PP GiainhanhHoa VoCo TN

8/9/2019 PP GiainhanhHoa VoCo TN

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Phm ngc sn

Phng php gii bi tptrc nghim

Dng cho hc sinh n luyn thi i hc nm 2008

H ni - 2008


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Phn mt : Ho hc v cChuyn 1

Phng php p dng nh lut bo ton khi lngI- Ni dung nh lut bo ton khi lng

Tng khi lng cc cht tham gia phn ng bng tng khi lng sn phm.

V d : trong phn ng A + B C + D

Ta c : mA + mB = mB

n

C + mD

- H qu 1 : Gi mT l tng khi lng cc cht trc phn ng, mS l tng khi

lng cc cht sau phn ng. D cho phn ng xy ra va hay c cht d, hiu

sut phn ng nh hn 100% th vn c mS = mT.

- H qu 2 : Khi cation kim loi kt hp vi anion phi kim to ra cc hp cht

(nhoxit, hiroxit, mui) th ta lun c :

Khi lng hp cht = khi lng kim loi + khi lng anion.

- H qu 3 : Khi cation kim loi thay i anion to ra hp cht mi, s chnh lch

khi lng gia hai hp cht bng s chnh lch v khi lng gia cc cation.

- H qu 4 : Tng khi lng ca mt nguyn t trc phn ng bng tng khi

lng ca nguyn t sau phn ng.

- H qu 5 : Trongphn ng kh oxit kim loi bng CO, H2, Al+ Cht kh ly oxi ca oxit to ra CO2, H2O, Al2O3. Bit s mol CO, H2, Al

tham gia phn ng hoc s mol CO2, H2O, Al2O3 to ra, ta tnh c lng oxi

trong oxit (hay hn hp oxit) v suy ra l

ng kim loi (hay hn hp kim loi).+ Khi kh oxit kim, CO hoc H2 ly oxi ra khi oxit. Khi ta c :

2 2O (trong oxit ) CO CO H On n n= = =

p dng nh lut bo ton khi lng tnh khi lng hn hp oxit ban

u hoc khi lng kim loi thu c sau phn ng.


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II- Bi tp minh hoBi 1. Cho 24,4 gam hn hp Na2CO3, K2CO3 tc dng va vi dung dchBaCl2. Sau phn ng thu c 39,4 gam kt ta. Lc tch kt ta, c cn dung dch

thu c m gam mui clorua. m c gi tr l

A. 2,66 B. 22,6

C. 26,6 D. 6,26

Hng dn gii.

2 3BaCl BaCOn = n = 0,2 (mol)

p dng nh lut bo ton khi lng : hh BaClm m+2

= mkt ta + m

=> m = 24,4 + 0,2.208 - 39,4 = 26,6 gam

p n C.

Bi 2. Ha tan 10,14 gam hp kim Cu, Mg, Al bng mt lng va dung dchHCl thu c 7,84 lt kh A (ktc) v 1,54 gam cht rn B v dung dch C. C cn

dung dch C thu c m gam mui, m c gi tr l :

A. 33,45 B. 33,25

C. 32,99 D. 35,58

Hng dn gii. Theo nh lut bo ton khi lng :

+= +

= + = + =

( Al Mg) Clm m m

(10,14 1,54) 0,7.35,5 6,6 24,85 33,45 (gam)

p n A

Bi 3. Ha tan hon ton 10 gam hn hp Mg v Fe trong dung dch HCl d thyto ra 2,24 lt kh H2 (ktc). C cn dung dch sau phn ng thu c gam mui

khan. Khi lng mui khan thu c l

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

C. 3,42 gam D. 34,2 gam

Hng dn gii.


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Theo phng trnh in li HCl H,n n n . , (mo

, += = = =

2

2 242 2 0 222 4

l)

=

=> 10 + 0,2.35,5 = 17,1 (gam)mui kimloi Clm m m = +

p n B.

Bi 4. Trn 5,4 gam Al vi 6,0 gam Fe2O3 ri nung nng thc hin phn ngnhit nhm. Sau phn ng ta thu c m gam hn hp cht rn. Gi tr ca m l

A. 2,24 gam B. 9,40 gam

C. 10,20 gam D. 11,40 gam

Hng dn gii. Theo nh lut bo ton khi lng :

mhh sau = mhh trc = 5,4 + 6,0 = 11,4 (gam)

p n C.

Bi 5. Cho 0,52 gam hn hp 2 kim loi Mg v Fe tan hon ton trong dung dchH2SO4 long, d thy c 0,336 lt kh thot ra (ktc). Khi lng hn hp mui

sunfat khan thu c l

A. 2 gam B. 2,4 gamC. 3,92 gam D. 1,96 gam

Hng dn gii. Ta c mui thu c gm MgSO4 v Al2(SO4)3. Theo nh lut

bo ton khi lng :

= + = = =

= + =

2 2-24 4

mui Kimloi HSO SO

mui

0,336m m m .Trong n n 0,015 (mol)

22,4

m 0,52 0,015.96 1,96 gam

p n D

Bi 6. Cho 2,81 gam hn hp A gm 3 oxit Fe2O3, MgO, ZnO tan va trong300 ml dung dch H2SO4 0,1M. C cn dung dch sau phn ng, khi lng hn

hp cc mui sunfat khan to ra l

A. 3,81 gam B. 4,81 gam

C. 5,21 gam D. 4,8 gam


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Hng dn gii. p dng nh lut bo ton khi lng :

2 4 2

2 4 2

2 2 4

oxit H SO mui H O

mui oxit H SO H O

H O H SO

mui

m + m = m + m

m = m + m - m

Trong : n = n = 0,3.0,1= 0,03 (mol)

m = 2, 81+ 0.03.98 - 0, 03.18 = 5, 21(gam)

p n C.

Bi 7. Thi mt lung kh CO d qua ng s ng m gam hn hp gm CuO,Fe2O3, FeO, Al2O3 nung nng thu c 2,5 gam cht rn. Ton b kh thot ra scvo nc vi trong d thy c 15 gam kt ta trng. Khi lng ca hn hp oxit

kim loi ban u l

A. 7,4 gam B. 4,9 gam

C. 9,8 gam D. 23 gam

Hng dn gii. Cc phng trnh ho hc :

MxOy + yCOot xM + yCO2

Ca(OH)2 + CO2 CaCO3 + H2O

Ta c :

oxit kim loi oxi

O CO CO CaCO

oxit

m m m

Trong n n n n , (mol)

m , , . , (gam)

= +

= = = = =

= + =

2 3

150 15

100

2 5 0 15 16 4 9

p n B

Bi 8. Chia 1,24 gam hn hp hai kim loi c ha tr khng i thnh hai phnbng nhau :

- Phn 1: b oxi ha hon ton thu c 0,78 gam hn hp oxit.

- Phn 2: tan hon ton trong dung dch H2SO4 long thu c V lt H2

(ktc). C cn dung dch thu c m gam mui khan.

1. Gi tr ca V l


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A. 2,24 lt B. 0,112 lt

C. 5,6 lt D. 0,224 lt

2. Gi tr ca m l

A. 1,58 gam B. 15,8 gam

C. 2,54 gam D. 25,4 gam

Hng dn gii.

1. Ta nhn thy, khi kim loi tc dng vi oxi v H2SO4, s mol O2- bng SO4

2-,

hay : O HSOn n n= =2 24 .

Trong = = =O oxit kimloi1,24

m m m 0,78 0,16 (gam2

)

= = = = =2H O

0,16n n 0,01 (mol). V 0,01.22, 4 0,224 (lt)

16

p n D

2.= + = + =

24mui Kimloi SO

1,24

m m m 0,01.96 1,58 (lt)2

Bi 9. Ha tan hon ton 20 gam hn hp Mg v Fe vo dung dch axit HCl dthy c 11,2 lt kh thot ra (ktc) v dung dch X. C cn dung dch X th khi

lng mui khan thu c l

A. 35,5 gam. B. 45,5 gam.

C. 55,5 gam. D. 65,5 gam

Hng dn gii.2H

11,2n =

22,4= 0,5 = = =

2HCl Hn 2n 2.0,5 1 mol

p dng nh lut bo ton khi lng, mKL+ mHCl = mMui + mHiro

= + 2mui kimloi HCl H

m m m m

mmui = 20 + 1.36,5 - 2.0,5 = 55,5 (gam).

p n A.


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Bi 10.Sc ht mt lng kh clo vo dung dch hn hp NaBr v NaI, un nngthu c 2,34 g NaCl. S mol hn hp NaBr v NaI phn ng l:

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

C. 0,02 mol D. 0,04 mol

Hng dn gii. p dng nh lut bo ton khi lng :

nNaBr + nNaI = nNaCl =2,34

58,5= 0,04 mol.

p n D

Bi 11.Ho tan ht 38,60 gam hn hp gm Fe v kim loi M trong dung dchHCl d thy thot ra 14,56 lt H2 (ktc). Khi lng hn hp mui clorua khan thu

c l

A. 48,75 gam B. 84,75 gam

C. 74,85 gam D. 78,45 gam

Hng dn gii. Ta c Mui Kimloi Clm m m = +

Trong 2HCl HCl

2.14,56n n 2n 1,3(mo22,4

= = = = l) . m = 38,6 + 1,3.35,5 = 84,75 (g).

p n B

Bi 12.Cho tan hon ton 8,0 gam hn hp X gm FeS v FeS2 trong 290 mldung dch HNO3, thu c kh NO v dung dch Y. tc dng ht vi cc cht

trong dung dch Y, cn 250 ml dung dch Ba(OH)2 1M. Kt ta to thnh em

nung ngoi khng kh n khi lng khng i c 32,03 gam cht rn Z.

