21 de Thi Hsg Co Da Chi Tiet(2)

S 1

S 1Mn : Ha hc - Lp 9

Thi gian lm bi : 150 pht

Cu 1 : ( 5 im)

a) Tm cc cht k hiu bng ch ci trong s sau v hon thnh s bng phng trnh phn ng :

A

Fe2O3 FeCl2

B b) Nung nng dy st trong khng kh, phn ng xong cho sn phm vo dung dch HCl va . Sau cho ton b vo NaOH. gii thch cc hin tng xy ra.

Cu 2 : ( 4 im) Nhit phn mt lng MgCO3 sau mt thi gian thu c cht rn A v kh B. Hp th ht kh B bng dung dch NaOH cho ra dung dch C. Dung dch C va tc dng vi BaCl2 va tc dng vi KOH. Ha tan cht rn A bng Axit HCl d thu c kh B v dung dch D. C cn dung dch D c mui khan E. in phn E nng chy c kim loi M. Hon thnh cc phng trnh phn ng trn.

Cu 3 : (6 im)

a) Bng phng php ha hc, hy tch ring tng kh ra khi hn hp gm CO2; SO2; N2.

b) Ha tan hon ton 3,78(g) mt kim loi X vo dung dch HCl, thu c 4,704(l) H2 ktc. Xc nh kim loi X.

Cu 4 : (5 im) Ha tan 1,42 (g) h n h p Mg ; Al ; Cu bng dung dch HCl th thu c dung dch A v kh B + cht rn D. Cho A tc dng vi NaOH d v lc kt ta nung nhit cao n lng khng i thu c 0,4 (g) cht rn E. t nng cht rn D trong khng kh n lng khng i thu c 0,8 (g) cht r n F. Tnh khi lng mi kim loi.

.......................................Ht.........................................

S 2 Cu1:

1. Ho tan hon ton 20,4 gam Al2O3 v 8 gam MgO trong 122,5 gam dung dch H2SO4. trung ho lng axit cn d phi dng 400 ml dung dch NaOH 0,5M. Nng phn trm dung dch H2SO4 ban u l: A: C.72%

2. Cho hn hp Al v Fe tc dng vi hn hp dung dch AgNO3 v Cu(NO3)2 thu c dung dch B v cht rn D gm 3 kim loi. Cho D tc dng vi dung dch HCl d c kh bay ln. Thnh phn cht rn D l:

A. Fe, Cu v Ag

Cu2:

1. C 4 l mt nhn A, B, C,D cha NaI, AgNO3, HI, K2CO3.

- Cho cht trong l A vo cc l: B,C,D u thy c kt ta

- Cht trong l B ch to 1 kt ta vi 1 trong 3 cht cn li

- Cht C to 1 kt ta v 1 kh bay ra vi 2 trong 3 cht cn li.

Xc nh cht cha trong mi l. Gii thch

2. Vit 6 phng trnh phn ng khc nhau thc hin phn ng.

PbCl2 + ? = NaCl + ? Cu3: 1. t hn hp C v S trong Oxi d _ hn hp A.

- Cho 1/2 A li qua dung dch NaOH thu c dung dich B v kh C.

- Cho kh C qua hn hp cha CuO, MgO nung nng thu c cht rn D v kh E.

- Cho kh E li qua dung dch Ca(OH)2 thu c kt ta F v dung dch G thm dung dch KOH vo G li thy c kt ta F xut hin. un nng G cng thy kt ta F.

Cho 1/2 kh A cn li qua xc tc nng thu c kh M. Dn M qua dung dch BaCl2 thy c kt ta N.

Xc nh thnh phn A,B,C,D,E,F,G,M,N v vit tt c cc phn ng xy ra.

2. Trnh by phng php tch ring tng cht nguyn cht t hn hp: vi, vi sng, thch cao v mui n.

Cu4:Trn 50ml dung dch Al2(SO4)3 4M vi 200ml Ba(OH)2 1,5M thu c kt ta A v dung dch B. Nung kt ta A trong khng kh n lng khng i thu c cht rn D. Thm BaCl2 d vo dung dch B th tch ra kt ta E.

a. Vit ptp. Tnh lng D v E

b. Tnh nng mol cht tan trong dung dch B ( coi th tch thay i khng ng k khi xy ra phn ng)

S 3Cu I: (5 im)1. T cc nguyn liu ban u l qung St Pirit FeS2, mui n, khng kh, nc, cc thit b v ho cht cn thit, c th iu ch c FeSO4, Fe(OH)3, NaHSO4. Vit cc phng trnh ho hc iu ch cc cht ?

2. Bng phng php ha hc, hy tch ring Al2O3 ra khi hn hp gm Al2O3,Fe2O3,SiO2.

Cu II: (5 im )

1. Ch c dng thm mt thuc th; hy phn bit 4 dung dch sau y bng phng php ho hc : KCl, NH4NO3, Ca(H2PO4)2, (NH4)2SO4.

2 . Cho s bin ho sau: Cu

Hy xc nh cc n cht A, B, C ri

hon thnh cc phng trnh phn ng? CuCl2 A

C B

Cu III: (5 im)

1. Cho 44,2 gam hn hp hai mui sunfat ca mt kim loi ho tr I v mt kim loi ho tr II tc dng va vi dung dch BaCl2, thu c 69,9gam mt cht kt ta. Tnh khi lng cc mui thu c trong dung dch sau phn ng?

2. Hai l Km c khi lng bng nhau, mt l c ngm trong dung dch Cu(NO3)2, mt l c ngm trong dung dch Pb(NO3)2. Sau mt thi gian phn ng, khi lng l th nht gim 0,05gam.

a. Vit phng trnh phn ng xy ra?

b. Khi lng l km th hai tng hay gim bao nhiu gam?

Bit rng trong c hai trng hp lng km b ho tan nh nhau.

Cu IV: (5 im)

1. Cho m gam bt St vo dung dch hn hp cha 0,16mol Cu(NO3)2 v 0,4mol HCl. Lc u cho phn ng xy ra hon ton. Sau phn ng thu c hn hp kim loi c khi lng bng 0,7m gam v V lt kh (ktc). Tnh V v m?

2. Nung n hon ton 30gam CaCO3 ri dn ton b kh thu c vo 800ml dung dch Ba(OH)2, thy thu c 31,08gam mui axt. Hy tnh nng mol ca dung dch Ba(OH)2?

Cho S = 32, O = 16, Ba = 137, Cl = 35,5, Zn = 65, Pb = 207, N = 14,

Fe = 56, Cu = 64, Ca = 40, H = 1. S 4Cu 1(2im): Vit 4 phn ng ho hc khc nhau iu ch trc tip ra:

a. dung dch NaOH b. dung dch CuCl2Cu 2( 4im): Hon thnh s phn ng bng cch thay cc cht thch hp vo cc ch ci A,B,C,D ,ghi r iu kin phn ng (nu c):

B (2) H (3) E

A (1) (5) (4) G

C (6) D (7) E

Bit A l mt hp cht ca Fe

Cu 3(4im): C 5 mu phn bn ho hc khc nhau dng rn b mt nhn gm :

NH4NO3 , Ca3(PO4)2 , KCl , K3PO4 v Ca(H2PO4)2 .Hy trnh by cch nhn bit cc mu phn bn ho hc ni trn bng phng php ho hc .

Cu 4(5im): Ho tan hon ton m1 gam Na vo m2 gam H2O thu c dung dch B c t khi d.

a. Vit phng trnh phn ng

b. Tnh nng % ca dung dch B theo m1 v m2c. Cho C% = 5% , d =1,2g/ml. Tnh nng mol ca dung dch thu c.

