BÀI TẬP VỀ XÁC ĐỊNH TÊN KIM LOẠI

7/31/2019 BI TP V XC NH TN KIM LOI

1/24

BI TP V XC NH TN KIM LOI 1) C th tnh c khi lng mol nguyn t kim loi Mtheo cc cch sau:

- T khi lng (m) v s mol (n) ca kim loi M = - T Mhp cht Mkim loi

- T cng thc Faraday M = (n l s electron trao i mi in cc)

- T a < m < b v < n < tm M tha mntrong khong xc nh - Lp hm s M = f(n) trong n l ha tr ca kim loi M (n = 1,2, 3), nu trong bi ton tm oxit kim loi MxOy th n = kimloi M - Vi hai kim loi k tip nhau trong mt chu k hoc phn nhm tm tn 2 kim loi 2) Mt s ch khi gii bi tp:

- Bit s dng mt s nh lut bo ton nh bo ton khi

lng, bo ton nguyn t, bo ton mol electron, Bit vit ccphng trnh ion thu gn, phng php ion electron

- Khi bi khng cho kim loi M c ha tr khng i th khi kimloi M tc dng vi cc cht khc nhau c th th hin cc s oxiha khc nhau t kim loi M c cc ha tr khc nhau

- Khi hn hp u c chia lm hai phn khng bng nhauthphn ny gp k ln phn kia tng ng vi s mol cc cht phn

ny cng gp k ln s mol cc cht phn kia 3) Mt s v d minh ha: V d 1: Cho 3,024 gam mt kim loi M tan ht trong dung dchHNO3 long, thu c 940,8 ml kh NxOy (sn phm kh duy nht, ktc) c t khi i vi H2 bng 22. Kh NxOy v kim loiM l:
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2/24

A. NO v Mg B. NO2 v Al C. N2O v Al D. N2O v FeHng dn : M(NxOy) = 44 nN2O = 0,042 mol

M Mn+ + ne 2NO3- + 8e + 10H+ N2O + 5H2O

Theo lbt mol electron: ne cho = ne nhn 3,024 No duy nht n = 3 v M = 27 Al p n C V d 2: Hn hp X gm Mg v kim loi M. Ha tan hon ton 8gam hn hp X cn va 200 gam dung dch HCl 7,3 %. Mtkhc cho 8 gam hn hp X tc dng hon ton vi kh Cl2 cndng 5,6 lt Cl2 ( ktc) to ra hai mui clorua. Kim loi M vphn trm v khi lng ca n trong hn hp X l: A. Al v 75 % B. Fe v 25 % C. Al v 30 % D. Fe v 70 % Hng dn : nHCl = 0,4 mol ; nCl2 = 0,25 mol ; nMg = x mol ; nM= y mol 24x + My = 8 (1)

- X tc dng vi dung dch HCl (M th hin ha tr n) 2x + ny =0,4 (2)

- X tc dng vi Cl2 (M th hin ha tr m) 2x + my = 0,5 (3)

- T (2) ; (3) y(m n) = 0,1 m > n No duy nht m = 3 v n= 2 x = y = 0,1 mol

- T (1) M = 56 Fe v % M = 70 % p n D V d 3: Hn hp X gm hai mui cacbonat ca 2 kim loi kimth hai chu k lin tip. Cho 7,65 gam X vo dung dch HCl d.Kt thc phn ng, c cn dung dch th thu c 8,75 gam muikhan. Hai kim loi l: A. Mg v Ca B. Ca v Sr C. Be v Mg D. Sr v BaHng dn :- t cng thc chung ca hai mui l CO3. Phng trnh phnng:

CO3 + 2HCl Cl2 + CO2 + H2O- T phng trnh thy: 1 mol CO3 phn ng th khi lngmui tng: 71 60 = 11 gam
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3/24

- Theo bi khi lng mui tng: 8,75 7,65 = 1,1 ga m c0,1 mol CO3 tham gia phn ng + 60 = 76,5 = 16,5 2 kim loi l Be v Mg p nC V d 4: Ha tan hon ton 6 gam hn hp X gm Fe v mt kimloi M (ha tr II) vo dung dch HCl d, thu c 3,36 lt kh H2( ktc). Nu ch ha tan 1,0 gam M th dng khng n 0,09 molHCl trong dung dch. Kim loi M l: A. Mg B. Zn C. Ca D. NiHng dn : nH2 = 0,15 mol- nX = nH2 = 0,15 mol X = 40

