Tài liệu Bồi dưỡng giáo viên đổi mới PPGD và KTĐG môn Hóa học

8/3/2019 Ti liu Bi dng gio vin i mi PPGD v KTG mn Ha hc

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b gio dc v o to

ti liubi dng gio vin

i mi ppdh v ktgMn Ha hc

h ni thng 10 nm 2009


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phng php gii mt s bi tp ha hcc ni dung nh lng

1. Phng php p dng s bo ton khi lng, s mol nguyn tC s

Trong cc qu trnh ho hc th :Tng khi lng ca cc cht trc phn ng lun bng tng khi lngca cc cht sau phn ng :

(tr c phn ng) (sau phn ng)m m= Tng s mol nguyn t ca nguyn t A trc phn ng lun bng tngs mol nguyn t ca nguyn t A sau phn ng.

= A(trc phn ng) A(sau phn ng)n nCch p dngKhi gii bi tp trc nghim ta nn lp s tm tt cc phn ng, ri pdng nhng s bo ton trn tm ra cc i lng khc nh : s mol, khi

lng cc cht trong s phn ng th bi ton s c gii nhanh hn.

Bi tp minh haBi 1. Ngi ta cho t t lung kh CO i qua mt ng s ng 5,44 g hn hp A

gm FeO, Fe3O4, Fe2O3, CuO nung nng, kt thc phn ng thu c hnhp cht rn B v hn hp kh C. Sc hn hp kh C vo dung dch nc vitrong d thy c 9 g kt ta v kh D bay ra. Khi lng cht rn B thu clA. 3g B. 4g C. 5g D. 3,4gLi giiS phn ng:

FeO

CO + Fe2O3ot

A + CO2

Fe3O4CuO

CO2 + Ca(OH)2 CaCO3 + H2O

0,099

0,09(mol)100

=

Theo nh lut BTKL th2CO A B CO

m m m m+ = +

0,09.28 + 5,44 = mB

+ 0,09.44 m = 4g


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Bi 2. Nung nng m g hn hp X gm ACO3 v BCO3 thu c m g hn hp rnY v 4,48 lt kh CO2. Nung nng Y n khi lng khng i thu thmc kh CO2 v hn hp rn Z. Cho ton b kh CO2 thu c khi nung Yqua dung dch NaOH d, sau cho dung dch BaCl2 d vo dung dch trnth thu c 19,7 g kt ta. Mt khc cho CO d qua hn hp Z nung nng

thu c 18,4 g hn hp Q v 4,48 lt kh CO2

(ktc). m c gi tr l

A. 34,8 g B. 25,7g C. 44,1g D. 19,8gLi giiS phn ng nhit phn :

3

3

ACO

BCO Y + CO2 (1)

Y 0t

Z + CO2 (2)

2BaClNaOH 22 3 3CO CO BaCO

19,70,1 0,1(mol)197

=

2CO Z Q CO+ + (3)

Bn cht ca s (3) l :

CO + O(trong Z) CO2

m(trong Z) =4,48

0,2(mol)22,4

=

Z Q Om m m 18,4 0,2.16 21,6(gam) = + = + =

2Y Z COm m m 21,6 0,1.44 26(gam) = + = + =

2X Y COm m m 26 0,2.44 34,8(gam) = + = + =

Bi 3. Ho tan hon ton hn hp gm 0,2 mol FeO, 0,3 mol Fe2O3, 0,4 molFe3O4vo dung dch HNO3 2M va , thu c dung dch mui v 5,6 ltkh hn hp kh NO v N2O4 (ktc) c t khi so vi H2 l 33,6. Th tch

dung dch HNO3 tham gia phn ng lA. 3,6 lt B. 2,4 lt C. 3,2 lt D. 4,8 ltLi giiS phn ng :

(FeO; Fe2O3; Fe3O4) + HNO3 Fe(NO3)3 + (NO ; N2O4) + H2O

t 2 4NO N On x(mol) ; n y (mol)= =


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5,6x + y = 0,25

x 0,1mol22,4Ta c h

30x 92y y 0,15 mol33,6

2(x y)

= =

+ = =

+

p dng s bo ton nguyn t Fe tnh s mol Fe(NO3)3 :

3 3 2 3 3 4

3 3 2 3 3 4

Fe(Fe(NO ) Fe(FeO,Fe O ,Fe O )

Fe(NO ) FeO Fe O Fe O

n n

n n 2n 3n 0,2 2.0,3 3.0,4 2 mol

=

= + + = + + =

p dng s bo ton nguyn t N :

3 3 3 2 4 3N(HNO ) N(Fe(NO ) NO N O ) HNOn n n 3.2 0,1 2.0,15 6, 4 mol+ += = + + =

Vy3HNO

6,4V 3,2 lt

2= =

Bi 4. Ha tan hon ton hn hp gm 0,12 mol FeS2 v a mol Cu2S vo dung

dch HNO3 (va ), thu c dung dch X (ch cha 2 mui sunfat) v mt

kh duy nht l NO. Gi tr ca a lA. 0,12 mol B. 0,04 mol C. 0,075 mol D. 0,06 molLi gii

S phn ng :

2

2

FeS

Cu S+ HNO3

2 4 3

4

Fe (SO )

CuSO+ NO + H2O

p dng nh lut bo ton nguyn t Fe, Cu, S

FeS2 Fe2(SO4)30,12 0,06

Cu2S CuSO4

a 2a2 2 2 4 3 4S(FeS ) S(Cu S) S(Fe (SO ) ) S(CuSO )

n n n n+ = +

2 2 2 4 3 4FeS Cu S Fe (SO ) CuSO2n n 3n n+ = +

2.0,12 + a = 3.0,06 + 2a a = 0,06 mol

Bi 5. Kh hon ton m g hn hp CuO, Fe3O4 bng kh CO nhit cao, thu

c hn hp kim loi v kh CO2. Sc kh CO2 vo dung dch Ca(OH)2 thu

c 20 g kt ta v dung dch A, lc b kt ta, cho Ba(OH)2 d vo dung

dch A thu c 89,1 g kt ta na. Nu dng H2 kh hon ton m g hn

hp trn th cn bao nhiu lt kh H2 (ktc) ?

A. 16,46 lt B. 19,72 lt C. 17,92 lt D. 16,45 ltLi giiS phn ng :


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3 4

CuO

Fe O+ CO

ot

Cu

Fe+ CO2 (1)

Cho CO2 vo dung dch Ca(OH)2 th

2

2

3

Ca(OH)2

Ba(OH)3 2 3 3

20CaCO 0,2(mol)

100

CO (2)

Ca(HCO ) CaCO BaCO

x x

=

+

100x 197x 89,1 x 0,3(mol)+ = =

p dng s bo ton nguyn t C

2 3 3

2 3 3

C(CO) C(CO ) C(CaCO ) C(BaCO )

CO CaCO BaCO

n n n n

n n n (0,2 0,3) 0,3 0,8(mol)

= = +

= + = + + =

Bn cht cc phn ng xy ra trong (1) l :

CO + O(oxit) CO20,8 0,8 0,8

Nu dng H2 kh m g hn hp CuO, Fe3O4 th bn cht cc phn ng l

H2 + O(oxit) H2OTng s mol nguyn t oxi trong hai qu trnh ny bng nhau nn

2 2H O H (ktc)n n 0,8(mol) V 0,8.22,4 17,92 (lit)= = = =

Bi tp vn dngBi 1. tc dng ht 5,44 g hn hp CuO, FeO, Fe2O3 v Fe3O4 cn dng va

90ml dung dch HCl 1M. Mt khc, nu kh hon ton 5,44 g hn hptrn bng kh CO nhit cao th khi lng st thu c lA. 3,20g B. 4,72 g C. 2,11 g D. 3,08 g

Bi 2. Cho hn hp gm : FeO (0,01 mol), Fe2O3 (0,02 mol), Fe3O4 (0,03 mol) tanva ht trong dung dch HNO3 thu c dung dch cha mt mui v 0,448lt kh N2O4 (ktc). Khi lng mui v s mol HNO3 tham gia phn ng lA. 32,8 g ; 0,4 mol B. 33,88 g ; 0,46 molC. 33,88 g ; 0,06 mol D. 33,28 g ; 0,46 mol

Bi 3. Cho 1,1 g hn hp Fe, Al phn ng vi dung dch HCl thu c dungdch X, cht rn Y v kh Z, ho tan ht Y cn s mol H2SO4 (long)bng 2 ln s mol HCl trn, thu c dung dch T v kh Z. Tng thtch kh Z (ktc) sinh ra trong c hai phn ng trn l 0,896 lt. Tng khilng mui sinh ra trong hai trng hp trn l


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A. 2,92 g B. 2,67 g C. 3,36 g D. 1,06gBi 4. Ho tan hon ton 3,72g hn hp 2 kim loi A, B trong dung dch HCl d

thy to ra 1,344 lt kh H2 (ktc). C cn dung dch sau phn ng thu cmui khan c khi lng lA. 7,12g B. 7,98g C. 3,42g D. 6,12g

Bi 5. Cho hn hp A gm 0,1 mol Cu, 0,2 mol Ag phn ng ht vi V lt dung

dch HNO3 1M thu c dung dch X v hn hp Y gm 2 kh NO, NO 2(

2NO NOn n 0,1 mol= = ). V c gi tr l

A. 1 lt B. 0,6 lt C. 1,5 lt D. 2 ltBi 6. t chy m g hp cht A (CnHn1ONa) vi mt lng va l 6,272 lt O2

(ktc) thu c 2,12 g Na2CO3 v hn hp X cha CO2, H2O. Nu cho hnhp X qua bnh ng H2SO4 c th khi lng bnh tng 1,8 g. Vy m cgi tr lA. 6,46 B. 4,64 C. 4,46 D. 6,44