Cu 5(5im): Ho tan hon ton 4gam hn hp gm 1 kim loi ho tr II v 1 kim loi ho tr III cn dng ht 170ml dung dch HCl 2M

a. Tnh th tch H2 thot ra ( KTC).

b. C cn dung dch thu c bao nhiu gam mui kh.

c. Nu bit kim loi ho tr III l Al v s mol bng 5 ln s mol kim loi ho tr II th kim loi ho tr II l nguyn t no .

( Cho Fe =56, Na =23, O =16, Cl =35,5, Cu =64, Zn =65 , Al =27 H =1, Ba =137)

5 thi hs gii mn : Ho 9Thi gian : 150 pht Cu 1 : (1,5 )

Nung hn hp gm 2 mui CaCO3 v MgCO3 thu c 76 h hn hp 2 oxt v 33,6 lt kh CO2 (ktc). Hiu sut ca phn ng l 96 %. Khi lng hn hp ban u l :

A. 147 (g)

Cu 2 : (4,5)1, Hy dng mt ho cht nhn bit 6 l ho cht b mt nhn ng cc dung dch sau : K2CO3 ; (NH4)2SO4 ; MgSO4 ; Al2(SO4)3; FeCl3

2, Tm cng thc ho hc ca cc ch ci A, B, C , D, E, G v vit cc phng trnh ho hc biu din cc bin ho sau :

a, Al A B C A NaAlO2

b, Fe D E Fe2O3 D F G FeO

Cu 3 : (3,5)

1, Hy cho bit cc hin tng c th xy ra v vit phng trnh phn ng xy ra trong nhng th nghim sau :

a, Cho t t dung dch Ba(OH)2vo dung dch (NH4)2SO4

b, Cho mu kim loi Na vo dung dch Al(NO3)3

c, Nh t t dung dch H2SO4 c vo ng glucz (C6H12O6)

2, Trong nc thi ca mt nh my c cha a xt H2SO4. Bng th nghim thy rng c 5 lt nc thi cn dng 1g Ca(OH)2 trung ho. Mi gi nh my thi ra 250 m3 nc thi a, Tnh khi lng Ca(OH)2 cn dng trung ho lng nc thi trong 1 gi.

b, Tnh khi lng CaO cn dng trong 1 ngy. Bit nh my hot ng 24gi/ngy.

Cu 4 (5): Ho tan 5,94 g kim loi ho tr III trong 564 ml dung dch HNO3 10% (d=1,05 g/ml) thu c dung dch A v 2,688 lt hn hp kh B (gm N2O v NO) ktc. T khi ca kh B i vi Hir l 18,5 .

a, Tm kim loi ho tr III . Tnh C % ca cht trong dd A .

b, Cho 800 ml dung dch KOH 1M vo ddA. Tnh khi lng kt ta to thnh sau phn ng .

Cu 5(4) : Nung 178 g hn hp gm cc mui Na2SO4, Na2CO3, NaHCO3 thu c hn hp cht rn A v 5.600 cm3 kh CO2 .Cho hn hp A vo 150 cm3 dung dch a xt HCl (d=1,08 g/cm3) thu c 12320 cm3 kh CO2 .

a,vit phng trnh ho hc xy ra .

b, Tnh thnh phn phn trm khi lng cc mui trong hn hp ban u.

6 CU I: (4 im)

1. Hon thnh s phn ng sau:

A

A Fe D G (Bit A + B D + G + H2O )

A

2. Tch cc cht ra khi hn hp gm CaO, NaCl, CaCl2.

CU II: (4,5im)

1. Nu hin tng v vit PTHH (nu c) cho mi th nghim sau:

a. Cho kim loi Natri vo dd CuCl2.

b. Sc t t n d kh CO2 vo nc vi trong.

c. Nh t t n d dd HCl c vo cc ng thuc tm.

d. Cho l kim loi ng vo dd st (III) sunfat.

2. Khng dng thuc th no khc hy phn bit cc dd ng trong cc l ring bit: NaHCO3, Na2CO3, BaCl2, Na3PO4, H2SO4.

CU III: (6 im)

1. Cho 0,2 mol Zn vo 100g dung dch X cha 0,1 mol CuSO4 v 0,2 mol FeSO4 c dung dch Y cha 2 mui tan. Tnh nng phn trm cc cht trong dung dch Y.

2. Ngi ta dng 200 tn qung c hm lng Fe2O3 l 30% luyn gang. Loi gang thu c cha 80% Fe. Tnh lng gang thu c bit hiu sut ca qu trnh sn xut l 96%.

CU IV: (5,5im)

Cho 14,8 gam gm kim loi ho tr II, oxit v mui sunfat ca kim loi tan vo dd H2SO4 long d thu c dd A v 4,48 lt kh ktc. Cho NaOH d vo dd A thu c kt ta B. Nung B n nhit cao th cn li 14 gam cht rn.

Mt khc, cho 14,8 gam hn hp vo 0,2 lt dd CuSO4 2M. Sau khi phn ng kt thc, tch b cht kt ta ri em c cn dd th thu c 62 gam cht rn. Xc nh kim loi.

7

Cu 1: ( 5,0 im )a- Vit cc phng trnh phn ng thc hin chuyn ho ho hc sau :

M N P Q

EMBED Equation.3 R T M

Cho bit A l kim loi thng dng c 2 ho tr thng gp l (II) v (III) kh bn .

b- Vit cc phng trnh phn ng xy ra trong qu trnh iu ch axit Sulfuric t qung Pirit .

Cu 2: ( 5,0 im ) Ha tan 115,3 g hn hp X gm MgCO3 v RCO3 bng 500ml dd H2SO4 thu c dd A , rn B v 4,48 lt kh CO2 (ktc). C cn dd A thu c 12g mui khan. Mt khc, nung B n khi lng khng i thu 11,2 lt CO2 (ktc) v rn C.

a. Tnh nng mol ca dd H2SO4, khi lng rn B v C.

b. Xc nh R bit trong X s mol RCO3 gp 2,5 ln s mol MgCO3.

Cu 3: ( 5,0 im )X l hn hp hai kim loi Mg v Zn. Y l dd H2SO4 cha r nng .

Th nghim 1 : Cho 24,3 g X vo 2 lt Y sinh ra 8,96 lt kh H2 (ktc).


a. Chng t rng trong th nghim 1 th X cha tan ht, trong th nghim 2 th X tan ht.

b. Tnh nng mol c dd Y v khi lng mi kim loi trong X.

Cu 4: ( 5,0 im )C 5,56 g hn hp A gm Fe v mot kim loi M (c ha tr khng i). Chia A lm hai phn bng nhau. Phn I ha tan ht trong dd HCl c 1,568 lt hydr. Ha tan ht phn II trong dd HNO3 long thu c 1,344 lt kh NO duy nht. Xc nh kim loi M v thnh phn phn trm khi lng mi kim loi trong A. (cc th tch kh ktc).

8 BI

Cu 1 : Chn phng n tr li ng. Khoanh trn v gii thch s la chn :

1- Khi lng tnh bng gam ca nguyn t Na l :

A. 3.10-23g. B. 2,82.10-23g. C. 3,82.10-23g. D. 4,5.10-23g.

2- Thnh phn cc nguyn t ca hp cht R c chim 58,5%C ; 4,1%H ; 11,4%N v oxi . Cng thc ho hc ca hp cht l :

A. C3H5NO2 ; B. C6H5NO2 ; C. C6H13NO2 ; D. C2H5NO2 Cu 2: 1- Lng cht cha trong 1 gam oxt ca nhng oxt no di y l nh nhau :

a. CO2 ; b. CO ; c. NO2 ; d. N2O.

