dch H2SO4 ti thiu phi dng (cc kh o ktc)Hng dn gii:a) Phn 1: Gm bt nhm v oxit st tc dng vi HCl, nH2= 3,36/22,4= 0,15 mol nAl(1/3 A)= 0,15*2/3= 0,1 molPhn 2: sn phm ca phn ng nhit nhm tc dng vi NaOH d sinh ra kh => Al d sau pnH2= 0,672/22,4= 0,03 mol => nAl d= 0,03*2/3= 0,02 mol nAl p = 0,1-0,02=0,08 molPhn 3: Sn phm ca phn ng nhit nhm tc dng vi H2SO4 th c Al d v Fe p sinh ra kh: nH2= 2,688/22,4=0,12 mol, nAl d= 0,02 mol => nFe= 0,12- 0,03= 0,09 molS mol oxi nguyn t c trong oxit st l: nO= nO(Al2O3)= 0,08/2*3= 0,12 molFexOy => x:y= 0,09: 0,12= 3:4Vy CT oxit st l Fe3O4.% cc cht trong hn hp cng chnh l % cc cht trong phn 1Phn 1 gm Al: 0,1 mol, Fe3O4: 0,03 mol% Al=27,95%, %Fe3O4= 72,05%Phn 2 gm: Al d: 0,02 mol, Al2O3: 0,04 mol, Fe: 0,09 mol%Al=5,59%, %Al2O3= 42,24%, %Fe= 52,17%b) 2Al+ 3H2SO4 Al2(SO4)3 + 3H20,02 0,03Fe + H2SO4 FeSO4 + H20,09 0,09Al2O3 + 3H2SO4 Al2(SO4)3 + 3H2O0,04 0,12Vy nH2SO4= 0,03+0,09+0,12= 0,24 mol => V=0,24/1,2= 0,2 lt.

Dng 5: ton v tnh lng tnh ca nhm v hp cht ca nhmPhng php tng qut:

Vi dng bi tp ny phng php ti u nht l pp i s: Vit tt c cc PTHH xy ra, sau da vo cc d kin cho v PTHH tnh ton- Mt s vn cn ch :+ Cn phi hiu th no l hp cht lng tnh( va tc dng vi axit, va tc dng vi bazo) bao gm mui HCO3-, HSO-3, cc oxit: Al2O3, ZnO, Cr2O3, cc hiroxit nh: Al(OH)3, Zn(OH)2, Cr(OH)3+ Bi ton v s lng tnh ca cc hidroxit c 2 dng nh sau: V d v Al(OH)3* Bi ton thun: Cho lng cht tham gia phn ng , hi sn phmVD: Cho dung dch mui nhm ( Al3+) tc dng vi dung dch kim (OH-). Sn phm thu c gm nhng cht g ph thuc vo t s k = nOH-/nAl3++ Nu k 3 th Al3+ phn ng va hoc d khi ch c phn ng Al3+ + 3OH- Al(OH)3 ( 1) (k= 3 c ngha l kt ta cc i)+ Nu k 4 th OH-phn ng (1) d v ha tan va ht Al(OH)3 theo phn ng sau: Al(OH)3 + OH- Al(OH)4- (2)+ Nu 3< k < 4 th OH- d sau phn ng (1) v ha tan mt phn Al(OH)3 (2)* Bi ton nghch: Cho sn phm , hi lng cht tham gia phn ngVD: Cho a mol OH- t t vo x mol Al3+, sau phn ng thu c y mol Al(OH)3 (x, y cho bit). Tnh aNhn xt: nu x=y th bi ton rt n gin, a= 3x=3yNu y< x Khi xy ra mt trong hai trng hp sau:

a = 3y+ Trng hp 1: Al3+ d sau phn ng (1) Vy

a= 4x-yTrng hp ny s mol OH- l nh nht+ Trng hp 2: Xy ra c (1) v (2) vy: Trng hp ny s mol OH- l ln nht+ Mun gii c nh bi ton trn chng ta cn quy v s mol Al3+ trong AlCl3, Al2(SO4)3.. v quy v s mol OH- trong cc dd sau: NaOH, KOH, Ba(OH)2, Ca(OH)2+ Cn ch n kt ta BaSO4 trong phn ng ca Al2(SO4)3 vi dung dich Ba(OH)2. Tuy cch lm khng thay i nhng khi lng kt ta thu c gm c BaSO4+ Trong trng hp cho OH- tc dng vi dung dch cha c Al3+ v H+ th OH- s phn ng vi H+ trc sau mi phn ng vi Al3++ Cn ch cc dung dch mui nh Na[Al(OH)4], Na2[Zn(OH)4]... khi tc dng vi kh CO2 d th lng kt ta khng thay i v:Na[Al(OH)4] + CO2 Al(OH)3 + NaHCO3, Cn khi tc dng vi HCl hoc H2SO4 long th lng kt ta c th b thay i ty thuc vo lng axit: HCl + Na[Al(OH)4] Al(OH)3 + NaCl + H2O Nu HCl d: Al(OH)3 + 3HCl AlCl3 + 3H2OBi tp: Chng ta s i vo tng bi ton c th sau:Bi 1: : Tnh th tch dung dch Ba(OH)2 0,01M cn thm vo 100 mL dung dch Al2(SO4)3 0,1M thu c 4,275 gam kt ta.HHHHHng dn gii:

v . Gi x l s mol Ba(OH)2 cn thm vo

=> v .Ba2+ + SO42- BaSO4(1)x (mol)0,03 (mol)Al3+ + 3OH- Al(OH)3(2)0,02 (mol)2x (mol)Al(OH)3 + OH- Al(OH)4-(3)1. Nu ch xy ra phn ng (1) v (2). Trong trng hp ny Al3+ tham gia phn ng va hoc d : , v nh vy Ba2+ phn ng ht phn ng (1).