Bi 7. Thu phn hon ton 1 este n chc A cn va 100ml NaOH 1M thuc ancol etylic v mui ca axit hu c B. Phn hu hon ton B thu c

5,6 lt kh CO2 (ktc), 4,5 g H2O v m g Na2CO3. Cng thc cu to ca A lA. C2H5COOC2H5 B. CH3COOC3H7

C. C3H7COOC2H5 D. C3H7COOCH3Bi 8. Cho 13,8g hn hp gm ancol etylic v glixerol tc dng va vi Na thu

c 4,48 lt H2 (ktc) v dung dch mui. C cn dung dch mui, khilng cht rn thu c lA. 22,6 g B. 22,4 g C. 34,2 g D. 25,0 g

Bi 9.un 13,8 g hn hp 3 ancol no, n chc vi H2SO4 c 1400C thu c

11,1g hn hp cc ete c s mol bng nhau. Tnh s mol mi ete.A. 0,025 mol B. 0,1 mol

C. 0,15 mol D. 0,2 molBi 10. t chy hon ton mt cht hu c A cha 1 nguyn t oxi thu c hn

hp sn phm B. Cho B i qua dung dch Ca(OH)2 d thy c 15 g kt ta vkhi lng dung dch gim 4,8 g. CTPT ca A l

A : CH4O B : C2H6O

C : C3H8O D : C4H10O

2. Phng php tng gim khi lngC sKhi mt nguyn t hay nhm nguyn t X trong cht tham gia phn ng(gi l cht u) c thay th bng mt nguyn t hay nhm nguyn t Y


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to ra cht mi (cht cui), th s chnh lch khi lng gia cht u v chtcui chnh bng hiu khi lng ca hai nhm nguyn t X v Y (|XY|).

Th d : CaCO3 CaSO4

Ta thy th s chnh lch khi lng gia hai mui CaCO3 v CaSO4 :M (40 96) (40 60) 36 g / mol = + + = ng bng s chnh lch khi

lng ca hai anion23CO (60g)

v24SO

(96g): M 96 60 36g / mol = = .

Cch p dngKhi mt cht thay anion c bng anion mi sinh ra cht mi th s chnhlch khi lng gia cht c v cht mi chnh l s chnh lch khi lngca anion c v anion mi.Khi mt cht thay cation c bng cation mi sinh ra cht mi th s chnhlch khi lng gia cht c v cht mi chnh l s chnh lch khi lngca cation c v cation mi.

Bi tp minh hoBi 1. Cho 41,2 g hn hp X gm Na2CO3, K2CO3 v mui cacbonat ca kim loi

ho tr 2 tc dng vi dung dch H2SO4 d. Kt thc phn ng thu c hn

hp Y gm ba mui sunfat v 8,96 lt kh CO2 (ktc). Khi lng ca Y lA. 58,6 g B. 55,6 g C. 45,0 g D. 48,5 gLi giiS phn ng :

2 3

2 3

3

Na CO

K CO

MCO

+ H2SO4 2 4

2 4

4

Na SO

K SO

MSO

+ H2O + CO2

1 mol X chuyn thnh 1 mol Y th tng khi lng lM 96 60 36(g / mol) = =

Theo nh lut bo ton nguyn t C : 2 23 COCOn n 0,4(mol) = = khi

lng Y ln hn khi lng ca X l 0,4.36 = 14,4 (g)

Vy mY = 41,2 + 14,4 =55,6 (g)

Bi 2. Cho 84,6 g hn hp A gm BaCl2 v CaCl2 vo 1 lt hn hp Na2CO30,3M v (NH4)2CO3 0,8 M. Sau khi cc phn ng kt thc ta thu c79,1 g kt ta A v dung dch B. Phn trm khi lng BaCl2 v CaCl2trong A ln lt l

A. 70,15% ; 29,25% B. 60,25% ; 39,75%C. 73,75%; 26,25% D. 75,50% ; 24,50%Li gii


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t2 2BaCl CaCl

n x(mol); n y(mol)= =

2

2

BaCl

CaCl+ 2 3

4 2 3

Na CO

(NH ) CO 3

3

BaCO

CaCO+

4

NaCl

NH Cl

C 2 mol Clmt i (71 g) c 1 mol mui 23CO thm vo (60 g)

chnh lch (gim) khi lng ca 1 mol mui l M = 71 60 =11(g)

gim khi lng mui : m = 84,6 79,1 = 5,5 (g)Vy s mol mui clorua bng s mol mui cacbonat phn ng

= =5,5

0,5 (mol)11

M s mol CO32 (theo gi thit) = 0,3 + 0,8 = 1,1(mol) > 0,5 mol (phn

ng). Vy mui cacbonat phn ng d.x + y = 0,5 (1)

208x + 111y = 84,6 (2)

x 0,3mol

y 0,2 mol

=

=

= =

= =

2

2

BaCl

CaCl

0,3.208%m .100% 73,75%

84.6%m 100 73,75 26,25%

Bi 3. Hn hp A gm 10 g MgCO3, CaCO3 v BaCO3 c ho tan bng HCl dthu c dung dch B v kh C. C cn dung dch B c 14,4 g mui khan.Sc kh C vo dung dch c cha 0,3 mol Ca(OH)2 thu c s g kt ta lA. 10g B. 20g C. 30g D. 40gLi gii

CO32 + 2H+ CO2 + H2O

p dng phng php tng gim khi lng khi chuyn 23CO thnh Cl ta

tnh c s mol A =

= = =2 23 COCO14, 4 10

n n 0, 4 (mol)11

CO2 + Ca(OH)2 CaCO3 + H2O0,4 0,3 0,3

CO2 + H2O + CaCO3 Ca(HCO3)20,1 0,1

3CaCOm 0,2.100 20 g= =

Bi 4. Nhng mt thanh kim loi A (ho tr II) vo dung dch CuSO4. Sau phn ngkhi lng thanh kim loi A gim 0,12g. Mt khc cng thanh kim loi A c nhng vo dung dch AgNO3 d th kt thc phn ng khi lng thanh

tng 0,26g. Bit s mol A tham gia hai phn ng bng nhau. Kim loi A lA. Zn B. Mg C. Cd D. Fe


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Li giiPhng trnh phn ng :

A + Cu2+ d A2+ + Cua a

A + 2Ag+ d A2+ + 2Aga 2a

Khi lng thanh kim loi tng = mA mCu = 0,12ga.MA 64a = 0,12 MA.a = 64a + 0,12 (1)

Mt khc khi lng thanh kim loi gim = mAg + mA = 0,26 g2a.108 MA.a = 0,26 MA.a = 2a.108 0,26 (2)

x = 2,5.103 mol MA =3

3

64.2,5.10 0,12112(g / mol)

2,5.10

+=

Cht X l Cd.Bi 5. C 2 dung dch FeCl2 v CuSO4 c cng nng mol.

Nhng thanh kim loi vo M (nhm IIA) vo V lt dung dch FeCl2, kt

thc phn ng khi lng thanh kim loi tng 16g. Nhng cng thanh kim loi y vo V lt dung dch CuSO4 kt thc phnng khi lng thanh kim tng 20g. Gi thit cc phn ng xy ra hon tonv kim loi thot ra bm ht vo M. Kim loi M lA. Zn B. Mg C. Cd D. Fe

Li giiCc phng trnh phn ng xy ra :

M + Fe2+ M2+ + Fex x x

M + Cu2+ M2+ + Cux x x

Theo gi thit cc phn ng u xy ra hon ton nn cc ion 2 2Fe v Cu+ +

phn ng ht 2 2Fe Cun n x(mol)+ += =

Khi lng thanh kim loi tng (1) l : m = mFe mM = 16g56x MM.x = 16 M.x = 56x 16

Khi lng thanh kim loi tng (2) l : m = mCu mM = 20 g64x M.x = 20 M.x = 64x 20

M = 24. Vy kim loi M l Mg.

Bi tp vn dngBi 1. Cho 20 g hn hp X gm hai axit cacboxylic no, n chc tc dng va

vi dung dch Na2CO3 thu c V lt kh CO2 (ktc) v dung dch mui. Ccn dung dch thu c 28,8 g mui. Gi tr ca V l


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A. 3,36 lt B. 4,48 lt C. 2,24 lt D. 6,72 ltBi 2. Cho 5,5 g hn hp 2 ancol n chc l ng ng k tip nhau tc dng va

vi Na kim loi to ra 8,8 g cht rn v V lt kh H2(ktc). Cng thc ca2 ancol lA. CH3OH v C2H5OH. B. C2H5OH v C3H7OH.