2- C mt hc sinh lm th nghim v thy hin tng xy ra nh sau :

Bn hc sinh dng mt ng nh thi vo ng nghim c cha nc vi trong, ban

u c hin tng nc vi trong vn c; bn tip tc thi vi hy vng nc s c trng xo nhng kt qu li khc l nc vi li trong dn li. Em hy gip bn gii thch hin tng trn v vit PTHH chng minh.

Cu 3 : Chn 2 cht v c tho mn cht R trong s sau :

A B C

R R R R

X Y Z

Cu 4 : Sc V(lt) CO2 vo 4lt dung dch Ca(OH)2 0,02M thu c 5g kt ta trng, tnh

9

Cu 1.(1,25 im)

Nguyn t ca nguyn t R c tng s ht proton, ntron v electron l 40. Trong s ht mang in nhiu hn s ht khng mang in l 12.

Xc nh R v s ht mi loi.

Cu 2.(1,75 im)

Nhn bit cc oxit ng ring bit trong mi l mt nhn sau ch dng hai ho cht khc: MgO, Na2O, P2O5 v ZnO.

Cu 3. (1 im) Vit 4 phng trnh phn ng iu ch O2 m em hc chng trnh lp 8, ghi iu kin phn ng (nu c).

Cu 4. (1,5 im) ha tan hon ton 8 gam oxit kim loi R cn dng 300ml dung dch HCl 1M. Hi R l kim loi g?Cu 5. (1,5 im)Cho bit tng s ht proton, ntron, electron trong 2 nguyn t ca nguyn t A v B l 78, trong s ht mang in nhiu hn s ht khng mang in l 26 ht. S ht mang in ca A nhiu hn s ht mang in ca B l 28 ht. Hi A, B l nguyn t g ?

Cho bit in tch ht nhn ca mt s nguyn t sau :

ZN = 7 ; ZNa = 11; ZCa = 20 ; ZFe = 26 ; ZCu = 29 ; ZC = 6 ; ZS = 16.

Cu 6.(3 im)Cho 7,73 gam hn hp gm km v st c t l nZn : nFe = 5 : 8 vo dung dch HCl d ta thu c V lt kh H2 (ktc). Dn ton b lng kh H2 ny qua hn hp E (gm Fe2O3 chim 48%, CuO chim 32%, tp cht cha 20%) c nung nng.

a. Tnh V

b. Tnh khi lng hn hp E va phn ng hon ton vi V lt kh H2 ni trn. Bit rng tp cht khng tham gia phn ng

(Cho Zn = 65; Fe = 56; O =16)

(Th sinh khng c s dng bt c ti liu no, cn b coi thi khng gii thch g thm)

10 Cu 1: (6,0 im)

a) Trnh by cc phng php iu ch Baz, mi phng php cho mt v d.

b) iu ch Cu(OH)2 th phng php no ph hp? Tm cc cht c th c ca phng php chn v vit tt c cc phn ng xy ra.

Cu 2: (5,0 im)

t chy mt di magi ri a vo y mt bnh ng kh lu hunh ioxit. Phn ng to ra mt cht bt A mu trng v mt cht bt mu vng B. Cht A phn ng vi dung dch H2SO4 long to ra cht C v nc. Cht B khng tc dng vi dung dch H2SO4 long, nhng B chy c trong khng kh to ra cht kh c trong bnh lc ban u.

a) Hy xc nh tn cc cht A, B, C

b) Vit cc phng trnh phn ng sau:

- Magi v kh lu hunh ioxit v cho bit phn ng ny thuc loi phn ng no? Vai tr ca Magi v lu hunh ioxit trong phn ng

- Cht A tc dng vi H2SO4 long

- Cht B chy trong khng kh.

Cu 3: (5,0 im)

a) C 4 l ha cht mt nhn ng ln lt cc cht: Nc, dung dch HCL, dung dch Na2CO3 v dung dch NaCl. Khng dng thm ha cht no khc. Hy nhn bit tng cht (c dng cc bin php k thut).

b) Cho 1g bt st tip xc vi oxi mt thi gian thy khi lng bt vt qu 1,41g.

Gi s ch to thnh 1 oxit st duy nht th l xit no?

A.FeO

B. Fe2O3

C.Fe3O4 D.Khng c xit no ph hp

Gii thch cho la chn ng.

Cu 4: (4,0 im)

Nhit phn hon ton 20g mui cacbonat kim loi ha tr II c cht rn A v kh B.

Dn ton b kh B vo 150ml dung dch Ba(OH)2 1M thu c 19,7g kt ta.

a) Tnh khi lng cht rn A

b) Xc nh cng thc mui cacbonat .

(Cho bit kim loi ha tr (II): Mg = 24; Ca = 40; Be = 9; Ba = 137)

11

Cu 1: ( 5,0 im )a- Vit cc phng trnh phn ng thc hin chuyn ho ho hc sau :

M N P Q

EMBED Equation.3 R T M

Cho bit A l kim loi thng dng c 2 ho tr thng gp l (II) v (III) kh bn .

b- Vit cc phng trnh phn ng xy ra trong qu trnh iu ch axit Sulfuric t qung Pirit .

Cu 2: ( 5,0 im ) Ha tan 115,3 g hn hp X gm MgCO3 v RCO3 bng 500ml dd H2SO4 thu c dd A , rn B v 4,48 lt kh CO2 (ktc). C cn dd A thu c 12g mui khan. Mt khc, nung B n khi lng khng i thu 11,2 lt CO2 (ktc) v rn C.

c. Tnh nng mol ca dd H2SO4, khi lng rn B v C.

d. Xc nh R bit trong X s mol RCO3 gp 2,5 ln s mol MgCO3.

Cu 3: ( 5,0 im )X l hn hp hai kim loi Mg v Zn. Y l dd H2SO4 cha r nng .



c. Chng t rng trong th nghim 1 th X cha tan ht, trong th nghim 2 th X tan ht.

d. Tnh nng mol c dd Y v khi lng mi kim loi trong X.

Cu 4: ( 5,0 im )C 5,56 g hn hp A gm Fe v mot kim loi M (c ha tr khng i). Chia A lm hai phn bng nhau. Phn I ha tan ht trong dd HCl c 1,568 lt hydr. Ha tan ht phn II trong dd HNO3 long thu c 1,344 lt kh NO duy nht. Xc nh kim loi M v thnh phn phn trm khi lng mi kim loi trong A. (cc th tch kh ktc).

12 bi :

Cu 1 (1) : Trn 100 ml dung dch H2SO4 1M vo 300 ml dung dch NaOH 1M , phn ng kt thc cho mu qu tm vo dung dch ta thy mu qu tm ha xanh. Ti sao?

Cu 2 (2) : Mun iu ch Canxi sunfat t Lu hynh v Canxi cn thm t nht nhng ha cht g ? Vit cc phng trnh phn ng.

Cu 3 () : Nu hin tng, vit phng trnh phn ng cho cc th nghim sau :

a/ Nhng inh st co sch g vo dung dch ng sunfat

b/ Sc kh SO2 vo dung dch Ca(HCO3)2

d/ Cho ng vo dung dch H2SO4 c nng

Cu 4 (1,5 ) : Nu hin tng xy ra trong mi trng hp sau v gii thch.