a. Khi lng mi cht trong X l

A. 3,6 gam FeS v 4,4 gam FeS2 B. 4,4 gam FeS v 3,6 gam FeS2

C. 2,2 gam FeS v 5,8 gam FeS2 D. 4,6 gam FeS v 3,4 gam FeS2

b. Th tch kh NO (ktc) thu c l

A. 1,12 lt B. 2,24 lt

C. 3,36 lt D. 6,72 lt


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c. Nng mol ca dung dch HNO3 dng l

A. 1 M B. 1,5 M

C. 2 M D. 0,5 M

Hng dn gii.

a. p dng nh lut bo ton khi lng i vi nguyn t Fe v S

Ta c : x mol FeS v y mol FeS2 0,5(x+y) mol Fe 2O3 v (x+2y) mol BaSO4

88x 120y 8

313x 546y 32,03

88x + 120y = 8

160.0,5(x+y) + 233(x+2y) = 32,03

+ =

+ =

Gii h c x = 0,05 v y = 0,03

Khi lng ca FeS = 88.x = 88.0,05 = 4,4 gam

Khi lng ca FeS2 : 8 - 4,4 = 3,6 gam.

p n B.

b. p dng nh lut bo ton electron

FeS - 9e Fe +3 + S+6

0,05 0,45 (mol)

FeS2 - 15e Fe+3 + 2S+6

0,03 0,45 (mol)

NO3- + 3e NO

3x ... x (mol)3x = 0,45 + 0,45 , x = 0,3 (mol). VNO = 0,3.22,4 = 6,72 (lit)

p n D

c. + = + =3Fen x y 0, 08 mol. lm kt ta ht lng Fe3+ cn 0,24 mol OH- hay

0,12 mol Ba(OH)2


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Kt ta (x + 2y) = 0,11 mol SO42- cn 0,11 mol Ba2+ hay 0,11 mol Ba(OH)2

S mol Ba(OH)2 dng = 0,12 + 0,11 = 0,23 < 0,25

Cn : 0,25 - 0,23 = 0,02 mol Ba(OH)2 trung ho vi 0,04 mol HNO3 d

3 33

3

HNO (p NO HNO (dNO

M(HNO )

n n n n

0,08.3 0,3 0,04 0,58 (mol)

0,58C 2M

0,29

) )= + +

= + + =

= =

p n C

Bi 13.Thi 8,96 lt CO (ktc) qua 16 gam FexOy nung nng. Dn ton b lngkh sau phn ng qua dung dch Ca(OH)2 d, thy to ra 30 gam kt ta. Khi

lng st thu c l

A. 9,2 gam B. 6,4 gam

C. 9,6 gam D. 11,2 gam

Hng dn gii.

yCO + FexOy xFe + yCO2 (1)y 1 x y

CO8,96

n = = 0,4(mol22,4

)

CO2 + Ca(OH)2 CaCO3 + H2O (2)

3 2CaCO CO30

n = = 0,3 (mol) n = 0,3 (mol)100

CO dv Fe2CO COn > n xOy ht

Theo nh lut bo ton khi lng c :

+ = +x y 2Fe O CO Fe CO

m m m m

16 + 28.0,3 = mFe + 0,3.44 mFe= 11,2 (gam).

p n D


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Bi 14.Thc hin phn ng nhit nhm vi 9,66 gam hn hp X gm FexOy vnhm, thu c hn hp rn Y. Cho Y tc dng vi dung dch NaOH d, thu c

dung dch D, 0,672 lt kh (ktc) v cht khng tan Z. Sc CO2 n d vo dung

dch D lc kt ta v nung n khi lng khng i c 5,1 gam cht rn.

a. Khi lng ca FexOy v Al trong X ln lt l

A. 6,96 v 2,7 gam B. 5,04 v 4,62 gam

C. 2,52 v 7,14 gam D. 4,26 v 5,4 gam

b. Cng thc ca oxit st l

A. FeO B. Fe2O3

C. Fe3O4 D. Khng xc nh c

Hng dn gii.

a. 2yAl + 3FexOy yAl2O3 + 3xFe (1)

Al + NaOH + H2O NaAlO2 + 3/2H2 (2)

0,02 ................................... 0,02 .......... 0,03

NaAlO2 + CO2 + 2H2O Al(OH)3 + NaHCO3 (3)

2Al(OH)3ot Al2O3 + 3H2O (4)

Nhn xt : Tt c lng Al ban u u chuyn ht v Al2O3 (4). Do

= = = = =2 3Al (ban u) Al O Al

5,1n 2.n 2. 0,1 (mol) m 0,1.27 2,7 (gam)

102

= =x yFe O

m 9,66 2,7 6,96 (gam)

p n A

b.

= = = = =2 3Al (ban u) Al O Al

5,1n 2.n 2. 0,1 (mol) m 0,1.27 2,7 (gam)

102

Theo nh lut bo ton khi lng nguyn t oxi, ta c :

= = =x y 2 3O(trongFe O ) O(trongAl O )

n n 1,5.0,08 0,12 (mol)


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= =Fe 6,96 0,12.16n 056,09 mol

. CTPT l FeFe On : n , : , := =0 09 0 12 3 4 3O4

p n C

Bi 15.Kh hon ton 32 gam hn hp CuO v Fe2O3 bng kh H2 thy to ra 9gam H2O. Khi lng hn hp kim loi thu c l

A. 12 gam B. 16 gam

C. 24 gam D. 26 gam

Hng dn gii. V H2 ly oxi ca oxit kim loi H2O

Ta c nO (trong oxit) = =9182H On = 0,5 (mol)

mO = 0,5.16 = 8 gam mKim loi = 32 - 8 = 24 (g)

p n C

Bi 16.Thi mt lung kh CO d i qua ng ng hn hp 2 oxit Fe3O4 v CuOnung nng n khi phn ng xy ra hon ton thu

c 2,32 gam hn hp kimloi. Kh thot ra c a vo bnh ng dung dch Ca(OH)2 d thy c 5 gam

kt ta trng. Khi lng hn hp 2 oxit kim loi ban u l

A. 3,12 gam B. 3,21 gam

C. 4 gam D. 4,2 gam

Hng dn gii. Cc phn ng

Fe3O4 + 4CO0t 3Fe + 4CO2

CuO + CO

0t

Cu + CO2

CO2+ Ca(OH)2 CaCO3 + H2O

CO ly oxi trong oxit CO2. nO (trong oxit) = nCO = 2 3CO CaCOn n 0,05(mol)= =

moxit = mKl + moxitrong oxit = 2,32 + 0,05.16 = 3,12 (g).

p n A


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Chuyn 2Phng php tng gim khi lngI - Ni dung

Da vo s tng gim khi lng khi chuyn t cht ny sang cht khc

xc nh khi lng hn hp hay mt cht.

- Da vo phng trnh ho hc tm s thay i v khi lng ca 1 mol

cht trong phn ng (A B) hoc x mol A y mol B. (vi x, y t l cn bng

phn ng).

- Tnh s mol cc cht tham gia phn ng v ngc li.

Phng php ny thng c p dng gii bi ton v c v hu c, trnh

c vic lp nhiu phng trnh, t s khng phi gii nhng h phng trnh

phc tp.

II - Bi tp minh ho

Bi 1. Ha tan 14 gam hn hp 2 mui MCO3 v N2(CO3)3 bng dung dch HCld, thu c dung dch A v 0,672 lt kh (ktc). C cn dung dch A th thu c

m gam mui khan. m c gi tr l

A. 16,33 gam B. 14,33 gam

C. 9,265 gam D. 12,65 gam

Hng dn gii.

Vn dng phng php tng gim khi lng.

Theo phng trnh ta c:

C 1 mol mui lng mui tng 71- 60 =11 gam3CO 2 mol Cl 1mol CO + 2

Theo s mol CO2 thot ra l 0,03 th khi lng mui tng 11.0,03 = 0,33 (g)

Vy mmui clorua = 14 + 0,33 = 14,33 (g).

p n B


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Bi 2. Nhng 1 thanh nhm nng 45 gam vo 400 ml dung dch CuSO4 0,5M.Sau mt thi gian ly thanh nhm ra cn nng 46,38 gam. Khi lng Cu thot ra

l

A. 0,64 gam B. 1,28 gam

C. 1,92 gam D. 2,56 gam

Hng dn gii.

C 2 mol Al 3 mol Cu khi lng tng 3.(64 - 54) = 138 gam

Theo n mol Cu khi l

ng tng 46,38 - 45 = 1,38 gamnCu = 0,03 mol. mCu = 0,03.64 = 1,92 gam

p n C

Bi 3. Ha tan 5,94 gam hn hp 2 mui clorua ca 2 kim loi A, B (u c hotr II) vo nc c dung dch X. lm kt ta ht ion Cl- c trong dung dch X

ngi ta cho dung dch X tc dng vi dung dch AgNO3 thu c 17,22 gam kt

ta. Lc b kt ta, thu c dung dch Y. C cn Y c m gam hn hp mui

khan. m c gi tr l

A. 6,36 gam B. 63,6 gam

C. 9,12 gam D. 91,2 gam

Hng dn gii. p dng phng php tng gim khi lng

C 1 mol MCl2 1 mol M(NO 3)2 v 2 mol AgCl th m tng 2.35,5 - 71 = 53 gam

0,12 mol AgCl khi lng tng 3,18 gam

m mui nitrat = mKl + m = 5,94 + 3,18 = 9,12 (g)

p n C

Bi 4. Mt bnh cu dung tch 448 ml c np y oxi ri cn. Phng in ozon ho, sau np thm cho y oxi ri cn. Khi lng trong hai trng hp

chnh lch nhau 0,03 gam. Bit cc th tch np u ktc. Thnh phn % v th

tch ca ozon trong hn hp sau phn ng l

A. 9,375 % B. 10,375 %

C. 8,375 % D.11,375 %


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Hng dn gii.

Th tch bnh khng i, do khi lng chnh l do s ozon ha.