Ta c : m(kt ta) =

VBa(OH)2=1. Nu xy ra cc phn ng (1), (2) v (3) th

(loi)Bi 2: Cho 240 ml dung dch Ba(OH)2 1M vo 200 ml dung dch hn hp AlCl3 aM v Al2(SO4)3 2aM thu c 51,3 gam kt ta. Gi tr ca a l bao nhiu?Hng dn gii:Kt ta y chc chn phi c BaSO4, c th c Al(OH)3.nBa(OH)2= 0,24*1=0,24 mol, nAlCl3= 0,2a mol, nAl2(SO4)3=0,4a mol nBaSO4= nSO42-=0,4a*3=1,2a molKhi cho OH- vo Al3+ th xy ra cc phn ng:Al3+ + 3OHAl(OH)3(1)Al(OH)3 + OH- AlO2- + H2O (2)Al3+ khng th d nn c th xy ra (1) hoc c th xy ra ng thi 2 p.Gi s xy ra 2 p, cc pt c vit li l:Al3+ + 3OHAl(OH)3a 3a aAl(OH)3 + OH- AlO2- + H2O (0,48-3a) (0,48-3a)Vy kt ta gm: BaSO4 (1,2a mol), Al(OH)3 (a-(0,48-3a))mol 1,2a*233+ (4a-0,48)*78=51,3 => a=0,15.Bi 3: Thm 150 ml dung dch NaOH 2M vo cc ng 100 ml dung dch AlCl3 xM, khuy u n khi p xy ra hon ton thy c 0,1 mol kt ta. Thm tip 100 ml dung dch NaOH 2M vo cc khuy u cho ti khi p hon ton thu c 0,14 mol kt ta. Tnh xHng dn gii:nNaOH= 0,15*2=0,3 mol, nAlCl3=0,1x mol, nkt ta=0,1 mol => c hai trng hp NaOH va hoc AlCl3 d. Nhng n khi thm NaOH vo th s mol kt ta l 0,14 mol suy ra trong trng hp u th AlCl3 dAl3+ + 3OH- Al(OH)30,1 0,3 0,1Gi a l s mol kt ta to thnh khi thm tip NaOH to thnh kt ta cc i.Al3+ + 3OH- Al(OH)3a 3a aAl(OH)3 + OH- AlO2- + 2H2O(0,2-3a) (0,2-3a)Ta c pt: 0,1+ a (0,2-3a)=0,14 => a=0,04Vy nAlCl3= 0,1+a= 0,1+ 0,04=0,14 mol =>x=0,14/0,1= 1,4M.Bi 4: Thchin 2 thnghim sau:TN1: Trn 100 ml dd Al2(SO4)3 vo 120 ml dd NaOH, sau p thu c 2,34 g kt taTN2: Trn 100 ml dd Al2(SO4)3 vo 200 ml dd NaOH, sau p cng thu c 2,34 g kt ta.Tnh nng mol ca Al2(SO4)3 v NaOH em dng.Hng dn gii:Gi x,y ln lt l nng mol ca Al2(SO4)3 v NaOH em dng.TN1: nAl2(SO4)3=0,1x mol, nNaOH=0,12y mol, nkt ta=2,34/78=0,03 molV trong TN2, gi nguyn s mol Al2(SO4)3 m tng s mol NaOH ln th lng kt ta vn khng i nn ta suy ra rng trong TN1 th Al3+ d, khi thm NaOH vo th lng kt ta tng n cc i, ri li b tan bng vi lng kt ta ban u. 0,12y= 0,03*3= 0,09 => y=0,75 (M)TN2: nNaOH= 0,2y =0,2*0,75=0,15 molS mol NaOH nhiu hn so vi TN1 l: 0,15-0,09= 0,06 molS mol kt ta cc i = nAl3+ban u= 0,2x mol Al3+ + 3OHAl(OH)3 : pt to kt ta ca Al3+ d sau TN1(0,2x-0,03) (0,6x-0,09)