C. C3H7OH v C4H9OH. D. C2H3OH v C3H5OH.Bi 3. Thy phn 0,01mol este ca 1 ancol a chc vi 1 axit n chc tiu tn

ht 1,2g NaOH. Mt khc khi thy phn 4,36g este th tiu tn ht 2,4gNaOH v thu c 4,92g mui. Cng thc ca este lA. (CH3COO)3C3H5 B. (C2H3COO)3C3H5

C. C3H5(COOCH3)3 D. C3H5(COOC2H3)3Bi 4. Thc hin phn ng este ho gia axit axetic (d) v hn hp gm 7,52 g 3

ancol k tip nhau trong cng mt dy ng ng. Sau phn ng thu c15,92 g 3 este. Gi s hiu sut phn ng l 100%. CTPT ca ba ancol lA. 3 2 5 3 5CH OH;C H OH;C H OH B. 2 5 3 5 4 7C H OH;C H OH;C H OH

C. 3 5 4 7 5 9C H OH;C H OH;C H OH D. 3 7 4 9 5 11C H OH;C H OH;C H OH

Bi 5. Ha tan hon ton 20,85 g hn hp X gm NaCl v NaI vo nc cdung dch A. Sc kh Cl2 d vo dung dch A. Kt thc th nghim, c cndung dch thu c 11,7 g mui khan. Khi lng NaCl c trong X lA. 5,85 g B. 7,55 g C. 2,95 g D. 5,10 g

Bi 6. Cho kh CO qua ng s cha 15,2g hn hp cht rn CuO v FeO nungnng. Sau mt thi gian thu c hn hp kh B v 13,6g cht rn C. Chohn hp kh B hp th hon ton vo dung dch Ca(OH) 2 d thy c kt ta.Lc ly kt ta v sy kh ri cn th khi lng kt ta thu c lA. 12g B. 11g C. 10g D. 9 g

Bi 7. Nhng thanh kim loi M (ha tr II) vo dung dch CuSO4, sau mt thi

gian ly thanh kim loi ra thy khi lng gim 0,1%. Mt khc cng nhngthanh kim loi trn vo dung dch AgNO3. Sau mt thi gian thy khi

lng tng 7,55%. Bit s mol CuSO4 v AgNO3 tham gia phn ng haitrng hp nhnhau. Kim loi M lA. Zn B. Mg C. Ni D. Ca

Bi 8. Ha tan 3,23 g hn hp mui CuCl2 v Cu(NO3)2 vo nc thu c dungdch X. Nhng thanh Mg vo dung dch X cho n khi mt mu xanh cadung dch, ly thanh Mg ra cn li, thy tng thm 0,8 g. C cn dung dchsau phn ng thu c m g mui khan. Gi tr m lA. 3,08 B. 4,03 C. 2,48 D. 2,84

Bi 9. Cho 16,2 g hn hp este ca ancol metylic v hai axit cacboxlic no, nchc tc dng va vi dung dch NaOH 1M thu c dung dch A. C


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cn dung dch A thu c 17,8 g hn hp hai mui khan, th tch dung dchNaOH 1 M dng lA. 0,2 lt B. 0,3 lt C. 0,4 lt D. 0,5 lt

Bi 10. un nng 3,188 g este ca glixerol vi ba axit cacboxylic no, n chcmch h X, Y, Z (X, Y l ng phn ca nhau v k tip vi Z) vi dungdch NaOH d, phn ng kt thc thu c 3,468 g hn hp mui. Cng

thc phn t ca cc axit lA. C2H4O2 v C3H6O2 B. C3H6O2 v C4H8O2

C. C4H8O2 v C5H10O2 D. C3H4O2 v C4H8O2


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Phng php s dng quy tc thng bng in tch, thng bngs mol electronC sTrong phn ng oxi ha kh th tng s mol e m cc cht kh cho i bngtng s mol e m cc cht oxi ho thu vo :

= e cho e nhnn nNu bi ton c nhiu cht oxi ho v nhiu cht kh tham gia trong s phn ng, hoc qu trnh phn ng phi i qua nhiu giai on th p dngphng php ny gii s rt nhanh v kt qu thu c chnh xc.Cc bc p dng quy tc thng bng s mol electron nh sau :

Phi xc nh c t cc cht ban u tham gia phn ng n cc chtsn phm c bao nhiu cht cho electron v s mol tng cht, c bao nhiucht nhn electron v s mol tng cht (c th phi t n s).

Vit cc qu trnh cho electron tnh tng s mol e m cc cht kh cho

i ( e chon ).

Vit cc qu trnh nhn electron tnh tng s mol e m cc oxi ho nhnvo ( e nhnn ).p dng nh lut bo ton electron :

= e cho e nhnn ni vi nhng h trung ho inNu trong h tn ti ng thi cc ht mang in th ta lun c tng s molin tch dng n t(+) bng tng s mol in tch m n t() :

n t(+) = n t()Vi nt = s mol ion s n v in tch ca ion .

Bi tp minh hoBi 1. Ho tan ht 7,5 g hn hp Al v Mg trong HNO3 long thu c dung dch

A gm 2 mui v 3,36 lt ( ktc) hn hp 2 kh NO v N2O, khi lng cahn hp kh l 5,2 g. Khi lng ca Al, Mg trong hn hp ln lt lA. 3,5g v 4,0g. B. 2,1g v 5,4g.C. 2,7g v 4,8g. D. 4g v 3,5g.Li gii

t = =2NO N O

n a (mol) ; n b (mol)

3,36a b 0,15 a 0,1mol

22,4b 0,2 mol30a 44b 5,2

+ = = =

= + =


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S phn ng :Al

Mg+ HNO3

3 3

3 2

Al(NO )

Mg(NO )+

2

NO

N O+ H2O

Cc cht cho electron : Al: x (mol); Mg: y (mol)

Al0 Al3+ + 3e

x 3x = + e chon 3x 2y (mol)Mg0 Mg2+ + 2ey 2y

Cht nhn electron l HNO3 c hai qu trnh nhn e:

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

0,3 0,1 0,1 e nhnn 0,7 (mol)= N+5 + 4e N+ (N2O)

0,4 0,1 0,05

p dng quy tc thng bng s mol electronta c: 3x + 2y = 0,7 (1)Phng trnh khi lng: 27x + 24y = 7,5 (2)

Gii h (1, 2) Al

Mg

m 2,7gx 0,1(mol)

m 4,8gy 0, 2(mol)

==

==

Bi 2. Ha tan hon ton hn hp gm 0,06 mol FeS2 v x mol Cu2S vo axit

HNO3 (va ) thu c dung dch A (ch cha hai mui sunfat) v kh duynht NO. Gi tr ca x lA. 0,04 B. 0,06 C. 0,12 D. 0,03Li gii

3 22 4

2 22 4

FeS Fe 2SO

0,06 0,06 0,12

Cu S 2Cu SO

a 2a a

+

+

+

+

Theo quy tc thng bng in tch n t(+) = n t()3.0,06 2.2a 2.0,12 2.a a 0,03(mol) + = + =

Bi 3. m g Fe ngoi khng kh mt thi gian nn b g (gi s g st ch ton loxit st) cn nng 10 g. Lng g st trn lm mt mu hon ton 200 ml

dung dch KMnO4 0,5M trong dung dch H2SO4 d. m c gi tr l

A. 17,2g B. 9,8g C. 9,0g D. 15,0gLi giiS phn ng :


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14

2 a 4 2 4 2 4 3 4 2Fe + O FeO KMnO H SO Fe (SO ) MnSO H O + + + +

Cht cho electron : Fe :m

(mol)56

Fe Fe3+ + 3em

56 3

m

56 =

e chom

n 3. 56

Cht nhn electron O :10 m

16

(mol) v KMnO4 : 0,1 (mol)

O + 2e O210 m

16

2

10 m

16

Mn+7 + 5e Mn+20,1 0,5

e nhn10 m

n 2. 0,516

= +

p dng quy tc thng bng s mol electron :10 m m

2. 0,5 3. m 9,8 (g)16 56

+ = =

Bi 4. Cho lung kh H2 i qua ng s ng m g oxit Fe2O3 nhit cao mtthi gian, ngi ta thu c 6,72 g hn hp A gm 4 cht rn khc nhau.

em ho tan hon ton hn hp ny vo dung dch HNO3 d thy tothnh 0,448 lt kh B ktc (duy nht) c t khi so vi hiro l 15 th mc gi tr lA. 7,5 g B. 7,2 g C. 8,0 g D. 8,4 g

Li giiS :

2 3 2 2Fe O H H O A+ + 3HNO+ Fe(NO3)3 + NO + H2O

Xt c qu trnh th : Fe+3 Fe+3hnh nhkhng c s cho v nhn e.

Cht cho electron : H2m 6,72

(mol)16

( do 2 (oxit) 2H O H O+ )

m 6,72 m 6,72

16 16

H2 2H+ + 2e

m 6,72

16

2

m 6,72

16

= e chonCht nhn electron : HNO3, kh B l NO.


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15

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

0,06 0,02 0,02 = e nhnn 0, 06

= =m 6,72

0,06 m 7,2 g8

Bi 5. Trn 60g bt Fe vi 30g bt lu hunh ri un nng (trong iu kin khng

c khng kh) thu c cht rn X. Ho tan X bng dung dch axit H2SO4long, d c dung dch B v kh C. t chy C cn V lt O2 (ktc). Ccphn ng xy ra hon ton th V c gi tr lA. 39,2 lt B. 32,928 lt C. 32,29 lt D. 38,292 ltLi giiS :

o2 4 2H SO + Ot 2 2

42 2

H H OFe FeFeSO +

S FeS H S SO

Xt c qu trnh phn ng th Fe v S cho electron, cn O2 nhn electron.

Cht cho electron : Fe :

60

(mol)56 ; S :

30

(mol)32 Fe Fe2+ + 2e60

562

60

56

S S+4 (SO2) + 4e30

324

30

32

Cht nhn electron : gi s mol O2 l x mol.