Cho CO2 li chm qua nc vi trong sau thm nc vi trong vo dung dch thu c

Cu 5 (2 ): Tnh lng Oxi trong ha cht A cha 98% H3PO4 tng ng vi lng Lu hynh c trong ha cht B cha 98% SO4. Bit lng Hyr A bng lng Hyr B

Cu 6 (2 ) : Trong mt ng nghim ngi ta ha tan 8 gam ng Sunfat ngm nc CuSO4.5H2O ri th vo mt ming km. C bao nhiu gam ng nguyn cht sinh ra sau phn ng, bit rng ly tha Km

Cu 7 (2 im) : Hy tm A, B, C, D hon thnh s phn ng

A B C D CuSO4Cu 8 (2,5 im) : 4,48 gam Oxit ca mt kim loi c ha tr 2 tc dng va vi 100 ml dung dch H2SO4 0,8 M, ri c cn dung dch th nhn c 13,76 gam tinh th mui ngm nc. Tm cng thc ca mui ngm nc trn.

Cu 9 (3 im) : Ha tan 10,8 gam hn hp gm Nhm, Magi v ng vo dung dch HCl 0,5 M ta c 8,96 lt Hyr ( ktc) v 3 gam mt cht rn khng tan.

a/ Tnh khi lng mi kim loi c trong hn hp

b/ Tnh th tch dung dch HCl cn dng

13

Cu 1: (1 im)

1) C 4 dung dch b mt nhn : AgNO3, NaOH, HCl, NaNO3Hy dng mt kim loi phn bit cc dung dch trn. Vit cc phng trnh ho hc minh ho.

2) Vit cc phng trnh ho hc xy ra cho cc th nghim sau:

a) Sc kh SO3 vo dung dch BaCl2

b) Nung nng Fe(OH)2 trong khng kh

c) in phn dung dch NaCl c mng ngn

Cu 2: ( 2,5 im )

Cho a gam Na vo 160 ml dung dch (D = 1,25 g/ml ) gm Fe2(SO4)3 0,125M v Al2(SO4)3 0,25M. Tch kt ta nung c 5,24 gam cht rn.

a) Tnh a?

b) Tnh C% cc cht trong dung dch sau phn ng?

Cu 3:(2 im) Ho tan 43,71 gam hn hp gm 3 mui Cacbonat, Hirocacbonat, Clorua ca mt kim loi kim ( ho tr I ) vo mt th tch dung dch HCl 10,52 % ( D = 1,05 g/ml ) ly d c dung dch A v 17,6 gam kh B

Chia dung dch A thnh hai phn bng nhau

Phn 1 : Cho tc dng vi AgNO3 d, c 68,88 gam kt ta

Phn 2 : Phn ng va vi 125 ml dung dch KOH 0,8 M sau phn ng c cn c 29,68 gam hn hp mui khan.

a) Tm tn kim loi kim ?

b) Tnh % khi lng mi kim loi ly ?

c) Tnh th tch dung dch HCl dng ?

Cu 4 : ( 3 im )

Cho 10,72 gam hn hp Fe v Cu tc dng vi 500 ml dung dch AgNO3 phn ng hon ton xong thu c dung dch A v 35,84 gam cht rn B.

a) Chng minh B khng phi hon ton l Ag

b) Cho dung dch A tc dng vi dung dch NaOH d ri lc kt ta nung trong khng kh n khi lng khng i thu c 12,8 gam cht rn. Tnh nng % v khi lng mi kim loi trong hn hp ban u v tnh nng mol / lit ca AgNO3 ban u ?

Hc sinh c s dng bng tun hon cc nguyn t ha hc

........................................................Ht.....................................................................

14 Cu 1: (2, 0 im )

Hy ch ra 3 cht n gin no nm trn cng mt dy ngang hay trn cng mt ct dc hoc trn cng mt ng cho ca hnh vung di y u l phi kim :

NaAlC

Fe

SiCa

P

S

Cu 2: ( 2, 0 im )

Trong thnh phn 3 mol lu hunh xt c 3,6 1024 nguyn t xy v

1,8 1024 nguyn t lu hunh . a ra cng thc phn t xt lu hunh ?

Cu 3: ( 6, 0 im )

Ngi ta un nng trong mt bnh cu 0,18 gam mt cht n gin A vi

Axt H2SO4 c d . Sn phm to thnh ca phn ng ngi ta cho i qua dung dch Can xi hyrxt , khi tch ra 5,1 gam kt ta . Hy xc nh cht A ( a ra cu tr li bng tnh ton v phng trnh chng minh ).

Cu 4: ( 4, 0 im )

Cht rn A mu xanh lam ,tan c trong nc to thnh dung dch , khi cho thm NaOH vo dung dch to ra kt ta B mu xanh lam . Khi nung nng ,cht B b ho en . Nu sau tip tc nung nng sn phm trong dng Hyr th to ra cht C mu . Cht C tng tc vi mt A xt v c m c to ra dung dch ca cht A ban u . Hy cho bit cht A l cht no , vit tt c cc phng trnh phn ng ho hc tng ng .

Cu 5 : ( 6, 0 im )

Ngi ta cho 5,60 lt hn hp xt Cc bon ( II ) v Cc bon ( IV ) khi nung nng i qua mt ci ng cha 20,0 gam xt ng ( II ) .Sau ngi ta s l ng cha trn bng 60,0 ml dung dch A xt H2SO4 nng 85 % ( t khi dung dch bng 1,80 g/ml ) .Khi 42,7 % A xt H2SO4 tham gia vo phn ng .

a/ Hy vit cc phng trnh phn ng xy ra .

b/ Hy tnh phn th tch ca cc xt cc bon trong hn hp u .

15 Cu 1: (5 im)

1- Trong hp cht kh vi Hir ca nguyn t R c ha tr IV, Hir chim 25% v khi lng. Xc nh nguyn t

2- Ngi ta dng mt dung dch cha 20 gam NaOH hp thu hon ton 22 gam CO2. Vit phng trnh phn ng v gi tn mui to thnh.

Cu 2: (5 im)

1- B tc chui phn ng v cho bit A, B, C, D, E, F l nhng cht g?

A + B

C + H2 C + Cl2

D

D + dd NaOHE + F

E

Fe2O3

H2O

Cu 2: ( 2, 0 im )

Trong thnh phn 3 mol lu hunh xt c 3,6 1024 nguyn t xy v 1,8 1024 nguyn t lu hunh . a ra cng thc phn t xt lu hunh ?

Cu 3: (5 im)

Cho a gam dung dch H2SO4 24,5% v b gam dung dch NaOH 8% th to c 3,6 gam mui axt v 2,84 gam mui trung ha.

1- Tnh a v b

2- Tnh thnh phn trm ca dung dch sau phn ng

Cu 4: ( 4, 0 im )

Cht rn A mu xanh lam ,tan c trong nc to thnh dung dch , khi cho thm NaOH vo dung dch to ra kt ta B mu xanh lam . Khi nung nng ,cht B b ho en . Nu sau tip tc nung nng sn phm trong dng Hyr th to ra cht C mu . Cht C tng tc vi mt A xt v c m c to ra dung dch ca cht A ban u . Hy cho bit cht A l cht no , vit tt c cc phng trnh phn ng ho hc tng ng

16 THI HSG Mn: Ha 9 - Thi gian lm bi: 90 pht

Cu 1: (3 im) 1- (1,5 im): Hy cho bit trong dung dch c th c ng thi cc cht sau y khng ?

a. KOH v HCl

d. HCl v AgNO3

b. Ca(OH)2 v H2SO4

e.Ca(OH)2 v Ca(HCO3)2

c. HCl v KNO3

g. KCl v NaOH

2- (1,5 im): Vit phng trnh ha hc thc hin chui bin ha sau:

Fe2O3 (a) FeCl3 (b) Fe(OH)3 (c) Fe2O3 (d) Fe.