- ha tan 1 gam M dng khng n 0,09 mol HCl 22,2 < M < 40 < 56 M l Mg p n A V d 5: ha tan hon ton 6,834 gam mt oxit ca kim loi Mcn dng ti thiu 201 ml dung dch HCl 2M. Kim loi M l: A. Mg B. Cu C. Al D. FeHng dn : Gi cng thc oxit l MxOy ; nHCl = nH+ = 0,402mol

- Ta c nO2 (trong oxit) = mol nMxOy = mol (Mx+

16y) = Mx = 18y M = No duy nht v M = 27 Al p n C II BI TON V KIM LOI TC DNG VI NC, KIM LOITC DNG VI DUNG DCH BAZ KIM 1) Mt s ch khi gii bi tp:

- Ch c kim loi kim, Ca, Sr, Ba mi tan trong nc nhit thng

- Cc kim loi m hiroxit ca chng c tnh lng tnh nh Al,Zn, Be, Sn, Pbtc dng c vi dung dch kim (c)
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4/24

- Nu bi cho nhiu kim loi tc dng vi nc to dung dchkim, ri sau ly dung dch kim tc dng vi dung dch hnhp axit th:

+ Gii bng cch vit phng trnh ion thu gn + nOH = 2nH2

- Nu bi cho hn hp kim loi kim hoc kim th v kim loiM ha tr n vo nc th c th c hai kh nng:

+ M l kim loi tan trc tip (nh kim loi kim, Ca, Sr, Ba)

+ M l kim loi c hiroxit lng tnh (nh Al, Zn)

M + (4 n)OH + (n 2)H2O MO2n 4 + H2 (da vo smol kim loi kim hoc kim th s mol OH ri bin lun xemkim loi M c tan ht khng hay ch tan mt phn) 2) Mt s v d minh ha: V d 1: Hn hp X gm Na, K, Ba ha tan ht trong nc d todung dch Y v 5,6 lt kh ( ktc). Tnh V ml dung dch H2SO42M ti thiu trung ha Y A. 125 ml B. 100 ml C. 200 ml D. 150 mlHng dn : nH2 = 0,25 mol

Ta c nOH = 2nH2 m nOH = nH+ nH2SO4 = =nH2 = 0,25 mol V = 0,125 lt hay 125 ml p n A V d 2: Thc hin hai th nghim sau: Th nghim 1: Cho m gam hn hp Ba v Al vo nc d, thuc 0,896 lt kh ( ktc) Th nghim 2: Cng cho m gam hn hp trn cho vo dungdch NaOH d thu c 2,24 lt kh ( ktc) Cc phn ng xy rahon ton. Gi tr ca m l: A. 2,85 gam B. 2,99 gam C. 2,72 gam D. 2,80 gamHng dn : nH2 th nghim 1 = 0,04 < nH2 th nghim 2 =0,1 mol th nghim 1 Ba ht, Al d cn th nghim 2 th cBa v Al u ht
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5/24

- Gi nBa = x mol v nAl = y mol trong m gam hn hp

- Th nghim 1:

Ba + 2H2O Ba2+ + 2OH+ H2

x 2x x

Al + OH + H2O AlO2+ H2

2x 3x

nH2 = 4x = 0,04 x = 0,01 mol

- Th nghim 2: tng t th nghim 1 ta c: x + = 0,1 y =0,06 mol m = 0,01.137 + 0,06.27 = 2,99 gam p n B V d 3: Ha tan hon ton 7,3 gam hn hp X gm kim loi Nav kim loi M (ha tr n khng i) trong nc thu c dung dchY v 5,6 lt kh hiro ( ktc). trung ha dung dch Y cn dng100 ml dung dch HCl 1M. Phn trm v khi lng ca kim loiM trong hn hp X l: A. 68,4 % B. 36,9 % C. 63,1 % D. 31,6 %Hng dn : nH2 = 0,25 mol ; nHCl = 0,1 mol

- Gi nNa = x mol v nM = y mol 23x + My = 7,3 (1)