O2 + 4e 2O2

x 4xp dng quy tc thng bng s mol electron : 4.

32

302.

56

604 +=x

Gii ra x =330

224mol

2O

330V 22,4 33

224= = (lt)

Bi 6: t chy hon ton 7,2 gam kim loi M (c ho trhai khngi trong hpcht) trong hn hp kh Cl2 v O2. Sau phn ng thu c 23,0 gam cht rn v thtch hn hp kh phn ng l 5,6 lt ( dktc). Xc nh Kim loi M

Li gii

Khi lng hn hp kh = 23 7,2 = 15,8 (g) M = 15,8 : 0,25 = 63,2


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2

2

Cl 71 31,2

63,2O 32 7,8

=

=

t l s mol 2

2

Cl 31,2 4 0,2

O 7,8 1 0,05= = =

M M+n + ne Cl2 + 2e 2Cl O2 + 4e 2O

2

x nx 0,2 0,4 0,05 0,2

Theo quy tc thng bng s mol e:

nx = 0,4 + 0,2 = 0,6 M =7,2

n0,6

= 12n vi n = 2 th M = 24 l Mg

Bi tp vn dngBi 1. Dung dch X c cha 5 ion : Cu2+, Ba2+

, Ca2+ v 0,1mol Cl v 0,2mol

3NO . Thm dn V lt dung dch K2CO3 1M vo dung dch X n khi c

kt ta cc i. V c gi tr lA. 150ml B. 300ml C. 200ml D. 250ml

Bi 2. Cho 0,04 mol Fe; 0,02 mol Al tc dng vi 100 ml dung dch X chaAgNO3 v Cu(NO3)2 thu c dung dch Y v 5,84 g cht rn D gm 3 kimloi. Cho D tc dng vi dung dch HCl d c 0,448 lt hiro (ktc). Nng mol cc mui AgNO3 v Cu(NO3)2 trong X ln lt lA. 0,4M; 0,2M B. 0,2M; 0,4M C. 0,4M; 0,6M D. 0,2M; 0,3M

Bi 3. Ho tan m g hn hp A gm Fe v kim loi R (c ho tr khng i) trongdung dch HCl d th thu c 1,008 lt kh (ktc) v dung dch cha 4,575gmui khan. Cng lng hn hp trn ha tan trong dung dch cha hn hpHNO3 c v H2SO4 nhit thch hp th thu c 0,063 mol kh NO2 v0,021 mol kh SO2. Kim loi R lA. Mg B. Al C. Ca D. Zn

Bi 4. 10,08 g bt st trong khng kh sau mt thi gian thu c hn hp Ac khi lng m g gm Fe, FeO, Fe3O4, Fe2O3. Cho A tc dng hon tonvi dung dch HNO3 thy gii phng ra 2,24 lt kh NO duy nht ( iukin tiu chun). Khi lng m ca hn hp A lA. 11 g B. 12g C. 13g D. 14g

Bi 5. Nung nng 5,6 g bt st trong bnh ng O2 thu c 7,36 g hn hp Xgm Fe, Fe2O3 v Fe3O4. Cho X tan hon ton trong dung dch HNO3 thuc V lt (ktc) hn hp kh Y gm NO v N2O4, t khi hi ca Y so viH2 l 25,33. V c gi tr lA. 22,4 lt B. 0,672 lt C. 0,372 lt D. 1,12 lt

Bi 6. Hn hp X gm FeS2 v MS c s mol nh nhau (M l kim loi ha tr II).Cho 6,51g X tc dng hon ton vi lng d dung dch HNO3 un nng,


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thu c dung dch Y (ch cha mui sunfat) v 13,216 lt (ktc) hn hp

kh Z gm NO v NO2 c khi lng 26,34g. Kim loi M lA. Mg B. Zn C. Mn D. Cu

Bi 7. Cho m g hn hp ba kim loi Al, Fe, Cu tan hon ton trong dung dch

HNO3 thu c V lt hn hp kh D (ktc) gm NO2 (gi thit tn ti NO2

ktc) v NO (khng sinh mui NH4NO3). T khi hi ca D so vi hirobng 18,2. Tng s g mui khan to thnh theo m v V l

A. m +V

22,4B. m

V

22,4 C.2m +

V

22,4 D. m +

V

7,75

4. Phng php ng cho trong bi ton trn ln hai dung dchhoc hn hp hai khC sp dng nh lut bo ton khi lng trong qu trnh trn ln cc dungdch ca cng mt cht tan, ta lun c :

Khi lng dung dch thu c bng tng khi lng ca cc dung dchthnh phn.

Khi lng cht tan thu c cng bng tng khi lng cht tan c trongtng dung dch thnh phn .Phm vi p dng

Pha long hay c cn dung dch Pha trn cc dung dch ca cng mt cht, cng loi nng Pha trn cc khKhi trn ln 2 dung dch c nng khc nhau hay cho thm cht tannguyn cht vo dung dch cha cht tan , hoc qu trnh c cn dungdch. tnh c nng dung dch trng thi cui ta c th gii bngphng php bo ton khi lng, tuy nhin ta nn dng phng php ngcho th gii bi ton s nhanh hn.

Sau y gii thiu mt s s hay c s dng :

Nu trn dung dch 1 c khi lng l m1(g) v nng C1% vi dung dch2 c khi lng m2(g) v nng C2% (gi s C1 < C2) thu c dung dch

mi c nng C% (vi C1 < C < C2) ta s dng s :

Ch :

Ta coi H2O c C% = 0.Ta coi cht tan nguyn cht c C = 100%.

=

m (g)......... C C C

m C CC

m C C

m (g).........C C C

1 1 2

1 2

2 1

2 2 1


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Nu trn dung dch 1 c th tch V1 (lt) v nng CM(1) vi dung dch 2c th tch V2 (lt) v nng CM(2) (gi s CM(1) < CM(2)) ta thu c dungdch mi c nng CM (vi CM(1) < C < CM(2)) ta s dng s sau :

Nu trn mt th tch V1 (lt) kh A c phn t khi MA vi mt th tch khB c phn t khi MB (gi s MA < MB) ta thu c hn hp kh c phn t

khi trung bnh l M (vi MA < M < MB) ta s dng s sau :

Bi ton minh hoBi 1. Cn cho s g H2O vo 100 g dung dch H2SO4 90% c dung dch

H2SO4 50% lA. 90 g B. 80 g C. 60 g D. 70 gLi gii

= =m 40

m 80(gam)100 50

Bi 2. Lm bay hi 500 ml dung dch cht A 20% (D = 1,2 g/ml) ch cn 300 gdung dch. Nng % ca dung dch ny lA. 30% B. 40% C. 50% D. 60%Li gii

mdd = 500.1,2 = 600 (g)y l bi ton c cn nn s :

600 xx 40%

300 x 20 = =

m 0 40

50

100 90 50

dung dch A : 600 20 - x

x

H2O: 300 x - 20

=

M( ) M( ) M

M( ) MM

M M( )

M( ) M M( )

V (lit)...... C C C

C CVC

V C CV (lit)...... C C C

1 1 2

21

2 1

2 2 1

A B

B

A

B A

V (lit)...... M M M

n V M MM

n V M M

V (lit)...... M M M

= =

1

1 1

2 2

2


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Bi 3. Trn V1 ml dung dch NaOH (d = 1,26 g/ml) vi V2 ml dung dch NaOH

(d = 1,06 g/ml) thu c 1lt dung dch NaOH (d = 1,16 g/ml). Gi tr V1, V2ln lt l

A. V1 = V2 = 500 B. V1 = 400, V2 = 600

C. V1 = 600, V2 = 400 D. V1 = 700, V2 = 300

Li gii

= = =1 1 22

V 0,1V V 500ml

V 0,1

Bi 4. Mt hn hp 104 lt (ktc) gm H2 v CO c t khi hi i vi metan bng1,5 th

2HV v VCO trong hn hp l

A. 16 lt v 88 lt. B. 88 lt v 16 lt.C. 14 lt v 90 lt. D. 10 lt v 94 lt.Li gii

1

2

V 2

V 11=

=

=

1

2

V 16 lt

V 88lt

Bi 5. Cho 6,12g Mg tc dng vi dung dch HNO3 thu c dung dch X ch cmt mui v hn hp kh Y gm NO v N2O c t khi hi i vi hirobng 16,75. Th tch NO v N2O ( ktc) thu c ln lt lA. 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.

Li gii

Qu trnh cho electron : Mg Mg2+ + 2e0,225 0,51

Qu trnh nhn electron : N+5 + 3e N+2 (NO)3x x

N+5 + 4e N+ (N2O)8y 2y y

= =2N O

NO

V 1 x

V 3 y

V1 H2 2 4

24

V2 CO 28 22

V1 NO 30 10,5

33,5

V2

N2O 44 3,5

V1 1,26 0,1

1,16

V2 1,06 0,1


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20

3x 8y 0,51 x 0,09

3x y 0 y 0,03

+ = =

= =

Bi tp vn dngBi 1. Trn hai th tch metan vi mt th tch hirocacbon X thu c hn hp

kh (ktc) c t khi so vi H2 bng 15. Cng thc phn t ca X l

A. C2H6 B. C3H8 C. C4H10 D. C5H12Bi 2. Cho hn hp X gm 2 este c CTPT l C4H8O2 v C3H6O2 tc dng vi

NaOH d thu c 6,14g hn hp 2 mui v 3,68g ancol B duy nht c tkhi so vi oxi l 1,4375. S g ca C4H8O2 v C3H6O2 trong A ln lt lA. 3,6g v 2,74g. B. 3,74g v 2,6g.C. 6,24g v 3,7g. D. 4,4g v 2,22g.