C 1mol oxi c thay bng 1mol ozon khi lng tng 16g

Vy khi lng tng 0,03 gam th s ml ozon (ktc) l0,03

16.22400 = 42 (ml).

%O3 =42

100%448

= 9,375 %.

p n ABi 5. Ho tan hon ton 4 gam hn hp MCO3 v M'CO3 vo dung dch HClthy thot ra V lt kh (ktc). Dung dch thu c em c cn thu c 5,1 gam

mui khan. Gi tr ca V l

A. 1,12 lt B. 1,68 lt

C. 2,24 lt D. 3,36 lt

Hng dn gii.

3 2 2MCO + 2HCl MCl + H O + CO

2 4 g 5,1 g x mol mtng = 5,1 - 4 = 1,1 (gam)

M+60 M+71 1 mol mtng = 11 gam

x =1,1

11= 0,1 (mol) V = 0,1 . 22,4 = 2,24 (lt).

p n C

Bi 6. Cho 1,26 gam mt kim loi tc dng vi dung dch H 2SO4 long to ra3,42 gam mui sunfat. Kim loi l

A. Mg B. Fe

C. Ca D. Al

Hng dn gii. p dng phng php tng - gim khi lng.

C 1 mol kim loi tc dng to thnh mui SO42- khi lng tng ln 96 gam.


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Theo khi lng tng 3,42 - 1,26 = 2,16 g.

Vy s mol kim loi M l 0,0225 mol. Vy1,26

M 56.0,0225

= = M l Fe

p n B

Bi 7. Ha tan hon ton 12 gam hn hp hai kim loi X v Y bng dung dchHCl ta thu c 12,71gam mui khan. Th tch kh H2 thu c (ktc) l

A. 0,224 lt B. 2,24 lt

C. 4,48 lt D. 0,448 ltHng dn gii. p dng phng php tng - gim khi lng:

C 1 mol Cl- sinh ra sau phn ng khi lng mui tng ln 35,5 gam.

Theo , tng 0,71 gam, do s mol Cl- phn ng l 0,02 mol.

2H Cl

1n n 0,01 (mo

2= = l) . V = 0,224 (l)

p n A.

Bi 8. Cho ho tan hon ton a gam Fe3O4 trong dung dch HCl, thu c dungdch D, cho D tc dng vi dung dch NaOH d, lc kt ta ngoi khng khn khi lng khng i na, thy khi lng kt ta tng ln 3,4 gam. em

nung kt ta n khi lng khng i c b gam cht rn. Gi tr ca a, b ln

lt l

A. 46,4 v 48 gam B. 48,4 v 46 gam

C. 64,4 v 76,2 gam D. 76,2 v 64,4 gam

Hng dn gii.

Fe3O4 + 8HCl 2FeCl 3 + FeCl2 + 4H2O

FeCl2 + 2NaOH Fe(OH) 2 + 2NaOH

FeCl3 + 3NaOH Fe(OH) 3 + 3NaOH

4Fe(OH)2 + O2 + 2H2O 4Fe(OH) 3

2Fe(OH)3ot Fe2O3 + 3H2O


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Nhn xt : Ta thy Fe3O4 c th vit dng Fe2O3.FeO. Khi cho D tc dng viNaOH kt ta thu c gm Fe(OH)2 v Fe(OH)3. ngoi khng kh Fe(OH)2

Fe(OH) 3.

1 mol Fe(OH)2 1 mol Fe(OH) 3 thm 1 mol OH khi lng tng ln 17 gam

0,2 mol 0,2 mol . 3,4 (gam)

2 3 2FeO Fe O Fe(OH)n n n 0, 2 (mo= = = l)

0,2 mol Fe3O4 0,3 mol Fe 2O3

a = 232.0,2 = 46,4 (gam), b = 160.0,3 = 48 (gam)

p n A

Bi 9. Cho 8 gam hn hp A gm Mg v Fe tc dng ht vi 200 ml dung dchCuSO4 n khi phn ng kt thc, thu c 12,4 gam cht rn B v dung dch D.

Cho dung dch D tc dng vi dung dch NaOH d, lc v nung kt ta ngoi

khng kh n khi lng khng i thu c 8 gam hn hp gm 2 oxit.

a. Khi lng Mg v Fe trong A ln lt l

A. 4,8 v 3,2 gam B. 3,6 v 4,4 gam

C. 2,4 v 5,6 gam D. 1,2 v 6,8 gam

b. Nng mol ca dung dch CuSO4 l

A. 0,25 M B. 0,75 M

C. 0,5 M D. 0,125 M

c. Th tch NO thot ra khi ho tan B trong dung dch HNO3 d l

A. 1,12 lt B. 3,36 lt

C. 4,48 lt D. 6,72 ltHng dn gii.

a. Cc phn ng :

Mg + CuSO4 MgSO4 + Cu

Fe + CuSO4 FeSO 4 + Cu

Dung dch D gm MgSO4 v FeSO4. Cht rn B bao gm Cu v Fe d


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MgSO4 + 2NaOH Mg(OH)2 + Na2SO4

FeSO4 + 2NaOH Fe(OH) 2 + Na2SO4

Mg(OH)2ot MgO + H2O

4Fe(OH)2 + O2ot 2Fe2O3 + 4H2O

Gi x, y l s mol Mg v Fe phn ng. S tng khi l ng t hn hp A (gm Mg

v Fe) hn hp B (gm Cu v Fe c th d) l

(64x + 64y) - (24x + 56y) = 12,4 - 8 = 4,4

Hay : 5x + y = 0,55 (I)

Khi lng cc oxit MgO v Fe2O3 m = 40x + 80y = 8

Hay : x + 2y = 0,2 (II)

T (I) v (II) tnh c x = 0,1, y = 0,05

mMg = 24.0,1 = 2,4 (g)

mFe = 8 - 2,4 = 5,6 (g)

p n C.

b.

CuSO

M

n x y ,

,C ,

,

= + =

= =

4

0 15

0 150 75

0 2M

p n B

c. Hn hp B gm Cu v Fe d. nCu = 0,15 mol; nFe = 0,1 - 0,05 = 0,05 mol. Khi

tc dng vi dung dch HNO3. Theo phng php bo ton eletron

- Cht kh l Fe v Cu

Fe - 3e Fe+3

0,05 .... 0,15

Cu - 2e Cu +2

0,15 . . . . 0,3


18/71

- Cht oxi ho l HNO3

N+5 + 3e N +2 (NO)

3a . . . . . . a . . ..a

Ta c 3a = 0,15 + 0,3 , a = 0,15 (mol). VNO = 0,15.22,4 = 3,36 lt

p n B

Bi 10.Cho 2,81 gam hn hp gm 3 oxit Fe2O3, MgO, ZnO tan va trong 300ml dung dch H2SO4 0,1M th khi lng hn hp cc mui sunfat khan to ra l

A. 3,81 gam B. 4,81 gamC. 5,21 gam D. 4,86 gam

Hng dn gii. p dng phng php tng - gim khi lng.

C 1 mol H2SO4 phn ng, thay th O (trong oxit) bng SO42- trong cc

kim loi, khi lng tng 96 - 16 = 80 gam.

Theo s mol H2SO4 phn ng l 0,03 th khi lng tng 0,24 gam.

Vy khi lng mui khan thu c l: 2,81 + 2,4 = 5,21 gam.

p n C


19/71

Chuyn 3Phng php s dng cc gi tr trung bnhI - Ni dung

- Dng khi lng mol trung bnh M l khi lng ca 1 mol hn hp.

hh 1 1 2 2 1 1 2 2

hh 1 2

m n .M n .M n .%V n .%V

M n n n 100

+ +

= = =+ vi M1 < M < M2

- Gi tr trung bnh dng bin lun tm ra nguyn t khi hoc phn t

khi hay s nguyn t trong phn t hp cht.

II Bi tp minh ho

Bi 1. Hn hp X gm hai kim loi A, B nm k tip nhau trong cng mt phnnhm chnh. Ly 6,2 gam X ho tan hon ton vo nc thu c 2,24 lt hiro

(ktc). A, B l

A. Li, Na B. Na, KC. K, Rb D. Rb, Cs

Hng dn gii.

t cng thc chung ca A v B l R

2R + 2H2O 2ROH + H2

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

6,2M = =31 (g/mol) Vy 2 kim loi l Na (23) v K (39)

0,2

p n B.

Bi 2. Ha tan 5,94 gam hn hp hai mui clorua ca hai kim loi A v B (cngthuc nhm IIA) vo nc c dung dch X. lm kt ta ht ion Cl- trong

dung dch X ngi ta cho tc dng vi dung dch AgNO3 thu c 17,22 gam kt

ta. Cng thc ha hc ca hai mui clorua ln lt l


20/71

A. BeCl2, MgCl2 B. MgCl2, CaCl2

C. CaCl2, SrCl2 D. SrCl2, BaCl2

Hng dn gii. t cng thc chung ca hai mui l RCl2

RCl2 + 2AgNO3 2AgCl + 2RCl

RCl AgCl

1 1 17, 22n n . 0,06 (m

2 2 143,5= = = ol)

2RCl

5,94M 99 R 99 71 28

0,06

= = = =

Vy 2 kim loi nhm IIA l Mg (24) v Ca (40).

p n B

Bi 3. Ho tan hon ton 4,68 gam hn hp mui cacbonat ca hai kim loi A vB k tip trong nhm IIA vo dung dch HCl thu c 1,12 lt CO2 (ktc). Kim

loi A v B l

A. Be v Mg B. Mg v Ca

C. Ca v Sr D. Sr v BaHng dn gii. Gi M l nguyn t khi trung bnh ca 2 kim loi A v B

M CO3 + 2HCl M Cl2 + CO2 + H2O

0,05 ........................................1,12

= 0,05 (mol)22,4

M CO3 = ;6,9305,0

68,4= M = 93,6 - 60 = 33,6

Bin lun: A < 33,6 A l Mg = 24

B > 33,6 B l Ca = 40.

p n B

Bi 4. X v Y l hai nguyn t halogen 2 chu k lin tip trong bng tun hon. kt ta ht ion X-, Y- trong dung dch cha 4,4 gam mui natri ca chng cn

150 ml dung dch AgNO3 0,4M. X v Y l


21/71

A. Flo, clo B. Clo, brom

C. Brom, iot D. Khng xc nh c.

Hng dn gii.

