tuyet chieu giai nhanh hoa hoc

8/8/2019 Tuyet Chieu Giai Nhanh Hoa Hoc

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Tuyt chiu s 1

y l 1 chiu thc dng x l cc bi ton hn hp phc tp (hn hp c t 3 cht trln) v dng rt n gin lm cho cc php tnh tr nn n gin, thun tin hn .Rt phhp vi hnh thc thi trc nghim

V d minh ha cho k thut 1 : Nung 8,4 gam Fe trong khng kh, sau phn ng thuc m(g) cht rn X gm: Fe, Fe2O3, Fe3O4, FeO. Ho tan m gam X vo dung dchHNO3 d thu c 2,24 lt NO2 (ktc) l sn phm kh duy nht. Gi tr m l:

A. 11,2 g. B. 10,2 g. C. 7,2g. D. 6,9 g.

Nhn xt:Vi cc bi ton hn hp phc tp c s cht trong hn hp ln hn 2 chtta u c th dng k thut 1 bin i v mt hn hp mi gm 2 cht bt k trongs cc cht trong hn hp. Trong bi ton trn X c 4 cht nn c 6 cch gii. Ta cth bin X thnh X gm (Fe; Fe2O3) hoc (Fe; FeO) hoc (FeO; Fe3O4) hoc (Fe;

Fe3O4) hoc (FeO; Fe2O3) hoc (Fe2O3; Fe3O4).

Hng dn gii: Ti ch lm 3 trong 6 cch trn, cc bn c th trin khai cc cch cn liu cho kt qu ging nhau.

Cch gii 1:

Quy hn hp X thnh X gm (FeO, Fe2O3) mX = mX = mFeO + mFe2O3

Theo bi ra ta c: nFe ban u = 8,4/56 = 0,15 Tng mol Fe trong X cng bng 0,15.

Mt khc:

FeO + 4HNO3 Fe(NO3)3 + NO2 + 2H2O.

0,1 mol 0,1 mol

Ta c nFe ban u = 0,15 mol

2Fe + O2 2FeO

0,1 0,1

4Fe + 3O2 2Fe2O3

(0,15 - 0,1) = 0,05 0,025

Vy m = 0,1. 72 + 0,025.160 = 11,2g p n A.

.Cch gii 2:

Quy hn hp X thnh X gm (Fe; Fe2O3) mX = mX = mFe + mFe2O3


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Theo bi ra ta c: Fe + 6HNO3Fe(NO3)3 + 3NO2 + 3H2O.

0,1/3 0,1

m nFe ban u = 8,4/56 = 0,15 S mol Fe nm trong Fe2O3 l: 0,15 0,1/3 = 0,35/3

nFe2O3 = 0,35/3.2mX = 0,1/3 . 56 + 0,35/6 . 160 = 11,2p n A.

Cch gii 3:

Quy hn hp X thnh X gm (Fe; FeO) -> mX = mX = mFe + mFeO

Theo bi ra ta c:

Fe + 6HNO3 Fe(NO3)3 + 3NO2 + 3H2O

a 3a

FeO + 4HNO3Fe(NO3)3 + NO2 + 2H2O

b b

Gi a, b l s mol ca Fe v FeO3a + b = 0,1 (1) v a + b = 0,15 (2)

T (1) v (2) ta c: a = -0,025 v b = 0,175.

mX = -0,025. 56 + 0,175.72 = 11,2g p n A

Nhn xt: Cc bn hc sinh thn mn! S dng chiu thc s 1 gip ta gii cc bi tonv hn hp cht rt nhTi; Lm gim s n s (v lm gim s lng cht trong hn hp).Khi s dng chiu thc ny i khi cc bn s thy xut hin s mol ca cc cht l sm, khi Ti mong cc bn hy bnh tnh. l s b tr khi lng ca cc cht cho cc nguyn t c bo ton. Kt qu cui cng ca ton bi s ko thay i. y lchiu thc s 1 Ti hng dn dng c bn. Nu cc bn bit vn dng chiu thc ny c 2 dng th li gii cn ngn gn hn rt nhiu. Dng nng caos gip cc bn gii c c hn hp cc cht hu c na. Ti s ging dy nng cao phn bi ging sau. Thn i cho tm bit.

Bi tp v nh thuc Chiu Thc 1

Cu 1: Nung 8,4gam Fe trong khng kh , sau phn ng thu c m gam cht rn X gmFe,Fe2O3,Fe3O4,FeO. Ho tan m gam hn hp X vo dung dch HNO3 d thu c 2,24lt kh NO2 (ktc) l sn phm kh duy nht. Gi tr ca m l

A:11,2 gam B: 10,2 gam

C:7,2 gam D:6,9 gam


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Cu 2: Ho tan ht m gam hn hp X gm Fe2O3,Fe3O4,FeO bng HNO3 c nng thuc 4,48 lt kh NO2(ktc).C cn dung dch sau phn ng thu c 145,2 gam muikhan. Gi tr ca m l

A:35,7 gam B: 46,4 gam

C:15,8 gam D:77,7 gam

Cu 3: Ho tan hon ton 49,6gam hn hp X gm Fe,Fe2O3,Fe3O4,FeO bng H2SO4 cnng thu c dung dch Y v 8,96 lt kh SO2(ktc).

a) Phn trm khi lng ca oxi trong hoonx hp X l

A:40,24 % B: 30,7 %

C: 20,97 % D: 37,5 %

b) Khi lng mui trong dung dch Y lA:160 gam B: 140 gam

C:120 gam D: 100 gam

Cu 4: kh hon ton 3,04 gam hnn hp X gm Fe2O3,Fe3O4,FeO th cn 0,05 molkh H2 .Mt khc ho tan hon ton 3,04 gam hn hp X trong dung dch H 2SO4 c nngth thu c V ml kh SO2(ktc).gi tr ca V l

A:224ml B: 448ml

C:336ml D:112ml

Cu 5: Nung m gam bt Fe trong oxi khng kh , sau phn ng thu c 3 gam hn hpcht rn X. Ho tan ht hn hp X vo dung dch HNO3 d thu c 0,56 lt kh NO(ktc) l sn phm kh duy nht. Gi tr ca m l

A:2,52 gam B: 2,22 gam

C:2,62 gam D:2,32 gam

Cu 6: Hn hp X gm Fe,Fe2O3,Fe3O4,FeO vi s mol moi cht l 0,1 mol . HO tanht vo dung dch Y gm (HCl v H2SO4 long) d thu c dung dch Z .Nh t t dungdch Cu(NO3)2 1M vo dung dch Z cho ti khi ngng thot ra kh NO .Th tch dungdch Cu(NO3)2 cn dng v th tch kh thot ra ktc thuc phng n no

A:25ml v 1,12 lt B: 500ml v 22,4 lt

C:50ml v 2,24 lt D: 50ml v 1,12 lt


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Cu 7: Nung 8,96 gam Fe trong khng kh , sau phn ng thu c hn hp cht rn Agm Fe2O3,Fe3O4,FeO. A Ho tan va trong dung dch cha o,5 mol HNO3 thu ckh NO(ktc) l sn phm kh duy nht. S mol kh NO l

A:0,01 mol B: 0,04 mol

C:0,03 mol D:0,02 mol

Cu 8: Cho 41,76 gam hn hp A gm FeO, Fe2O3 v Fe3O4 trong s mol FeO = smol Fe2O3 tc dng va vi

V lt dung dch cha HCl 1M v H2SO4 0,5M (long). Gi tr ca V l:

A. 0,6 lt B. 0,7 lt

C. 0,8 lt. D. Mt kt qu khc.

Tuyt Chiu S 2

Nu nh tuyt chiu s 1 cc bn c tip cn vi mt phng php kh mnh vgii ton hn hp, th vi tuyt chiu s 2, cc bn s c tip cn mt ngh thut giiton rt su sc, gip hc sinh nhm ra kt qu mt cch nhTi nht.

c im ca cc bi ton c gii bng tuyt chiu s 2 l cho mt hn hp gm cnhiu cht (tng t cc bi tp thuc tuyt chiu s 1) nhng v mt bn cht ch gm 2hoc 3 nguyn t. V vy, dng tuyt chiu s 2 quy i thng v cc nguyn t tngng.

V d 1: cho hn hp X gm Fe, FeO, Fe2O3, Fe3O4. Khi ta i thnh 1 hn hpmi X' ch gm Fe v O.

V d 2: cho hn hp X gm Cu2S, CuS, CuO. Khi ta i thnh 1 hn hp mi X'ch gm Cu, S, O.

V d 3: cho hn hp X gm CuO, Cu, Cu2O. Khi ta i thnh 1 hn hp mi X'ch gm Cu v O.

.................

V d minh ha 1: Nung m gam bt st trong oxi, thu c 6 gam hn hp cht rn X.Ho tan ht hn hp X trong dung dch HNO3 (d), thot ra 1,12 lt ( ktc) NO (l snphm kh duy nht). Gi tr ca m l:

A. 5,04. B. 4,44. C. 5,24. D. 4,64.

Hng dn gii:


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Tm tt:

Fe + O2 X (Fe; FeO; Fe2O3; Fe3O4) + dd HNO3 Fe3+ + NO + H2O

m gam 6 gam 1,12 lt

S ha bng tuyt chiu s 2.

Ta c th quy i hn hp X thnh hn hp X' gm Fe v O vi s mol ln lt l x, y.

Fe + O2 (Fe; O) + HNO3 Fe3+ + N2+ + O2-

. x y 0,05 mol y

Theo nguyn l bo ton nguyn t v bo ton khi lng ta c: Khi lng Fe ban ulun bng s lng Fe nm trong X'. V vy m = 56x.

Mt khc: 56x + 16y = 6 (I)Cc qu trnh nhng v nhn e:

Fe - 3e Fe+3

x 3x x

O0 + 2e O-2

y 2y y

N+5

+ 3e N+2

. 0,15 0,05

Theo LBT electron ta c: 3x = 2y + 0,15 (II).

T (I), (II) x = 0,09; y = 0,06

m = 0,09 . 56 = 5,04 p n A.

V d minh ha 2: Ho tan hon ton 60,8 gam cht rn X gm Cu, CuS, Cu2S v Sbng HNO3 d, thot ra 40,32 lt kh NO duy nht (ktc) v dung dch Y. Thm Ba(OH)2

d vo Y thu c m gam kt ta. Gi tr ca m l:

A. 163,1. B. 208,4. C. 221,9. D. 231,7.

Hng dn gii:



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Ta c th quy i hn hp X thnh hn hp X' gm Cu v S vi s mol ln lt l x, y.

X (Cu ; S ) + HNO3 d dd Y (Cu2+ + SO42-) + NO + H2O

60,8 x mol y mol x y 1,8 mol

dd Y (Cu2+ + SO42-) + Ba(OH)2 d (Cu(OH)2 + BaSO4)

. x mol y mol x mol y mol

Tnh khi lng kt ta (Cu(OH)2 + BaSO4).

tnh c khi lng kt ta, ta ch cn xc nh x v y.

Tht vy, 64x + 32y = 60,8 (I)

Cc qu trnh nhng v nhn e:

Cu0 - 2e Cu+2

x 2x

S - 6e S+6

y 6y

N+5 + 3e N+2

. 5,4 1,8

Theo nh lut bo ton e: 2x + 6y = 5,4 (II)

T (I), (II) ta c: x = 0,6 v y = 0,7

m = 0,6 . 98 + 0,7 . 233 = 221,9g p n C.

V d minh ha 3: Nung m gam bt Cu trong oxi thu c 49,6 gam hn hp cht rn Xgm Cu, CuO v Cu2O. Ho tan hon ton X trong H2SO4 c nng thot ra 8,96 lt SO2duy nht (ktc). Gi tr ca m l:

A. 19,2. B. 29,44. C. 42,24. D. 44,8.

Hng dn gii:


Ta c th quy i hn hp X thnh hn hp X' gm Cu v O vi s mol ln lt l x, y.

Cu + O2 X' ( Cu; O ) + H2SO4 .n Cu2+ + S+4 + O2-


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. m(g) 49,6 x mol y mol x mol 0,4 y mol

Theo bi ra ta c: 64x + 16y = 49,6 (I)

Cc qu trnh nhng v nhn e:

Cu - 2e Cu+2

x 2x

O0 + 2e O-2

y 2y

S+6 + 2e S+4

. 0,8 0,4

Theo LBT e ta c: 2x = 2y + 0,8 (II)

T (I), (II) ta c: x = 0,7 v y = 0,3

Theo nguyn l bo ton nguyn t v bo ton khi lng, m(g) Cu ban u bin htthnh Cu nm trong X'.

m = 64 . x = 64 . 0,7 = 44,8 n n D.

Bi tp v nh thuc tuyt chiu s 2

Cu 1: Nung m gam bt Cu trong oxi thu c 24,8 gam hn hp cht rn X gm Cu,CuO v Cu2O. Ho tan hon ton X trong H2SO4 c nng thot ra 4,48 lt SO2 duy nht(ktc). Gi tr ca m l:

A. 9,6. B. 14,72. C. 21,12. D. 22,4

Cu 2: Ho tan hon ton 30,4 gam cht rn X gm Cu, CuS, Cu2S v S bng HNO3 d,thot ra 20,16 lt kh NO duy nht (ktc) v dung dch Y. Thm Ba(OH)2 d vo Y thuc m gam kt ta. Gi tr ca m l:

A. 81,55. B. 104,2. C. 110,95. D. 115,85.

Cu 3: Hn hp X c t khi so vi H2 l 21,2 gm propan, propen v propin. Khi tchy hon ton 0,1 mol X, tng khi lng ca CO2 v H2O thu c l:

A. 18,6 gam. B. 18,96 gam. C. 19,32 gam. D. 20,4 gam.

Cu 4: Cho 9,12 gam hn hp gm FeO, Fe2O3, Fe3O4 tc dng vi dung dch HCl (d).Sau khi cc phn ng xy ra hon ton, c dung dch Y; c cn Y thu c 7,62 gamFeCl2 v m gam FeCl3. Gi tr ca m l:


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A. 4,875. B. 9,75. C. 14,625. D. 19,5.