Bi 3. T 1 tn qung hematit (A) iu ch c 400kg st. T 1 tn qungmanhetit (B) iu ch c 500kg st. c 1 tn qung hn hp m t 1tn qung hn hp ny iu ch c 460kg st thphi trn 2 qung A. Bvi t l v khi lng lA. 2 : 3 B. 3 : 5 C. 3 : 4 D. 1 : 3

Bi 4. Mt hn hp kh X gm SO2 v O2 c t khi so vi metan bng 3. Thm V

lt O2 vo 20 lt hn hp X thu c hn hp Y c t khi so vi metan bng2,5. Gi tr ca V lA. 20 B. 30 C. 5 D. 10

Bi 5.S ml H2O cn thm vo 1 lt dung dch HCl 2M thu c dung dchmi c nng 0,8M lA. 1,5 lt B. 2 lt C. 2,5 lt D. 3 lt

Bi 6. Trn 1 lt dung dch KCl C1 M (dung dch A) vi 2 lt dung dch KCl C2 M(dung dch B) c 3 lt dung dch KCl (dung dch C). Cho dung dch C tcdng va vi dung dch AgNO3 thu c 86,1 g kt ta. Nu C1 = 4C2 th

C1 c gi tr lA. 1 M B. 1,2 M C. 1,4 M D.1,5 M

Bi 7. T khi hi ca hn hp kh C3H8 v C4H10 i vi hiro l 25,5. Thnhphn % th tch ca hn hp lA. 50% ; 50%. B. 25% ; 75%.C. 45% ; 55%. D. 20% ; 80%.

5. Phng php dng phng trnh ion rt gnC s

Bn cht ca cc phn ng trao i xy ra trong dung dch cht in li l

phn ng ca cc ion vi nhau to ra cht kt ta, bay hi hay cht inli yu,...


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Cch p dngKhi cho dung dch hn hp (X) phn ng vi dung dch hn hp (Y) thay vvic vit nhiu phng trnh phn ng gia cc phn t ta vit cc phngtrnh dng ion rt gn. Sau y l mt s s minh ha :

Cho s : 2 4

3

HCl

H SO

HNO

+2

NaOH

Ba(OH)

hn hp 6 mui + H2O

Bn cht l: H+(axit) + OH

(baz) H2O v2 2

4 4Ba SO BaSO+ +

Khi mi trng trung tnh th :H (axit )

n + = OH ( )n baz

Cho s

2 3

4 2 3

2 3

Na CO

(NH ) CO

K CO

+

2

2

BaCl

CaCl

Bn cht l :

+

+

+

+

2 23 3

2 23 3

Ca CO CaCO

Ba CO BaCO

Cho s :

Fe

Mg

Zn

+2 4

HCl

H SO (l)

Hn hp mui + H2

Bn cht l

Fe

Mg

Al

+ H+ Hn hp mui + H2

Bi tp minh haBi 1. Cho dung dch X cha ng thi 2 axit H2SO4 1M v HCl 2M vo 200ml

dung dch Y cha NaOH 1,5 M v KOH 1M. Khi mi trng dung dchtrung tnh th th tch dung dch X cn lA. 120 ml B. 125 ml C. 200 ml D. 150 mlLi gii

Bn cht cc phn ng trn l 2H OH H O+ +

+

= + =

= + =

H

OH

n V.(2.1 2) 4V (mol)

n 0,2.(1,5 1) 0,5 (mol)

Khi mi trng trung tnh : 4V = 0,5 V= 125 mlBi 2. Cho 100ml dung dch A cha ng thi 2 axit HCl 1M v HNO3 2M vo

200ml dung dch B cha NaOH 0,8 M v KOH x M thu c dung dch C.


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Bit rng trung ho 100ml dung dch C cn 60ml dung dch HCl 1M. x cgi tr lA. 1,2M B. 2,2M C. 3,3M D. 2,5MLi giiC 3 axit phn ng vi 2 baz. Bn cht cc phn ng l

2H OH H O+ +

+

= + + =

= +

H

OH

60.500n 0,1(1 2) .1 0,6(mol)100.1000

n 0,2(0,8 x)(mol)

Mi trng trung tnh: 0,6 = 0,2(0,8+x) x = 2,2M.

Bi 3. Trn 100ml dung dch X gm KHCO3 1M v K2CO3 1M vi 100ml dung

dch Y gm NaHCO3 1M v Na2CO3 1M thu c dung dch Z. Nh t t

100ml dung dch T gm H2SO4 2M v HCl 1M vo dung dch Z thu c V

lt CO2 (ktc). Gi tr ca V lA. 2,24 lt B. 8,96 lt C. 6,72 lt D. 4,48 lt

Li giiBn cht ca cc phn ng gia cc cht trong T v Z l :

23 3

3 2 2

CO 2H HCO

HCO H H O CO

+

+

+

+ +

Khi cho dung dch X vo Y thu c dung dch Z c3HCO 0,2 (mol) v

23CO 0,2 (mol).

+ = + =2 4HCl H SOH (dd T)

n n 2n 0,5(mol)

Nh t t dung dch T vo dung dch Z, phn ng xy ra theo th t : +

+ 23 3CO H HCO

0,2 0,2 0,2

Tng s mol = + =3HCO

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

+ = = < =3H (cn li) HCO

n 0,5 0,2 0,3(mol) n 0,4(mol)

3H (d) HCOn 0,5 0,2 0,3 n 0,2 0,2 0, 4(mol)+ = = < = + =

3 2 2HCO H CO H O

0,3 0,3 0,3

++ +

Bi 4. Tnh th tch dung dch cha HCl 0,5M v H2SO4 0,75M cn thit hotan hon ton 23,2g Fe3O4.


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23

A. 200 ml B. 300 ml C. 350 ml D. 400 mlLi giiBn cht phn ng gia hai axit v Fe3O4 l :

Fe3O4 + 8H+

Fe2+ + 2Fe3+ + 4H2O.0,1 0,8Gi th tch dung dch l V : 0,5V + 2V.0,75 = 0,8 V = 400 ml

Bi 5.Cho 2 kim loi Fe, Mg tc dng vi 200ml dung dch A gm HCl 0,1M,H2SO4 0,2M thu c dung dch B v kh C. Cho t t dung dch D gmNaOH 0,3M, KOH 0,1M vo B tc dng va vi cc cht trong B thth tch dung dch D cn dng lA. 0,15 lt. B. 0,25 lt. C. 0,35 lt. D. 0,45 lt.Li gii

+

+ +

+

+ + +

22

2 2

22

Fe Fe Fe(OH)

H H H OH H O

Mg Mg Mg(OH)

Quy tc thng bng in tch :+ ( )n trong B = + Hn trong A = OHn trong D.

Dung dch trung tnh khi :

+ = + = + = H OHn n 0,2(0,1 0,2.2) V(0,3 0,1) V 0,25 (lt)

Bi tp vn dngBi 1. tc dng va vi 0,96g hiroxit ca 2 kim loi kim hai chu k lin

tip trong bng tun hon, phi dng 20ml dung dch HCl 0,4M v H2SO40,3M. Cc kim loi kim lA. Na, K B. Li, Na C. K, Rb D. Na, Rb

Bi 2. Ha tan hn hp hai kim loi Ba v Na (dng ht rt nh) vo nc thuc dung dch A v 672 ml kh (ktc). Nh t t dung dch FeCl3 vo dung dchA cho n d, lc kt ta, ra sch, sy kh v nung n khi lng khng i thuc m g cht rn. Gi tr m l

A. 3,2 g B. 6,4 g C. 1,6 g D. 4,8 gBi 3. Cho 4,64g hn hp A gm FeO, Fe2O3, Fe3O4 ( 2 3FeO Fe On : n 1: 1= ) ho

tan hon ton trong V lt dung dch H2SO4 0,2M v HCl 0,6M. V c gi trlA. 1,60 lt B. 1,22 lt C. 1,90 lt D. 1,56 lt

Bi 4. Hn hp cht rn X gm Fe, Fe2O3, Fe3O4 v FeO (c s mol bng nhau l

0,1 mol). Ha tan ht X vo dung dch Y gm HCl v H2SO4 long (d), thuc dung dch Z v 1,12 lt kh H2 (ktc). Nh t t dung dch Cu(NO3)2


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0,5M vo dung dch Z cho ti khi ngng kh NO thot ra th dng li. Thtch dung dch Cu(NO3)2 dng lA. 158,3 ml. B. 140,0 ml. C. 100,0 ml D. 160,5 ml.