Cu 2: (2im)

Cho 4 l dung dch NaCl, CuSO4, MgCl2, NaOH thuc th ch c phe nolph talein. Lm th no nhn bit chng?

Cu 3: (2 im)

Phi ly 2 ming nhm c t l vi nhau nh th no v khi lng khi cho mt ming vo dung dch axt v 1 ming kia vo dung dch baz, th ta c th tch kh Hir thot ra bng nhau?

Cu 4: (3 im)

Ha tan 1,68 gam hn hp gm Ag v Cu trong 29,4 gam dung dch A(H2SO4 c, nng) thu c ch mt loi kh v dung dch B. Cho kh thot ra hp th ht vo nc brm, sau thm Ba(NO3)2 d th thu c 2,796 gam kt ta.

a. Tnh khi lng Ag v Cu trong hn hp u.

b. Tnh nng % H2SO4 trong dung dch A, bit lng H2SO4 phn ng vi Ag v Cu ch bng 10% lng ban u. 17 THI HC SINH GII LP 9Cu 1:

Khi cho dung dch H3 PO4 Tc dng vi dung dch NaOH to c dung dch M.

a/ Hi M c th cha nhng mui no?

b/ Phn ng no c th xy ra khi thm KOH vo dung dch M

c/ Phn ng no c th xy ra khi thm H3 PO4 ( hoc P2 O5) vo dung dch M?

Vit phng trnh phn ng.

Cu 2:

C th c hin tng g xy ra khi cho kim loi A vo dung dch mui B? Vit phng trnh phn ng.

Cu 3:

Hy nu mt mui va tc dng vi HCl, va tc dng vi NaOH , tho mn iu kin:

a/ C hai phn ng u c kh thot ra.

b/ Phn ng vi HCl c kh bay ln v phn ng vi NaOH c kt ta.

c/ C hai phn ng u to kt ta.

Cu 4:

A + O2 ( B + C

B + O2 ( D

D + E ( F

D + BaCl2 + E ( G ( + H

F + BaCl2 ( G ( + H

H + AgNO3 ( Ag Cl( + I

I + A ( J + F + NO ( + E

I + C ( J + E

J + NaOH ( Fe(OH)3 + K

Cu 5:

Nung x1 gam Cu vi x2 gam O2 thu c cht rn A1. un nng A1 trong x3 gam H2SO4 98% , sau khi tan ht thu c dung dch A2 v kh A3 . Hp th ton b A3 bng 200ml NaOH 0,15 M to ra dung dch cha 2,3g mui . Bng phng php thch hp tch CuSO4 ra khi dung dch A2 s thu c 30g tinh th CuSO4. 5 H2O . Nu cho A2 tc dng vi dung dch NaOH 1M th to ra lng kt ta nhiu nht phi dng ht 300ml NaOH. Vit phng trnh phn ng . Tnh x1, x2, x3.

18 THI HC SINH GII LP 9Cu1:

Cu2: Dung dch A c cha CuSO4 v FeSO4

a/ Thm Mg vo dung dch A ( dung dch B c 3 mui tan.

b/ Thm Mg vo dung dch A ( dung dch C c 2 mui tan.

c/ Thm Mg vo dung dch A ( dung dch D ch c 1 mui tan.

Gii thch mi trng hp bng phng trnh phn ng.

Cu3:

Ho tan mt hn hp gm Mg v mui Magi cacbonnal bng axit HCl th thu c mt hn hp kh c th tch l 6,72 lt ( o KTC) . Sau khi t hn hp kh ny v lm ngng t ht hi nc th th tch hn hp kh ch cn 1,12 lt ( KTC).

a/ Vit phng trnh phn ng xy ra.

b/ Tnh thnh phn v khi lng ca mi cht trong hn hp?

Cu 4:

a/ Cho 1,625 g st Clorua ( cha r ho tr ca st) tc dng vi dung dch AgNO3 d thu c 4,305 g AgCl kt ta. Xc nh cng thc ca st Clorua.

b/ Cn bao nhiu mililt dung dch NaOH cha 0,02 g NaOH trong 1ml dung dch chuyn 1,25 g FeCl3 . 6 H2O thnh Fe(OH)3 .

Cu 5:

Cho A l mt hn hp bt gm : Ba , Al , Mg.

- Ly m gam A cho vo nc ti khi ht phn ng thy thot ra 6,94 lt H2 ( KTC) .

- Ly m gam A cho vo dung dch xt d ti ht phn ng thy thot ra 6,72 lt H2 ( KTC) .

- Ly m gam A ho tan bng mt lng va dung dch axit HCl c mt dung dch v 9,184 lt H2 (o KTC) . Hy tnh m v % khi lng cc kim loi trong A.

19Cu 1: (2 im)

T nhng cht c sn l K2O, BaO, H2O v cc dung dch CuCl2, FeCl3. Hy vit cc phng trnh ha hc iu ch baz tan v baz khng tan.

Cu 2: ( 4 im)

Vit cc phng trnh phn ng ghi r iu kin (nu c) theo s :

a. A to B + H2O C + CO2 A +HCl D + 1NaOH E

b. AlCl3

Al2(SO4)3 Al(OH)3

Al2O3Cu 3: (3 im)

a. Bng phng php ha hc hy nhn bit 3 l ng cht rn khng nhn: NaOH, NaCl, Ba(OH)2.

b. Ch c dng qu tm lm th no nhn bit cc dung dch cht cha trong cc l mt nhn ring bit: KCl, K2SO4, KOH v Ba(OH)2.

Cu 4: (1.5 im) T cc ha cht bit. Hy vit 3 PTHH iu ch clo

Cu 5: (1.5 im)

Ha tan M2O3 trong mt lng va dung dch H2SO4 20%. Ngi ta thu c dung dch mui c nng 21,756%. Xc nh cng thc oxit.

Cu 6: (1.5 im)

Cho 24 g hn hp CuO v Fe2O3 ha tan va vo 146g dung dch HCl 20%. Tnh thnh phn trm theo khi lng c trong hn hp.

Cu 7: (4.5 im) Kh 15.2g hn hp FeO v Fe2O3 bng hidro nhit cao, thu c st kim loi. ha tan ht lng st ny cn dng 100ml dung dch H2SO4 2M.

a. Xc nh phn trm khi lng mi oxit.

b. Tnh th tch H2 ktc cn dng kh hn hp trn.

c. Nu c cn cn thn dung dch sau phn ng s thu c bao nhiu gam tinh th FeSO4. 7H2O

Th sinh c s dng bng h thng tun hon cc nguyn t ha hc.

20

Cu 1: (6,0 im)

a) Trnh by cc phng php iu ch Baz, mi phng php cho mt v d.

b) iu ch Cu(OH)2 th phng php no ph hp? Tm cc cht c th c ca phng php chn v vit tt c cc phn ng xy ra.

Cu 2: (5,0 im)

t chy mt di magi ri a vo y mt bnh ng kh lu hunh ioxit. Phn ng to ra mt cht bt A mu trng v mt cht bt mu vng B. Cht A phn ng vi dung dch H2SO4 long to ra cht C v nc. Cht B khng tc dng vi dung dch H2SO4 long, nhng B chy c trong khng kh to ra cht kh c trong bnh lc ban u.

a) Hy xc nh tn cc cht A, B, C

b) Vit cc phng trnh phn ng sau:

- Magi v kh lu hunh ioxit v cho bit phn ng ny thuc loi phn ng no? Vai tr ca Magi v lu hunh ioxit trong phn ng

- Cht A tc dng vi H2SO4 long

- Cht B chy trong khng kh.