Al(OH)3 + OH- AlO2- + H2O(0,2x-0,03) (0,2x-0,03)Ta c: (0,6x-0,09) + (0,2x-0,03)= 0,06 => x=0,225 (M).Bi 5: Ha tan hon ton m gam hn hp gm Na2O v Al2O3 vo nc thu c dung dch X trong sut. Thm t t dung dch HCl 1M vo X, khi ht 100 ml th bt u xut hin kt ta; khi ht 300 ml hoc 700 ml th u thu c a gam kt ta. Gi tr ca a v m ln lt l bao nhiu?Hng dn gii:1. S mol HCl p NaOH d trong ddX=s mol NaOH=0,1. 1. NaAlO2 trong ddX khi p 0,2mol HCl to 0,2mol Al(OH)3 nn a=15,6g. Nhng khi dng 0,6mol HCl s to 0,2mol Al(OH)3 v mui Al3+. Vy c 0,4mol HCl p sau: AlO2- + 4H+ Al3+ + 2H2O tng mol AlO2- l 0,2 + 0,1=0,3 nn Al2O3=0,15. Na2O s l 0,15 + 0,1/2= 0,2. vy m= 27,7.Bi 6: Cho 500 ml dd Ba(OH)2 0,1 M vo V (l) dd Al2(SO4)3 0,1M. Sau khi cc phn ng kt thc thu c 12,045g kt ta. Tnh V.Hng dn gii:Ta c: nBa(OH)2= 0,5*0,1=0,05 mol, nAl2(SO4)3= 0,1V mol => nAl3+=0,2V molng thi xy ra 2 p sau:Ba2+ + SO42- BaSO4Al3+ + 3OH- Al(OH)3Nu OH_ d th:Al(OH)3 + OH- AlO2- + H2OKt ta gm BaSO4, c th c Al(OH)3Nu kt ta ch gm BaSO4 th khi lng kt ta ti a l: 0,05*233=11,65g < 12,045g Kt ta gm hai cht trnAl3+ khng th d nn c hai trng hp: p va to kt ta nhm hiddroxit hoc kt ta nhm hiddroxit b tan mt phn.Ta s xt trng hp tng qut nht l kt ta nhm hiddroxit b tan mt phn nBaSO4= nSO42-=0,3V (v Ba2+ cn nm trong hp cht aluminat)Al3+ + 3OH- Al(OH)3 0,2V 0,6V 0,2V Al(OH)3 + OH- AlO2- + H2O (0,1-0,6V) (0,1-0,6V) 0,3V*233 + (0,2V-(0,1-0,6V))*78=12,045 => V=0,15 lt.Bi 7: Trn 150 ml dd Ba(OH)2 xM vi 100 ml dd Al2(SO4)3 yM, sau phn ng thu c 85,5 g kt ta. Nu trn 250 ml dd Ba(OH)2 trn vi lng Al2(SO4)3 th thu c 120,45 g kt ta. Tnh x v y.Hng dn gii:Kt ta thu c chc chn phi c BaSO4, c th c Al(OH)3Nu trong trng hp u hai cht phn ng va vi nhau hoc l Ba(OH)2 d ha tan tip Al(OH)3 th trong trng hp 2 nu dng lng Ba(OH)2 nhiu hn th lng kt ta BaSO4 khng i, cn lng Al(OH)3 t i v b ha tan th khi lng kt ta s gim ch khng tng. Theo gi thit th khi lng kt ta tng => ban u Al2(SO4)3 d. TN1: nBa(OH)2= 0,15x mol, nAl2(SO4)3= 0,1y mol Ba2+ + SO42- BaSO4 (tnh theo s mol ca Ba2+) 0,15x 0,15x 0,15x Al3+ + 3OH- Al(OH)3 (tnh theo s mol ca OH-) 0,1x 0,1x 0,1x 0,15x*233 + 0,1x*78=85,5 => x=2(M) TN2: Ta s xt hai trng hp: TH1: Al2(SO4)3 vn cn d sau p nBaSO4= nBa2+= 0,25*2= 0,5 mol, nAl(OH)3= nOH-= 0,5*2/3=1/3 molsuy ra: 0,5*233 + 1/3*78 =142,5 g > 120,45 g : loi TH2: p to kt ta nhm hirit cc i, ri sau kt ta c th b tan. Trong trng hp ny ta xt kt ta tan l trng hp tng qut nht Ba2+ + SO42- BaSO4 0,3y 0,3y 0,3y (tnh theo s mol SO42- v Ba2+ cn nm trong hp cht aluminat) Al3+ + 3OH- Al(OH)3 0,2y 0,6y 0,2y Al(OH)3 + OH- AlO2- + H2O 1-0,6y 1-0,6yVy ta c: 0,3y*233 + (0,2y-(1-0,6y))*78=120,45 g y=1,5 (M).