Cu 5: Cho 11,36 gam hn hp gm Fe, FeO, Fe2O3 v Fe3O4 phn ng ht vi dungdch HNO3 long (d), thu c 1,344 lt kh NO (sn phm kh duy nht, ktc) vdung dch X. C cn dung dch X thu c m gam mui khan. Gi tr ca m l:

A. 49,09. B. 38,72. C. 35,5. D. 34,36.

Cu 6: Oxi ha chm m gam Fe ngoi khng kh thu c 12 gam hn hp X gm FeO,Fe3O4, Fe2O3 v Fe d. Ho tan hon ton X trong HNO3 thu c 2,24 lt NO (cht khduy nht, o ktc). Gi tr m l:

A. 7,57. B. 7,75. C. 10,08. D. 10,8.

Cu 7: Cho 13,92 gam hn hp X gm Cu v mt oxit st vo dung dch HNO3 long dthu c 2,688 lt kh NO (sn phm kh duy nht, o ktc). C cn dung dch sauphn ng thu c 42,72 gam mui khan. Cng thc ca oxit st l:

A. FeO. B. Fe3O4. C. Fe2O3. D. Khng xc nh c.

Cu 8: t chy 6,72 gam bt Fe trong khng kh thu c m gam hn hp X gm FeO,Fe3O4, Fe2O3 v Fe d. ho tan X cn dng va ht 255ml dung dch cha HNO3 2M,thu c V lt kh NO2 (sn phm kh duy nht, o ktc). Gi tr ca m, V ln lt l:

A. 8,4 v 3,36. B. 8,4 v 5,712. C. 10,08 v 3,36. D. 10,08 v 5,712.

Cu 9: Hn hp bt X gm Zn, ZnS v S. Ho tan hon ton 17,8 gam X trong HNO3nng, d thu c V lt kh NO2 duy nht (ktc) v dung dch Y. Thm t t Ba(OH)2vo Y thy lng kt ta ti a thu c l 34,95 gam. Gi tr ca V l:

A. 8,96. B. 20,16. C. 22,4. D. 29,12.

Cu 10:Nung m gam bt st trong oxi, thu c 3 gam hn hp cht rn X. Ho tan hthn hp X trong dung dch HNO3 (d), thot ra 0,56 lt ( ktc) NO (l sn phm khduy nht). Gi tr ca m l:

A. 2,52. B. 2,22. C. 2,62. D. 2,32.

Cu 11: Cho 18,5 gam hn hp gm Fe v Fe3O4 vo 200ml HNO3 un nng. Sau phnng thu c 2,24 lt kh NO duy nht (ktc), dung dch D v cn li 1,46 gam kim loi

cha tan. Nng mol ca dung dch HNO3 dng l:A. 2,7M. B. 3,2M. C. 3,5M. D. 2,9M.

Cu 12: Ho tan hon ton 34,8 gam mt oxit st dng FexOy trong dung dch H2SO4 c,nng. Sau phn ng thu c 1,68 lt kh SO2 (sn phm kh duy nht, o ktc). OxitFexOy l:

A. FeO. B. FeO hoc Fe3O4. C. Fe3O4. D. Khng xc nh c.


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Cu 13: Hn hp X gm Mg, MgS v S. Ho tan hon ton m gam X trong HNO3 c,nng thu c 2,912 lt kh N2 duy nht (ktc) v dung dch Y. Thm Ba(OH)2 d vo Yc 46,55 gam kt ta. Gi tr ca m l:

A. 4,8. B. 7,2. C. 9,6. D. 12.

Cu 14: Ho tan hon ton 25,6 gam cht rn X gm Fe, FeS, FeS2 v S bng dung dchHNO3 d, thot ra V lt kh NO duy nht (ktc) v dung dch Y. Thm Ba(OH)2 d vo Ythu c 126,25 gam kt ta. Gi tr ca V l:

A. 17,92. B. 19,04. C. 24,64. D. 27,58.

Cu 15: Cho hn hp X gm FeO, Fe2O3, Fe3O4 vi s mol bng nhau. Ly a gam X chophn ng vi CO nung nng, sau phn ng trong bnh cn li 16,8 gam hn hp rn Y.Ho tan hon ton Y trong H2SO4 c, nng thu c 3,36 lt kh SO2 duy nht (ktc).Gi tr ca a v s mol H2SO4 phn ng ln lt l:

A. 19,2 v 0,87. B. 19,2 v 0,51. C. 18,56 v 0,87. D. 18,56 v 0,51.

Cu 16: Hn hp X c t khi so vi H2 l 27,8 gm butan, metylxiclopropan, but-2-en,etylaxetilen v ivinyl. Khi t chy hon ton 0,15 mol X, tng khi lng ca CO2 vH2O thu c l:

A. 34,5 gam. B. 36,66 gam. C. 37,2 gam. D. 39,9 gam.

Cu 17: ho tan hon ton 2,32 gam hn hp gm FeO, Fe3O4 v Fe2O3 (trong smol FeO bng s mol Fe2O3), cn dng va V lt dung dch HCl 1M. Gi tr ca V l:

A. 0,08. B. 0,16. C. 0,18. D. 0,23.

Cu 18: Ho tan hon ton 14,52 gam hn hp X gm NaHCO3, KHCO3 v MgCO3trong dung dch HCl d, thu c 3,36 lt kh CO2 (ktc). Khi lng mui KCl tothnh trong dung dch sau phn ng l:

A. 8,94 gam. B. 16,17 gam. C. 7,92 gam. D. 12 gam.

Cu 19: Ho tan hon ton m gam hn hp X gm Fe, FeCl2, FeCl3 trong H2SO4 cnng, thot ra 4,48 lt kh SO2 duy nht (ktc) v dung dch Y. Thm NH3 d vo Y thuc 32,1 gam kt ta. Gi tr ca m l:

A. 16,8. B. 17,75. C. 25,675. D. 34,55.

Tuyt Chiu S 3Th by, 09 Thng 5 2009 16:16 Ti quang dung

c im nhn dng : Vi tt c cc bi ton m trong c xy ra nhiugiai on oxi ha khc nhau (thng l 2 giai on) bi cc cht oxi ha khc


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nhau. Khi y, ta c th thay i vai tr oxi ha ca cht oxi ha ny cho cht oxiha kia bi ton tr nn n gin hn.

S ca chiu thc:

Cht kh X + Cht oxi ha 1 Sn phm trung gian + Cht oxi ha 2 Snphm cui.

Ta i cht oxi ha 2 bng cht oxi ha 1.

* C s ca tuyt chiu s 3 l:

S mol electron cht oxi ha c nhn = s mol electron cht oxi ha mi nhn

Do s thay i tc nhn oxi ha nn c s thay i sn phm sao cho ph hp.

V d minh ha 1: Nung m gam bt st trong oxi, thu c 12 gam hn hp

cht rn X. Ho tan ht hn hp X trong dung dch HNO3 (d), thot ra 2,24 lt (ktc) NO (l sn phm kh duy nht). Gi tr ca m l:

A. 10,08. B. 8,88. C. 10,48. D. 9,28.

Hng dn gii:

Tm tt:

Fe + O2 X (Fe; FeO; Fe2O3; Fe3O4) + dd HNO3 Fe3+ + NO + H2O

m gam 12 gam 2,24 lt


Fe + O2 X + O2 Fe2O3.

m gam 2 a (mol)

Gi a l s mol Fe c trong m (g). Theo nguyn l bo ton nguyn t Fe ta c:S mol ca Fe nm trong Fe2O3 l 2a.

y ta thay vai tr nhn e ca N+5 bng Oxi. Gi y l s mol nguyn t Oxitrong Fe2O3.

M : N+5 + 3e N+2.

0,3 0,1

O + 2e O-2.

y 2y y


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Do s mol electron cht oxi ha c nhn = s mol electron cht oxi ha mi nhn

nn 2y = 0,3 y = 0,15.

Mt khc, khi lng Fe2O3 = mX + mO = 12 + 0,15 . 16 = 14,4.

S mol Fe2O3 = 14,4/160 = 0,09.

Vy s mol Fe nm trong Fe2O3 = 0,09 . 2 = 0,18 m = 0,18 . 56 = 10,08 (g) p n A.

V d minh ha 2: Nung m gam bt Cu trong oxi thu c 74,4 gam hn hpcht rn X gm Cu, CuO v Cu2O. Ho tan hon ton X trong H2SO4 c nngthot ra 13,44 lt SO2 duy nht (ktc). Gi tr ca m l:

A. 28,8. B. 44,16. C. 42,24. D. 67,2.

Hng dn gii:S ha bng tuyt chiu s 3.

Cu + O2 X (Cu; CuO; Cu2O) + O2 CuO

m(g) 74,4g a (mol)

Thay vai tr oxi ha ca H2SO4 bng Oxi.

y ta thay vai tr nhn e ca S+6 bng Oxi. Gi y l s mol nguyn t Oxitrong CuO.

M : S+6 + 2e S+4.

1,2 0,6

O + 2e O-2.

y 2y y

Do s mol electron cht oxi ha c nhn = s mol electron cht oxi ha mi nhn

nn 2y = 1,2 y = 0,6.

Mt khc, khi lng CuO = mX + mO = 74,4 + 0,6 . 16 = 84.

S mol CuO = 84/80 = 1,05.

mCu = 1,05 . 64 = 67,2(g) p n D.

Tuyt Chiu S 4 (Tuyt Chiu 3 Dng)


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Th by, 09 Thng 5 2009 17:19 Ti quang dung

* C s ca tuyt chiu s 4 (Tuyt chiu 3 dng) l:

S dng nh lut bo ton nguyn t v khi lng.

Nhn xt:

Trong cc phng trnh phn ng ca kim loi, oxit kim loi... vi HNO 3 hocH2SO4 c nng ta lun c 2 h thc:

- Nu l HNO3: S mol ca H2O = 1/2 s mol ca HNO3 phn ng.

- Nu l H2SO4: S mol ca H2O = s mol ca H2SO4 phn ng.

V d minh ha 1: Cho m gam bt stra ngoi khng kh sau mt thi gianngi ta thu c 12 gam hn hp B gm Fe; FeO; Fe2O3; Fe3O4. Ho tan hn

hp ny bng dung dch HNO3 ngi ta thu c dung dch A v 2,24 lt kh NO(ktc). Tnh m.

Hng dn gii:


Fe + O2 Cht rn B + HNO3 Fe(NO3)3 + NO + H2O.

m gam 12 gam 0,1mol

x mol x mol

Gi x l s mol ca Fe c trong m gam. Theo nguyn l bo ton th s mol Fec trong Fe(NO3)3 cng l x mol.

Mt khc, s mol HNO3 phn ng = (3x + 0,1) s mol ca H2O = 1/2 s molHNO3 = 1/2 (3x + 0,1)

Theo nh lut bo ton khi lng ta c: 12 + 63(3x + 0,1) = 242 . x + 0,1 . 30 +18. 1/2(3x + 0,1)

x = 0,18 (mol). m = 10,08 (g).

Tuyt chiu s 4 ny c tm p dng rt tng qut, c th x l ht c tt ccc bi ton thuc cc chiu 1, 2, 3. Trn y Ti ch trnh by mt kha cnh rtnh b ca tuyt chiu ny. Ti s phn tch k hn cho cc bn trn lp luynthi ti cc trung tm. Cc bn ch theo di.

Cc bi tp c th gii bng tuyt chiu ny:


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Bi 1: Ho tan hon ton 4,431 gam hn hp Al v Mg trong HNO3 long thuc dung dch A v 1,568 lt (ktc) hn hp hai kh u khng muc khilng 2,59 gam trong c mt kh b ho nu trong khng kh.

1. Tnh phn trm theo khi lng ca mi kim loi trong hn hp.

2. Tnh s mol HNO3 phn ng.

3. Khi c cn dung dch A th thu c bao nhiu gam mui khan.

Bi 2:Cho m gam bt stra ngoi khng kh sau mt thi gian ngi ta thuc 12 gam hn hp B gm Fe; FeO; Fe2O3; Fe3O4. Ho tan hn hp ny bngdung dch HNO3 ngi ta thu c dung dch A v 2,24 lt kh NO (ktc). Vitphng trnh phn ng xy ra v tnh m.

Bi 3: Mt hn hp A gm Fe v kim loi R ho tr n khng ic khi lng14,44 gam. Chia hn hp A thnh 2 phn bng nhau. Ho tan htphn 1 trong

dung dch HCl thu c 4,256 lt kh H2. Ho tan htphn 2trong dung dchHNO3 thu c 3,584 lt kh NO.

1. Xc nh kim loi R v thnh phn % khi lng mi kim loi trong hn hp A.

2. Cho 7,22gam A tc dng vi 200ml dung dch B cha Cu(NO3)2 v AgNO3.Sau phn ng thu c dung dch C v 16,24 gam cht rn D gm 3 kim loi.Cho D tc dng vi dung dch HCl thu c 1,344 lt H2. Tnh nng mol/l caCu(NO3)2 v AgNO3 trong B; (cc th tch o ktc, phn ng xy ra hon ton).

Bi 4: Nung M gam bt st trong khng kh sau mt thi gian ngi ta thu c104,8 gam hn hp rn A gm Fe, FeO, Fe2O3, Fe3O4. Ho tan hon ton Atrong dung dch HNO3 d thu c dung dch B v 12,096 lt hn hp kh NO vN2O ( ktc) c t khi hi so vi H2 l 20,334.

1. Tnh gi tr ca M

2. Cho dung dch B tc dng vi dung dch NaOH d thu c kt taC. Lc kt ta ri nung n khi lng khng i c cht rn D. Tnh khilng ca D.

Bi 5: Ho tan hon ton 24,3 gam nhm vo dung dch HNO3 long d thuc hn hp kh NO v N2O c t khi hi so vi H2 l 20,25 v dung dch B

khng cha NH4NO3. Tnh th tch mi kh thot ra ktc)

Bi 6: Cho 200 ml dung dch HNO3 tc dng vi 5 gam hn hp Zn v Al. Phnng gii phng ra 0,896 lt (ktc) hn hp kh gm NO v N2O. Hn hp kh c t khi hi so vi H2 l 16,75. Sau khi kt thc phn ng bn lc, thu c2,013 gam kim loi. Hi sau khi c cn dung dch A th thu c bao nhiu gammui khan? Tnh nng dung dch HNO3 trong dung dch ban u.


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Bi 7: Ho tan hon ton 2,43 gam kim loi A va vo Z ml dung dch HNO30,6M c dung dch B c cha A (NO3)3 ng thi to ra 672 ml hn hp khN2O v N2 c t khi hi so vi O2 l 1,125.