S mol AgNO3 = s mol X- v Y- = 0,4.0,15 = 0,06 (mol)

Khi lng mol trung bnh ca hai mui l M =4,4

0,0673,3

M X,Y = 73,3 - 23=50,3, hai halogen l Clo (35,5) v Brom (80).

p n B.

Bi 5. Hn hp X gm hai kim loi A, B nm k tip nhau trong nhm IA. Ly7,2 gam X ho tan hon ton vo nc thu c 4,48 lt hiro ( ktc). A, B l

A. Li, Na B. Na, K

C. K, Rb D. Rb, Cs

Hng dn gii. Dng phng php phn t khi trung bnh

X + H2O XOH + 1/2H2

2X H

4,48n 2n 2. 0,4 (mol)

22,4

7,2M 18. Hai kim loi l Li (9)v Na (23)

0,4

= = =

= =

p n A


22/71

Chuyn 4Phng php ng choI - Ni dung

c s dng trong cc bi ton trn ln dung dch c cng cht tan, cng loi

nng hoc trn ln cc cht kh khng tc dng vi nhau.

1. Cc cht cng nng C%m ............ C C C

m C CC

m C C

m ............ C C C

=

1 1 2

1 2

2 1

2 2 1

Trong :

m1 l khi lng dung dch c nng C1 (%)

m2 l khi lng dung dch c nng C2 (%)

C (%) l nng dung dch thu c sau khi trn ln. Vi C1 < C < C2

2. Cc cht cng nng mol

Trong :

V1 l th tch dung dch c nng CM (1)

V2 l th tch dung dch c nng CM (2)

CM l nng mol dung dch thu c sau khi trn ln.

Vi CM (1) < CM < CM (2)

M( ) M( ) M

M( ) MM

M M(

M( ) M M( )

)

............ C C C

C CVC

V C C

V ............ C C C

V

=

1 1 2

21

2 1

2 2 1


23/71

3. Cc cht kh khng tc dng vi nhau.V ............ M M M

V M MM

V M M

V ............ M M M

=

1 1 2

1 2

2 1

2 2 1

Trong :

V1 l th tch cht kh c phn t khi M1

V2 l th tch cht kh c phn t khi M2

M l khi lng mol trung bnh thu c sau khi trn ln. Vi M1 < M < M2

II - Bi tp minh ho

Bi 1. Mt dung dch HCl nng 45% v mt dung dch HCl khc c nng 15%. c mt dung dch mi c nng 20% th cn phi pha ch v khi

lng gia 2 dung dch theo t l l

A. 1 : 3 B. 3 : 1

C. 1 : 5 D. 5 : 1

Hng dn gii.p dng qui tc ng cho ta c

m ............

m

m

m ............

= =

1

1

2

2

45 20 15

5 125

25 5

15 45 20

p n C.

Bi 2. iu ch c hn hp 26 lt H2 v CO c t khi hi i vi metanbng 1,5 th th tch H2 v CO cn ly lA. 4 lt v 22 lt

B. 22 lt v 4 lt

C. 8 lt v 44 lt

D. 44 lt v 8 lt


24/71

Hng dn gii. p dng qui tc ng cho

=

2

2

H

H

CO

CO

V ............ 2 4

V 424

V 2

V ............ 28 22

2

Mt khc + =2H CO

V V 26

Vy cn 4 lt H2 v 22 lt CO.

p n ABi 3. Khi lng dung dch NaCl 15% cn trn vi 200 gam dung dch NaCl 30% thu c dung dch NaCl 20 % l

A. 250 gam B. 300 gam

C. 350 gam D. 400 gam

Hng dn gii. Dng phng php ng cho

= =

m ............ 15 10

m 10

20 m 400200 5200 ............ 30 5

Nhvy khi lng NaCl 15 % cn trn l 400 gam.

p n D

Bi 4. Th tch H2O v dung dch MgSO4 2M cn pha c 100 ml dung dchMgSO4 0,4M ln lt l

A. 50 ml v 50 ml B. 40 ml v 60 ml

C. 80 ml v 20 ml D. 20 ml v 80 mlHng dn gii. Gi V l th tch H2O cn cho vo, khi th tch dung dch

MgSO4 2M l 100 - V.

V .

= =

........... 0 1,6

V 1,60,4 V 80

100 V 0,4

100 V.... 2 0,4


25/71

Vy pha 80 ml H2O vi 20 ml dung dch MgSO4 2M th thu c 100 mldung dch MgSO4 0,4 M.

p n C

Bi 5. Ha tan 4,59 gam Al bng dung dch HNO3 thu c hn hp kh NO vN2O c t khi hi i vi hiro bng 16,75. Th tch NO v N 2O (ktc) thu c

l

A. 2,24 lt v 6,72 lt

B. 2,016 lt v 0,672 lt

C. 0,672 lt v 2,016 lt

D. 1,972 lt v 0,448 lt

Hng dn gii. S dng phng php bo ton electron

- Al l cht kh

4,590,17.......0,51 mol

27

3+Al - 3e Al

=

- Cht oxi ho

5 2

5 12

N 3e N (NO)

3x .................. x

N 2.4e 2N (N

8

O)

y .............. 2y....... y

+ +

+ +

+

+

Theo phng php ng chox ......... ,

x ,,y ,

y.............. ,

= =

30 10 5

10 5 333 53 5 1

44 3 5

+ =

=

3x 8y 0,51

x 3y

= =

= =2

NO

N O

x 0,09 V 2,016 (l)

y 0,03 V 0,671 (l)

p n B


26/71

Bi 6. Mt dung dch NaOH nng 2M v mt dung dch NaOH khc c nng 0,5M. c mt dung dch mi c nng 1M th cn phi pha ch v th tch

gia 2 dung dch theo t l l

A. 1 : 2 B. 2 : 1

C. 1 : 3 D. 3 : 1

Hng dn gii. Dng phng php ng cho, gi V1 l th tch ca dung dch

NaOH 2M, V2 l th tch ca dung dch NaOH 0,5M.

= =

1

1

2

2

V ......... 2 0,5

V 0,5 11V 1

V .............. 0,5 1

2

p n A

Bi 7. Hn hp gm NaCl v NaBr. Cho hn hp tc dng vi dung dch AgNO3d th to ra kt ta c khi lng bng khi lng ca AgNO3 tham gia phn

ng. Thnh phn % theo khi lng ca NaCl trong hn hp u l

A. 25,84% B. 27,84%

C. 40,45% D. 27,48%

Hng dn gii.

NaCl + AgNO3 AgCl + NaNO3 (1)

NaBr + AgNO3 AgBr + NaNO3 (2)

Khi lng kt ta (gm AgCl v AgBr) bng khi lng AgNO3, do

khi lng mol trung bnh ca hai mui kt ta + = =3AgCl AgBr AgNOM M 170 v

= =Cl , BrM 170 108 62 . Hay khi lng mol trung bnh ca hai mui ban u

l = + =NaCl,NaBrM 23 62 85 .p dng phng php ng cho, ta c

NaBr ........ 103 26, 5

85

NaCl ...........58, 5 18


27/71

NaCl

NaBr NaCl

m

m +m= 18.58,5 100%

(26,5.103) (18.58,5)+= 27,84 %

p n B

Bi 8. Cho hn hp gm N2, H2 v NH3 c t khi so vi hiro l 8. Dn hn hpi qua dung dch H2SO4 c, d th th tch kh cn li mt na. Thnh phn phn

trm (%) theo th tch ca mi kh trong hn hp ln lt l

A. 25% N2, 25% H2 v 50% NH3.

B. 25% NH3, 25% H2 v 50% N2.C. 25% N2, 25% NH3v 50% H2.

D. 15% N2, 35% N2v 50% NH3

Hng dn gii.

Khi i qua dung dch H2SO4 c, d ton b NH3 b hp th, do thnh

phn ca NH3 l 50%.

p dng phng php ng cho, = =hn hp banuM 8.2 16 ta c:

= =

+

3

2 2

N . . 1H ....... 7 16 M16 M 1

16 M 151 1

N H ........M 1

M = 15 l khi lng mol trung bnh ca hn hp ca N2 v H2. Tip tc p

dng phng php ng cho ta c:

= = =

2

22 2

2

2

N ............28 13

N 115 %N %H 25%

H 1H ................2 13

p n A


28/71

Chuyn 5Phng php p dng nh lut bo ton in tchI - Ni dung

- nh lut bo ton in tch c p dng trong cc trng nguyn t,

phn t, dung dch trung ha in.

- Trong phn ng trao i ion ca dung dch cht in li trn c s canh lut bo ton in tch ta thy c bao nhiu in tch dng hoc m ca cc

ion chuyn vo trong kt ta hoc kh tch ra khi dung dch th phi tr li cho

dung dch by nhiu in tch dng hoc m.

II - Bi tp p dng

Bi 1. Chia hn hp 2 kim loi A, B c ha tr khng i thnh 2 phn bngnhau :

- Phn 1 tan ht trong dung dch HCl, to ra 1,792 lt H2 (ktc).

- Phn 2 nung trong khng kh n khi lng khng i thu c 2,84gam cht rn. Khi lng hn hp 2 kim loi trong hn hp u l

A. 2,4 gam B. 3,12 gam

C. 2,2 gam D. 1,8 gam

Hng dn gii.