- Nu M tc dng trc tip vi nc nH2 = nOH = 0,5 > nHCl = 0,1 loi

- Nu M l kim loi c hiroxit lng tnh (n = 2 hoc 3):

M + (4 n)OH + (n 2)H2O MO2n 4 + H2

y (4 n)y ny/2
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6/24

- Do OH d nn kim loi M tan ht v nOH d = x (4 n)ymol x (4 n)y = 0,1 (2) v x + ny = 0,5 (3) y = 0,1 mol

- Thay ln lt n = 2 hoc 3 vo (1) ; (2) ; (3) ch c n = 3 ; x =

0,2 ; M = 27 l tha mn %M = 36,9 % p n B III BI TON V KIM LOI TC DNG VI DUNG DCHAXIT 1) Kim loi tc dng vi dung dch axit: a) i vi dung dch HCl, H2SO4 long :M + nH+ Mn+ + n/2H2

(M ng trc hiro trong dy th in cc chun)

b) i vi H2SO4 c, HNO3 (axit c tnh oxi ha mnh): - Kim loi th hin nhiu s oxi ha khc nhau khi phn ng viH2SO4 c, HNO3 s t s oxi ha cao nht

- Hu ht cc kim loi phn ng c vi H2SO4 c nng (trPt, Au) v H2SO4 c ngui (tr Pt, Au, Fe, Al, Cr), khi S+6trong H2SO4 b kh thnh S+4 (SO2) ; So hoc S-2 (H2S)

- Hu ht cc kim loi phn ng c vi HNO3 c nng (trPt, Au) v HNO3 c ngui (tr Pt, Au, Fe, Al, Cr), khi N+5trong HNO3 b kh thnh N+4 (NO2)

- Hu ht cc kim loi phn ng c vi HNO3 long (tr Pt, Au), khi N+5 trong HNO3 b kh thnh N+2 (NO) ; N+1 (N2O) ;No (N2) hoc N-3 (NH4+)

c) Kim loi tan trong nc (Na, K, Ba, Ca,) tc dng vi axit:c2 trng hp

- Nu dung dch axit dng d: ch c phn ng ca kim loi viaxit
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7/24

- Nu axit thiu th ngoi phn ng gia kim loi vi axit (xy ratrc) cn c phn ng kim loi d tc dng vi nc ca dungdch

2) Mt s ch khi gii bi tp: - Kim loi tc dng vi hn hp axit HCl, H2SO4 long (H+ ngvai tr l cht oxi ha) th to ra mui c s oxi ha thp v giiphng H2: M + nH+ Mn+ + n/2H2 (nH+ = nHCl + 2nH2SO4)

- Kim loi tc dng vi hn hp axit HCl, H2SO4 long, HNO3 vit phng trnh phn ng di dng ion thu gn (H+ ng vaitr mi trng, NO3 ng vai tr cht oxi ha) v so snh cc t

s gia s mol ban u v h s t lng trong phng trnh xemt s no nh nht th cht s ht trc ( tnh theo)

- Cc kim loi tc dng vi ion NO3 trong mi trng axit H+xem nh tc dng vi HNO3

- Cc kim loi Zn, Al tc dng vi ion NO3trong mi tr ngkim OH gii phng NH3

4Zn + NO3 + 7OH 4ZnO22+ NH3 + 2H2O

(4Zn + NO3 + 7OH + 6H2O 4[Zn(OH)4]2+ NH3)

8Al + 3NO3 + 5OH + 2H2O 8AlO2+ 3NH3

(8Al + 3NO3 + 5OH + 18H2O 8[Al(OH)4]+ 3NH3

- Khi hn hp nhiu kim loi tc dng vi hn hp axit th dngnh lut bo ton mol electron v phng php ion electron

gii cho nhanh. So snh tng s mol electron cho v nhn bin lun xem cht no ht, cht no d

- Khi hn hp kim loi trong c Fe tc dng vi H2SO4 cnng hoc HNO3 cn ch xem kim loi c d khng. Nu kimloi (Mg Cu) d th c phn ng kim loi kh Fe3+ v Fe2+. Vd: Fe + 2Fe3+ 3Fe2+ ; Cu + 2Fe3+ Cu2+ + 2Fe2+