Cu 3: (5,0 im)

a) C 4 l ha cht mt nhn ng ln lt cc cht: Nc, dung dch HCL, dung dch Na2CO3 v dung dch NaCl. Khng dng thm ha cht no khc. Hy nhn bit tng cht (c dng cc bin php k thut).

b) Cho 1g bt st tip xc vi oxi mt thi gian thy khi lng bt vt qu 1,41g.

Gi s ch to thnh 1 oxit st duy nht th l xit no?

A.FeO

B. Fe2O3

C.Fe3O4

D.Khng c xit no ph hp

Gii thch cho la chn ng.

Cu 4: (4,0 im)

Nhit phn hon ton 20g mui cacbonat kim loi ha tr II c cht rn A v kh B.

Dn ton b kh B vo 150ml dung dch Ba(OH)2 1M thu c 19,7g kt ta.

a) Tnh khi lng cht rn A

b) Xc nh cng thc mui cacbonat .

(Cho bit kim loi ha tr (II): Mg = 24; Ca = 40; Be = 9; Ba

21

Cu 1 :(3 25 ) Vit nhng phng trnh phn ng , ghi iu kin (nu c ) thc hin nhng bin ha theo s di y :

a. Natri Natri oxit Natri hidroxit Natri nitrat

b. Bari Bari oxit Bari hidroxit bari clorua

c. photpho anhidric photphoric axit photphoric can xi photphat

d. ng hydroxit ng oxit ng clorua

e. St (III)hydroxit St (III) oxit St (III) sunfat

Cu 2 : ( 2 25 ) Cho cc cht : Axit clohiric , dung dch Natri hydroxit ,Bari sunfat ,magie cacbonat , kali cacbonat ,ng nitrat . Hi :

- Nhng cht no tc dng c vi nhau to thnh cht tn ti ?

-Vit cc phng trnh phn ng tng ng .

Cu 3 : ( 1 ) Mt loi duyra c thnh phn khi lng nh sau : 94% Al ,4% Cu ,0,5% mi nguyn t Mg , Mn ,Fe ,Si . Nu c 1 tn nhm nguyn cht th phi ly bao nhiu kg mi nguyn t cn li luyn thnh duyra nh ni trn .

Cu 4 : (2 ) Cho cc cht sau :

Kali clorua , Canxi clorua ,Mangandioxit , axitsunfuric m c . em trn ln hai hoc ba cht vi nhau . Trn nh th no th to thnh hidro clorua ? trn nh th no th to thnh clor ? Vit cc phng trnh phn ng tng ng .

Cu 5 :(4 50 ) Mt hn hp X gm Al , Fe2 O3 ,c khi lng l 234 gam .Thc hin phn ng nhit nhm (Al kh Fe2 O3 cho ra Fe v Al2 O3 ) thu c cht rn Y .Cho Y tc dng vi dung dch NaOH d cn li cht rn Z c khi lng l 132 gam (trong phn ng khng c kh Hidro bay ra ).

a.Tnh khi lng Al ,Fe , Al2O3 , Fe2O3 trong hn hp Y

b.Tnh khi lng ca Al v Fe trong hn hp X .

Cu 6 : (4 ) Chia lm hai phn bng nhau 1 lt dung dch ( nhn c sau khi ho tan 31 gam Na2O vo nc ).

a. Cho phn (1) phn ng va vi x (ml ) dung dch Fe2(SO4)3 0,5 M . Tnh x v nng M cc cht tan trong dung dch sau phn ng .

b.Cho phn (2) tc dng vi y (ml ) dung dch H2SO4 20% (D=1,14 g/ml ). Tnh y cn dng thu c mui trung ha v khi lng ca mui .

Cu 7 :(3 ) B 27,05 gam tinh th FeCl3 .6H2O vo 100 gam dung dch NaOH 20%

a.Tnh khi lng ca cht kt ta to thnh ?

b.Tnh nng phn trm ca cc cht c trong dung dch sau phn ng ?

( Cho bit Na :23 ; Fe: 56 ; Al : 27 ; O : 16 ; H : 1 ; S :32 ; Cl : 35,5 )

AP N 1 CuNi dungim

1aFe2O3 + 6HCl 2FeCl3 + 3H2OFe2O3 + 3H2 2Fe + 3H2O2FeCl3 + Fe 3FeCl2Fe + 2HCl FeCl2 + H2 0.5

0.5

0.5

0.5

bFe + O2 FeO ; Fe2O3 . Fe3O4 FeO + 2HCl FeCl2 + H2O Fe2O3 + 6HCl 2FeCl3 + 3H2O Fe3O4 + 8 HCl FeCl2 . + 2 FeCl3 + 4H2O FeCl2 + 2NaOH Fe( OH )2 + 2NaCl FeCl3 + 3NaOH Fe ( OH )3 + 3NaCl FeCl2 . 2FeCl3 + 8NaOH Fe( OH )2 . 2Fe( OH )3 + 8NaCl 0,5 0,25 0,25 0,5 0,5 0,5 0,5

2MgCO3 MgO + CO2 . Kh B l CO2 , cht rn A ( MgO + MgCO3 )

CO2 + NaOH Na2CO3 + H2O

CO2 + NaOH NaHCO3

Dung dch cha 2 mui Na2CO3 v NaHCO3 vy mui Na2CO3 tc dng vi BaCl2 , cn NaHCO3 tc dng vi KOH .

Na2CO3 + BaCl2 BaCO3 + NaCl

2 NaHCO3 + 2KOH K2CO3 + Na2CO3 + 2H2O

MgO + 2HCl MgCl2 + H2O

MgCO3 + 2HCl MgCl2 + CO2 + H2O

Mui khan E l MgCl2 .

MgCl2 Mg + Cl2

kim loi ( M ) l Mg0,5 0,5 0,5 0,5 0,5 0,5 0,5 0,5

3a Cho hn hp i qua bnh ng dd NaOH d th kh CO2 v SO2 b gi li , kh thot ra l N2

CO2 + 2NaOH Na2CO3 + H2O

SO2 + 2NaOH Na2SO3 + H2O

Cho dd H2SO3 vo dung dch va thu c trn cho n d ta s thu c CO2 .

Phn ng :

H2SO3 + Na2CO3 Na2SO3 + CO2 + H2O

Cho tip dd va to thnh trn 1 lng dd HCl ta s thu c SO2 do phn ng

P/ ng : Na2SO3 + 2HCl 2NaCl + SO2 + H2O 0,5 0,5

0,5 0,750,25

bGi n l ho tr ca kim loi v a l s mol X dng ta c phn ng X + HCl XCln + n/2 H2 1 ( mol ) ( mol )

a ( mol ) ( mol ) Suy ra ta c h : a.X = 3,78 ( 1 )

= ( 2 )

an = 0,42 ( 3 )

T ( 1 ) , ( 2 ) , ( 3 ) => = 9

=> X= 9n

V ho tr ca kim loi c th 1 , 2 , hoc 3

Do xt bng sau :

n

1

2

3

4

X

9

18

27

36

- Trong cc kim loi bit Al c ho tr 3 , vi nguyn t lng 27 l ph hp

0,5 0,5 0,5 0,5 1

4 Mg + 2HCl MgCl2 + H2

2Al + 6 HCl 2AlCl3 +3H2

Cht rn D l Cu khng tan .