Nhn xt : S mol in tch ca hai kim loi A v B trong hai phn l

khng thay i, do s mol in tch m trong hai phn l nhnhau.

V nn

2

O 2Cl

O(trong oxit ) Cl( trong mu H2

1 1,796n n n 0,08 (

2 22,4i)= = = = mol)

mKim loi = m oxit - mO = 2,84 0,08.16 = 1,56 gam

Khi lng trong hn hp ban u m = 2.1,56 = 3,12 gam

p n B


29/71

Bi 2. Dung dch A c cha 5 ion : Mg2+, Ba2+ , Ca2+, 0,1 mol Cl- v 0,2mol. Thm dn V lt dung dch K3NO

2CO3 1M vo A n khi c lng kt ta ln

nht. V c gi tr l

A. 150 ml B. 300 ml

C. 200 ml D. 250 ml

Hng dn gii. Phng trnh ion rt gn

Mg2+ + MgCO23CO

3

Ba2+ + BaCO23CO

3

Ca2+ + CaCO23CO

3

Khi phn ng kt thc, cc kt ta tch khi dung dch, phn dung dch

cha Na+, Cl- v . trung ha in th3NO

3Na Cl NOn n n 0,3(mo+ = + = l)

2 3

Na

dd Na CO

n 0,3

V 0,15 (l) 150 ml2Na

+

+ = = = =

p n A

Bi 3. Dung dch A cha cc ion CO32-, SO32-, SO42- v 0,1 mol HCO3-, 0,3 molNa+. Thm V (lt) dung dch Ba(OH)2 1M vo dung dch A th thu c lng kt

ta ln nht. Gi tr ca V l

A. 0,15 lt B. 0,2 lt

C. 0,25 lt D. 0,5 lt

Hng dn gii.

Nng cc ion [Ba2+] = 1M, [OH-] = 2M. thu c lng kt ta ln

nht, cn 0,1 mol OH- tc dng ht vi HCO3-

HCO3- + HO- CO3

2- + H2O

Mt khc cn 0,3 mol OH- trung ho Na+. Vy tng s mol OH- cn l 0,1

+ 0,3 = 0,4 (mol)


30/71

Th tch dung dch Ba(OH)2 l0,4V 0,22

= = (lit)

l)

l)

p n B

Bi 4. Cho tan hon ton 15,6 gam hn hp gm Al v Al 2O3 trong 500 ml dungdch NaOH 1M thu c 6,72 lt H2 (ktc) v dung dch D. Th tch HCl 2M cn

cho vo D 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 gii.

Trong dung dch D c cha AlO2- v OH- (nu d). Dung dch D trung ho

v in nn2AlO OH Na

n n n 0,5 (mo ++ = =

Khi cho HCl vo D :

H+ + OH- H2O

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

thu c lng kt ta ln nht th2H AlO OH

n n n 0,5 (mo+ = + =

Th tch dung dch HCl l0,5

V 0,252

= = (l)

p n B

Bi 5. Cho tan hon ton 10 gam hn hp Mg v Fe trong dung dch HCl 4M thuc 5,6 lt H2 (ktc) v dung dch D. kt ta hon ton cc ion trong D cn

300 ml dung dch NaOH 2M. Th tch dung dch HCl dng l

A. 0,1 lt B. 0,12 ltC. 0,15 lt D. 0,2 lt

Hng dn gii.

Khi cho 0,6 mol NaOH vo dung dch D cha Mg2+, Fe2+ v H+ (nu d)

tch ra khi dung dch D. Dung dch to thnh cha Cl- phi trung ho in vi 0,6

mol Na+


31/71

Cl Na

HCl

n n 0,6 mo

0,6V 0,1

4t

+= =

= =

l

5 l

p n C

Bi 6. Cho 20 gam hn hp X gm Fe, FeO, Fe3O4, Fe2O3 tan va ht trong 700ml dung dch HCl 1M thu c 3,36 lit H2 (ktc) v dung dch D. Cho dung dch

D tc dng vi NaOH d, lc kt ta v nung trong khng kh n khi lng

khng i thu c cht rn Y. Khi lng Y l

A. 16 gam B. 32 gam

C. 8 gam D. 24 gam


Fe + 2HCl FeCl 2 + H2

FeO + 2HCl FeCl 2 + H2O

Fe3O4 + 8HCl 2FeCl 3 + FeCl2 + 4H2O

Fe2O3 + 6HCl 2FeCl 3 + 3H2O

FeCl2 + 2NaOH Fe(OH)2 + 2NaCl




Vi cch gii thng thng, ta t n s l s mol cc cht ri tnh ton theo

phng trnh phn ng. gii nhanh bi ton ny, ta p dng phng php bo

ton in tch.

S mol HCl ho tan Fe l = = =2HCl H

3,36n 2n 2. 0,3 m

22,4ol

S mol HCl ho tan cc oxit = 0,7 - 0,3 = 0,4 (mol)

Theo nh lut bo ton in tch ta cO (trongoxit) Cl

,n n , = = =2

1 0 40 2

2 2mol


32/71

= = =oxit oxiFe(trongX)m m 20 0,2.16n 0

56 56,3 mol

0,3 mol Fe 0,15 mol Fe2O3 ; = =Fe Om , . gam2 3 0 15 160 24

p n D

Bi 7. Trn 100 ml dung dch AlCl3 1M vi 200 ml dung dch NaOH 1,8M thuc kt ta A v dung dch D.

a. Khi lng kt ta A l

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

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

b. Nng mol ca cc cht trong dung dch D l

A. NaCl 0,2M v NaAlO2 0,6M B. NaCl 1M v NaAlO2 0,2M

C. NaCl 1M v NaAlO2 0,6M D. NaCl 0,2M v NaAlO2 0,4M

Hng dn gii. Ta c th s dng nh lut bo ton in tch

3

Al Cl

Na OH

n 0,1mol, n 3.0,1 0,3 mol

n n 0, 2.1,8 0,36 mol

+

+

= = =

= = =

Sau khi phn ng kt thc, kt ta tch ra, phn dung dch cha 0,3 mol Cl-

trung ho in vi 0,3 mol Na+ cn 0,06 mol Na+ na phi trung ho in vi mt

anion khc, ch c th l 0,06 mol AlO2- (hay [Al(OH)4]

-). Cn 0,1 - 0,06 = 0,04

mol Al3+ tch ra thnh 0,04 mol Al(OH)3. Kt qu trong dung dch cha 0,3 mol

NaCl v 0,06 mol NaAlO2 (hay Na[Al(OH)4])

a. .3Al(OH)

m 0, 04.78 3,12 gam= =

p n A

b.2M(NaCl) M(NaAlO )

0,3 0,06C 1M, C

0,3 0,3= = = = 0,2M. .

p n B


33/71

Chuyn 6Phng php p dng nh lut bo ton electronI - Ni dung

Trong phn ng oxi ha - kh, s mol electron m cht kh cho i bng s

mol electron m cht oxi ha nhn v.

enhn enhngn n=

- S dng cho cc bi ton c phn ng oxi ha - kh, c bit l cc bi

ton c nhiu cht oxi ha, nhiu cht kh.

- Trong mt phn ng hoc mt h phn ng, cn quan tm n trng thi

oxi ha ban u v cui ca mt nguyn t m khng cn quan tm n cc qu

trnh bin i trung gian.

- Cn kt hp vi cc phng php khc nh bo ton khi lng, bo ton

nguyn t gii bi ton.

- Nu c nhiu cht oxi ha v nhiu cht kh cng tham gia trong bi

ton, ta cn tm tng s mol electron nhn v tng s mol electron nhng ri mi

cn bng.

II - bi tp p dng

Bi 1. m gam bt st ngoi khng kh mt thi gian thu c11,8 gam hnhp cc cht rn FeO, Fe3O4, Fe2 O3, Fe. Ha tan hon ton hn hp bng dung

dch HNO3 long thu c 2,24 lt kh NO duy nht (ktc). Gi tr ca m l

A. 5,02 gam B. 10,04 gam

C. 15,12 gam D. 20,16 gam

Hng dn gii.

nFe =m56 ;

=

2O phn ng

11,8 mn

32; nNO gii phng = 0,1 mol

- Cht kh l Fe :


34/71

3

56 56

0 +Fe - 3e Fem m

..........

3

- Cht oxi ha gm O2 v HNO3 :

+

0 -2O 4e 2O

11,8 m 11,8 m.......

32 8

2

3

+5 +2

N e N (NO)0,3 ................ 0,1+

mol e- Fe nhng = ne- cht oxi ha (O2, 3NO ) nhn:

3 11 80 3

56 8

= +

m , m, m = 10,04 (g).

p n B.

Bi 2.Ha tan hon ton 17,4 gam hn hp 3 kim loi Al, Fe, Mg trong dungdch HCl thy thot ra 13,44 lt kh. Nu cho 34,8 gam hn hp trn tc dng vidung dch CuSO4 d, lc ly ton b cht rn thu c sau phn ng tc dng vi

dung dch HNO3 nng d th thu c V lt kh NO2 (ktc). Gi tr V l

A. 11,2 lt B. 22,4 lt

C. 53,76 lt D. 76,82 lt

Hng dn gii. Al, Mg, Fe nhng e, s mol electron ny chnh bng s mol e

Cu nhng khi tham gia phn ng vi HNO3. S mol electron m H+ nhn cng

chnh l s mol electron m HNO3 nhn.

13 441 2 0 6

22 4

+ 22H + 2e H,

, ......... ,,

=

17,4 gam hn hp H+ nhn 1,2 mol e. Vy 34,8 gam s mol e m H+ nhn l 2,4

mol.