1. Xc nh kim loi A v tnh gi tr ca Z

2. Cho vo dung dch B 300ml dung dch NaOH 1M. Sau khi phn ng song lcly kt ta, ra sch, un nng n khi lng khng i c mt cht rn.Tnh khi lng ca mt cht rn . Cc V o ktc

Bi 8: Cho a gam hn hp A gm 3 oxit FeO, CuO, Fe3O4c s mol bng nhautc dng hon ton vi lng va 250ml dung dch HNO3 khi un nng nhthu c dung dch B v 3,136 lt (ktc) hn hp kh C gm NO2 v NO c tkhi so vi H2 l 20,143. Tnh a v nng mol ca dung dch HNO3 dng.

Bi 9: Cho mt hn hp gm 2,8 gam Fe v 0,81 gam Al vo 200 ml dung dchC cha AgNO3 v Cu(NO3)2. Khi cc phn ng kt thc c dung dch D v

8,12g cht rn E gm ba kim loi. Cho E tc dng vi dung dch HCl d c0,672 lt H2 (kc). Tnh nng mol ca Ag(NO3)2 trong dung dch C

Bi 10: t chy x mol Fe bi oxi thu c 5,04g hn hp A gm cc oxt st.Ho tan hon ton A trong HNO3 thu c 0,035 mol hn hp Y gm NO vNO2.T khi hi ca Y i vi H2 l 19. Tnh x.

Bi 11: Nung nng 16,8g bt st ngoi khng kh, sau mt thi gian thu c mgam hn hp X gm oxt st. Ho tan ht hn hp X bng H2SO4 c nng thuc 5,6 lt SO2 (kc).

a) Vit tt c phn ng xy ra)b) Tm m.

c) Nu ho tan ht X bng HNO3 c nng th th tch NO2 (kc) thu c l baonhiu?

Bi 12: Nung nng m gam bt st ngoi khng kh. Sau mt thi gian thu c10g hn hp (X) gm Fe, FeO, Fe2O3 v Fe3O4.Ho tan ht (X) bng HNO3 thuc 2,8 lt (kc) hn hp Y gm NO v NO2. cho dY/H2 = 19. Tnh m ?

Bi 13: Cho mt lung CO i qua ng s ng m gam Fe2O3 nung nng mt thi

gian, thu c 13,92 gam cht rn X gm Fe, Fe3O4, FeO v Fe2O3. Ho tan htX bng HNO3 c nng thu c 5,824 lt NO2 (kc). Tnh m?

Bi 14 Cho mt lung kh CO i qua ng s ng m gam Fe2O3 nung nng. Saumt thi gian thu c hn hp X nng 44,64g gm Fe3O4, FeO, Fe v Fe2O3d. Ho tan ht X bng HNO3 long thu c 3,136 lt NO (kc). Tnh m ?


15/85

Tuyt chiu s 5(Bo ton Electron)Th t, 13 Thng 5 2009 18:05 Ti quang dung

Bi 1: ho tan ht mt hn hp gm 0,02 mol kim loi A (ho tr II) v 0,03mol kim loi B (ho tr III) cn m gam dung dch HNO3 21%. Sau phn ng thu

c 0,896 lt (kc) hn hp NO v N2O. Vit cc phng trinh phn ng xy rav tnh M.

Hng dn gii

Cc phn ng xy ra:

3A + 8 HNO3 = 3A(NO3)2 + 2NO + 4H2O

4A + 10HNO3 = 4A(NO3)2 + N2O + 4H2O

B + 4HNO3 = B(NO3)3 + NO + 2H2O

8B +30HNO3 = 8B(NO3)3 + 3N2O + 15H2O

Gi a, b l s mol NO v N2O thu c, ta c cc qu trnh cho nhn electron.

Cho

A - 2e = A2+

0,02mol 0,04mol

B - 3e = B3+

0,03mol 0,09mol

Nhn

NO3- + 3e + 4H+ = NO + 2H2O

3a 4a a

2NO3- + 8e + 10H+ = N2O + 5H2O

8b 10b b

3a + 8b = 0,04 + 0,09 = 0,13 (I)

a + b = 0,896/22,4 = 0,04 (II)

T (I), (II) : a = 0,038 v b = 0,02

S mol HNO3 = S mol H+ = 4a + 10b = 0,172


16/85

S mol dd HNO3 21% = (0,172 . 63 . 100) / 21 = 21,6(g)

Bi 2: Hn hp A gm 3 kim loi X, Y c ho tr ln lt l 3; 2; 1 v t l mol lnlt l 1:2:3, trong s mol ca X l x. Ho tan hon ton A bng dung dch ccha y gam HNO3 (ly 25%). Sau phn ng thu c dung dch B khng

cha NH4NO3 v V lt (kc) hn hp kh G gm NO v NO2. Lp biu thc tnh ytheo x v V.

Hng dn gii

Gi a, b l s mol NO v NO2 sinh ra, ta c cc qu trnh cho, nhn electron:

Cho

X - 3e = X3+

x 3x

Y - 2e = Y2+

2x 4x

Z - e = Z+

3x 3x

Nhn

NO3- + 3e + 4H+ = NO + 2H2O

3a 4a a

NO3- + e + 2H+ = NO2 + H2O

b 2b b

3a + b = 3x + 4x + 3x = 10x (I)

a + b = V / 22,4 (II)

T (I), (II) a = 1/2 (10x - V / 22,4) v b = 1/2 (3V / 22,4 - 10x)

S mol HNO3 = S mol H+ = 4a + 2b = 10x + V / 22,4

y = 63 (10x + V / 22,4) + 25/100 . 63 (10x + V / 22,4) = 78,75 (10x + V / 22,4)

Bi 3: Cho mt hn hp gm 2,8g Fe v 0,81g Al vo 200ml dung dch C chaAgNO3 v Cu(NO3)2. Khi cc phn ng kt thc c dung dch D v 8,12 gam


17/85

cht rn E gm ba kim loi. Cho E tc dng vi dung dch HCl d c 0,672 ltH2 (kc). Tnh nng mol ca AgNO3 v Cu(NO3)2 trong dung dch C.

Hng dn gii

Do Al u tin phn ng trc Fe nn ba kim loi trong E phi l Fe, Cu, Ag. Tac:

nFe ban u = 2,8 / 56 = 0,05 mol

nAl ban u = 0,81 / 27 = 0,03 mol

Khi cho E tc dng vi HCl, ch xy ra phn ng:

Fe + 2HCl = FeCl2 + H2

nFe cn d = S mol H2 = 0,672 / 22,4 = 0,3

Dung dch C (gm x mol AgNO3 v y mol Cu(NO3)2) tc dng va vi0,03 mol Mg v (0,05 - 0,03) = 0,02 mol Fe

Ta c cc qu trnh cho, nhn electron:

Cho

Al - 3e = Al3+

0,03mol 0,09mol

Fe - 2e = Fe2+

0,02mol 0,04mol

Nhn

AgNO3 + e = Ag + NO3-

x x x

Cu(NO3)2 + 2e = Cu + 2NO3-

y 2y y

x + 2y = 0,04 + 0,09 = 0,013 (I)

108x + 64y + 0,03 . 56 = 8,12 (II)

T (I), (II) : x = 0,03 v y = 0,05


18/85

CM AgNO3 = 0,03 / 0,2 = 0,15M.

CM Cu(NO3)2 = 0,05 / 0,2 = 0,25MBi 4: Ho tan 62,1 gam kim loi M trong dung dch HNO3 long c 16,8 lt(kc) hn hp X gm 2 kh khng mu, khng ho nu ngoi khng kh. Bit d

x /H2 = 17,2. a. Tm tn M. b. Tnh th tch dung dch HNO3 2M dng, bit rng ly d 25% so

vi lng cn thit.

Hng dn gii

a. Ta c: MX = 17,2 2 = 34,4

Hai kh khng mu, khng ho nu ngoi khng kh v tho iu kin M1 < 34,4< M2 y ch c th l N2 v N2O.

Gi x l s mol M dng v n l ha tr ca M. Gi a, b l s mol N2 v N2O ctrong X, ta c cc qu trnh cho nhn e:

Cho

M - ne = Mn+

x nx

Nhn

2NO3- + 10e + 12H+ = N2 + 6H2O

10a 12a a

2NO3- + 8e + 10H+ = N2O + 5H2O

8b 10b b

x . M = 62,1 (I)

n . x = 10a + 8b (II)

a + b = 16,8/22,4 = 0,75 (III)

(28a + 44b) / 0,75 = 34,4 (IV)

T (I), (II), (III), (IV) :

a = 0,45


19/85

b = 0,3

x . M = 62,1

n . x = 6,9

Rt ra M = 9n. Ch c n = 3, ng vi M = 27 l ph hp. Vy M l Al

b. Ta c:

S mol HNO3 = S mol H+ = 12a + 10b = 8,4

Th tch dd HNO3 = 8,4 / 2 + 25/100 . 8,4 / 2 = 5,25 lt

Bi 5: Cho 12,45 gam hn hp X (Al v kim loi M ho tr II) tc dng vi dungdch HNO3 d c 1,12 lt hn hp N2O v N2, c t khi i vi H2 l 18,8 vdung dch Y. Cho Y tc dng vi dung dch NaOH d c 0,448 lt NH 3. Xc

nh kim loi M v khi lng mi kim loi trong X. Cho nx = 0,25 mol v cc thtch o kc.

Hng dn gii

Gi a, b l s mol ca Al v M c trong X

Gi c, d, e l s mol N2O, N2 v NH4NO3 c to ra, ta c cc qu trnh cho,nhn electron.

Cho

Al - 3e = Al3+

a 3a

M - 2e = M2+

b 2b

Nhn

2NO3- + 8e + 10H+ = N2O + 5H2O

8c 10c c

2NO3- + 10e + 12H+ = N2 + 6H2O

10d 12d d

2NO3- + 8e + 10H+ = NH4NO3 + 3H2O


20/85

8e e

Phn ng ca dung dch Y vi NaOH:

NH4NO3 + NaOH = NH3 + H2O + NaNO3

e e

suy ra :

27a + b.M = 12,45

a + b = 0,25

3a + 2b = 8c + 10 d + 8e

c + d = 1,12/ 22,4 = 0,05

(44c + 28d)/ (c + d) = 18,8 . 2 = 37,6

e = 0,448/22,4 = 0,02

a = 0,1

b = 0,15

c = 0,03

d = 0,2

e = 0,02

M = 65 M l Zn

Bi 6: t chy x mol Fe bi oxi thu c 5,04g hn hp A gm cc oxit st.Ho tan hon ton A trong HNO3 thu c 0,035 mol hn hp Y gm NO vNO2. T khi hi ca Y i vi H2 l 19. Tnh x.

Hng dn gii

Cn c vo s phn ng:

x mol Fe + O2 Cc oxi st + HNO3 Fe(NO3)3 + NO + NO2 +H2O

Ta c cc qu trnh cho nhn electron:

Cho


21/85

Fe - 3e = Fe3+

X 3x

Nhn

O2 + 4e = 2O2-

(5,04 - 56x)/32 4(5,04 - 56x)/32

NO3- + 3e + 4H+ = NO + 2H2O

3a a

NO3- + e + 2H+ = NO2 + H2O

b b

Suy ra:

a + b = 0,035

(30a + 46b) / (a + b) = 19 . 2 = 38

4(5,04 - 56x)/32 + 3a + 3b = 3x

a = 0,0175; b = 0,0175; x = 0,07

Bi 7: m gam phi bo st (A) ngoi khng kh, sau mt thi gian c hn

hp (B) nng 12g gm Fe, FeO, Fe2O3, Fe3O4. Ho tan ht B bng HNO3 thygii phng 2,24 lt NO (kc) duy nht

a. Vit phng trnh phn ng

b. nh m.

Hng dn gii

a. Cc phn ng xy ra:

2Fe + O2 = 2FeO

3Fe + 2O2 = Fe3O4

4Fe + 3O2 = 2Fe2O3

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

3FeO + 10HNO3 = 3Fe(NO3)3 + NO + 5H2O


22/85

Fe2O3 + 6HNO3 = Fe(NO3)3 + 3H2SO4

3Fe3O4 + 28HNO3 = 9Fe(NO3)3 + NO + 14H2O

b. Cn c vo s phn ng:

a mol Fe + O2 Fe, FeO, Fe2O3, Fe3O4 + HNO3 Fe(NO3)3 + NO + H2O.

Ta c cc qu trnh cho, nhn electron:

Cho

Fe - 3e = Fe3+

a 3a

Nhn

O2 + 4e = 2O2-

(12 - 56a)/32 4(12 - 56a)/32

NO3- + 3e + 4H+ = NO + 2H2O

0,03 mol 0,1 mol

3a = 4(12 - 56a)/32 + 0,3 a = 0,18

m = 56a = 10,08g

Tuyt chiu s 6 (Bo ton khi lng)Ch nht, 17 Thng 5 2009 17:00 Ti quang dung

P DNG PHNG PHP BO TON KHI LNG

Nguyn tc ca phng php ny kh n gin, da vo nh lut bo ton khilng (LBTKL): "Tng khi lng cc cht tham gia phn ng bng tng khilng cc cht to thnh sau phn ng".

Cn lu l: khng tnh khi lng ca phn khng tham gia phn ng cngnh phn cht c sn, v d nc c sn trong dung dch. Khi c cn dung dchth khi lng mui thu c bng tng khi lng cc cation kim loi v aniongc axit.

V d 1: Hn hp X gm Fe, FeO v Fe2O3. Cho mt lung kh CO i qua ng sng m gam hn hp X nung nng. Sau khi kt thc th nghim thu c 64


23/85

gam cht rn A trong ng s v 11,2 lt kh B (ktc) c t khi so vi H2 l 20,4.Gi tr ca m l:

A. 105,6 gam. B. 35,2 gam.

C. 70,4 gam. D. 140,8 gam.

Hng dn gii:

Cc phn ng kh st oxit c th c:

Nh vy, cht rn A c th gm 3 cht Fe, FeO, Fe3O4 hoc t hn, iu khng quan trng v vic cn bng cc phng trnh trn cng khng cn thitcho vic xc nh p n, qua trng l s mol CO phn ng bao gi cng bngs mol CO2 to thnh.

nB = 11,2/22,5 = 0,5 (mol)

Gi x l s mol ca CO2, ta c phng trnh v khi lng ca B: 44x + 28(0,5 -x) = 0,5 20,4 2 = 20,4

Nhn c x = 0,4 mol v cng chnh l s mol CO tham gia phn ng.