Bi 5. Cho 12,15 g bt Al vo 100 ml dung dch hn hp NaNO3 1,5M v NaOH3M, khuy u cho n khi ngng kh thot ra th dng li. Th tch khthot ra ktc l

A. 5,04 lt B. 7,56 lt C. 6,72 lt D. 4,48 ltBi 6. Cho 6,4 g Cu tc dng vi 60 ml dung dch hn hp gm HNO3 2M v

H2SO4 1M, thu c V lt kh NO duy nht (ktc), phn ng xy ra honton. Gi tr ca V lA. 0,672 B. 0,896 C. 1,344 D. 2,24

Bi 7. Dung dch A th tch 200ml cha ng thi hai mui MgCl2 0,4M vCu(NO3)2 0,2M. Dung dch B cha ng thi KOH 0,16M v Ba(OH)20,02M. Th tch dung dch B cn lm kt ta ht hai ion Mg2+, Cu2+ lA. 1 lt B. 1,2 lt C. 1,5 lt D. 1,7 lt

Bi 8. Cho 3,75g hn hp A gm Mg v Al vo 250 ml dung dch X cha axit HCl

1M v H2SO4 0,5M, c dung dch B v 3,92 lt H2 (ktc). Thnh phn %khi lng Mg, Al trong A lA. 65% ; 35%. B. 64% ; 36%.B. 55% ; 45%. D. 50% ; 50%.

Bi 9. Cho hn hp A gm Mg v Al vo 500 ml dung dch X cha axit HCl 1Mv H2SO4 0,5M, c dung dch B v 3,92 lt H2 (ktc). Th tch dung dchC gm NaOH 0,2M v Ba(OH)2 0,1M cn phn ng ht vi cc chttrong B lA. 0,125 lt B. 1,112 lt C. 1,875 lt D. 1,235 lt

Bi 10. Cho m g hn hp Mg, Al vo 250 ml dung dch X cha hn hp axit

HCl 1 M v axit H2SO4 0,5 M, thu

c 5,32 lt H2 ( ktc) v dung dch Y(coi th tch dung dch khng i). Dung dch Y c pH lA. 7 B. 1 C. 2 D. 6

6. Phng php gii cc bi ton cc i - cc tiu Cch tm khong gii hn ca mui : Hn hp kim loi (A, B) tc dng vi hn

hp axit (HNO3 v H2SO4) to ra hn hp mui sunfat v mui nitrat.

+) Do 1 mol 24SO (nng 96 gam) tng ng vi 2 mol 3NO

(nng 124

gam). Vi cng mt hn hp kim loi nu to mui nitrat th khi l ng snng hn mui sunfat. Khi lng mui cc i khi hn hp ch to ra muinitrat v cc tiu khi hn hp ch to mui sunfat. Vy khi lng thc t l

2- -4 3

thc tmuiSO mui NOm m m< <


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+) Do 1 mol ion kim loi A+ (nng MA g) tng ng vi 1 mol B+ (nng MB).

Nu MA < MB th mui ca A nh hn mui ca B (cng gc axit). Khilng mui cc i khi hn hp ch to ra mui B v cc tiu khi hn hpch to mui A. Vy khi lng thc t l

muiA thc t mui Bm m m< < Bi tp minh haBi 1. Ha tan 1,02 gam Al2O3 vo 300 ml dung dch H2SO4 0,2M ta thu c

dung dch A. Rt tip vo A 200ml dung dch NaOH th thu c mt ktta, em nung n khi lng khng i th c 0,51 gam cht rn. Coi thtch dung dch thay i khng ng k. Nng mol ca dung dch NaOHban u lA. 0,55M ; 0,65M B. 0,65M ; 0,75MC. 0,45M ; 0,7M D. 0,85M ; 1,5MLi gii

2 3 2 4Al O H SOn 0,01mol ; n 0,06 mol= =

2 3 2 4 2 4 3 2Al O 3H SO Al (SO ) 3H O+ +

Dung dch A2 4(d)

2 4 3

H SO 0,06 0,03 0, 03mol

Al (SO ) : 0,01mol

= =

2H OH H O

0,06 0, 06

+ +

TH1 : NaOH ch kt ta mt phn Al3+.3

3 2 3

NaOH M(NaOH)

Al 3OH Al(OH) Al O

0,01 0, 01 0,005

0,09n 0,06 0,03 0,09(mol) C 0,45M0,2

+ +

= + = = =

TH2 : NaOH ch kt ta ht Al3+ v ho tan mt phn kt ta ny

o

33

3 4

t3 2 3 2

Al 3OH Al(OH)

0,02 0,06 0,02

Al(OH) OH [Al(OH) ]

0,01 0,01

2Al(OH) Al O H O

0,510,01 0,005102

+

+

+

+

=


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26

NaOH M0,14

n 0,14 mol C 0,7 M0,2

= = =

Bi 2: Cho 200ml dung dch AlCl3 tc dng vi 400 ml dung dch NaOH 2M tathu c mt kt ta em nung nng n khi lng khng i th thu c5,1gam cht rn.

a) Nng mol ca dung dch AlCl3 lA. 0,155M B. 1,125M C. 0,175M D. 1,185Mb) Lc tch kt ta thu c dung dch nc lc. Cho dung dch nc lc tcdng vi dung dch HCl 1M. Th tch dung dch HCl kt ta thu c lnnht lA. 62,5 ml B. 25,5 ml C. 35,0 ml D. 45,0 mlLi gii

NaOHn 0,4.2 0,8mol= = ot3

3 2 3Al 3OH Al(OH) Al O

0,1 0,3 0,1 0,05

+ +

a) TH1 : Al3+ cha b kt ta ht 3 MAl

0,1n 0,1mol C 0,50,2

+ = = =

TH2 : Al3+ b kt ta ht, OH- d

33

3 4

Al 3OH Al(OH)

a 3a a

Al(OH) OH [Al(OH) ]

a 0,1 a 0,1

+

+

+

3 MAl

0,225OH 4a 0,1 0,8 a 0,225 mol n C 1,125 M

0,2

+ = = = = = =

b) Dung dch nc lc4Na[Al(OH) ]

n 0,125 mol=

4 3 2Na[Al(OH) ] HCl Al(OH) NaCl H O

0,125 0,125 0,125

+ + +

M(HCl)0,125

C 0,062 lit 62,5ml2

= = =

Bi 3. Chia 3,64 gam hn hp (A) gm 3 kim loi Fe, Al, Mg ( dng bt mn)thnh hai phn bng nhau. Ho tan ht phn th nht trong dung dch HCl

va thu c 1,568 lt H2 v dung dch (A). Cho phn th hai vo dung

dch NaOH 0,5 M (ly d), thu c dung dch (B) v cht rn C. Cho C


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27

phn ng vi dung dch HNO3 un nng thu c 2,016 l kh NO2 ktc(khng c mui amoni) v dung dch (D).a) Khi lng Al trong hn hp A lA. 1,08g B. 1,28g C. 0,28g D. 1,18gb) Khi lng mui nitrat to thnh trong dung dch (D) lA. 6,68g B. 5,68g C. 4,28g D. 3,18g

c) Th tch dung dch NaOH 0,5 M cn phi cho vo dung dch A kt tathu c l cc i v cc tiu ln lt lA. 200ml v 300ml B. 280 ml v 320 mlC. 320 ml v 280 ml D. 250 ml v 350 mlLi gii

Gi Al Fe Mgn x mol;n y mol;n z mol= = = trong 1/2 hn hp A

Ta c 27x + 56y + 24z = 1,82 (I)S phn ng vi HCl :

2

32

2

Fe Fe

Al H Al H

Mg Mg

+

+ +

+

+ +

3x + 2y + 2z = 0,14 (II)Phn 2 tc dng vi dung dch NaOH

2 4 23

Al NaOH 3H O Na[Al(OH) ] H2

+ + +

S phn ng ca C vi HNO3:Fe

Mg+ HNO3

3

2

Fe

Mg

+

++ NO2 + H2O

3y + 2z = 0,09 (III)Gii h (I), (II), (III) ta c x = 0,02 mol ; y = 0,01 mol ; z =0,03 mol

a.Al

Fe

Mg

m 0,02.2.27 1,08gam

m 0,02.56 1,12 gam

m 1,44 gam

= =

= = =

b. mmui = mFe, Mg+3NO

m = 0,01.56+0,03.24 + (0,01.3+0,03.2)62 = 6,68 gam

c. Dung dch A gm

3

2

2

Al : 0,02 mol

Fe : 0, 01mol

Mg : 0, 03 mol

+

+

+


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28

kt ta thu c cc i khi ton b ba kim loi b kt ta va ht :3

3

22

22

Al 3OH Al(OH)

0,02 0,06

Fe 2OH Fe(OH)

0,01 0,02Mg 2OH Mg(OH)

0,03 0,06

+

+

+

+

+

+

NaOHOH

0,14n 0,06 0,06 0,02 0,14 mol V 0,28lit 280 ml

0,5 = + + = = = =

Kt ta cc tiu khi Al(OH)3 b tan ht :

3 4Al(OH) NaOH Na[Al(OH) ]

0,02 0,02

+

NaOHOH

0,16

n 0,14 0, 02 0,16 mol V 0,32 lit 320 ml0,5= + = = = =

Bi 4. Ho tan hon ton 11,9 g hn hp 2 kim loi Al v Zn vo dung dch HNO3

2M (va ) thu c dung dch A v 3,64 lt hn hp 2 cht kh N2O v NO( ktc) c khi lng 5,75 gam.a) Thnh phn phn trm ca Al v Zn c trong hn hp lA. 30,00% ; 70,00% B. 43,38% ; 54,62%C. 35,00% ; 65,00% D. 54,62% ; 43,38%

b) Th tch NH3 (ktc) cho vo dung dch A kt ta ln nht, nh nht lA. 17,92 lt v 26,88 lt B. 15,52 lt v 25,68 ltC. 13,96 lt v 27,24 lt D. 12,52 lt v 15,68 lt