8/24

- Khi ha tan hon hon hn hp kim loi trong c Fe bngdung dch HNO3 m th tch axit cn dng l nh nht muiFe2+

- Kim loi c tnh kh mnh hn s u tin phn ng trc - Nu bi yu cu tnh khi lng mui trong dung dch, ta pdng cng thc sau:

mmui = mcation + manion to mui = mkim loi + manion tomui

(manion to mui = manion ban u manion to kh)

- Cn nh mt s cc bn phn ng sau: 2H+ + 2e H2 NO3- + e + 2H+ NO2 + H2O

SO42 + 2e + 4H+ SO2 + 2H2O NO3- + 3e + 4H+ NO +2H2O

SO42 + 6e + 8H+ S + 4H2O 2NO3- + 8e + 10H+ N2O +5H2O

SO42 + 8e + 10H+ H2S +4H2O 2NO3- + 10e + 12H+ N2+ 6H2O

NO3- + 8e + 10H+ NH4+ + 3H2O

- Cn nh s mol anion to mui v s mol axit tham gia phnng:

nSO42 to mui = . nX (a l s electron m S+6 nhn tosn phm kh X)

nH2SO4 phn ng = 2nSO2 + 4nS + 5nH2S

nNO3 to mui = a.nX (a l s electron m N+5 nhn to rasn phm kh X)
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9/24

nHNO3 phn ng = 2nNO2 + 4nNO + 10nN2O + 12nN2

3) Mt s v d minh ha V d 1: Cho 3,68 gam hn hp gm Al v Zn tc dng vi mt

lng va dung dch H2SO4 10 %, thu c 2,24 lt kh H2 (ktc). Khi lng dung dch thu c sau phn ng l: A. 101,68 gam B. 88,20 gam C. 101,48 gam D. 97,80 gamHng dn : nH2 = nH2SO4 = 0,1 mol m (dung dch H2SO4)= 98 gam m (dung dch sau phn ng) = 3,68 + 98- 0,2 =101,48 gam p n C V d 2: Ho tan hon ton 14,6 gam hn hp X gm Al v Snbng dung dch HCl (d), thu c 5,6 lt kh H2 ( ktc). Thtch kh O2 ( ktc) cn phn ng hon ton vi 14,6 gam hnhp X l: A. 2,80 lt B. 1,68 lt C. 4,48 lt D. 3,92 lt Hng dn : Gi nAl = x mol ; nSn = y mol 27x + 119y = 14,6(1) ; nH2 = 0,25 mol

- Khi X tc dng vi dung dch HCl:

V d 3: Cho 7,68 gam hn hp X gm Mg v Al vo 400 ml dungdch Y gm HCl 1M v H2SO4 0,5M. Sau khi phn ng xy rahon ton thu c 8,512 lt kh ( ktc). Bit trong dung dch,cc axit phn li hon ton thnh cc ion. Phn trm v khi lngca Al trong X l: A. 56,25 % B. 49,22 % C. 50,78 % D. 43,75 %Hng dn : nH+ = 0,8 mol ; nH2 = 0,38 mol nH+phn ng= 0,76 mol < 0,8 mol axit d, kim loi ht
http://www.moon.vn/Images/Teachers/binhnguyentrang83/Baigiang/Haohoc/PP_KimLoai/image048.GIF


10/24

- Gi nMg = x mol ; nAl = y mol % Al

= %

p n A V d 4: Cho 0,10 mol Ba vo dung dch cha 0,10 mol CuSO4v 0,12 mol HCl. Sau khi cc phn ng xy ra hon ton, lc lykt ta nung nhit cao n khi lng khng i thu c mgam cht rn. Gi tr ca m l: A. 23,3 gam B. 26,5 gam C. 24,9 gam D. 25,2 gamHng dn : Cc phn ng xy ra l:

Ba + 2HCl BaCl2 + H2 BaCl2 + CuSO4 BaSO4 + CuCl2

0,06 0,12 0,06 0,06 0,06 0,06

Ba + 2H2O Ba(OH)2 + H2 Ba(OH)2 + CuSO4 BaSO4 +Cu(OH)2

0,04 0,04 0,04 0,04 0,04 0,04

Cu(OH)2 CuO + H2O

0,04 0,04 m (cht rn) = mBaSO4 + mCuO = (0,06 + 0,04).233 + 0,04.80= 26,5 gam p n B V d 5: Th tch dung dch HNO3 1M (long) t nht cn dng ho tan hon ton 18 gam hn hp gm Fe v Cu trn theo t lmol 1 : 1 l: (bit phn ng to cht kh duy nht l NO) A. 1,0 lt B. 0,6 lt C. 0,8 lt D. 1,2 ltHng dn : nFe = nCu = 0,15 mol