Theo LBTKL, ta c: mX + mCO = mA + mCO2 m = 64 + 0,4 . 44 - 0,4 . 28 =70,4(gam) (p n C).

V d 2: un 132,8 gam hn hp 3 ancol no, n chc vi H2SO4 c 1400Cthu c hn hp cc ete c s mol bng nhau v c khi lng l 111,2 gam.S mol ca mi ete trong hn hp l:

A. 0,1 mol. B. 0,15 mol.

C. 0,4 mol. D. 0,2 mol.

Hng dn gii:

Ta bit rng c 3 loi ancol tch nc iu kin H2SO4 c, 1400C th tothnh 6 loi ete v tch ra 6 phn t H2O.

Theo LBTKL ta c: mH2O = mru - mete = 132,8 - 111,2 = 21,6 (gam)

nH2O = 21,6/18 = 1,2(mol)


24/85

Mt khc, c hai phn t ancol th to ra mt phn t ete v mt phn t H2O.Do s mol H2O lun bng s mol ete, suy ra s mol mi ete l 1,2/6=0,2(mol).(p n D).

Nhn xt: Chng ta khng cn vit 6 phng trnh ca phn ng t ancol tch

nc to thnh 6 ete, cng khng cn tm CTPT ca cc ancol v cc ete trn.Nu sa vo vic vit phng trnh phn ng v t n s mol cho cc ete tnh ton th vic gii bi tp rt phc tp, tn nhiu thi gian.

V d 3: Cho 12 gam hn hp hai kim loi Fe, Cu tc dng va vi dung dchHNO3 63%. Sau phn ng thu c dung dch A v 11,2 lt kh NO2 duy nht(ktc). Nng % cc cht c trong dung dch A l:

A. 36,66% v 28,48%.

B. 27,19% v 21,12%.

C. 27,19% v 72,81%.

D. 78,88% v 21,12%.

Hng dn gii:

Fe + 6HNO3 Fe(NO3)3 + 3NO2 + 3H2O

Cu + 4HNO3 Fe(NO3)3 + 2NO2 + 3H2O.

nNO2 = 0,5mol

nHNO3 = 2nNO2 = 1 mol

p dng LBTKL ta c:

mdd mui = mhh k.loi + mddHNO3 - mNO2

= 12 + (1. 63 . 100) /63 - (46 . 0,5) = 89(gam)

t nFe = x mol, nCu = y mol, ta c:

56x + 64y = 12

3x = 2y = 0,5

x = 0,1 v y = 0,1

C% Fe(NO3)3 = (0,1 . 242 /89) . 100% = 27,19%

C% Cu(NO3)2 = (0,1 . 188/89) . 100% = 21,12%. (p n B)


25/85

V d 4: Ho tan hon ton 23,8 gam hn hp mt mui cacbonat ca cc kimloi ho tr I v mui cacbonat ca kim loi ho tr II trong dung dch HCl. Sauphn ng thu c 4,48 lt kh (ktc). bn c cn dung dch thu c khilng mui khan l:

A. 13 gam. B. 15 gam.C. 26 gam. D. 30 gam.

Hng dn gii

M2CO3 + 2HCl 2MCl + CO2 + H2O

RCO3 + 2HCl RCl2 + CO2 + H2O

nCO2 = 4,88/22,4 = 0,2 (mol)

Tng nHCl = 0,4 mol v nH2O = 0,2 molp dng LBTKL ta c:

23,8 + 0,4 . 36,5 = mmui + 0,2 44 + 0,2 18

mmui = 26 gam (p n C).

V d 5: Hn hp A gm KClO3; Ca(ClO2)2; Ca(ClO3)2; CaCl2 v KCl nng 83,68gam. Nhit phn hon ton A, thu c cht rn B gm CaCl2; KCl v 17,472 ltkh ( ktc). Cho cht rn B tc dng vi 360 ml dung dch K2CO3 0,5M (va )thu c kt ta C v dung dch D. Khi lng KCl trong dung dch D nhiu gp

22/3 ln lng KCl c trong A. % khi lng KClO3 c trong A l:

A. 47,83%. B. 56,72%.

C. 54,67%. D. 58,55%.

Hng dn gii


26/85

(p n D).

V d 6: t chy hon ton 1,88 gam cht hu c A (cha C, H, O) cn 1,904lt O2 (ktc) thu c CO2 v hi nc theo t l th tch 4:3. CTPT ca A l (Bitt khi ca A so vi khng kh nh hn 7).

A. C8H12O5. B. C4H8O2.

C. C8H12O3. D. C6H12O6.

Hng dn gii

1,88 gam A + 0,085 mol O2 4a mol CO2 + 3a mol H2O.


27/85

p dng LBTKL, ta c:

mCO2 + mH2O = 1,88 + 0,085 . 32 = 46 (gam)

Ta c: 44 . 4a + 18 . 3a = 46 a = 0,02 mol

Trong cht A c:

nC = 4a = 0,08 (mol)

nH = 3a . 2 = 0,12 (mol)

nO = 4a . 2 + 3a - 0,085 . 2 = 0,05 (mol)

nC : nH : nO = 0,08 : 0,12 : 0,05 = 8 : 12 : 5

Vy cng thc ca cht hu c A l C8H12O5 c MA < 203 gam. (p n A).

V d 7: Cho 0,1 mol este to bi 2 ln axit v ancol mt ln ancol tc dng honton vi NaOH thu c 6,4 gam ancol v mt lng mui c khi lng nhiuhn lng este l 13,56% (so vi lng este). CTCT ca este l:

A. CH3 - COO - CH3.

B. CH3OCO - COO - CH3.

C. CH3COO - COOCH3.

D. CH3COO - CH2 - COOCH3.

Hng dn gii

R(COOR')2 + 2NaOH R(COONa)2 + 2R'OH

0,1 0,2 0,1 0,2 mol

MR'OH = 6,4/0,2 = 32 (gam) Ancol CH3OH.

p dng LBTKL, ta c:

meste + mNaOH = mmui + mancol

mmui - meste = 0,2 . 40 - 64 = 16 (gam)

m mmui - meste = 13,56/100 meste

meste = 1,6 . 100/ 13,56 = 11,8 (gam)

Meste = 118 gam


28/85


29/85

- Phn 2: Tc dng vi H2 d (Ni, t0) th thu c hn hp A. bn t chyhon ton hn hp A th th tch kh CO2 (ktc) thu c l:

A. 1,434 lt. B. 1,443 lt.

C. 1,344 lt. D. 0,672 lt.

Hng dn gii

Phn 1: V andehit no n chc nn nCO2 = nH2O = 0,06 mol

nCO2 (phn 1) = nC (phn 2) = 0,06 mol.

Theo nh lut bo ton nguyn t v LBTKL, ta c:

nC (phn 2) = nC (A) = 0,06 mol.

nCO2 (A) = 0,06 molTh tch CO2 = 22,4 . 0,06 = 1,344 (lt). (p n C).

V d 10: Cho mt lung kh CO i qua ng s ng 0,04 mol hn hp A gmFeO v Fe2O3 t nng. Sau khi kt thc th nghim thu c B gm 4 cht nng4,784 gam. Kh i ra khi ng s cho hp th vo dung dch Ba(OH)2 d th thuc 9,062 gam kt ta. Phn trm khi lng Fe2O3 trong hn hp A l:

A. 86,96%. B. 16,04%.

C. 13,04%. D. 6,01%.

Hng dn gii

0,04 mol hn hp A (FeO v Fe2O3) + CO 4,784 gam hn hp B + CO2

CO2 + Ba(OH)2 d BaCO3 + H2O.

nCO2 = nBaCO3 = 0,046 mol.

v nCO (p.) = nCO2 = 0,046 mol.

p dng LBTKL, ta c:

mA + mCO = mB + mCO2.

mA = 4,784 + 0,046 . 44 - 0,046 . 28 = 5,52 (gam)

t nFeO = x mol, nFe2O3 = y mol trong hn hp B, ta c:

x + y = 0,04


30/85

72 x + 160 y = 5,52

x = 0,01 mol v y = 0,03 mol.

%mFeO = 0,01 . 72 / 5,52 . 100% = 13,04%.

%Fe2O3 = 86,96% (p n A).

Bi tp vn dng

Bi 1: Ho tan 9,14 gam hp kim Cu, Mg, Al bng mt lng va dung dchHCl thu c 7,84 lt kh X (ktc). 2,54 gam cht rn Y v dung dch Z. Lc bcht rn Y, c cn cn thn dung dch Z thu c khi lng mui khan l:

A. 31,45 gam. B. 33,99 gam.

C. 19,025 gam. D. 56,3 gam.

Bi 2: Cho 15 gam hn hp 3 amin n chc, bc mt tc dng va vi dungdch HCl 1,2M th thu c 18,504 gam mui. Th tch dung dch HCl dng l:

A. 0,8 lt B. 0,08 lt.

C. 0,4 lt. D. 0,04 lt.

Bi 3: Trn 8,1 gam bt Al vi 48 gam bt Fe2O3 ri cho tin hnh phn ngnhit nhm trong iu kin khng c khng kh. Kt thc th nghim, khi lng

cht rn thu c l:

A. 61,5 gam. B. 56,1 gam.

C. 65,1 gam. D. 51,6 gam.

Bi 4: Ho tan hon ton 10 gam hn hp X gm 2 kim loi (ng trc H trongdy in ho) bng dung dch HCl d thu c 2,24 lt kh H2 (ktc). C cndung dch sau phn ng thu c khi lng mui khan l:

A. 1,71 gam. B. 17,1 gam

C. 13,55 gam. D. 34,2 gam.

Bi 5: Nhit phn hon ton m gam hn hp X gm CaCO3 v Na2CO3 thu c11,6 gam cht rn v 2,24 lt kh (ktc). Hm lng % CaCO 3 trong X l:

A. 6,25%. B. 8,62%.

C. 50,2%. D. 62,5%.


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Bi 6: Cho 4,4 gam hn hp hai kim loi nhm IA hai chu k lin tip tc dngvi dung dch HCl d thu c 4,48 lt H2 (ktc) v dung dch cha m gam muitan. Tn hai kim loi v khi lng m l:

A. 11 gam; Li v Na.

B. 18,6 gam; Li v Na.

C. 18,6 gam; Na v K.

D. 12,7 gam; Na v K.

Bi 7: t chy hon ton 18 gam FeS2 v cho ton b lng SO2 vo 2 lt dungdch Ba(OH)2 0,125M. Khi lng mui to thnh l:

A. 57,4 gam. B. 56,35 gam.

C. 59,17 gam. D. 58,35 gam.Bi 8:Ho tan 33,75 gam mt kim loi M trong dung dch HNO3 long , d thuc 16,8 lt kh X(ktc) gm hai kh khng mu ho nu trong khng kh c tkhi hi so vi hiro bng 17.8.

a. Kim loi l:

A.Cu. B. Zn.

C. Fe. D. Al.

b. Nu dng dung dch HNO3 2M v ly d 25% th th tch dung dch cn ly l

A. 3,15 lt B. 3,00 lit

C. 3,35 lt D. 3.45 lt

Bi 9: Ho tan hon ton 15,9 gam hn hp gm 3 kim loi Al, Mg v Cu bngdung dch HNO3 thu c 6,72 lt khi NO v dung dch X. bn c cn dung dchX thu c s gam mui khan l

A.77,1 gam B.71,7 gam

C. 17,7 gam D. 53,1 gam

Bi 10: Ho tan hon ton 2,81 gam hn hp gm Fe2O3, MgO, ZnO trong 500ml axit H2SO4 0,1 M (va ). Sau phn ng, hn hp mui sunfat khan thuc khi c cn dung dch c khi lng l

A.6,81 gam B.4,81 gam


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C. 3,81 gam D. 4.81 gam

p n cc bi tp vn dng:

1. A 2. B 3. B. 4. B. 5. D

6. B.7. D.8. a-D, b-B9. B. 10. A

Tuyt chiu s 7Th ba, 19 Thng 5 2009 18:00 Ti quang dung

KIM LOI PHN NG VI MUI

I. PHNG PHP

Dng I: Mt kim loi y mt ion kim loi khc.

iu kin kim loi X y c kim loi Y ra khi dung dch mui ca Y:

- X phi ng trc Y trong dy in ha.

V d: Xt phn ng sau:

Cu + 2Ag+ Cu2+ + 2Ag

Phn ng trn lun xy ra v: Cu c tnh kh mnh hn Ag v Ag+ c tnh oxiha mnh hn Cu2+.

Fe + Al3+: Phn ng ny khng xy ra v Fe ng sau Al trong dy in ha.

- Mui ca kim loi Y phi tan trong nc.


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Zn + Fe(NO3)3 Zn(NO3)2 + Fe

phn ng ny xy ra v Zn ng trc Fe v mui st nitrat tan tt trong nc.

Al + PbSO4: Phn ng ny khng xy ra v mui ch sunfat khng tan trongnc.

Ch : Khng c ly cc kim loi kim (Na, K, ...) v kim th (Ca, Sr, Ba)mc d chng ng trc nhiu kim loi nhng khi cho vo nc th s tcdng vi nc trc to ra mt baz, sau s thc hin phn ng trao i vimui to hiroxit (kt ta).

V d: Cho kali vo dung dch Fe2(SO4)3 th c cc phn ng sau:

K + H2O KOH + 1/2 H2.

Fe2(SO4)3 + 6KOH 2Fe(OH)3 + 3K2SO4

Dng II: Cho mt kim loi X vo dung dch cha hai mui ca hai ion kim loi Yn+

v Zm+.

- n gin trong tnh ton, ta ch xt trng hp X ng trc Y v Z, nghal kh c c hai ion Yn+ v Zm+ (Y ng trc Z).

- Do Zm+ c tnh oxi ha mnh hn Yn+ nn X phn ng vi Zm+ trc:

mX + qZm+ mXq+ + qZ (1) (q l ha tr ca X)

Nu sau phn ng (1) cn d X th c phn ng:

nX + qYn+ nXq+ + qY (2)

Vy, cc trng hp xy ra sau khi phn ng kt thc:

+ Nu dung dch cha 3 ion kim loi (Xq+, Yn+ v Zm+) th khng c phn ng (2)xy ra, tc l kim loi X ht v ion Zm+ cn d.

+ Nu dung dch cha hai ion kim loi (Xq+, Yn+) th phn ng (1) xy ra xong (tcht Zm+), phn ng (2) xy ra cha xong (d Yn+), tc l X ht.