Li giiGi

2N O NOn x mol ; n y mol= = . Ta c

x y 0,1625 x 0, 0625 mol

44a 30b 5,75 y 0,1mol

+ = =

+ = =

t Al Znn a mol ; n b mol 27a 65b 11,9 (I)= = + =

Al

Zn+ HNO3

3 3

3 2

Al(NO )

Zn(NO )+ 2

N O

NO+ H2O

Dng phng php bo ton electron ta c :Cc qu trnh cho e :


30/39

29

3

2

Al Al 3e

a 3a

Zn Zn 2e

b 2b

+

+

+

+ echon 3x 2y = +

Cc qu trnh nhn e :5

2

5 2

N 4e N N O

0,5 0,125 0,0625

N 3e N NO

0,3 0,1 0,1

+ +

+ +

+

+ enhnn 0,8 =

Ta c : 3x + 2y = 0,8 (II)Gii h (I), (II) thu c :

a) Al

Zn

m 0,2.27 5,4 g %Al 43,38%x 0,2 mol

y 0,1mol m 6,5g %Zn 54,62%

= = ==

= = =

b) Dung dch A gm3

2

Al :0,2 mol

Zn :0,1mol

+

+

Cc phn ng

Kt ta thu c cc i khi 3 2Al , Zn+ + b kt ta va ht :3

3 2 3 4

23 2 2 4

Al 3NH 3H O Al(OH) 3NH

0,2 0,6

Zn 2NH 2H O Zn(OH) 2NH

0,1 0,2

+ +

+ +

+ + +

+ + +

3NHV 0,8.22,4 17,92 lit= = Kt ta thu c cc tiu khi Zn(OH)2 b ho tan ht

2 3 3 4 2Zn(OH) 4NH [Zn(NH ) ](OH)+

3NHV 1,2.22,4 26,88 lit= =

Bi tp vn dngBi 1. Ho tan hon ton 0,81 g Al vo dung dch NaOH va thu c dung

dch mui A v kh H2. Th tch dung dch HCl 0,1 M kt ta thu ccc i v cc tiu ln lt lA. 0,3 lt v 1,5 lt B. 0,3 lt v 1,2 lt

C. 0,5 lt v 1,5 lt D. 0,2 lt v 1,6 ltBi 2. Cho 100 ml dung dch A gm MgCl2 0,3M; AlCl3 0,5M; HCl 0,50M. Thmt t V lt dung dch X gm NaOH 0,2M v Ba(OH)2 0,1M vo dung dch A


31/39

30

thu c kt ta Y. lng kt ta Y ln nht ri nh nht th V ln ltnhn cc gi tr lA. 1,75 lt v2,24 lt B. 0,65 lt v 0,775 ltC. 0,75 lt v 0,25 lt D. 1,50 lt v 1,75 lt

Bi 3. Ho tan hon ton m g Zn vo lng va 300ml HCl 1M thu c dung

dch A v kh H2. Th tch dung dch NaOH 0,01 M vo dung dch A thu

c 7,425 g kt ta cc i ri cc tiu ln lt lA. 150 ml v 200 ml B. 250 ml v 500 mlC. 150 ml v 750 ml D. 250 ml v 750 ml

Bi 4. Nu ho tan hon ton 28,1g hn hp MgCO3 v BaCO3 c thnh phn thay

i trong cha a% MgCO3 bng dung dch HCl v cho tt c kh thot rahp th ht vo dung dch cha 0,2 mol Ca(OH)2 th thu c kt ta D. lng kt ta D nhiu nht v t nht th a c gi tr ln lt lA. 29,89% v 100,00% B. 30,00% v 100,00%C. 29,89% v 60,00% D. 39,89% v 79,89%

Bi 5. Ho tan 15,3 g BaO vo nc c dung dch A. Cho 12,3 gam hn hp

CaCO3 v MgCO3 (c thnh phn thay i) ho tan ht vo dung dch HCld th thu c kh B. Nu cho kh B hp th ht vo dung dch A th sauphn ng khi lng kt ta to thnh trong khongA.

3BaCO10,638(g) m 15,169(g)< <

B.3BaCO

12,680(g) m 15,690(g)< <

C.3BaCO

10,000(g) m 15,000(g)< <

D.3BaCO

15,638(g) m 25,169(g)< <

Bi 6. Thm t t dung dch cha HCl 1M vo dung dch c cha 3,82 gam hn

hp mui Na2CO3 v K2CO3 ( 2 3 2 3Na CO K COn : n 1 : 2)= thu c dung dch

X v kh Y. th tch kh Y thu c c gi tr cc tiu, cc i th th tchdung dch HCl 1M ln lt lA. 30 ml ; 40 ml B. 40 ml ; 70 mlC. 30 ml ; 60 ml D. 40 ml ; 60 ml

Bi 7. Ho tan 1,29 gam Zn, Cu (c =Zn Cun : n 1 :1 ) vo dung dch cha HNO3

v H2SO4 (c, nng) thu c hn hp kh gm NO2 v kh SO2. Nu t l

kh NO2 v kh SO2 thay i th khi lng mui khan thu c trong khongA. mui3,20(gam) m 3,70(gam)< <

B. <


32/39

31

Bi 8. Ho tan 0,89 gam Zn, Mg vo dung dch cha HCl v H2SO4 thu c

0,448 lt kh H2 (ktc). Nu thay i t l s mol hai axit th khi lng muikhan thu c trong khongA. <


33/39

32

1 A 2 B K K

A B K

y .n y .n ... y .ny

n n ... n

+ + +=

+ + +(Vi y1 < y < yK)

(y1, y2, y3,.. l s nguyn t H ca A, B, C,...)Cc gi tr trung bnh khc nh s nguyn t oxi trung bnh, phn t khica gc hirocacbon trung bnh, s nhm chc trung bnh,... cch thit lptng t.

Phm vi p dngDng gii nhanh cc loi bi hu c, v c nh: bi tp xc nh cngthc phn t ca cht ho hc, tnh th tch, tnh s mol hay tnh % s mol,% th tch vi cc cht kh,...Bi tp minh ha

Bi 1. t chy mt hn hp X gm 2 hirocacbon A, B k tip nhau trong dyng ng thu c 96,8g CO2 v 57,6g H2O. Cng thc phn t ca A, Bv thnh phn % mi cht trong hn hp X l

4 2 6A. CH , C H 2 6 3 8B. C H , C H

3 8 4 10C. C H , C H 2 4 3 6D. C H , C H

Li gii

2 2CO H O96,8 57,6

n 2,2 (mol) ; n 3,2 (mol)44 18

= = = =

2 2CO H On n A, B thuc dy ng ng ankan. <

Gi n l s nguyn t C trung bnh ca hn hp ( n n m 4 ), khi thay A, B bng 1 cht duy nht

n 2n 2C H

+c s mol = (a+b) mol.

2 2 2n 2n 2

3n 1C H O nCO (n 1)H O (1)

2

2,2 3,2

+

++ + +

n n 1n 2,2. Vy : n 1 n 2,2 m 2

2,2 3,2

+= = = = =

Vy : hai hirocacbon l C2H6 v C3H8

Theo quy tc ng cho:

2 3 - 2,2

2,23 2,2 - 2

2 6

3 8

C H 0,8

C H 0,2= 80% v 20%

Bi 2. t chy ht 0,5mol hn hp X gm 2 hirocacbon mch h A, B thuccng dy ng ng (phn t khi hn km nhau 28vC) cn 40,32 lt O

2

to ra 26,88 lt CO2. Tm cng thc phn t ca A, B.Li gii


34/39

33

2 2O p CO40,32 26,88

n 1,8 (mol) ; n 1,2 (mol)22.4 22, 4

= = = =

2 2 2 2

2

2 2 2

O(O p) O(CO ) O(H O) O(H O)

O(H O)

H O O(H O) CO

Theo s bo ton nguyn t O:

n n n n 2.1,8 2.1,2

n 1,2(mol)

Ta c: n n 1,2(mol) n A,B thuc anken

= + =

=

= = =

Gi n l s nguyn t C trung bnh ca hn hp ( n n m ), khi thay A, B bng 1 cht duy nht

n 2nC H c:

2 2 2n 2n

2 4 4 8

3nC H O nCO nH O

20,5 1,2

1,2n 2,4 n n m 4 n 2 A : C H ; m 4 B: C H

0,5

+ +

= = = =

Bi 3. Hn hp kh A ( ktc) gm 2 anken. t chy 7 th tch A cn 31 thtch kh O2 ( cng iu kin nhit v p sut). Trong A, anken chanhiu C hn, chim khong 4050% th tch hn hp. Cng thc phn tca 2 anken lA. C2H4 ; C3H6 B. C2H4 ; C4H8 C. C2H4 ; C5H10 D. C3H6 ; C4H8Li gii

Gi n l s nguyn t C trung bnh ca hn hp (2 n 4)< CTPTTB ca

A ln 2n

C H .

o

2

t2 2 2n 2n

A O

3nC H O nCO nH O

27V 31V

+ +

t s

3n.V1V 2:

7V 31V=

1 3n

7 2.31=

2.31n 2,95

21= =

1 anken c s nguyn t C l 2 l C2H4 olefin cn li c s nguynt C l 3 hoc 4Gi x l %V ca CnH2n ; (1x) l %V ca C2H4 (phn trm th tch chnhl % v s mol trong cng iu kin).Theo s bo ton nguyn t C :