Dng 6: ton tng hpBi 1: : Cho kh Cl2 tc dng vi Ca(OH)2 ta c clorua vi l hn hp CaCl2, CaClO2, Ca(OCl)2 v nc m. sau khi loi b nc nh un nh v ht chn khng kh th thu c 152,4 gam hn hp A cha (% khi lng)50% CaOCl2; 28,15% Ca(ClO)2 v phn cn li l CaCl2. Nung nng hn hp A thu c 152,4 gam hn hp B ch cha canxi clorua v canxi clorat.1. Vit cc phng trnh phn ng xy ra.1. Tnh th tch kh Cl2 ( iu kin tiu chun) phn ng1. Tnh % khi lng CaCl2 trong hn hp B1. Nung hn hp B nhit cao ti phn ng hon ton v ly tt c cc kh thot ra cho vo bnh kn dung tch khng i ch cha 16,2 gam kim loi M ha tr n duy nht (th tch cht rn khng ng k). nhit v p sut ban u trong bnh l t0C v p atm. Nung nng bnh mt thi gian sau a nhit bnh v t0C, p sut trong bnh lc ny l 0,75 p atm. Ly cht rn cn lai trong bnh ha tan hon ton bng dd HCl d thy bay ra 13,4 lt kh (ktc). Hi M l kim loi g Hng dn gii:1. Cc phn ng :2Ca(OH)2 + Cl2 CaCl2 + Ca(ClO)2 + 2 H2O(1)Ca(OH)2 + Cl2 CaOCl2 + H2O(2)6CaOCl2 t 5CaCl2 + Ca(ClO3)2(3)3Ca(ClO)2 t 2CaCl2 + Ca(ClO3)2(4)Ca(ClO3)2 t CaCl2 + 3O2(5)1. nCaOCl2 = = 0,6 molnCa(ClO)2 = 0,3 molnCaCl2 = 0,3 molTng s mol Cl2 tham gia phn ng:nCl2 = 0,6 + 0,3 + 0,3 = 1,2 molVCl2 = 26,88 lt1. %CaCl2 == 72,84 %1. Tng s mol O2 thot ra bng 0,6. + 0,3 = 0,6 molV nhit bnh khng i, m p sut gim 25% ng vi lng O2 phn ng vi kim loi:4 M + n O2 2 M2OnnO2 = 0,6 . 0,25 = 0,15 molnH2 = 0,6 molGi x l s mol kim loi Mp dng nh lut bo ton electron: nx = 0,15. 4 + 0,6. 2 = 1,8X = M = 9n n= 3 M = 27 (Al)Bi 2: : Ngi ta tin hnh phn tch nh lng mt hn hp Cr, Al, Cu nh sau:1. Cho axit HCl d phn ng ht hn hp ni trn trong mt bnh kn khng c khng kh thu c 5,6 lt (ktc) v tch ra c 16 gam b rn1. Lc ly dung dch, cho vo dung dch mt lng d dung dch NaOH v nc Clo. Sau thm d dung dch BaCl2, thu c 25,3 gam kt ta mu vng.1. Vit phng trnh phn ng qua cc giai on?Tnh thnh phn phn trm cc kim loi trong hn hp?Hng dn gii:B rn: Cu = 16 gamKh H2 5,6 litClo ha CrCl2: CrCl2 + Cl2 CrCl3CrCl3 + 3/2 Cl2 + 8 NaOH Na2CrO4 + 6 NaCl + 4 H2OAlCl3 + 4 NaOH NaAlO2 + 3 NaClNa2CrO4 + BaCl2 BaCrO4 + 2 NaClKhi lng Cu trong hn hp l 16 gamKhi lng Crom trong hn hp chnh l khi lng Crom trong 25,3 gam BaCrO4 l 5,2 gammAl = 2,7 gam% Cu = 66,95 % ; %Cr = 21,76 % ; % Al = 11,3 %Bi 3: Cho t t b mol dung dch HCl vo a mol Na2CO3, sau p thy thot ra V lt kh. Cn nu cho t t dung dch cha a mol Na2CO3 vo dung dch cha b mol HCl th kt thc phn ng thu c 2V lt kh cng iu kin. Tm t l a:b?Hng dn gii:Khi cho t t dung dch HCl vo dung dch Na2CO3 th xy ra th t phn ng nh sau:H+ + CO32- HCO3-(1)H+ + HCO3- CO2 + H2O (2)iu kin TN1 c kh thot ra l: b>a. Ta s c 2 trng hp sau:TH1: b>2a :trong trng hp ny axit u d c 2 TN, v vy s mol CO2 thu c 2 TN u bng nhau v bng a: loiTH2: a s mol khng kh = 3x mol1. nO2(A2)= x+0,2*3x=1,6x mol, nN2= 0,8*3x=2,4x molHn hp A3 gm 3 kh nn ta c cc trng hp sau:TH1: N2, CO2, O2: ngha l oxi d => nCO2= nC= 0,528/12=0,044 molnhn hp kh= 0,044:22,92*100=0,192 mol1. 