+ Nu dung dch ch cha ion kim loi (Xq+) th phn ng (1), (2) xy ra honton, tc l cc ion Yn+ v Zm+ ht, cn X ht hoc d.

Ch :

- Nu bit s mol ban u ca X, Yn+ v Zm+ th ta thc hin th t nh trn.

- Nu bit c th s mol ban u ca Yn+ v Zm+ nhng khng bit s mol banu ca X, th:


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+ Khi bit khi lng cht rn D (gm cc kim loi kt ta hay d), ta ly hai mc so snh:

Mc 1: Va xong phn ng (1), cha xy ra phn ng (2). Z kt ta ht, Y chakt ta, X tan ht.

mCht rn = mZ = m1

Mc 2: Va xong phn ng (1) v phn ng (2), Y v Z kt ta ht, X tan ht.

mCht rn = mZ + mY = m2

Ta tin hnh so snh khi lng cht rn D vi m1 v m2

Nu mD < m1: Z kt ta mt phn, Y cha kt ta.

Nu m1 < mD < m2 : Z kt ta ht, Y kt ta mt phn

Nu mD > m2 : Y v Z kt ta ht, d X.

+ Khi bit khi lng chung cc oxit kim loi sau khi nung kt ta hidroxit to rakhi thm NaOH d vo dung dch thu c sau phn ng gia X vi Yn+ v Zm+,ta c th s dng 1 trong 2 phng php sau:

Phng php 1: Gi s ch c phn ng (1) (Z kt ta ht, X tan ht, Yn+ chaphn ng) th:

m1 = m cc oxit

Gi s va xong phn ng (1) v (2) (Y v Z kt ta ht, X tan ht) th:

m2 = mcc oxit

xc nh im kt thc phn ng, ta tin hnh so snh mcht rn vi m1, m2 nh:

m2 < mcht rn < m1: Z kt ta ht, Y kt ta mt phn, X tan ht.

mcht rn > m1: Z kt ta mt phn, Y cha kt ta, X tan ht.

Phng php 2: Xt 3 trng hp sau:

D X, ht Yn+ v Zm+.

Ht X, d Yn+ v Zm+.

Ht X, ht Zm+ v d Yn+.


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Trong mi trng hp, gii h phng trnh va lp. Nu cc nghim udng v tha mn mt iu kin ban u ng vi cc trng hp kho st thng v ngc li l sai.

Dng 3: Hai kim loi X,Y vo mt dung dch cha mt ion Zn+.

- Nu khng bit s mol ban u ca X, Y, Zn+, th ta vn p dng phng phpchung bng cch chia ra tng trng hp mt, lp phng trnh ri gii.

- Nu bit c s mol ban u ca X, Y nhng khng bit s mol ban u caZn+, th ta p dng phng php dng 2 mc so snh.

Nu ch c X tc dng vi Zn+ mcht rn = m1.

Nu c X, Y tc dng vi Zn+ (khng d Zn+) mcht rn =m2

Nu X tc dng ht, Y tc dng mt phn m1 < mcht rn < m2.

Dng 4: Hai kim loi X, Y cho vo dung dch cha 2 ion kim loi Zn+, Tm+ (X, Yng trc Z, T).

Gi s X > Y, Zn+ > Tm+, ta xt cc trng hp sau:

Trng hp 1: Nu bit s mol ban u ca X, Y, Zn+, Tm+, ta ch cn tnh smol theo th t phn ng.

X + Tm+ ...

X + Zn+ ... (nu d X, ht Tm+)

Y + Tm+ ... (nu ht X, d Tm+)

Trng hp 2: Nu khng bit s mol ban u, da trn s ion tn ti trongdung dch sau phn ng d on cht no ht, cht no cn.

V d: Nu dung dch cha ba ion kim loi (Xa+, Yb+, Zn+) Ht Tm+, ht X, Y (cnd Zc+), ... th ta s dng phng php tnh sau y:

Tng s electron cho bi X, Y = tng s electron nhn bi Zn+, Tm+.

V d: Cho a mol Zn v b mol Fe tc dng vi c mol Cu2+.

Cc bn phn ng.

Zn Zn2+ + 2e

(mol) a 2a

Fe Fe2+ + 2e


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(mol) b 2b

Cu Cu2+ + 2e

(mol) c 2c

Tng s mol electron cho: 2a + 2b (mol)

Tng s mol electron nhn: 2c (mol)

Vy: 2a + 2b = 2c a + b = c.

II. V D P DNG

V d 1: Cho 0,387 gam hn hp A gm Zn v Cu vo dung dch Ag2SO4 c smol l 0,005 mol. Khuy u ti phn ng hon ton thu c 1,144gam chtrn. Tnh khi lng mi kim loi.

Hng dn gii:

- Phn ng:

Zn + Ag2SO4 = ZnSO4 + 2Ag

Cu + Ag2SO4 = CuSO4 + 2Ag

- V tng s mol Zn v Cu nm trong gii hn:

0,387/65 < nhh < 0,387/64

0,0059 < nhh < 0,00604

nhh ln hn 0,005 mol, chng t Ag2SO4 ht.

- Gi s Zn phn ng mt phn, Cu cha tham gia phn ng.

Gi s mol Zn ban u l x; s mol Zn phn ng l x'

Gi s mol Cu ban u l y.

Khi lng kim loi tng:

108.2x' - 65.x' = 1,144 - 0,387 = 0,757 (gam)

151x' = 0,757 x' = 0,00501.

S mol ny ln hn 0,005 mol, iu ny khng ph hp vi bi, do Znphn ng ht v x = x'.


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- Zn phn ng ht, Cu tham gia phn ng mt phn.

Gi s mol Cu tham gia phn ng l y.

Ta c phng trnh khi lng kim loi tng:

108.2x - 65.x + 108 . 2y' - 64 . y' = 0,757 (*)

Gii phng trnh (*) kt hp vi phng trnh:

x + y' = 0,005

Ta c: x = 0,003 v y = 0,002

Vy: mZn = 0,003 . 65 = 0,195 (gam)

mCu = 0,387 - 0,195 = 0,192 (gam)

V d 2: Cho 4,15 gam hn hp Fe, Al phn ng vi 200ml dung dch CuSO40,525M. Khuy k hn hp phn ng xy ra hon ton. bn lc kt ta (A)gm hai kim loi nng 7,84 gam v dung dch nc lc (B). ha tan kt ta(A) cn t nht bao nhiu mililit dung dch HNO3 2M, bit phn ng to NO?

Hng dn gii:

Phn ng xy ra vi Al trc, sau n Fe. Theo gi thit, kim loi sinh ra lCu (kim loi II).

Gi x l s mol Al, y l s mol Fe phn ng v z l mol Fe d:

2Al + 3CuSO4 Al2(SO4)3 + 3Cu

x 1,5x 1,5x (mol)

Fe + CuSO4 FeSO4 + Cu

y y y (mol)

Ta c: 27x + 56(y + z) = 4,15 (1)

1,5x + y = 0,2 . 0,525 = 0,105 (2)

64(1,5x + y) + 56z = 7,84 (3)

Gii h (1), (2), (3)

x = 0,05, y = 0,03 v z = 0,02.

Phn ng vi HNO3:


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

z 4z (mol)

3Cu + 8HNO3 3Cu(NO3)2 + 2NO + 4H2O

(1,5x + y) 8/3(1,5x +y) (mol)

nHNO3 = 8,3(1,5x + y) + 4z = 0,36 (mol)

Vy V dd HNO3 = 0,36 /2 = 0,18 (lt)

V d 3: Cho hn hp (Y) gm 2,8 gam Fe v 0,81 gam Al vo 200ml dung dch(C) cha AgNO3 v Cu(NO3)2. Kt thc phn ng thu c dung dch (D) v 8,12gam cht rn (E) gm ba kim loi. Cho (E) tc dng vi dung dch HCl d, ta thuc 0,672 lt H2 (ktc). Tnh nng mol/l AgNO3, Cu(NO3)2 trc khi phnng.

Hng dn gii:

V phn ng gia Al v AgNO3 xy ra trc nn kim loi sau phn ng phi cAg, k n l CuSO4 c phn ng to thnh Cu. Theo gi thit, c ba kim loi kim loi th ba l Fe cn d.

Ta c: nFe = 2,8/5,6 = 0,05 (mol);

nAl = 0,81/27 = 0,03 (mol)

v nH2 = 0,672/22,4 = 0,03 (mol)

Phn ng: Fed + 2HCl 2FeCl2 + H2

(mol) 0,03 0,03

S mol Fe phn ng vi mui:

0,05 - 0,03 = 0,02 (mol)

Ta c phn ng sau (c th xy ra):

Al + 3AgNO3 3Ag + Al(NO3)2

Al + 3Ag+ 3Ag + Al3+.

2Al + 3Cu(NO3)2 2Al(NO3)2 + Cu

2Al + 3Cu2+ 2Al3+ + 3Cu

Fe + 2AgNO3 Fe(NO3)2 + 2Ag


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Fe + 2Ag+ Fe2+ + 2Ag

Fe + Cu(NO3)2 Fe(NO3)2 + Cu

Fe + Cu2+ Fe2+ + Cu

Ta c s trao i electron nh sau:

Al Al3+ + 3e

0,03 0,09 (mol)

Fe Fe2+ + 2e

0,02 0,04 (mol)

Ag+ + 1e Ag

x x x (mol)

Cu2+ + 2e Cu

y 2y y (mol)

Tng s electron nhng = Tng s electron nhn

x + 2y = 0,09 + 0,04 = 0,13 (1)

108x + 64y + 56 . 0,03 = 8,12 (2)

Gii h phng trnh (1) v (2), ta c x = 0,03; y = 0,05.

Vy: CM AgNO3 = 0,03 / 0,2 = 0,15M

CM Cu(NO3)2 = 0,05/0,2 = 0,25M.

V d 4: Cho 9,16 gam bt A gm Zn, Fe, Cu vo cc ng 170ml dung dchCuSO4 1M. Sau phn ng thu c dung dch B v kt ta D, nung D trongkhng kh nhit cao n khi lng khng i c 12 gam cht rn. Thmdung dch NaOH vo mt na dung dch B, lc kt ta, ra v nung trong khngkh n khi lng khng i thu c 5,2 gam cht rn E. Cc phn ng xy ra

hon ton.

Tnh khi lng mi kim loi trong hn hp ban u.

Hng dn gii

- Theo u bi cc phn ng xy ra hon ton nn Zn phn ng trc, sau nFe v ion Cu2+ c th ht hoc cn d.


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nCuSO4 = 0,17 . 1 = 0,17 (mol)

- Gi s 9,16 gam A hon ton l Fe (khi lng nguyn t nh nht) th nhn hp= 9,16/56 = 0,164 (mol). V vy nA < nCuSO4. Do phn ng CuSO4 cn d,hn hp kim loi ht.

- Phng trnh phn ng:

Zn + CuSO4 ZnSO4 + Cu

x x x x (mol)

Fe + CuSO4 FeSO4 + Cu

y y y y (mol)

Gi nCu ban u l z mol

Ta c: 65x + 56y + 64z = 9,16 (1)

Cht rn D l Cu: Cu + 1/2O2 = CuO

(x + y + z) = 12/80 = 0,15(mol) (2)

- Khi cho 1/2 dung dch B + NaOH s xy ra cc phn ng:

Ta c phng trnh:

(0,25y . 160) + 0,5(0,17 - x - y) . 80 = 5,2 (3)

Gii h phng trnh (1), (2), (3) ta c:

x = 0,04 (mol) Zn; y = 0,06 (mol) Fe v z = 0,05 mol Cu

T tnh c khi lng ca tng kim loi

Tuyt chiu s 8 (bo ton in tch)Th su, 22 Thng 5 2009 03:52 Ti quang dung


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Bo Ton in Tch

I. C S Ca Phng Php

1. C s: Nguyn t, phn t, dung dch lun trung ha v in- Trong nguyn t: s proton = s electron

- Trong dung dch:

tngs mol x in tch ion = | tng s mol x in tch ion m |

2. p dng v mt s ch a) khi lng mui (trong dung dch) = tng khi lng cc ion m b) Qu trnh p dng nh lut bo ton in tch thng kt hp: - Cc phng php bo ton khc: Bo ton khi lng, bo ton nguyn

t - Vit phng trnh ha hc ng ion thu gn

II. CC DANG BI TP THNG GP

Dng 1: p dng n thun nh lut bo ton in tch

V D 1: Mt dung dch c cha 4 ion vi thnh phn : 0,01 mol Na+, 0,02

mol Mg2+ , 0,015 mol SO42- ,

x mol Cl- . Gi tr ca x l:

A. 0,015. C. 0,02.

B. 0,035. D. 0,01.

Hng dn:

p dng nh lut bo ton in tch ta c:

0,01x1 + 0,02x2 = 0,015x2 + Xx1 x = 0,02 p n


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Dng 2: Kt hp vi nh lut bo ton khi lng

V D 2: Dung dch A cha hai cation l Fe2+: 0,1 mol v Al3+ : 0,2 mol v

hai anion l Cl-: x mol v SO42- : y mol. bn c cn dung dch A thu c 46,9

gam hn hp mui khan.

Gi tr ca x v y ln lt l:

A. 0,6 v 0,1 C. 0,5 v 0,15

B. 0,3 v 0,2 D. 0,2 v 0,3

Hng dn:

- p dng nh lut bo ton in tch ta c:

0,1x2 + 0,2x3 = Xx1 + y x 2 X + 2y = 0,8 (*)

- Khi c cn dung dch, khi lng mui = tng khi lng cc ion to mui

0,1x56 + 0,2x27 + Xx35,5 + Yx 96 = 46,9

35,5X + 96Y = 35,9 (**)

T (*) v (**) X = 0,2 ; Y = 0,3 p n D

V d 3: Chia hn hp X gm 2 kim loi c ha tr khng i thnh 2 phn bngnhau.

Phn 1: Ha tan haonf ton bng dung dch HCl d thu c 1,792 lt H2 (ktc).

Phn 2: Nung trong khng kh d, thu c 2,84 gam hn hp rn ch gm ccoxit.

Khi lng hn hp X l:A. 1,56 gam. C. 2,4 gam.

B. 1,8 gam. D. 3,12 gam.

Hng dn:


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Nhn xt: Tng s mol x in tch ion dng (ca 2 kim loi) trong 2 phn l

Bng nhau Tng s mol x in tch ion m trong 2 phn cng bng nhau.