- Do th tch dung dch HNO3 cn dng t nht mui Fe2+ ne cho = 2.(0,15 + 0,15) = 0,6 mol

- Theo lbt mol electron nH+ = nHNO3 = mol VHNO=0,8 lt p n C
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11/24

V d 6: Ha tan 9,6 gam Cu vo 180 ml dung dch hn hpHNO3 1M v H2SO4 0,5M, kt thc phn ng thu c V lt (ktc) kh khng mu duy nht thot ra, hanu ngoi khng kh.Gi tr ca V l: A. 1,344 lt B. 4,032 lt C. 2,016 lt D. 1,008 ltHng dn : nCu = 0,15 mol ; nNO3 = 0,18 mol ; nH+ = 0,36mol

3Cu + 8H+ + 2NO3 3Cu2+ + 2NO + 4H2O Do H+ ht ; Cu d

0,36 0,09

VNO = 0,09.22,4 = 2,016 lt p n C V d 7: Cho hn hp gm 1,12 gam Fe v 1,92 gam Cu vo 400ml dung dch cha hn hp gm H2SO4 0,5M v NaNO3 0,2M.Sau khi cc phn ng xy ra hon ton, thu c dung dch X vkh NO (sn phm kh duy nht). Cho V ml dung dch NaOH 1Mvo du ng dch X th lng kt ta thu c l ln nht. Gi tr tithiu ca V l: A. 360 ml B. 240 ml C. 400 ml D. 120 mlHng dn : nFe = 0,02 mol ; nCu = 0,03 mol ne cho =0,02.3 + 0,03.2 = 0,12 mol ; nH+ = 0,4 mol ; nNO3 = 0,08 mol(Ion NO3 trong mi trng H+ c tnh oxi ha mnh nh HNO3) - Bn phn ng: NO3 + 3e + 4H+ NO + 2H2O

Do kim loi kt v H+ d

0,12 0,16

nH+ d = 0,4 0,16 = 0,24 mol nOH (to kt ta max) =0,24 + 0,02.3 + 0,03.2 = 0,36 V = 0,36 lt hay 360 ml p nA V d 8: Cho 24,3 gam bt Al vo 225 ml dung dch hn hpNaNO3 1M v NaOH 3M khuy u cho n khi kh ngng thotra th dng li v thu c V lt kh ( ktc).Gi tr ca V l: A. 11,76 lt B. 9,072 lt C. 13,44 lt D. 15,12 lt


12/24

Hng dn : nAl = 0,9 mol ; nNO3 = 0,225 mol ; nOH = 0,675mol8Al + 3NO3 + 5OH + 18H2O 8[Al(OH)4]+ 3NH3 (1)

Do NO3 ht B: 0,9 0,225 0,675

P: 0,6 0,225 0,375 0,225

D: 0,3 0 0,3

Al + OH (d) + H2O AlO2+ H2 (2)0,3 0,3 0,45

T (1) ; (2) V = (0,225 + 0,45).22,4 = 15,12 lt p n D V d 9: Ha tan hon ton 100 gam hn hp X gm Fe, Cu , Agtrong dung dch HNO3 (d). Kt thc phn ng thu c 13,44 lthn hp kh Y gm NO2, NO, N2O theo t l s mol tng ng l3 : 2 : 1 v dung dch Z (khng cha mui NH4NO3). C cndung dch Z thu c m gam mui khan. Gi tr ca m v s molHNO3 phn ng ln lt l:

A. 205,4 gam v 2,5 mol B. 199,2 gam v 2,4 mol

C. 205,4 gam v 2,4 mol D. 199,2 gam v 2,5 molHng dn : nY = 0,6 mol nNO2 = 0,3 mol ; nNO = 0,2 mol ;nN2O = 0,1 mol- nNO to mui = nNO + 3.nNO + 8.nN O = 0,3 + 3.0,2 + 8.0,1= 1,7 mol mZ = mKl + mNO to mui = 100 + 1,7.62 = 205,4gam (1)- nHNO phn ng = 2.nNO + 4.nNO + 10.nN O = 2.0,3 + 4.0,2 +10.0,1 = 2,4 mol (2)- T (1) ; (2) p n C
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13/24

V d 10: Cho 6,72 gam Fe vo 400 ml dung dch HNO3 1M, nkhi phn ng xy ra hon ton, thu c kh NO (sn phm khduy nht) v dung dch X. Dung dch X c th ho tan ti a mgam Cu. Gi tr ca m l: A. 1,92 gam B. 3,20 gam C. 0,64 gam D. 3,84 gam

Hng dn: nFe = 0,12 mol ne cho = 0,36 mol; nHNO3 = 0,4mol ne nhn = 0,3 mol

- Do ne cho > ne nhn Fe cn d dung dch X c Fe2+ vFe3+

- Cc phn ng xy ra l:

Fe + 4HNO3 Fe(NO3)3 + NO + 2H2O

0,1 0,4 0,1

Fe (d) + 2Fe3+ 3Fe2+

0,02 0,04

Cu + 2Fe3+ (d) Cu2+ + 2Fe2+

0,03 0,06

mCu = 0,03.64 = 1,92 gam p n A V d 11: Ho tan hon ton 12,42 gam Al bng dung dch HNO3long (d), thu c dung dch X v 1,344 lt ( ktc) hn hpkh Y gm hai kh l N2O v N2. T khi ca hn hp kh Y so vikh H2 l 18. C cn dung dch X, thu c m gam cht rn khan.Gi tr ca m l:

A. 38,34 gam B. 34,08 gam C. 106,38 gam D. 97,98 gamHng dn : nAl = 0,46 mol ne cho = 1,38 mol ; nY = 0,06 mol; Y = 36- D dng tnh c nN2O = nN2 = 0,03 mol ne nhn =0,03.(8 + 10) = 0,54 mol < ne cho dung dch X cn cha mui

NH4NO3 nNH4+ = NO3= mol


14/24

- Vy mX = mAl(NO) + mNH NO = 0,46.213 + 0,105.80 =106,38 gam p n C (Hoc c th tnh mX = mKl + mNO to mui + mNH = 12,42 +(0,03.8 + 0,03.10 + 0,105.8 + 0,105).62 + 0,105.18 = 106,38 gam)III BI TP V KIM LOI TC DNG VI DUNG DCH MUI 1) Kim loi tc dng vi dung dch mui:

- iu kin kim loi M y c kim loi X ra khi dung dchmui ca n:

xM (r) + nXx+ (dd) xMn+ (dd) + nX (r)

+ M ng trc X trong dy th in cc chun

+ C M v X u khng tc dng c vi nc iu kinthng

+ Mui tham gia phn ng v mui to thnh phi l mui tan

- Khi lng cht rn tng: m = mX to ra mM tan

- Khi lng cht rn gim: m = mM tan mX to ra

- Khi lng cht rn tng = khi lng dung dch gim - Ngoi l:

+ Nu M l kim loi kim, kim th (Ca, Sr, Ba) th M s kh H+ca H2O thnh H2 v to thnh dung dch baz kim. Sau lphn ng trao i gia mui v baz kim

+ trng thi nng chy vn c phn ng: 3Na + AlCl3 (khan) 3NaCl + Al

+ Vi nhiu anion c tnh oxi ha mnh nh NO3-, MnO4- ,thkim loi M s kh cc anion trong mi trng axit (hoc baz)

- Hn hp cc kim loi phn ng vi hn hp dung dch muitheo th t u tin: kim loi kh mnh nht tc dng vi cation
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15/24

oxi ha mnh nht to ra kim loi kh yu nht v cation oxiha yu nht

- Th t tng dn gi tr th kh chun (Eo) ca mt s cp oxi

ha kh: Mg2+/Mg < Al3+/Al < Zn2+/Zn < Cr3+/Cr < Fe2+/Fe < Ni2+/Ni

BÀI TẬP VỀ XÁC ĐỊNH TÊN KIM LOẠI

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