MgCl2 + 2NaOH Mg ( OH ) 2 + 2NaCl

Do NaOH d nn Al( Cl)3 tan

AlCl3 + 4NaOH NaAlO2 + 3NaCl + 2 H2O

Mg( OH )2 MgO + H2O

Cht rn E l MgO = 0,4 ( g )

2Cu + O2 2CuO

Cht rn F l CuO = 0,8 ( g )

Theo PT :

m Mg = . 24 ( g )

m Cu = . 64 ( g )

m Al = 1,42 ( 0,64 + 0,24 ) = 0,54 ( g )

0,5 0,5 0,5 0,5 0,5 0,250,5 0,250,5 0,5 0,5

P N S 2CuNi dungim

11 A to kt ta vi B,C,D nn A l AgNO3

AgNO3 + NaI = AgI + NaNO3

AgNO3 + HI = AgI + HNO3

2AgNO3 +K2CO3 = Ag2CO3 + 2KNO3

C to kt ta vi A v to kh vi HI C l K2CO3

B ch to kt ta vi 1 trong 3 cht cn li B l NaI

D l HI

2HI + K2CO3 = 2KI + CO2 k + H2O

0,25 0,25 0,25 0,25 0,25 0,25 0,25 0,25

2 Mi phng trnh ng c 0,5 im

1. PbCl2 + Na2CO3 = PbCO3 + 2NaCl

2. PbCl2 + Na2S = PbS + 2NaCl

3. PbCl2 + Na2SO3 = PbSO3 + 2NaCl

4. PbCl2 + Na2SO4 = PbSO4 + 2NaCl

5. 3PbCl2 + 2Na3PO4 = Pb3(PO4)2 + 6NaCl

6. PbCl22+ Na2SiO3 = PbSiO3 + 2NaCl

3

21 2C + O2 = 2 CO

C + O2 = CO2

S + O2 = SO2

Kh A:, CO2 , SO2, O2d, CO

Cho A qua dung dch NaOH

CO2 + 2NaOH = Na2CO3 + H2O

SO2 + 2NaOH = Na2SO3 + H2O

Dung dch B cha Na2CO3, Na2SO3 cn kh C cha: CO2, O2, CO

C qua CuO, MgO nng.

CuO + CO = Cu + CO2

Cht rn D ( MgO, Cu ) v kh E c: CO2, O2, CO d

E li qua Ca(OH)2

CO2 + Ca(OH)2 = CaCO3 + H2O

2CO2 + Ca(OH)2 = Ca(HCO3)2

Kt ta F l CaCO3

Dung dch G: Ca(HCO3)2

Ca(HCO3)2+ 2KOH = CaCO3 + K2CO3 + H2O

Ca(HCO3)2 = CaCO3+ CO2 + H2O

A qua xc tc nng

2SO3 + O2 = 2SO3 ( kh M)

M qua dung dch BaCl2

SO3 + H2O + BaCl2 = BaSO4 + 2HCl

(Kt ta N)( 0,25( 0,25)( 0,25)( 0,25) ( 0,25)( 0,25)( 0,25)( 0,25)( 0,25)( 0,25) ( 0,25)( 0,25)( 0,25)( 0,25)

32Ho tan trong nc

CaO + H2O = Ca(OH)2

Ra nhiu ln thu c cht rn A c CaCO3+ CaSO4v nc lc B c NaCl v Ca(OH)2 Thm Na2CO3 vo nc lc

Na2CO3 + Ca(OH)2= CaCO3+ 2 NaOH

Lc kt ta c nc lc C. em un nng kt ta

CaCO3= CaO + CO2

Trung ho nc lc C ri c cn c NaCl

Ngm cht rn A trong dung dch HCl

CaCO3 + 2HCl = CaCl2 + CO2 + H2O

Lc sn phm khng tan l CaSO4

Thm Na2CO3 vo nc lc thu li CaCO3

CaCl2 + Na2CO3= CaCO3+ 2 NaCl

( 0,5)

(0,25)

( 0,5)( 0,5)( 0,5)( 0,25)( 0,5)

4S mol Al2(SO4)3 = 0,2mol

nBa(OH)2 = 0,3mol

Pt: Al2(SO4)3 + 3Ba(OH)2 = 3 BaSO4 + 2 Al(OH)3

Khi nung BaSO4 c BaSO4 khng i

2Al(OH)3 = Al2O3 + 3H2O

Cht rn D gm BaSO4 v Fe2O3, dung dch B c Al2(SO4)3 d

Al2(SO4)3 + 3BaCl2 = 3 BaSO4 + 2FeCl3

Kt qu: mD= 80,1gam

mE = 69,9gam

CM = 0,4M

Ghi ch:

- HS lm cch khc m ng vn cho im ti a

- Cc phng trnh ho hc khng cn bng hoc khng ghi r trng thi tr 1/2 s im.( 0,5)( 0,5)( 0,5)( 0,5)( 0,25)( 0,5)( 0,5)( 0,25)( 0,5)

3 THI HSG CP HUYNcu hip nim

Cu I1. 5

(3)

- Nung qung St Pirit trong khng kh:

4FeS2 + 11O2 2Fe2O3 + 8SO2(

0,5

- in phn dung dch NaCl c mng ngn xp:

2NaCl + 2 H2O 2 2NaOH + 2Cl2(+ H2(

0,5

- iu ch Fe:

Fe2O3 + 3H2 2Fe + 3H2O

0,25

- iu ch H2SO4:

2SO2 + O2 2SO3( SO3 + H2O H2SO4

0,5

- iu ch FeSO4:

Fe + H2SO4 FeSO4 + H2(0,25

- iu ch Fe(OH)3:

2Fe + 3Cl2 2FeCl3 FeCl3 + 3NaOH Fe(OH)3(+ 3NaCl

0,5

- iu ch NaHSO4:

NaOH + H2SO4 NaHSO4+ H2O0,5

2.(2)

- Ho tan hn hp bng dung dch NaOH d, un nng ri lc b Fe2O3 khng tan:

Al2O3 + 2NaOH 2NaAlO2 + H2O

SiO2 + 2NaOH Na2SiO3 + H2O1

- Sc kh CO2 d i qua nc lc:

NaAlO2 + CO2 + 2H2O Al(OH)3( + NaHCO30,5

- Lc ly kt ta em nung nhit cao:

2Al(OH)3 0 Al2O3 + 3H2O0,5

Cu II15

(2)

- Chn dung dch Ba(OH)20,25

- Lp bng ghi cc hin tng nhn bit0,25

- Vit 3 phng trnh ng, mi phng trnh 0,5 dim1,50

2(3)

- Xc nh ng: A l Cu(OH)2; B l CuSO4; C l CuO0,5

- Vit ng 5 phng trnh, mi phng trnh 0,5 dim2,5

Cu III5

1(3)

- Gi A, B ln lt l k hiu ho hc ca kim loi tr I v II.

a, b ln lt l s mol ca 2 mui sunfat tng ng.