O2 2 Cl-

Mt khc: nCl- = nH+ = 2nH2 = 1,792/ 22,4 = 0,08 (mol)

Suy ra: nO (trong oxit) = 0,04 (mol)

Suy ra: Trong mt phn: mKim Loi - m oxi = 2,84 - 0,08.16 = 1,56 gam

Khi lng hn hp X = 2.1,56 = 3,12 gam

p n D

Dng 3: Kt hp vi bo ton nguyn t

V D 4: Cho hn hp X gm x mol FeS2 v 0,045 mol Cu2S tc dng va vi

HNO3 long, un nng thu c dung dch ch cha mui sunfat ca cc kim loi

V gii phng kh NO duy nht, Gi tr ca x l

A. 0,045. B. 0,09.

C. 0,135. D. 0,18.

Hng dn:

- p dng bo ton nguyn t:

Fe3+ : x mol ; Cu2+ : 0,09 ; SO42- : ( x + 0,045) mol

- p dng nh lut bo ton in tch (trong dung dch ch cha mui sunfat)

Ta c : 3x + 2.0,09 = 2(x + 0,045)

x = 0,09

p n B

V D 5: Dng dch X c cha 5 ion : Mg2+ , Ba2+ , Ca2+ , 0,1 mol Cl- v

0,2 mol NO3-. Thm dn V lt dung dch K2CO3 1M vo X n khi c lng

Kt ta ln nht thig gi tr ti thiu cn dng l:A. 150ml. B. 300 ml.


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C. 200ml. D. 250ml.

Hng dn:

C th qui i cc ion Mg2+, Ba2+, Ca2+ thnh M2+

M2+ + CO32- MCO3

Khi phn ng kt thc, phn dung dch cha K+, Cl-, v NO3-

p dng nh lut bo ton ton in tch ta c:

nk+ = nCl- + nNO3- = 0,3 (mol) suy ra: s mol K2CO3 = 0,15 (mol)

suy ra th tch K2CO3 = 0,15/1 = 0,15 (lt) = 150ml

p n A

Dng 4: Kt hp vi vic vit phng trnh dng ion thu gn

V D 6: Cho ha tan hon ton 15,6 gam hn hp gm Al v

Al2O3 trong 500 dung dch NaOH 1M thu c 6,72 lt H2 (ktc)

V dung dch X. Th tch HCl 2M ti thiu cn cho vo X thu c

lng kt ta ln nht l:

A. 0,175 lt. B. 0,25 lt.

C. 0,25 lt. D. 0,52 lt.

Hng dn :

Dung dch X cha cc ion Na+ ; AlO2- ; OH- d (c th). p dng nh lut

Bo ton in tch:

n AlO2- + n OH- = n Na+ = 0,5

Khi cho HCl vaof dung dch X:

H+ + OH H2O (1)

H+ + AlO2- + H2O Al(OH)3 (2)

3H+ + Al(OH)3 Al3+ + 3H2O (3)


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kt ta l ln nht, suy ra khng xy ra (3) v

n H+ = n AlO2- + n OH- = 0,5

Suy ra th tch HCl = 0,5/2 = 0,25 (lt)

p n B

Dng 5 : Bi ton tng hp

V d 7: Hon ton 10g hn hp X gm Mg v Fe bng dung dch HCl 2M.

Kt thc th nghim thu c dung dch Y v 5,6l kh H2 (ktc). kt ta

hon ton cc cation c trong Y cn va 300ml NaOH 2M.Th tch dung

dch HCl dng l:

A: 0,2 lt B: 0,24 lt

C: 0,3 lt D: 0,4 lt

Hng dn:

nNa+ = nOH- = nNaOH = 0,6M

Khi cho NaOH vo dung dch Y(cha cc ion :Mg2+;Fe2+;H+ d;Cl-) cc ion

dng s tc dng vi OH- to thnh kt ta .Nh vy dung dch thu

c sau phn ng ch cha Na+ v Cl-.

=>nCl- = nNa+=0,6 =>VHCl=0,6/2= 0,3 lt ==> p n C.

V d 8: ha tan hon ton 20 gam hn hp X gm Fe,FeO,Fe3O4,

Fe2O3 cn va 700ml dung dch HCl 1M thu c dung dch X v

3,36 lt kh H2 (ktc). Cho NaOH d vo dung dch X ri ly ton b

kt ta thu c bn nung trong khng kh n khi lng khng i

th lng cht rn thu c l :

A: 8 gam B: 16 gam


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C: 24 gam D:32 gam

Hng dn:

Vi cch gii thng thng ,ta vit 7 phng trnh ha hc,sau

t n s,thit lp h phng trnh v gii

Nu p dng nh lut bo ton din tch ta c :

Fe + 2HCl FeCl2 + H2

S mol HCl ha tan Fe l : nHCl = 2nH2 =0,3 mol

S mol HCl ha tan cc oxit =0,7- 0,3 = 0,4 mol

Theo nh lut bo ton din tch ta c

nO2-(oxit) =1/2 nCl- = 0,2 mol ==>

nFe (trong X) =moxit - moxi /56 =(20-0,2 x 16)/56 = 0,3 mol

C th coi : 2Fe (trong X ) Fe2O3

nFe2O3 =1,5 mol ==> mFe2O3 = 24 gam ==> p n C

III . BI TP T LUYN

Cu 1: Dung dch X c cha a mol Na+ ,b mol Mg2+ ,C mol Cl- v

d mol SO42-.. Biu thc lin h gia a,b,c,d l

A: a+2b=c+2d B:a+2b=c+d

C:a+b=c+ D : 2a+b=2c+d

Cu 2:C 2 dung dch,mi dung dch u cha 2 cation v 2 anion khng

trng nhau trong cc ion sau


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K+ :0,15 mol, Mg2+ : 0,1 mol,NH4+:0,25 mol,H+ :0,2 mol, Cl- :0,1 mol SO42- :0,075

mol NO3- :0,25 mol,NO3- :0,25 mol v CO32- :0,15 mol. Mt trong 2 dung dch trncha

A: K+,Mg2+,SO42- v Cl-; B : K+,NH4+,CO32- v Cl-

C :NH4+,H+,NO3-, v SO42- D : Mg2+,H+,SO42- v Cl-

Cu 3: Dung dch Y cha Ca2+ 0,1 mol ,Mg2+ 0,3 mol,Cl- 0,4 mol,HCO3- y mol.

Khi c cn dung dch Y th c mui khan thu c l :

A: 37,4 gam B 49,8 gam

C: 25,4 gam D : 30,5 gam

Cu 4 : Mt dung dch cha 0,02 mol Cu2+;0,03 mol K+,x mol Cl- v y mol

SO42-.Tng khi lng cc mui tan c trong dung dch l 5,435 gam. Gi tr

ca x v y ln lt l:

A:0,03 v 0,02 B: 0,05 v 0,01

C : 0,01 v 0,03 D:0,02 v 0,05

Cu 5: Ha tan hon ton hn hp gm 0,12 mol FeS2 v x mol Cu2S vo

dung dch HNO3 va , thu c dung dch X ch cha 2 mui sunfat ca

cc kim loi v gii kh NO duy nht. Gi tr l :

A :0,03 B :0,045

C:0,06 D:0,09

Cu 6: Cho m gam hn hp Cu,Zn,Mg tc dng hon ton vi dung dch

HNO3 long,d. C cn dung dch sau phn ng thu c (m+62). Gam

mui khan. Nung hn hp mui khan trn n khi lng khng i thu


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c cht rn c khi lng l:

A: (m+4) gam B: (m+8) gam

C: (m+16) gam D: (m+32)gam

Cu 7:Cho 2,24 gam hn hp Na2CO3,K2CO3 tc dng va vi dung

dch BaCl2.Sau phn ng thu c 39,4 gam kt ta.Lc tch kt ta,c

cn dung dch th thu c bao nhiu gam mui clorua khan ?

A: 2,66 gam B 22,6 gam

C: 26,6 gam D : 6,26 gam

Cu 8: Trn dung dch cha Ba2+;OH- 0,06 mol v Na2+ 0,02 mol vi

dung dch cha HCO3- 0,04 mol; CO32- 0,03 mol va Na+. Khi lng kt

ta thu c sau khi trn l

A: 3,94 gam B 5,91 gam

C: 7,88 gam D : 1,71 gam

Cu 9:Ha tan hon ton 5,94 gam hn hp hai mui clorua ca 2 kim

loi nhm IIA vo nc c 100ml dung dch X. lm kt ta ht

ion Cl- c trong dung dch X,ngi ta cho dung dch X trn tc

dng va vi dung dch AgNO3. Kt thc th nghim, thu c

dung dch Y v 17,22 gam kt ta. Khi lng mui khan thu c khi

kt ta dung dch Y l:

A: 4,86 gam B: 5,4 gam

C: 7,53 gam D : 9,12 gam


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Cu 10: Dung dch X cha 0.025 mol CO32-;0,1 mol Na+;0,25 mol

NH4+ v 0,3 mol Cl-. Cho 270ml dung dch Ba(OH)2 0,2M vo v

un nng nh (gi s H2O bay hi khng ng k). Tng khi lng

dung dch X v dung dch Ba(OH)2 sau qu trnh phn ng gim i l :

A: 4,125 gam B: 5,296 gam

C: 6,761 gam D : 7,015 gam

Cu 11: Trn 100ml dung dch AlCl3 1M vi 200ml dung dch

NaOH1,8M n phn ng hon ton th lng kt ta thu c l :

A: 3,12 gam B: 6,24 gam

C: 1,06 gam D : 2,08 gam

Cu 12: Dung dch B cha ba ion K+;Na+;PO43-. 1 lt dung dch

B tc dng vi CaCl2 d thu c 31 gam kt ta. Mt khc nu c

cn 1 lt dung dch B thu c 37,6 gam cht rn khan. Nng

ca 3 ion K+;Na+;PO43- ln lt l:

A:0,3M;0,3M v 0,6M B: 0,1M;0,1M v 0,2M

C: 0,3M;0,3M v 0,2M D : 0,3M;0,2M v 0,2M

Cu 13: Cho dung dch Ba(OH)2 n d vo 100ml dung dch X

gm cc ion: NH4+, SO42-,NO3-, ri tin hnh un nng th c 23,3

gam kt ta v 6,72 lt(ktc) mt cht duy nht. Nng mol ca

(NH4)2SO4 v NH4NO3 trong dung dch X ln lt l :

A: 1M v 1M B: 2M v 2M


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C: 1M v 2M D : 2M v 1M

Cu 14:Dung dch X cha cc ion : Fe3+,SO42-,NH4+,Cl-. Chia

dung dch X thnh 2 phn bng nhau:

-Phn 1 tc dng vi lng d dung dch NaOH,un nng thu c

0.672 lt kh (ktc) v 1,07 gam kt ta

-Phn 2 tc dng vi lng d dung dch BaCl2, thu c 4,66gam kt ta

Tng khi lng cc mui khan thu c khi c cn dung dch X l

(qu trnh c cn ch c nc bayhi)

A:3,73 gam B: 7,04 gam

C: 7,46 gam D : 3,52 gam

Tuyt chiu s 9 (p dng phng trnh ion - electron)Th ba, 26 Thng 5 2009 17:19 Ti quang dung

gii tt cc bi ton bng vic p dng phng php ion, iu u tin ccbn phi nm chc phng trnh phn ng di dng phn t t suy raphng trnh ion. i khi c mt s bi tp khng th gii theo cc phng trnhphn t c m phi gii da theo phng trnh ion. Vic gii bi ton ho hcbng cch p dng phng php ion gip chng ta hiu k hn v bn cht cacc phng trnh ho hc. T mt phng trnh ion c th ng vi rt nhiuphng trnh phn t. V d phn ng gia hn hp dung dch axit vi dungdch baz u c chung mt phng trnh ion l:

H+ + OH- H2O

hoc phn ng ca Cu kim loi vi hn hp dung dch HNO3 v dung dch H2SO4l:

3Cu + 8H+ + 2NO3- 3Cu2+ + 2NO + 4H2O ...

Sau y l mt s v d:

V d 1: Hn hp X gm (Fe, Fe2O3, Fe3O4, FeO) vi s mol mi cht l 0,1 mol,ho tan ht vo dung dch Y gm (HCl v H2SO4 long) d thu c dung dch


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Z. Nh t t dung dch Cu(NO3)2 1M vo dung dch Z cho ti khi ngng thot khNO. Th tch dung dch Cu(NO3)2 cn dng v th tch kh thot ra ( ktc) l:

A. 25 ml; 1,12 lt. B. 500ml; 22,4 lt.

C. 50ml; 2,24 lt. D. 50ml; 1,12 lt.

Hng dn gii

Quy hn hp 0,1 mol Fe2O3 v 0,1 mol FeO thnh 0,1 mol Fe3O4.

Hn hp X gm: (Fe3O4: 0,2 mol; Fe: 0,1 mol) tc dng vi dung dch Y.

Fe3O4 + 8H+ Fe2+ + 2Fe3+ + 4H2O.

0,2 0,2 0,4 mol

Fe + 2H

+

Fe

2+

+ H2

0,1 0,1 mol

Dung dch Z: (Fe2+: 0,3 mol; Fe3+: 0,4 mol) + Cu(NO3)2:

3Fe2+ + NO3- + 4H+ 3Fe3+ + NO + 2H2O

0,3 0,1 0,1

VNO = 0,1 . 22,4 = 2,24 (lt)

n Cu(NO3)2 = 1/2 n NO3-

= 0,05 (mol) V dd Cu(NO3)2 = 0,05 / 1 = 0,05 (lt) (hay 50ml)

p n C.

V d 2: Ho tan 0,1 mol Cu kim loi trong 120ml dung dch X gm HNO3 1M vH2SO4 0,5M. Sau khi phn ng kt thc thu c V lt kh NO duy nht (ktc).

A. 1,344 lt. B. 1,49 lt.

C. 0,672 lt. D. 1,12 lt.

Hng dn gii

n HNO3 = 0,12 mol; n H2SO4 = 0,06 mol

Tng n H+ = 0,24 mol v n NO3- = 0,12 mol.

Phng trnh ion:


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3Cu + 8H+ + 2NO3- 3Cu2+ + 2NO + 4H2O.

Ban u: 0,1 0,24 0,12 mol

Phn ng: 0,09 0,24 0,06 0,06 mol

Sau phn ng: 0,01 (d) (ht) 0,06 (d)

VNO = 0,06 . 22,4 = 1,344 (lt)

p n A.