35/71

2 4 2 4

+5 +4 2N + 1e N (NO ), ......... , mol

2NOV 2,4.22,4 53,76 lt= = .

p n C

Bi 3.Ha tan hon ton 43,2 gam kim loi Cu vo dung dch HNO3 long, tt ckh NO thu c em oxi ha thnh NO2 ri sc vo nc c dng oxi chuyn

ht thnh HNO3. Th tch kh oxi ktc tham gia vo qu trnh trn l

A. 5,04 lt B. 7,56 lt

C. 6,72 lt D. 8,96 lt

Hng dn gii.

Ta nhn thy, Cu nhng electron cho HNO3 to thnh NO2, sau NO2 li

nhng cho O2. Vy trong bi ton ny, Cu l cht nhng, cn O2 l cht nhn

electron.

Cu - 2e Cu2+

0,675 ...... 1,35O2 + 4e 2O

2-

x ........ 4x

4x = 1,35 x = 0,3375 = 0,3375.22,4 = 7,56 lt2O

V

p n B

Bi 4.Chia m gam hn hp 2 kim loi A, B c ha tr khng i thnh 2 phnbng nhau :

- Phn 1 tan ht trong dung dch HCl, to ra 1,792 lt H2 (ktc).

- Phn 2 nung trong oxi thu c 2,84 g hn hp oxit.

Gi tr ca m l

A. 1,56 gam B. 2,64 gam

C. 3,12 gam D. 4,68 gam


36/71

Hng dn gii. A, B l cht kh, H+ ( phn 1) v O2 ( phn 2) l cht oxi ha.

S mol e- H+ nhn bng s mol e- O2 nhn

2H+ + 2.1e- H2

0,16 ............ 0,08

O2 + 4e 2O2-

0,04 ...... 0,16

mkl phn 2 = moxit - mOxi = 2,84 - 0,04.32 = 1,56 gam. m = 1,56.2 = 3,12 gam.

p n C

Bi 5.Chia 44 gam hn hp gm Fe v kim loi M c ha tr duy nht thnh 2phn bng nhau:

- Phn 1: Tan va trong 2 lt dung dch HCl thy thot ra 14,56 lt H2 (ktc).

- Phn 2: Tan hon ton trong dung dch HNO3 long nng thy thot ra 11,2

lt kh NO duy nht (ktc)

a. Nng mol ca dung dch HCl l

A. 0,45 M B. 0,25 M

C. 0,55 M D. 0,65 M

b. Khi lng hn hp mui clorua khan thu c khi c cn dung dch sau phn

ng phn 1 l

A. 65,54 gam B. 68,15 gam

C. 55,64 gam D. 54,65 gam

c. % khi lng ca Fe trong hn hp ban u l

A. 49,01 % B. 47,97 %

C. 52,03 % D. 50,91 %

d. Kim loi M l

A. Mg B. Zn

C. Al D. Cu


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Hng dn gii.

a. = 0,65 (mol) n2H

n HCl = 2nH = 2.0,65 = 1,3 mol

CM =1,3

2= 0,65 M.

p n D

b. mmui = mKl + Clm

Trong 1 3HClCln n , m= =

olmmui = 22 + 1,3.35,5 = 68,15 gam

p n B

c. p dng phng php bo ton e

- Phn 1:

Fe - 2e Fe2+

x ........ 2.x

M - ae Ma+

y ........ a.y

2H+ + 2e H2

1,3 ........... 0,65

- Phn 2:

Fe - 3e Fe3+

M - ae Ma+

N+5 + 3e N+2 (NO)

1,5 0,5

2x ay 1,3

3x ay 1,5

+ =

+ =


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x = 0,2, ay = 0,9

nFe = 0,2 % mFe = =0,2.56

.100% 50,91%22

p n D

d. mM = 22 0,2.56 = 10,8 gam

nM = y =0,9

a; = = =

m 10,8.aM 12.a

n 0,9

Vy a = 2, M = 24 (Mg) l ph hp

p n A

Bi 6.Mt hn hp gm 3 kim loi Al, Fe, Mg c khi lng 26,1 gam c chialm 3 phn u nhau.

- Phn 1, cho tan ht trong dung dch HCl thy thot ra 13,44 lt kh.

- Phn 2, cho tc dng vi dung dch NaOH d thu c 3,36 lt kh.

- Phn 3, cho tc dng vi dung dch CuSO4 d, lc ly ton b cht rn thu

c sau phn ng em ho tan trong dung dch HNO3nng d th thu c V lt

kh NO2. Cc kh u c o iu kin tiu chun.

Th tch kh NO2 thu c l

A. 26,88 lt B. 53,70 lt

C. 13,44 lt D. 44,8 lt

Hng dn gii.

2Al + 6HCl AlCl3 + 3H2

Mg + 2HCl MgCl2 + H2

Fe + 2HCl FeCl2 + H2

Khi lng ca mi phn m = =26,1

8,7 gam3

t s mol Al, Mg, Fe trong 17,4 gam hn hp l x, y, z


39/71

+ + = = + + = =

= =

27x 24y 56z 8,7 x 0,11,5x y z 0,3 y 0.075

1,5x 0,15 z 0,075

Trong 34,7 gam hn hp : nAl = 0,4; nMg = 0,3; nFe = 0,3

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

Mg + CuSO4 MgSO4 + Cu

Fe + CuSO4 FeSO4 + Cu

Cu + 4HNO3 Cu(NO3)2 + 2NO2 + H2O

phn 3, khi cc kim loi tc dng vi dung dch CuSO4 to thnh Cu,

lng Cu ny tc dng vi HNO3 to ra Cu2+, do :

- Al, Mg, Fe l cht kh, nhng electron

enhn 3.0,1 2.0,075 2.0,075 0,6 mole= + + = ng

- HNO3 l cht oxi ho, nhn electron

N

+5

+ 1e N

+4

(NO2)a .................... a

a = 0,6

=2NO

n 0,6 mol = =2NO

V 0,6.22, 4 13, 44 lt

p n C

Bi 7.Cho tan hon ton 3,6 gam hn hp gm Mg v Fe trong dung dch HNO 32M, thu c dung dch D, 0,04 mol kh NO v 0,01 mol N2O. Cho dung dch D

tc dng vi dung dch NaOH ly d, lc v nung kt ta n khi lng thu cm gam cht rn.

a. Gi tr ca m l

A. 2,6 gam B. 3,6 gam

C. 5,2 gam D. 7,8 gam

b. Th tch HNO3 phn ng l


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A. 0,5 lt B. 0,24 lt

C. 0,26 lt D. 0,13 lt

Hng dn gii.

a. - HNO3 l cht oxi ho

N+5 + 3e NO

0,12 ........ 0,04 (mol)

2N+5 + 8e 2N +1 (N2O)

0,08 ....... 0,02 ... 0,01 (mol)

enhn 0,12 0,08 0,2 mn = + = ol

3y

- Mg v Fe l cht kh. Gi x, y l s mol Mg v Fe trong hn hp

Mg - 2e Mg +2

x .........2x (mol)

Fe - 3e Fe +3

y ...... 3y (mol)

enhn 2xng = +

Ta c h phng trnh :24x 56y 3,6

2x 3y 0,2

+ =

+ =

Gii h ra x = 0,01 mol Mg 0,01 mol MgO

y = 0,06 mol Fe 0,03 mol Fe 2O3

m = khi lng MgO + Fe2O3 = 0,01.40 + 0,03.160 = 5,2 (gam)

p n C

Ta c th tnh theo cch sau : Ta c s hp thc Mg MgO, Fe Fe 2O3.

Trong Mg v Fe l cht kh, oxi l cht oxi ho, s mol e nhn vn l 0,2 mol.

O + 2e O2-

0,1....... 0,2


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m = mMg, Fe + mO = 3,6 + 16.0,1 = 5,2 (gam)

b. Theo nh lut bo ton nguyn t N ta c

3 23

3 3 2 3 3 2

3

N(HNO ) N(NO) N(N O)N(NO )

HNO Mg(NO ) Fe(NO ) NO N O

HNO

n n n n

hay n 2n 3n n 2n

2.0,01 3.0,06 0,04 2.0,01 0,26 mol

0,26V 0,13 (lit)

2

= + +

= + + + =

= + + + =

= =

p n DBi 8.Cho mt lung kh CO qua m gam bt Fe2O3 nung nng, thu c 14 gamhn hp X gm 4 cht rn. Cho hn hp X tan hon ton trong dung dch HNO 3

thu c 2,24 lit kh NO (ktc). Gi tr ca m l

A. 16,4 gam B. 14,6 gam

C. 8,2 gam D. 20,5 gam

Hng dn gii.

- CO l cht kh (ta coi Fe2O3 khng tham gia vo phn ng oxi ho kh, do Fe2O3

Fe(NO 3)3 c s oxi ho khng thay i)

m oxi trong oxit = m - 14 (gam)

CO Otrong oxit

m 14n n

16

= =

m 14 m 14............

16 8

+2 +4C + 2e C

- HNO3 l cht oxi ho

N+5 + 3e N+2

0,3 ......... 0,1 (mol)

Ta cm 14

0,3 m 16, 4 gam8

= =

p n A.


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Bi 9.Cho tan hon ton 58 gam hn hp A gm Fe, Cu, Ag trong dung dchHNO3 2M thu c 0,15 mol NO, 0,05 mol N2O v dung dch D. C cn dung

dch D, khi lng mui khan thu c l

A. 120,4 gam B. 89,8 gam

C. 116,9 gam D. kt qu khc

Hng dn gii.