35/39

34

4 8

0,95n mx 2(1-x) 2,95 x(m-2) 2 2,95 x (1)

m 2M 0,4 < x 0,5 Thay vo (1) 3,9 m 4,375 m 4 C H

= + = + = =

< < < =

Bi 4. Mt hn hp X gm 2 hirocacbon A, B thuc cng 1 dy ng ng u lkh ktc, cn 20,16 lt O2 t chy ht X v phn ng to 7,2g H2O. Khi

cho l

ng hn hp X trn tc dng vi dung dch 3 3AgNO / NH d

thu

c62,7g kt ta. CTCT ca A, B lA. C2H2 ; C3H4 B. C2H2 ; C4H6 C. C3H4 ; C4H6 D. C2H2 ; C5H8Li gii

2 2O H O20,16 7,2

n 0,9 (mol) ; n 0,4 (mol)22,4 18

= = = =

p dng s bo ton nguyn t O :2 2 2O(O ) O(CO ) O(H O)

n n n= +

2O(CO )

n 1,8 0,4 1,4 (mol)= =

2CO

n 0,7mol= 2 2CO H O

n n ankin>

Gi n 2n 2

m 2m 2

A : C H : a(mol)B : C H : b(mol)

CTTB : n 2n 2C H (a b) mol +

n 2n 2 2 2 23n 1

C H O nCO (n 1)H O2

0,7 0,4

+ +

Ta c: 2COn 0,7

n 2,33(a b) 0,3

= = =+

( 2 n n m 4 < < )

2 2n 2 A : C H HC CH=

3 4 3

3 24 6

3 3

m 3 B : C H CH C CHCH CH C CH

m 4 B : C HCH C C CH

=

=

Xt 2 2 3 4C H v C H :

a b 0,3mol+ =

2a 3b

n 2,33 2a 3b 0,7a b

+= = + =

+

gii h2a 3b 0,7 a 0,2

a b 0,3 b 0,1

+ = =

+ = =

Phn ng vi 3 3AgNO NH


36/39

35

3 2 2 3CH CH 2[Ag(NH ) ]OH AgC CAg 2H O 4NH + + +

0,2 0,2

3 3 2 3 2 3CH C CH [Ag(NH ) ]OH CH C CAg H O 2NH + + +

0,1 0,1

m 0,2.240 0,1.147 62,7g m (gt)= + = =

Cch lm tng t vi hai trng hp cn li, kt qu thu c la khngph hp.Vy A l 2 2 3 4C H v B l C H .

Bi 5. Tch nc hon ton 10,6 g hn hp hai ancol thu c hn hp A gm 2olefin l ng ng k tip nhau. Cho hn hp A ( ktc) qua bnh ngdung dch brom d, ngi ta thy khi lng ca bnh tng thm 7g. Cngthc phn t ca 2 olefin lA. C2H4 v C3H6 B. C3H6 v C4H8C. C4H8 v C5H10 D. C5H10 v C6H12Li giit cng thc chung ca hai ancol no n chc, mch h l ng ng ktip l

n 2n 1C H OH

+.

n 2n 1C H OH

+ 2 4o

H SO

170 C

n 2nC H 2Br

+ 2n 2nC H Br

Khi chuyn :n 2n 1

C H OH+

n 2n

C H . Th M 18 =

hh3,6

m 10,6 7 3,6 (g) n 0,2 (mol)18

= = = =

7M 35

0,2= = M1 < 35 < M2 ; m M1, M2 l ng ng k tip.

M1 = 28 C2H4

M2 = 42 C3H6Bi tp vn dng

Bi 1. A, B l 2 ancol no, n chc k tip nhau trong dy ng ng. Cho 3,9ghn hp tc dng ht vi Na thu c 1,12 lt H2(ktc). Cng thc phn tca 2 ancol l

A. CH3OH, C2H5OH B. C2H5OH, C3H7OHC. C3H7OH, C4H9OH D. C4H9OH, C5H11OH

Bi 2. t chy hon ton 0,03 mol hn hp X gm 1 axit cacboxylic n chc Av 1 ancol no B, u mch h vi O2 th thu c 1,12 lt kh CO2 (ktc) v1,08 g H2O. Bit A, B hn km nhau 1 nguyn t cacbon trong phn t.Cng thc cu to ca A, B ln lt lA. C2H5COOH , C3H5(OH)3 B. HCOOH , C2H5OHC. C2H5COOH , C3H7OH D. C2H3COOH , C3H6(OH)3


37/39

36

Bi 3. Hn hp 3 ancol n chc A, B, C c tng s mol l 0,08 v khi lng l

3,387. xc nh CTPT ca A, B, C, bit rng B v C c cng s nguyn t

cacbon v khng phn ng vi AgNO3/ NH3, s mol A bng3

5tng s mol

ca B v C. Cng thc cu to ca A, B, C ln lt lA. 3 3 5 3 7CH OH ; C H OH; C H OH B. 2 5 3 5 3 7C H OH ; C H OH; C H OH

C. 2 5 4 9 4 9C H OH ; C H OH; C H OH D. 3 4 9 4 9CH OH ; C H OH; C H OH

Bi 4. t chy hon ton a g hn hp 2 ancol no, n chc mch h lin tiptrong dy ng ng thu c 3,584 lt CO2 ktc v 3,96g H2O. CTPT cacc ancol lA. CH3OH; C2H5OH B. C2H5OH; C3H7OH

C. C3H7OH; C4H9OH D. C4H9OH; C5H11OHBi 5. t chy hon ton 0,56 lt hn hp X gm hai hirocacbon c cng s

nguyn t cacbon, thu c2H O

m 1,9125 gam= ,2CO

m 4,4gam= . Trong

X khng c cht no cha qu 1 lin kt pi. Cng thc ca hai hirocacbon

lA. C2H4; C2H6 B. C3H6 ; C3H8 C. C4H8 ; C4H10 D. C5H10 ; C5H12

Bi 6. t chy hon ton 0,672 lt hn hp A gm hai kh hirocacbon (ktc) ccng s nguyn t cacbon, thu c 2,64g CO2 v 1,26 g H2O. Mt khc khicho A qua dung dch [Ag(NH3)2]OH ng trong ng nghim thy c kt tabm vo ng nghim. Cng thc phn t cc cht trong A lA. C2H2; C2H4 B. C2H2; C2H6 C. C3H4; C3H8 D. C2H2; C3H4

Bi 7. Mt hn hp hi ca 3 ancol n chc A, B, C v 13,44 g kh O2 c chatrong mt bnh kn dung tch 16 lt p sut 0,92 atm, 109,2oC. t chyhon ton hn hp, sn phm thu c

2 2H O COm 3,78g ; m 6,16g= = . Bit

rng B v C c cng s nguyn t cacbon v u khng phn ng vi dung

dch AgNO3/ NH3, s mol A bng3

5tng s mol ca B v C. Cng thc cu

to ca A, B, C ln lt lA. 3 3 5 3 7CH OH; C H OH ; C H OH B. 3 2 5 3 7CH OH; C H OH ; C H OH

C. 2 5 3 5 3 7C H OH; C H OH ; C H OH D. 3 3 5 4 9CH OH; C H OH ; C H OH

Bi 8. t chy hon ton 0,896 lt hn hp kh X (ktc) gm hai hirocacbon A,B thu c 1,12 lt kh CO2 (ktc) v 1,26 g H2O. Cng thc phn t ca A,B l

A. C2H2 ; C2H4 B. CH4 ; C2H4 C. CH4; C2H6 D. CH4; C2H2


38/39

37

Bi 9. t chy hon ton 3,24g hn hp X gm hai cht hu c A, B trong Ahn B mt nguyn t cacbon (MA


39/39

38

(11,2 gam) (1,8 gam)Th t li vn theo cch nh sau:

Cho 11,2 gam Fe tc dng vi 1,8 gam H2O thu c hn hp gm Fe v ccoxit ca n. Cho hn hp ny tc dng vi HNO 3 th thu c bao nhiu lt NO2 iu kin tiu chun.

Fe +

)gam8,1(OH2

32

43

OFeOFe

FeO

3HNO+

NO2

(11,2 gam)* p dng phng php bo ton e

Cht kh Cht oxi hoFe Fe3+ + 3e0,2 0,6

H2O + 2e H2 + O2-

0,1 0,2

NO 3 + e + 4H+ NO2 + 2H2O

a a

ta c: 0,6 = 0,2 + a

a = 0,4 mol

2NOV = 8,96 lt2. Phng php t n s ph

V d: Cho 18,5 gam hn hp Z gm Fe, Fe3O4tc dng vi 200 ml dung dchHNO3long un nng v khuy u. Sau khi phn ng xy ra hon ton thuc 2,24 lt kh NO duy nht (ktc), dung dch Z

1v cn li 1,46 gam kim

loi.1. Vit cc phng trnh ha hc cho cc phn ng xy ra.2. Tnh nng mol/l ca dung dch HNO

3.

Gii:t cng thc chung ca Fe (tham gia phn ng) v Fe3O4 trong hn hp l FexOy,s mol l a, ta c: 56ax + 16ay = 18,5-1,46 = 17,04 (1)

N+5 + 3e N+2(trong NO)0,3 0,1

FexOy xFe+2 + yO-2

+ (2x -2y)ea (2x -2y)a2ax 2ay = 0,3 (2)

Gii h hai phng trnh (1), (2): ax = 0,27; ay = 0,12nFe =

23 )(NOFen = ax nn

3HNOn 0,27 2 0,1 0,64(mol)= + =

MCHNO

2,32,0

64,03

==

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