0,044+ 2,4x+ (1,6x- 0,044)= 0,192 => x= 0,048 molmA= 11 + 0,048*32= 12,536 gam, nKClO3= 0,012 mol%KClO3= 0,012*122,5/12,536*100=11,73%, %KMnO4= 88,27%TH2: N2, CO2, CO: cacbon chy trong oxi thiu to ra hn hp CO2 v COGi y l s mol ca CO2 => nCO= 0,044-yBo ton oxi ta c: y+ =1,6x => y= 3,2x-0,044(*)Theo gi thit ta c: *100=22,92 Thay (*) vo, ta suy ra: x= 0,02mA= 11+ 0,02*32=11,64 gam%KClO3=12,63%, %KMnO4= 87,37%Bi 5: Hn hp A gm KClO3, Ca(ClO3)2, Ca(ClO)2, CaCl2 v KCl nng 83,68 gam. Nhit phn hon ton A ta thu c cht rn B gm CaCl2, KCl v mt th tch kh O2 va oxi ha SO3 iu ch 191,1 gam H2SO4 80%. Cho cht rn B tc dng vi 360ml dung dch K2CO3 0,5M (va ) thu c kt ta C v dung dch D. Lng KCl c trong dung dch D nhiu gp 22/3 ln lng KCl c trong A3. Tnh khi lng kt ta C3. Tnh % khi lng ca KClO3 trong AHng dn gii:Cc pt p xy ra l:2KClO3 2KCl + 3O2(1)Ca(ClO3)2 CaCl2 + 3O2(2)Ca(ClO)2 CaCl2 + O2 (3)S mol kh oxi sinh ra =0,5 nH2SO4= 0,5*1,56= 0,78 molCht rn B gm CaCl2 v KCl tc dng vi Na2CO3 th:Ca2+ + CO32- CaCO3, vi nK2CO3= 0,36*0,5= 0,18 mol0,18 0,18 0,18Vy kt ta C l kt ta CaCO3 vi khi lng l: 0,18*100= 18 gamKhi lng cht rn B l: 83,68- 0,78*32= 58,72 gam(p dng nh lut BTKL)1. mKCl(trong B)= 58,72-0,18*111= 38,74 gam => nKCl(trong B)= 0,52 mol1. nKCl(trong D)= 0,52+ 0,18*2= 0,88 mol => nKCl(trong A)= 0,88/22*3= 0,12 mol1. nKClO3(trong A)= 0,52-0,12= 0,4 mol (bo ton nguyn t K)Vy %KClO3 = 58,56%Bi 6: Thm 5,64 gam hn hp K2CO3 v KHCO3 vo 600ml dung dch hn hp Na2CO3 v NaHCO3, c dung dch A, gi s th tch dung dch A vn l 600ml. Chia dung dch A thnh 3 phn bng nhau:Cho rt t t 100 ml dung dch HCl vo phn th nht, thu c dung dch B v 448 ml kh(ktc) bay ra. Thm nc vi trong (c d) vo dung dch B thy to thnh 2,5 gam kt ta. Phn th 2 tc dng va vi 150 ml dung dch NaOH 0,1 MCho kh HBr d i qua phn th 3, sau c cn th thu c 8,125 gam mui khan.a) Vit cc pt p xy ra dng ionb) Tnh nng M ca cc mui trong dung dch A v ca dung dch HCl dng.Hng dn gii:Phn 1: Xy ra cc phn ng theo th t sau: H+ + CO32- HCO3- H+ + HCO3- CO2 + H2OS mol CO2 sinh ra: 0,448/22,4= 0,02 mol. Dd B ch gm mui HCO3-Khi cho nc vi trong d vo th: HCO3- + OH- CO32- + H2O 0,025 0,0251. S mol nguyn t Cacbon trong 1/3 dung dch A l: 0,02+0,025= 0,045 molPhn 2: HCO3- + OH- CO32- + H2O 0,015 0,015Suy ra trong 1/3 dung dch A th: nCO32-= 0,045-0,015= 0,03 molPhn 3: Khi cho axit d vo th: nH+= 2*0,03 +0,015= 0,075 mol => nBr-= 0,075 molKhi lng cation c trong 1/3 dung dch A l: 8,125-0,075*80= 2,125 gamKhi lng mui c trong 1/3 dung dch A l: 2,125+0,03*60+0,015*61=4,84 gam (cation+anion) => mmui natri(1/3A)= 4,84- 5,64/3=2,96 gamGi a, b,x,y ln lt l s mol ca K2CO3, KHCO3, Na2CO3, NaHCO3 c trong 1/3 dung dch ATa c cc hpt sau: a+x= 0,03(1) b+y= 0,015(2) 138a+100b=1,88(3) 106x+ 84y= 2,96(4)T (1) => a=0,03-x thay vo (3), ta c:138x- 100b= 2,26(5)T (2,4,5) ta gii ra