Nhn xt : Nu ch dng phng php bo ton electron thng thng, ta cng

ch lp c 2 phng trnh 3 n s v s gp kh khn trong vic gii. tnh

khi lng mui NO3- trong bi ton trn ta c cng thc

3XNO (trongmu

n ai)

= .n

Trong a l s electron m N+5 nhn to thnh X

Nhvy :

23NO N ONO

n 3.n 8.n 3.0,15 8.0, 05 0,95 mol

-3

mui khan Fe, Cu, Ag NOm = m + m

= + = + =

mmui khan = 58 + 0,95.62 = 116,9 gam

p n C

Bi 10. Kh Fe2O3 bng CO nhit cao, c hn hp X gm 4 cht rn.Chia X thnh 2 phn bng nhau. Phn mt tc dng vi dung dch HNO3 d, thu

c 0,02 mol NO v 0,03 mol N2O. Phn hai cho tan hon ton trong dung dch

H2SO4 c nng, thu c V lt (ktc) SO2. Gi tr ca V l

A. 2,24 B. 3,36

C. 4,48 D. 6,72Hng dn gii.

- HNO3 l cht kh


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5 2

5 1

2

N 3e N

0,06........0,02 mol

N 8e 2N (N O)

0,24........0,06....0,03 mol

+ +

+ +

+

+

enhn 0,06 0,24 0,3 mn = + = ol

- Cht kh hai phn l nh nhau, do s mol electron H2SO4 nhn bng s mol

electron HNO3 nhn, hay

S+6 + 2e S+4 (SO2)

0,3 ..................... 0,15

2SOV 0,15.22,4 3,36 (l)= =

p n B

Bi 11. Chia hn hp X gm Al, Al2O3, ZnO thnh hai phn bng nhau. Phnmt cho tc dng vi dung dch NaOH d, thu c 0,3 mol kh. Phn hai tan

hon ton trong dung dch HNO3 thu c 0,075 mol kh Y duy nht. Y l

A. NO2 B. NOC. N2O D. N2

Hng dn gii.

- Trong X ch c Al c tnh kh, nc b nhm kh theo phng trnh :

2H2O + 2e H 2 + 2OH-

0,6 ....... 0,3 (mol)

- Khi tc dng vi HNO3, cht oxi ho l HNO3

N+5 + ne Y

0,075.n .... 0,075 (mol)

Ta c 0,075.n = 0,6, vi n l s electron m N+5 nhn to thnh Y, n = 8.

Vy Y l N2O

p n C


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Bi 12. Cho tan hon ton 3,76 gam hn hp X dng bt gm S, FeS v FeS 2trong dung dch HNO3 thu c 0,48 mol NO2 v dung dch D. Cho dung dch D

tc dng vi dung dch Ba(OH)2 d, lc v nung kt ta n khi lng khng i,

c m gam hn hp rn. Gi tr ca m l

A. 11,650 gam B. 12,815 gam

C. 13,980 gam D. 15,145 gam

Hng dn gii.

Fe+2S2-1 tng ng vi Fe+2S-2.S0. V vy c th coi hn hp X gm hai cht S

v FeS, c s mol l a v b, ta c :

S gam X = 32a + 88b = 3,76 (I)

- Cht kh

0 6

0 3

S 6e S

a ........ 6a

FeS 9e Fe S

b .............9b

+

6+ +

+

- Cht oxi ho

N+5 + 1e N +4 (NO2)

0,48 ............. 0,48

Ta c 6a + 9b = 0,48 (II)

Gii h (I), (II) c : a = 0,035 mol S v b = 0,03 mol FeS

24 4

4

BaSO S FeSSO

BaSO

n n n n 0,035 0,03 0,065 (mol)

m 0,065.233 15,145 gam

= = + = + =

= =

p n D

Bi 13. Cho tan hon ton 7,2 gam FexOy trong dung dch HNO3 thu c 0,1mol NO2. Cng thc phn t ca oxit l

A. FeO B. Fe3O4

C. Fe2O3 D. c FeO v Fe3O4 u ng


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Hng dn gii.

- Cht oxi ho l HNO3

N+5 + 1e N +4 ..... NO2

0,1 ....... 0,1 .... 0,1

- Cht kh l FexOy

2y

3x2y

xFe x.(3 )e xFex

7,2 7,2............. .( 3x 2y)56x 16y 56x 16y

++

+ +

Ta c7, 2

.(3x 2y) 0,156x 16y

=+

72.(3x - 2y) = 56x + 16y

160x = 160y x = y : FeO

p n A


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Chuyn 7Lp s hp thc ca qu trnh chuyn ho,tm mi quan h gia cht u v cht cui

I - Ni dung

i vi cc bi ton hn hp bao gm nhiu qu trnh phn ng xy ra, tach cn lp s hp thc, sau cn c vo cht u v cht cui, b qua cc

phn ng trung gian.

V d.

- Cho hn hp A gm cc cht rn Fe, FeO, Fe3O4, Fe2O3 tan hon tontrong dung dch HCl, dung dch thu c cho tc dng vi dung dch NaOH d,lc kt ta, nung trong khng kh n khi lng khng i thu c m gam chtrn. Tnh m.

Ta thy, cht cui cng l Fe2O3, vy nu tnh c tng s mol Fe c trongA ta s tnh c s mol Fe2O3.

- Cho hn hp Fe, Zn, Mg tc dng ht vi dung dch HCl, cho t t dungdch NaOH vo dung dch thu c n kt ta ln nht, lc kt ta, nung trongkhng kh n khi lng khng i thu c m gam cht rn, tnh m

Ta thy, nu bit c s mol cc kim loi ban u, ta lp c s hpthc gia cht u v cui Fe Fe 2O3, Zn ZnO, Mg MgO ta s tnhc khi lng cc oxit.

II - Bi tp p dng

Bi 1. Cho 11,2 gam Fe v 2,4 gam Mg tc dng vi dung dch H2SO4 long d.Sau phn ng thu

c dung dch A v V lt kh H2 ( ktc). Cho dung dch NaOHd vo dung dch A thu c kt ta B. Lc B nung trong khng kh n khilng khng i c m (gam) cht rn.

a. V c gi tr l

A. 2,24 lt B. 3,36 lt

C. 5,6 lt D. 6,72 lt


47/71

b. Gi tr ca m l

A. 18 gam B. 20 gam

C. 24 gam D. 36 gam

Hng dn gii

2

2

H Mg Fe

H

2,4 11,2a. n n n 0,3 mol

24 56V 0,3.22,4 6,72 lt

= + = + =

= =

p n D

b. Da vo s thay i cht u v cui, ta lp c s hp thc :

2Fe Fe2O3 ; Mg MgO

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

m = 0,1.160 + 0,1.40 = 20 (g)

p n B

Bi 2. Ha tan hon ton hn hp gm 0,2 mol Fe v 0,1 mol Fe 2O3 vo dungdch HCl d thu c dung dch A. Cho dung dch A tc dng vi NaOH d thuc kt ta. Lc ly kt ta, ra sch, sy kh v nung trong khng kh n khi

lng khng i c m gam cht rn, m c gi tr l

A. 23 gam B. 32 gam

C. 24 gam D. 42 gam



Fe2O3 + 6HCl 2FeCl3 + 3H2O

HCl + NaOH NaCl + H2O



4Fe(OH)2 + O2 + 2H2O 4Fe(OH)3


48/71


Trong m gam cht rn c 0,1 mol Fe2O3 (16 gam) ban u. Vy ch cn tnh

lng Fe2O3 to ra t Fe theo mi quan h cht u (Fe) v cui (Fe2O3)

2Fe Fe2O3.2 3Fe O Fe

1n n 0,1

2= = mol

0,2

2 3Fe O (thun 0,1 0,1

m 0,2.160 32 gam

c) = + =

= =

p n B

Bi 3. Hn hp Al, Fe c khi lng 22 gam c chia thnh 2 phn bng nhau.- Phn 1 tc dng vi HCl d thu c dung dch A v 8,96 lt H2 (ktc).

Cho dung dch A tc dng dung dch NaOH d c kt ta B, lc kt ta B nung

trong khng kh n khi lng khng i c m1 cht rn.

- Phn 2 cho vo dung dch CuSO4 d n khi phn ng hon ton thu c

m2 gam cht rn khng tan.

a. m1 c gi tr l

A. 8 gam B. 16 gam

C. 32 gam D. 24 gam

b. m2 c gi tr l

A. 12,8 gam B. 16 gam

C. 25,6 gam D. 22,4 gam

Hng dn gii.

a. 2Al + 6HCl 2AlCl3 + 3H2


HCl + NaOH NaCl + H2O

AlCl3 + 3NaOH Al(OH)3 + 3NaCl

Al(OH)3 + NaOH NaAlO2 + 2H2O


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4Fe(OH)2 + O20t 2Fe2O3 + 4H2O

- Khi tc dng vi HCl, gi x, y ln lt l s mol Al v Fe ta c:

+ = =

+ = =

27x 56y 11 x 0,2 : Al

1,5x y 0,4 y 0,1 : Fe

- Sau cc phn ng cht rn thu c ch cn l Fe2O3.

2Fe Fe2O3

0,1....... 0,05 m1 = 8 (g)

p n A

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

Fe + CuSO4 FeSO4 + Cu

phn 2, Cu2+ nhn electron chnh bng H+ nhn phn 1, do

nCu = = 0,4 m2H

n Cu = 25,6 (g)

p n C

Bi 4. Cho tan hon ton 13,6 gam hn hp gm Fe v Fe2O3 trong dung dchHCl thu c 2,24 lt H2 (ktc) v dung dch D. Cho D tc dng vi dung dchNaOH d, lc, nung kt ta trong khng kh n khi lng khng i c a gamcht rn, gi tr ca a l

A. 8 gam B. 12 gam

C. 16 gam D. 24 gam

Hng dn gii. Fe + 2HCl FeCl 2 + H2

2 2 3Fe H Fe O (bann n 0,1 mol. m 13,6 0,1.56 8gamu)= = = =

Ta c s hp thc 2Fe Fe 2O3.2 3Fe O Fe

1n n 0,05 m

2= = ol.

Vy a = 8 + 0,05.160 = 16 gam

Cng c th dng phng php tng - gim khi lng.