V d 3: Dung dch X cha dung dch NaOH 0,2M v dung dch Ca(OH) 2 0,1M.Sc 7,84 lt kh CO2 (ktc) vo 1 lt dung dch X th khi lng kt ta thu cl:

A. 15 gam. B. 5 gam.

C. 10 gam. D. 0 gam.

Hng dn gii

n CO2 = 0,35 mol; nNaOH = 0,2 mol;

n Ca(OH)2 = 0,1 mol.

Tng: n OH- = 0,2 + 0,1 . 2 = 0,4 (mol)

v n Ca2+ = 0,1 mol.

Phng trnh ion rt gn:


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p n B.

V d 4: Ho tan ht hn hp gm mt kim loi kim v mt kim loi kim thtrong nc c dung dch A v c 1,12 lt H2 bay ra ( ktc). Cho dung dchcha 0,03 mol AlCl3 vo dung dch A. Khi lng kt ta thu c l:

A. 0,78 gam. B. 1,56 gam.

C. 0,81 gam. D. 2,34 gam.

Hng dn gii

Phn ng ca kim loi kim v kim loi kim th vi H2O:

M + nH2O M(OH)n + n/2H2

T phng trnh ta c:

n OH- = 2n H2 = 0,1 (mol)

Dung dch A tc dng vi 0,03 mol dung dch AlCl3:


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Tip tc ho tan kt ta theo phng trnh:

Al(OH)3 + OH- AlO2- + 2H2O

0,1 0,01 mol

Vy: m Al(OH)3 = 78.0,02 = 1,56 (gam)

p n B.

V d 5: Dung dch A cha 0,01 mol Fe(NO3)3 v 0,15 mol HCl c kh nng hotan ti a bao nhiu gam Cu kim loi? (Bit NO l sn phm kh duy nht)

A. 2,88 gam. B. 3,92 gam.

C. 3,2 gam. D. 5,12 gam.

Hng dn gii

Phng trnh ion:

p n C

V d 6: Cho hn hp gm NaCl v NaBr tc dng vi dung dch AgNO3 d thuc kt ta c khi lng ng bng khi lng AgNO3 phn ng. Phntrm khi lng NaCl trong hn hp u l:

A. 23,3%. B. 27,84%.

C. 43,23%. D. 31,3%.


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Hng dn gii

Phng trnh ion:

p n B.

V d 7: Trn 100ml dung dch A (gm KHCO3 1M v K2CO3 1M) vo 100mldung dch B (gm NaHCO3 1M v Na2CO3 1M) thu c dung dch C. Nh t t100ml dung dch D (gm H2SO4 1M v HCl 1M) vo dung dch C thu c V lt

CO2 (ktc) v dung dch E. Cho dung dch Ba(OH)2 ti d vo dung dch E th thuc m gam kt ta. Gi tr ca m v V ln lt l:

A. 82,4 gam v 2,24 lt. B. 4,3 gam v 1,12 lt.

C. 2,33 gam v 2,24 lt. D. 3,4 gam v 5,6 lt.

Hng dn gii

Dung dch C cha: HCO3- : 0,2mol; CO32- = 0,2 mol

Dung dch D c tng: n H+ = 0,3 mol.

Nh t t dung dch C v dung dch D:


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Tng khi lng kt ta:

m = 0,3 . 197 + 0,1 . 233 = 82,4 (gam)

p n A.

V d 8: Ho tan hon ton 7,74 gam mt hn hp gm Mg, Al bng 500ml dungdch gm H2SO4 0,28M v HCl 1M thu c 8,736 lt H2 (ktc) v dung dch X.

Thm V lt dung dch cha ng thi NaOH 1M v Ba(OH)2 0,5M vo dung dchX thu c lng kt ta ln nht.

a. S gam mui thu c trong dung dch X l:

A. 38,93 gam. B. 38,95 gam.

C. 38,97 gam. D. 38,91 gam.

b. Th tch V l:

A. 0,39 lt. B. 0,4 lt.

C. 0,41 lt. D. 0,42 lt.

c. Khi lng kt ta l:


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A. 54,02 gam. B. 53,98 gam.

C. 53,62 gam. D. 53,94 gam.

Hng dn giia. Xc nh lng mui thu c trong dung dch X:


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lng kt ta t gi tr ln nht th s lng OH - phi kt ta ht ccion Mg2+ v Al3+. Theo cc phng trnh phn ng (1), (2), (4), (5) ta c:

n H+ = n OH- = 0,78 mol

2V = 0,78 V = 0,39 lt.

p n A.

c. Xc nh khi lng kt ta:

p n C.

V d 9: Cho m gam hn hp Mg v Al vo 250ml dung dch X cha hn hpaxit HCl 1M v axit H2SO4 0,5M thu c 5,32 lt H2 (ktc) v dung dch Y (coith tch dung dch khng i). Dung dch Y c pH l:

A. 7. B. 1. C.

2. D. 6.

Hng dn gii

Khi cho Mg, Al tc dng vi hn hp 2 axit HCl v H2SO4 ta c s phn ng:


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p n B.

V d 10:Thc hin 2 th nghim:

- TN1: Cho 3,84g Cu phn ng vi 80ml dung dch HNO3 1M thot ra V1 lt NO.

- TN2: Cho 3,84g Cu phn ng vi 80ml dung dch HNO3 1M v H2SO4 0,5Mthot ra V2 lt NO. Bit NO l sn phm kh duy nht, cc th tch kh o cngiu kin. Quan h gia V1 v V2 l:

A. V2 = V1. B. V2 = 2 V1. C. V2 =2,5V1. D. V2 = 1,5V1.

Hng dn gii

TN1:

V2 tng ng vi 0,04 mol NO.

Nh vy V2 = 2V1


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p n B.

V d 11: Trn 100ml dung dch gm Ba(OH)2 0,1M v NaOH 0,1M vi 400mldung dch gm H2SO4 0,0375M v HCl 0,0125M thu c dung dch X. Gi tr pHca dung dch X l:

A. 1. B. 2. C.6. D. 7.

Hng dn gii

Ta c

Khi trn hn hp dung dch baz vi hn hp dung dch axit ta c phng trnhion rt gn:

p n B.

V d 12: Cho 1 mu hp kim Na - Ba tc dng vi nc d thu c dung dchX v 3,36 lt H2 (ktc). Th tch dung dch axit H2SO4 2M cn dng trung hodung dch X l:

A. 150ml. B. 75ml. C. 60ml. D. 30ml.

Hng dn gii


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Na + H2O NaOH + 1/2 H2

Ba + 2H2O Ba(OH)2 + H2

p n B.

Tuyt chiu s 10 (s dng cng thc kinh nghim)Th nm, 28 Thng 5 2009 10:06 Ti quang dung

PHNG PHP S DNG CNG THC KINH NGHIM

I. PHNG PHP GII

1. Ni dung phng php

* Xt bi ton tng qut quen thuc:

+ O2 +HNO3(H2SO4 c, nng)

m gam m1gam (n: max)


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Gi:

S mol kim loi a

S oxi ho cao nht (max) ca kim loi l n

S mol electron nhn (2) l t mol

Ta c:

Mt khc:

ne nhn = n e (oxi) + ne(2)

+ ng vi M l Fe (56), n = 3 ta c: m = 0.7.m1 + 5,6.t (2)

+ ng vi M l Cu (64), n = 2 ta c: m = 0.8.m1 + 6,4.t (3)

T (2,3) ta thy:

+ Bi ton c 3 i lng: m, m1 v ne nhn (2) (hoc V kh (2)).


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Khi bit 2 trong 3 i lng trn ta tnh c ngay i lng cn li.

+ giai on (2) bi c th cho s mol, th tch hoc khi lng ca 1

kh hay nhiu kh; giai on (1) c th cho s lng cht rn c th l

cc oxit hoc hn hp gm kim loi d v cc oxit.

2. Phm vi p dng v mt s ch

+ Ch dng kh HNO3 (hoc (H2SO4 c, nng) ly d hoc va .

+ Cng thc kinh nghim trn ch p dng vi 2 kim loi Fe v Cu.

3. Cc bc gii

+ Tm tng s mol electron nhn giai on kh N+5hoc S+6

+ Tm tng khi lng hn hp rn (kim loi v oxit kim loi): m1

+ p dng cng thc (2) hoc (3)

II. TH D MINH HO

Th d 1. t chy hon ton 5,6 gam bt Fe trong bnh O2 thu c

7,36 gam hn hp X gm Fe2O3, Fe3O4 v mt phn Fe cn d.

Ho tan hon ton lng hn hp X trn vo dung dch HNO3 thu

c V lt hn hp kh Y gm NO2 v NO c t khi so vi H2 bng 19.

Gi tr ca V l:

A. 0,896. B. 0,672. C. 1,792. D.0,448.

Hng dn gii:

p dng cng thc (1): ne nhn (2) => ne nhn (2) = 0,08

T dy/H2 =19 => nNO2 = nNO = x


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Vy: V = 22,4.0,02.2 = 0,896 lt -> p n A.

Th d 2. m gam bt Fe trong khng kh mt thi gian thu c

11,28 gam hn hp X gm 4 cht. Ho tan ht X trong lng d

dung dch HNO3 thu c 672ml kh NO (sn phm kh duy nht,

ktc). Gi tr ca m l:

A. 5,6. B.11.2. C.7,0 D. 8.4.

Hng dn gii:

p dng cng thc (2):

-> p n D.

Th d 3. Cho 11,36 gam hn hp X gm Fe, FeO, Fe2O3 v Fe3O4

phn ng ht vi dung dch HNO3 long, d thu c 1,344 lt kh NO

(sn phm kh duy nht, o ktc) v dung dch Y. C cn dung dch

Y thu c m gam mui khan.Gi tr ca m l:

A. 49,09 B. 35,50 C. 38,72. D.34,36.

Hng dn gii:

p dng cng thc (2):

=> m = 38,72 gam

p n C.

Th d 4. Cho 11,6 gam hn hp X gm Fe, FeO, Fe2O3 vo dung


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A. 5,6 gam.

B. 20,08 gam.

C. 11,84 gam.

D. 14,95 gam.

Cu 2: Ha tan hon ton 10 gam hn hp X (Fe, Fe2O3) trong dung

dch HNO3 va thu c 1,12 lt NO ( ktc, sn phm kh duy

nht) v dung dch Y. Cho Y tc dng vi dung dch NaOH d c

kt ta Z. Nung Z trong khng kh n khi khi lng khng i c

m gam cht rn. Gi tr ca m l:

A. 12 gam.

B. 16 gam.

C. 11,2 gam.

D. 19,2 gam.

Cu 3: Ha tan ht m gam hn hp Fe, Fe2O3, Fe3O4 trong dung

dch HNO3 c,nng d c 448 ml kh NO2 ( ktc). C cn

dung dch sau phn ng c 14,52 gam mui khan. Gi tr ca m l:

A. 3,36 gam.

B. 4,28 gam.

C. 4,64 gam.

D. 4,80 gam.

Cu 4: t chy hon ton 5,6 gam bt Fe trong bnh oxi thu c 7,36

gam hn hp X gm Fe2O3, Fe3O4 v mt phn Fe d. Ha tan hon

ton hn hp X bng dung dch HNO3 thu c V lt kh Y gm NO2 v

NO c t khi so vi H2 bng 19.

Gi tr ca V l:


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A. 0,896 lt.

B. 0,672 lt.

C. 0,448 lt.

D. 1,08 lt.

Cu 5: Cho lung kh CO i qua ng s ng m gam Fe2O3 nung nng.

Sau mt thi gian thu c 13,92 gam hn hp X gm 4 cht. Ha tan

ht X bng HNO3 c, nng d c 5,824 lt NO2 (sn phm kh duy

nht, ktc). Gi tr ca m l:

A. 16 gam.

B. 32 gam.

C. 48 gam.

D. 64 gam.

Cu 6: Cho 11,6 gam hn hp X gm Fe, FeO, Fe2O3 vo dung dch

HNO3 long, d c V lt kh Y gm NO v NO2 c t khi hi so vi

H2 l 19. Mt khc, nu cho cng lng kh hn hp X trn tc dng vi

kh CO d th sau phn ng hon ton c 9,52 gam Fe.

Gi tr ca V l:

A. 2,8 lt.

B. 5,6 lt.

C. 1,4 lt.

D. 1,344 lt.

Cu 7: Nung m gam bt ng kim loi trong oxi thu c 24,8 gam hn

hp rn X gm Cu, CuO v Cu2O. Ha tan hon ton X trong H2SO4

c nng thot ra 4,48 lt kh SO2 (sn phm kh duy nht, ktc).

Gi tr ca m l:


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A. 9,6 gam.

B. 14,72 gam.

C. 21,12 gam.

D. 22,4 gam.

Cu 8: Ha tan hon ton 18,16 gam hn hp X gm Fe v Fe3O4

trong 2 lt dung dch HNO3 2M thu c dung dch Y v 4,704 lt kh

NO (sn phm kh duy nht, ktc).

Phn trm khi lng Fe trong hn hp X l:

A. 38,23%.

B. 61,67%.

C. 64,67%.

D. 35,24%.

Cu 9: Cho m gam hn hp X gm Fe, Fe3O4 tc dng vi 200 ml dung

dch HNO3 3,2M. Sau khi phn ng hon ton c 0,1 mol kh NO (sn

phm kh duy nht) v cn li 1,46 gam kim loi khng tan.

Gi tr ca m l:

A. 17,04 gam.

B. 19,20 gam.

C. 18,50 gam.

D. 20,50 gam.

Cu 10: m gam Fe trong khng kh 1 thi gian c 7,52 gam hn

hp X gm 4 cht. Ha tan ht X trong dung dch H2SO4 c, nng d

c 0,672 lt kh SO2 (sn phm kh duy nht, ktc) v dung dch Y.

C cn cn thn dung dch Y c m1 gam mui khan.

Gi tr ca m v m1 ln lt l:


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B. 32 gam.