c : b=0,005, x=0,02, y=0,01 => a=0,01Nng ca cc mui c trong dung dch A l:K2CO3= 0,01: 0,2= 0,05(M), KHCO3= 0,005:0,2= 0025(M)Na2CO3= 0,02:0,2= 0,1(M), NaHCO3= 0,01:0,2= 0,05(M)S mol HCl dng l: 0,03+0,02= 0,05 mol => CM(HCl)= 0,05:0,1= 0,5(M)Bi 7: Mt hn hp A gm M2CO3, MHCO3, MCl (M l kim loi kim). Cho 43,71 gam A tc dng ht vi V ml d dung dch HCl 10,52% (d= 1,05g/ml) thu c dung dch B v 17,6 gam kh C.Chia B lm 2 phn bng nhau:Phn 1 phn ng va vi 125 ml dung dch KOH 0,8M, c cn dung dch thu c m gam mui khan.Phn 2 tc dng hon ton vi AgNO3 d thu c 68,88 gam kt ta trng.1. Tnh khi lng nguyn t M v tnh % v khi lng cc cht trong A1. Tnh gi tr ca V v m.Hng dn gii:Phn 1: Xy ra cc phn ng theo th t sau: H+ + CO32- HCO3- H+ + HCO3- CO2 + H2OS mol CO2 sinh ra: 0,448/22,4= 0,02 mol. Dd B ch gm mui HCO3-Khi cho nc vi trong d vo th: HCO3- + OH- CO32- + H2O 0,025 0,0251. S mol nguyn t Cacbon trong 1/3 dung dch A l: 0,02+0,025= 0,045 molPhn 2: HCO3- + OH- CO32- + H2O 0,015 0,015Suy ra trong 1/3 dung dch A th: nCO32-= 0,045-0,015= 0,03 molPhn 3: Khi cho axit d vo th: nH+= 2*0,03 +0,015= 0,075 mol => nBr-= 0,075 molKhi lng cation c trong 1/3 dung dch A l: 8,125-0,075*80= 2,125 gamKhi lng mui c trong 1/3 dung dch A l: 2,125+0,03*60+0,015*61=4,84 gam (cation+anion) => mmui natri(1/3A)= 4,84- 5,64/3=2,96 gamGi a, b,x,y ln lt l s mol ca K2CO3, KHCO3, Na2CO3, NaHCO3 c trong 1/3 dung dch ATa c cc hpt sau: a+x= 0,03(1) b+y= 0,015(2) 138a+100b=1,88(3) 106x+ 84y= 2,96(4)T (1) => a=0,03-x thay vo (3), ta c:138x- 100b= 2,26(5)T (2,4,5) ta gii ra c : b=0,005, x=0,02, y=0,01 => a=0,01Nng ca cc mui c trong dung dch A l:K2CO3= 0,01: 0,2= 0,05(M), KHCO3= 0,005:0,2= 0025(M)Na2CO3= 0,02:0,2= 0,1(M), NaHCO3= 0,01:0,2= 0,05(M)S mol HCl dng l: 0,03+0,02= 0,05 mol => CM(HCl)= 0,05:0,1= 0,5(M)Bi 8: Mt hn hp A gm M2CO3, MHCO3, MCl (M l kim loi kim). Cho 43,71 gam A tc dng ht vi V ml d dung dch HCl 10,52% (d= 1,05g/ml) thu c dung dch B v 17,6 gam kh C.Chia B lm 2 phn bng nhau:Phn 1 phn ng va vi 125 ml dung dch KOH 0,8M, c cn dung dch thu c m gam mui khan.Phn 2 tc dng hon ton vi AgNO3 d thu c 68,88 gam kt ta trng.1. Tnh khi lng nguyn t M v tnh % v khi lng cc cht trong A1. Tnh gi tr ca V v m.Hng dn gii:1.A tc dng vi HCl d th: 2H+ + CO32- CO2 + H2O H+ + HCO3- CO2 + H2OS mol kh CO2 sinh ra l: 17,6/44= 0,4 mol. Dung dch B gm MCl v HClPhn 1: nHCl d= nKOH= 0,125*0,8= 0,1 mol. Mui khan by gi l MCl v KClPhn 2: Ag+ + Cl- AgCl vi nAgCl= 68,88/143,5= 0,48 mol 0,48 0,481. S mol ca cation M+ trong dung dch B l: (0,48-0,1)*2=0,76 molGi x,y,z ln lt l s mol ca M2CO, MHCO3 v MCl trong hn hp A ban u.Ta c hpt sau: x+ y= 0,4 (1) nM= 2x +y +z =0,76(2)0,76*M + 60x+ 61y+ 35,5z= 43,71(3)T(1) suy ra: y= 0,4-xT (2) suy ra z= 0,36-x. Thay y v z vo (3), ta c:M= + D thy M=f(x) lin tc v tng trong khong 0 V=297,4 mlM= mNaCl +m KCl= 0,38*58,5 + 0,1*74,5=29,68 gam