50/71

2Fe Fe 2O3. Khi lng tng ln 48 gam0,1 mol Fe 0,05 mol Fe 2O3 khi lng tng 2,4 gam

a = 13,6 + 2,4 = 16 gam

p n C

Bi 5. 7,68 gam hn hp gm FeO, Fe3O4 v Fe2O3 tc dng va ht vi 260 mldung dch HCl 1M thu c dung dch X. Cho X tc dng vi dung dch NaOH

d, lc kt ta, nung trong khng kh n khi lng khng i thu c m gam

cht rn, gi tr ca m l

A. 8 gam B. 12 gam

C. 16 gam D. 24 gam

Hng dn gii.

FeO + 2HCl FeCl2 + H2O

Fe3O4 + 8HCl 2FeCl3 + FeCl2 + 4H2O

Fe2O3 + 6HCl 2FeCl3 + 3H2O

FeCl2 + 2NaOH Fe(OH)2 + 2NaClFeCl3 + 3NaOH Fe(OH)3 + 3NaCl



- p dng phng php bo ton in tch tnh s mol Fe c trong cc oxit

O(trongoxit) Cl HCl

Fe (trongoxit)

1 1 1n n n .0,26 0,13 mol

2 2 2

7,68 0,13.16n 0,1

56

= = = =

= = mol

- Lp s hp thc 0,1 mol Fe 0,05 mol Fe2O3. m = 0,05 .160 = 8 gam

p n A


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Chuyn 8Phng php giI bI tp v stI ni dung

- St l mt trong nhng nguyn t quan trng, c nhiu ng dng trong i

sng v c mt v tr quan trng trong chng trnh Ho hc ph thng cng nh

trong cc k thi Tt nghip THPT, Cao ng, i hc.

- Ngoi cc phng php nu cc chuyn trn, cc bi tp v st v

hp cht ca st cn c th s dng thm mt s cch gii nhanh sau y :

+ Khi Fe3O4 tc dng vi cc cht oxi ho, ta coi Fe3O4 l hn hp ca Fe2O3

v FeO. Trong ch c FeO tham gia phn ng oxi ho - kh vi =3 4FeO Fe O

n n

+ V tr ca Fe trong dy in ho+ +

+>

2 3

2

Fe Fe

Fe Fe. Do trong cc phn ng

c th xy ra theo nhiu trng hp khc nhau.

+ Trong bi ton tm cng thc phn t ca oxit st, cn tm s mol Fe v s

mol oxi c trong oxt ri lp t l mol Fe : O, t suy ra cng thc phn t.+ S dng phng php bo ton electron vi bi ton cho mt oxit st

FexOy tc dng vi dung dch HNO3 to ra sn phm kh do s kh N+5.

II Bi tp p dngBi 1.C mt loi oxit st dng luyn gang. Nu kh oxit st ny bngcacbon oxit nhit cao ngi ta thu c 0,84 gam st v 0,448 lt kh

cacbonic(ktc). Cng thc ho hc ca loi oxit st ni trn l

A. Fe2O3. B. Fe3O4C. FeO D. Khng xc nh c

Hng dn gii. Ta thy, CO ly O ca oxit to ra CO2, do

= = = =2O CO CO

0,448n n n 0,02 mo

22,4l


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= =Fe 0,84n 0,015 mol56, nFe : nO = 0,015 : 0,02 = 3 : 4.

Vy cng thc ca oxit l Fe3O4.

p n B

Bi 2. ha tan hon ton 10,8 gam oxit st cn va 300 ml dung dch HCl1M. Oxit st l

A. FeO B. Fe2O3

C. Fe3O4 D. C FeO v Fe3O4 u ng

Hng dn gii. Theo nh lut bo ton in tch, Cl- thay th O trong oxit nn

= = =Otrongoxit HClCl1 1

n n n 0,152 2

mol

= = =oxit oxiFe

m m 10,8 0,15.16n 0

56 56,15 mol

nFe : nO = 1 : 1. Vy CTPT l FeO

p n A

Bi 3.Ho tan ht m gam hn hp gm FeO, Fe3O4 v Fe2O3 c s mol bng nhautrong dung dch HNO3 thu c 2,688 lt NO (ktc). Gi tr ca m l

A. 70,82 gam B. 83,52 gam

C. 62,64 gam D. 41,76 gam

Hng dn gii.

Gi s mol ca mi oxit l x mol. Coi Fe3O4 l hn hp FeO v Fe2O3. Do

, hn hp gm FeO v Fe2O3 u 2x mol.

Khi tc dng vi HNO3 ch c FeO tham gia phn ng oxi ho kh to NO.

Fe+2 - 1e Fe +3

2x .......... 2x

N+5 + 3e N+2

0,36 ..... 0,12

2x = 0,36 x = 0,18 mol


53/71

= + = + =2 3FeO Fe Om m m 2.0,18. (72 160) 83,52 gam

p n B

Bi 4.Ho tan hon ton 8,64 gam mt oxit st trong dung dch HNO3 thu c0,896 lt NO (ktc) duy nht. Oxit st l

A. FeO B. Fe2O3

C. Fe3O4 D. C FeO v Fe3O4 u ng

Hng dn gii.

Trong oxit FexOy , s oxi ho ca Fe l +2y/x. p dng phng php bo

ton electron, ta c

- FexOy l cht kh

2y3x

2yFe x.(3 ) Fe

x8,64 8,64.(3x 2y)

..........56x 16y 56x 16y

++

+ +

- HNO3

l cht oxi ho

N+5 + 3e N +2

0,12 . 0,04

Ta c

8, 64.(3x 2y)0,12 72(3x 2y) 56x 16y

56x 16y

x y

= = +

+

=

Vy oxit l FeO.

p n A.

Bi 5.Cho ming st nng m gam vo dung dch HNO 3, sau phn ng thy c6,72 lt kh NO2 (ktc) thot ra v cn li 2,4 gam cht rn khng tan. Gi tr ca

m l

A. 8,0 B. 5,6


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C. 10,8 D. 8,4

Hng dn gii.

Sau phn ng st cn dnn c cc phn ng

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

0,1 ................................ 0,1 ............... 0,3

Fe + 2Fe(NO3)3 3Fe(NO 3)2

0,05 ............. 0,1

Lng st c hai phn ng l nFe = 0,01 + 0,005 = 0,015 mol

m = 0,15.56 + 2,4 = 10,8 gam

p n C


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Chuyn 9Phng php giI bI tp v nhmI ni dung

Vi cc bi ton ho hc v nhm, hp cht ca nhm cng nh cc bi ton

hn hp. Ngoi vic s dng cc phng php gii nh bo ton electron, bo ton

khi lng, tng gim khi lng ... trnh by cc chuyn trc, cn c

mt s dng bi tp c trng ring ca nhm, l :

1. Mui Al3+ tc dng vi dung dch kim to kt ta

Khi cho mt lng dung dch cha OH- vo dung dch cha Al3+ thu c

kt ta Al(OH)3. Nu 33Al(OH) Al

n n +< s c hai trng hp ph hp xy ra. Bi ton

c hai gi tr ng.

- Trng hp 1. Lng OH- thiu, ch to kt ta theo phn ng

Al3+ + 3OH- Al(OH) 3

Lng OH- c tnh theo kt ta Al(OH)3, khi gi tr OH- l gi tr nh nht.

- Trng hp 2. Lng OH- xy ra hai phn ng :

Al3+ + 3OH- Al(OH) 3 (1)

Al(OH)3 + OH- AlO2

- + 2H2O (2)

Trong , phn ng (1) hon ton, phn ng (2) xy ra 1 phn. Lng OH-

c tnh theo c (1) v (2), khi gi tr OH- l gi tr ln nht.

2. Dung dch H+ tc dng vi dung dch AlO2- to kt ta

Khi cho t t dung dch cha OH- vo dung dch cha Al3+ thu c kt ta

Al(OH)3. Nu s c hai trng hp ph hp xy ra. Bi ton c hai

gi tr ng.

33Al(OH) Aln n + 7 D. pH = 14

Bi 47.Cho 200 ml dung dch hn hp CuSO4 1M v Al2(SO4)3 1M tc dng vidung dch NaOH d, lc ly kt ta em nung n khi lng khng i c cht

rn c khi lng l

A. 8 gam B. 18 gam

C. 19,8 gam D. 36,4 gam

Bi 48.Nu hm lng Fe l 70% th l oxit no trong s cc oxit sau :A. FeO B. Fe2O3

C. Fe3O4 D. Khng c oxit no ph hp

Bi 49.Trn 40 gam Fe2O3 vi 10,8 gam Al ri nung nhit cao, hn hp sauphn ng ha tan vo dung dch NaOH d thu c 5,376 lt kh (ktc). Hiu sut

ca phn ng nhit nhm l

A. 12,5 % B. 60 %


71/71

C. 40 % D. 16,67 %

Bi 50.Cho 2,06 gam hn hp gm Fe, Cu tc dng vi dung dch HNO 3 long,d thu c 0,896 lt NO (ktc). Khi lng mui nitrat sinh ra l

A. 9,50 gam B. 7,44 gam

C. 7,02 gam D. 4,54 gam

p n phn bi tp t luyn

1 B, C, A / 2 B, B / 3 C / 4 A / 5 D / 6 B, C / 7 A, B / 8 D / 9 B / 10 A / 11 A / 12 C

/ 13 B / 14 A / 15 C / 16 B / 17 D / 18 A / 19 C / 20 C / 21 C / 23 B / 24 C / 25 B /

26 A / 27 A / 28 B / 29 C / 30 A / 31 D / 32 A / 33 C / 34 B / 35 C / 36 B / 37 C /

38 C / 39 A / 40 B / 41 A / 42 A / 43 A / 44 D / 45 D / 46 A / 47 D / 48 B / 49 C /

50 A

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