C. 40 gam.

D. 48 gam.

Cu 14: Ha tan 11,2 gam kim loi M trong dung dch HCl d thu c

4,48 lt H2 ( ktc). Cn nu ha tan hn hp X gm 11,2 gam kim loi M

v 69,6 gam oxit MxOy trong lng d dung dch HNO3 th c 6,72 lt

kh NO (sn phm kh duy nht, ktc). Cng thc ca oxit kim loi l:

A. Fe3O4.

B. FeO

C. Cr2O3.

D. CrO.

Cu 15: Cho 37 gam hn hp X gm Fe, Fe3O4 tc dng vi 640 ml

dung dch HNO3 2M long, ung nng. Sau khi cc phn ng xy ra

hon ton thu c V lt kh NO (sn phm kh duy nht, ktc),

dung dch Y v cn li 2,92 gam kim loi. Gi tr ca V l:

A. 2,24 lt.

B. 4,48 lt.

C. 3,36 lt.

D. 6,72 lt.

Cu 16: Cho lung kh CO i qua ng s cha 0,12 mol hn hp

gm FeO v Fe2O3 nung nng, phn ng to ra 0,138 mol CO2.

Hn hp cht rn cn li trong ng nng 14,325 gam gm 4 cht.

Ha tan ht hn hp 4 cht ny vo dung dch HNO3 d thu c

V lt kh NO (sn phm kh duy nht, ktc).

Gi tr ca V l:


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b lng kh NO thot ra bn trn vi lng O2 va hn hp

hp th hon ton trong nc c dung dch HNO3. Bit th tch Oxi

tham gia vo qu trnh trn l 336 ml ( ktc). Gi tr ca m l:

A. 34,8 gam.

B. 13,92 gam.

C. 23,2 gam.

D. 20,88 gam.

Cu 20: Thi t t V lt hn hp kh CO v H2 c t khi hi so vi H2

l 7,5 qua mt ng s ng 16,8 gam hn hp 3 oxit CuO, Fe3O4, Al2O3

nung nng. Sauk hi phn ng thu c hn hp kh v hi c t khi so

vi H2 l 15,5; dn hn hp kh ny vo dung dch Ca(OH)2 d thy c

5 gam kt ta. Th tch V ( ktc) v khi lng cht rn cn li trong

ng s ln lt l:

A. 0,448 lt; 16,48 gam.

B. 1,12 lt; 16 gam.

C. 1,568 lt; 15,68 gam.

D. 2,24 lt; 15,2 gam.

p n:

1B - 2C - 3C - 4A - 5A - 6A - 7D - 8B - 9C - 10D

11A - 12C - 13C - 14A - 15B - 16C - 17B - 18C - 19B - 20D.

Tuyt chiu s 11 (Phng Php ng cho)Th hai, 01 Thng 6 2009 06:54 Ti quang dung
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I. PHNG PHP GII

1. Ni dung phng php: Trn ln 2 dung dch

Khi lng Th tich Nng

(C% hoc CM)

Dung dch 1 m1 V1 C1

Dung dch 2 m2 V2 C2

Dung dch

Cn pha ch

m = m1+m2 V = V1+V2 C

S ng cho ng vi mi trng hp:

a. i vi nng % v khi lng:

b. i vi nng mol:

2. Cc dng ton thng gp


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Dng 1. Pha ch dung dch

Pha dung dch vi dung dch: xc nh C1, C2, C v p dng cc cng thc

(1) v (2). Pha ch dung dch vi dung mi (H2O): dung mi nguyn cht c C = 0%. Pha ch cht rn c tng tc vi H2O to cht tan vo dung dch: lc

ny, do c s tng tc vi H2O to cht tan nn ta phi chuyn cht rnsang dung dch c nng tng ng C > 100%.

Pha ch tinh th mui ngm nc vo dung dch: tinh th c coi nhdung dch c

C < 100%, y gi tr ca C chnh l hm lng % ca cht tan trong tinh thmui ngm ngc.

Ch :

- Khi lng ring ca H2O l 1g/ml.

- Phng php ny khng p dng c khi trn ln 2 dung dch cxy ra phn ng gia cc cht tan vi nhau (tr phn ng vi H2O) nnkhng p dng c vi trng hp tnh ton pH.

Dng 2: Tnh t l mol cc cht trong hn hpi vi hn hp gm 2 cht, khi bit khi lng phn t cc cht v khi lngphn t trung bnh ca hn hp, ta d dng tnh c t l mol ca cc cht theocng thc s (2) v ngc li.

Ch :

- y cc gi tr ca C c thay bng cc gi tr KLPT tng ng.

- T phng php ng cho ta rt ra cng thc tnh nhTi thnh phn% s mol ca hn hp 2 cht c khi lng phn t M1, M2 v khi lng trung

bnh l:


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II. CC BC GII

- Xc nh tr s cn tm t bi

- Chuyn cc s liu sang dng i lng % khi lng

- Xy dng ng cho => Kt qu bi ton

III. CC TRNG HP P DNG V TH D MINH HO

Dng 1. Pha ch dung dch

Th d 1. thu c dung dch HNO3 20% cn ly a gam dung dch HNO3 40%pha vi b gam dung dch HNO3 15%. T l a/b l:

A. 1/4. B.1/3.

C.3/1. D.4/1.

Hng dn gii:

p dng cng thc (1): a / b = (15 - 20) / (40 - 20) = 1 /4 => p n A

Th d 2. Ho tan hon ton m gam Na2O nguyn cht vo 75,0 gam dung dchNaOH 12,0% thu c dung dch NaOH 58,8%. Gi tr ca m l

A. 66,0. B.50,0. C.112,5. D.85,2.

Phn ng ho tan: Na2O + H2O -> 2NaOH

62 gam 80 gam

Coi Na2O nguyn cht nh dung dch NaOH c nng C = (80 / 62)100 =

129,0%

Theo (1): m / 75 = ( | 12,0 - 58,8| ) / ( |129,0 - 58,8| ) = 46,8 / 70,2 = 50 gam

p n B


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A. vinylaxetilen. B. buten. C.ivinyl D.butan

Hng dn gii:

S ng cho:

=> X l CH2 = CH - CH=CH2

-> p n C.

Th d 6. t chy hon ton 12,0 lt hn hp hai hp cht hu c k tip nhautrong dy ng ng thu c 41,4 lt CO2. Thnh phn % th tch ca hp chtc khi lng phn t nh hn l (cc th tch kh o cng iu kin).

A.55,0%. B.51,7%. C.48,3%. D.45,0%.

Hng dn gii:

Dng 3. Bi ton hn hp cc cht c tnh cht ho hc tng t nhau.

Th d 7. Nung hn hp X gm CaCO3 v CaSO3 ti phn ng hon ton ccht rn Y c khi lng bng 50,4% khi lng ca X. Thnh phn % khilng ca CaCO3 trong X l:

A.60%. C.45,5%


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B.54,5%. D.40%.

Hng dn gii:

Dng 4. Bi ton trn ln hai cht rn

Th d 8. X l qung hbnatit cha 60% Fe2O3. Y l qung mTietit cha 69,6%Fe3O4. Trn a tn qung X vi b tn qung Y thu c qung Z, m t 1 tnqung Z c th iu ch c 0,5 tn gang cha 4% cacbon. T l a/b l:

A.5/2. B.4/3.

C.3/4. D.2/5.

Hng dn gii:

"Cht tan" y l Fe. % khi lng Fe trong cc qung ln lt l:

Trong qung X: C1 = 60(112/160) = 42%.

Trong qung Y: C2 = 69,6(168/1232) = 50,4%

Trong qung Z: C = (100 - 4) / 2 = 48%

Theo (1): a/b = ( | 50,4 - 48,0 | ) / ( | 42,0 - 48,0 | ) = 2/5

=> n n D

Th d 9. Nhit phn hon ton a gam hn hp X gm Al(OH)3 v Cu(OH)2 thuc hn hp cht rn Y c khi lng 0,731a gam. Thnh phn % v khilng ca Al(OH)3 trong X l.


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A. 47,5%. B.50,0% C.52,5% D.55,0%

Hng dn gii:

Ta xbn nh y l bi ton trn ln 2 "dung dch" vi "cht tan" tng ng lnlt l Al2O3 v CuO.

i vi Al(OH)3: 2Al(OH)3 => Al2O3 c C1 = (102 / 2.78)100 = 65,4%

i vi Cu(OH)2: Cu(OH)2 => CuO c C2 = (80 / 98)100 = 81,6%

Tng hm lng Al2O3 v CuO trong hn hp X:

C = (0,731a / a)100 = 73,1%

Theo (1): m Al(OH)3 / m Cu(OH)2 = ( | 81,6 - 73,1 | ) / ( | 65,4 - 73,1 | )

=> %m Al(OH)3 = (8,5.100) / ( 8,5 + 7,7 ) = 52,5% => p n C

IV. BI TP P DNG

Cu 1: thu c dung dch HCl 30% cn ly a gam dung dch HCl 55% phavi b gam dung dch HCl 15%. T l a/b l:

A. 2/5.

B. 3/5.

C. 5/3.

D. 5/2.

Cu 2: pha c 100 ml dung dch nc mui c nng mol 0,5M cn lyV ml dung dch NaCl 2,5M. Gi tr ca V l:

A. 80,0.

B. 75,0.

C. 25,0.

D. 20,0.


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Cu 3: Ha tan 10 gam SO3 vo m gam dung dch H2SO4 49,0% ta c dungdch H2SO4 78,4%. Gi tr ca m l:

A. 6,67.

B. 7,35.

C. 13,61.

D. 20,0.

Cu 4: thu c 100 gam dung dch FeCl3 30% cn ha tan a gam tinh thFeCl3.6H2O vo b gam dung dch FeCl3 10%. Gi tr ca b l:

A. 22,2.

B. 40,0.

C. 60,0.

D. 77,8.

Cu 5: Mt hn hp gm CO v CO2 iu kin tiu chun c t khi i vihidro l 18,2. Thnh phn % v th tch ca CO2 trong hn hp l:

A. 45,0%.

B. 47,5%.C. 52,5%.

D. 55,0%.

Cu 6: Cn tren 2 th tch metan vi 1 th tch hidrocacbon X thu c hnhp kh c t khi hi so vi hidro bng 15. X l:

A. C4H10.

B. C3H8.

C. C4H8.

D. C3H6.

Cu 7: Mt loti kh l cc (thnh phn chnh l CH4 v H2) c t khi so vi He l1,725. Th tch H2 c trong 200,0 ml kh l cc l:


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A. 20,7 ml.

B. 179,3 ml.

C. 70,0 ml.

D. 130,0 ml.

Cu 8: Thm 150 ml dung dch KOH 2M vo 120 ml dung dch H3PO4 1M. Khilng cc mui thu c trong dung dch l:

A. 9,57 gam K2HPO4; 8,84 gam KH2PO4.

B. 10,44 gam K2HPO4; 12,72 gam K3PO4.

C. 10,24 gam K2HPO4; 13,50 gam KH2PO4.

D. 13,05 gam K2HPO4; 10,60 gam K3PO4.Cu 9: Ha tan 2,84 gam hn hp 2 mui CaCO3 v MgCO3 bng dung dch HCld, thu c 0,672 lt kh iu kin tiu chun. Thnh phn % s mol caMgCO3 trong hn hp l:

A. 33,33%.

B. 45,55%.

C. 54,45%.

D. 66,67%.Cu 10: X l khong vt cuprit cha 45% Cu2O. Y l khong vt tenorit cha70% CuO. Cn trn X v Y theo t l khi lng t = mx/my c qung C, mt 1 tn qung C c th iu ch c ti a 0,5 tn ng nguyn cht. Gi trca t l:

A. 5/3.

B. 5/4.

C. 4/5.

D. 3/5.

Cu 11: Nhit phn hon ton 108 gam hn hp X gm Na2CO3 v NaHCO3c cht rn Y c khi lng bng 75,4% khi lng ca X. Khi lngnaHCO3 c trong X l:

A. 54,0 gam.


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B. 27,0 gam.

C. 72,0 gam.

D. 36,0 gam.

Cu 12: t chy hon ton 21,0 gam dy st trong khng kh thu c 29, 4gam hn hp cc oxit Fe2O3 v Fe3O4. Khi lng Fe2O3 to thnh l:

A. 12,0 gam.

B. 13,5 gam.

C. 16,5 gam.

D. 18,0 gam.

Cu 13: t chy hon ton 15,68 lt hn hp kh (ktc) gm 2 hidrocacbonthuc cng dy dng ng, c khi lng phn t hn km nhau 28 vC, thuc n CO2 / n H2O = 24/31. CTPT v % khi lng tng ng vi cchidrocacbon ln lt l:

A. C2H6 (28,57%) v C4H10 (71,43%).

B. C3H8 (78,57%) v C5H12 (21,43%).

C. C2H6 (17,14%) v C4H10 (82,86%).

D. C3H8 (69,14%) v C5H12 (30,86%).

Cu 14: Cho 6,72 gam Fe vo dung dch cha 0,3 mol H2SO4 c, nng (githit SO2 l sn phm kh duy nht). Sau khi phn ng xy ra hon ton, thuc:

A. 0,03 mol Fe2(SO4)3 v 0,06 mol FeSO4.

B. 0,12 mol FeSO4.

C. 0,02 mol Fe2(SO4)3 v 0,08 mol FeSO4.

D. 0,06 mol Fe2(SO4)3.

Cu 15: thu c dung dch HCl 25% cn ly m1 gam dung dch HCl 45%pha vi m2 gam dung dch HCl 15%. T l m1/m2 l:

A. 1:2.


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B. 1:3.

C. 2:1.

D. 3:1.

Cu 16: pha 500 ml dung dch nc mui sinh l (C = 0,9%) cn ly V mldung dch NaCl 3%. Gi tr ca V l:

A. 150.

B. 214,3.

C. 285,7.

D. 350.

Cu 17: Mt hn hp gm O2, O3 ktc c t khi hi i vi hidro l 18. Thnhphn % v th tch O3 trong hn hp l:

A. 15%.

B. 25%.

C. 20%.

D. 30%.

Cu 18: S lt nc nguyn cht cn thm vo 1 lt dung dch H2SO4 98% (d =

1,84 g/ml) c dung dch mi c nng 10% l:A. 14,192.

B. 15,192.

C. 16,192.

D. 17,192.

Cu 19: Ha tan 2,84 gam hn hp 2 mui CaCO3 v MgCO3 bng dung dchHCl d, thu c 0,672 lt kh ktc. Thnh phn % s mol ca MgCO3 trong

hn hp l:

A. 66,67%.

B. 54,45%.

C. 45,55%.


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D. 33,33%.

p n:

1B - 2D - 3D - 4C - 5C - 6A - 17D

8B - 9A - 10D - 11C - 12A - 13C - 14A - 15A - 16A - 17B - 18C - 19D

tuyet chieu giai nhanh hoa hoc

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