hhhhjhhsahh

Mc lc

A.C S L THUYT2B. CC DNG BI TP5Dng 1: Tm tn kim loi hoc cng thc ca hp cht5Phng php tng qut5Bi tp5DNG 2: Dng ton v CO2 hoc SO2 tc dng vi dung dch kim15Phng php tng qut:15Bi tp:16Dng 3: Hn hp nhm vi kim loi kim, kim th hoc vi kim loi khc20Phng php tng qut:20Bi tp:21Dng 4: phn ng nhit nhm28Phng php tng qut:28VN 1: bi cho sn phm sau phn ng tc dng vi NaOH v cho kh thot ra th sn phm sau phn ng c Al d29VN 2: bi cho sn phm sau phn ng nhit nhm ch tc dng vi dung dch H+ cho kh H2 bay ra29VN 3: Thc hin phn ng nhit nhm gm hn hp cc oxit kim loi km hot ng vi nhm. Sn phm sau phn ng tc dng vi dd H+ cho kh H2 v cho tc dng vi dung dch OH- v cho khi lng cht rn khng tc dng.29VN 4: bi cho sn phm sau phn ng nhit nhm c chia thnh hai phn, mt phn cho tc dng vi dung dch OH- (th nghim 1) v mt phn cho tc dng vi dung dch H+ (th nghim 2). Tnh khi lng mi cht trong hn hp.29Bi tp:29Dng 5: ton v tnh lng tnh ca nhm v hp cht ca nhm37Phng php tng qut:37Bi tp:38Dng 6: ton tng hp42

52Sv: v th nh hng