C phng trnh:

A2SO4 + BaCl2 BaSO4 + 2 ACl (1)

amol amol amol

BSO4 + BaCl2 BaSO4 + BCl2 (2)

bmol bmol bmol

1

- Ta c

0,5

- Theo phng trnh phn ng (1) v (2):

nBaCl2 = nBaSO4 = 0,3(mol) mBaCl2 = 0,3x208 = 62,4(gam)0,5

Theo nh lut bo ton khi lng:

m(A2SO4; BSO4) + mBaCl2 = mBaSO4 + m(ACl; BCl2)

suy ra: 44,2 + 62,4 = 69,9 + m (ACl; BCl2)

Vy, hai mui tan trong dung dch thu c l ACl v BCl2 c khi lng bng 36,7gam1

2(2)

- Phng trnh phn ng:

Zn + Cu(NO3)2 Zn(NO3)2 + Cu (1)

amol amol

Zn + Pb(NO3)2 Zn(NO3)2 + Pb (2)

amol amol0,5

- V khi lng hai l km bng nhau nn s mol bng nhau

Gi a l mol mi l km: nZn(1) = nZn(2) = a0,5

- Theo PT (1): mZn(1) gim: 65a 64a = 0,05.

suy ra: a = 0,05(mol)0,5

- Theo PT (2): mZn tng: 207a 65a = 142a

V a = 0,05 nn l km th 2 tng 142 x 0,05 = 7,1(gam)0,5

Cu IV15

(2)

Fe + Cu (NO3)2 Fe(NO3)2 + Cu (1)

0,16 mol 0,16 mol 0,16 mol

Fe + 2HCl FeCl2 + H2( (2)

0,2 mol 0,4 mol 0,2 mol0,5

- V phn ng xy ra hon ton m sau thu c hn hp kim loi, suy ra Fe cn d; Cu(NO3) 2 v HCl phn ng ht0,25

- Theo PT (2): nH2 = 1/2nHCl = 0,2 (mol)

Th tch H2 sinh ra ktc = 0,2 x 22,4 = 4,48 (lt)0.5

- Theo PT (1): nFe = nCu = nCu(NO3)2 = 0,16 (mol)

- Theo PT(2): nFe = 1/2nHCl = 0,2 (mol)

suy ra, khi lng Fe d = m (0,16 + 0,2) x 56 = (m 20,16)

- Khi lng Cu sinh ra = 0,16 x 64 = 10,24 (gam)0,5

- V hn hp hai kim loi thu c c khi lng = 0,7m (gam) nn ta c PT: (m 20,16) + 10,24 = 0,7m

Gii PT c m = 33,067(gam)0,25

2 CaCO3 CaO + CO2( (1)

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

2CO2 + Ba(OH)2 Ba(HCO3)2 (3)

Mi phng trnh vit ng cho 0,25 im

0,75

nCaCO3 = 0,3 (mol); nBa(HCO3)2 = 31,08/259 = 0,12 (mol)0,25

Nu ch to mui axit th CM ca Ba(OH)2 = 0,12/0,8 = 0,15(M)0,5

Nu to ra hn hp hai mui th CM ca Ba(OH)2 = 0,18/0,8 = 0,225(M)1,5

Ch :

- Cc PT vit thiu iu kin; cn bng sai tr 1/2 s im ca PT.

- Cc cch gii khc vn ng th cho im tng ng./.

P N 4Cup nim

1aiu ch NaOH: 1. 2Na + 2H2O ( 2NaOH + H2 2. Na2O + H2O ( 2NaOH 3. 2NaCl + 2H2O 2NaOH + Cl2 + H2 4. Na2CO3 + Ca(OH)2 ( 2NaOH + CaCO3 ( Mi phn ng ng cho 0,25 im)

biu ch CuCl2:

1. CuSO4 + BaCl2 ( CuCl2 + BaSO4

2. CuO + 2HCl ( CuCl2 + H2O

4. Cu(OH)2 + 2HCl ( CuCl2 + H2O

3. Cu + Cl2 ( CuCl2Nu hc sinh vit phn ng khc m ng vn cho im ti a.Nu thiu iu kin hoc khng cn bng th tr mt na s im ca phng trnh

2 FeCl2 (2) Fe(OH)2 (3) Fe(OH)3 Fe3O4 (1) (5) (4) Fe2O3 FeCl3 (6) Fe2(SO4)3 (7) Fe(OH)30,5

1. Fe3O4 + 8HCl ( FeCl2 + 2FeCl3 + 4 H2O

2. FeCl2 + 2KOH ( Fe(OH)2 + 2KCl

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

4. 2Fe(OH)3 Fe2O3 + 3H2O

5. 2FeCl2 + Cl2 2FeCl3

6. 2FeCl3 + 3H2SO4 ( Fe2(SO4)3 + 6HCl

7. Fe2(SO4)3 + 6NaOH ( 2Fe(OH)3 + 3Na2SO4 Nu hc sinh vit s khc m ng th vn cho im ti a.Nu khng cn bng hoc thiu iu kin th tr mt na s im ca phng trnh 0,5

0,5

0,5

0,5

0,5

0,5

0,5

3Trch cc mu th t cc mu phn bn v nung nng nu mu no c mi khai thot ra th l: NH4NO3 v NH4NO3 b phn hy theo phng trnh :

2NH4NO3 2NH3 + H2O + N2O5

KhaiCc cht cn li cho vo nc nu cht no khng tan trong nc l Ca3(PO4)2 .

Cc cht cn li tan to thnh dung dch .Ta cho 1 t dung dch AgNO3 vo 3 cht cn li nu c kt ta trng(AgCl) l mu phn bn KCl cn c kt ta vng(Ag3PO4) l K3PO4 khng c hin tng g l Ca(H2PO4)2.

PTP: KCl + AgNO3 ( AgCl (Trng) + KNO3 K3PO4 + 3AgNO3 ( Ag3PO4 (Vng) + 3KNO3 1

1

1

0,5

0,5

4)a m1nNa =

23

a. PTP: 2Na + 2H2O ( 2NaOH + H2

0,5

0,75

0,25

b.

= ; x2 = =

= =

1,5

c.

p dng cng thc : CM(NaOH) = = = 1,5M2

Cu 5

(5)aGi A v B ln lt l kim loi ho tr II v ho tr III ta c :

PTP: A + 2HCl ( ACl2 + H2 (1)

2B + 6HCl ( 2BCl3 + 3H2 (2)

nHCl = V.CM = 0,17x2 = 0,34 (mol)

T (1) v (2) ta thy tng s mol axit HCl gp 2 ln s mol H2 to ra

nH2 = 0,34: 2 = 0,17 (mol)

VH2 = 0,17. 22,4 3,808 (lit) 0,25

0,5

0,5

0,25

0,5

0,25

0,5

0,5

bnHCl = 0,34 mol => nCl = 0,34 mol

mCl = 0,34.35,5 = 12,07g

Khi lng mui = m(hn hp) + m(Cl) = 4+ 12,07 = 16,07g

c. gi s mol ca Al l a => s mol kim loi (II) l a:5 = 0,2a (mol)

t (2) => nHCl = 3a. v t (1) => nHCl = 0,4a

3a + 0,4a = 0,34

a = 0,34: 3,4 = 0,1 mol => n(Kimloai) = 0,2.0,1 = 0,02mol

mAl = 0,1.27 = 2,7 g

m(Kimloi) = 4 2,7 = 1,3 g

Mkimloi = 1.3 : 0,02 = 65 => l : Zn 0,25

0,5

0,25

0,25

0,25

0,5

0,25

0,25

0,25

p n 5 v biu im Mn ho 9Cup nim

21Cho dung dch NaOH vo c 6 l dung dch .

+ Nu khng c phn ng l dung dch K2CO3

Nu c cht mi khai bc ln l ( NH4)2SO4

PTHH: ( NH4)2SO4 + 2NaOH ( Na2SO4 + 2 NH3 + 2H2O

+ Nu c cht kt ta trng hi xanh l FeCl2

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

Trng hi xanh

+ Nu c cht kt ta nu l FeCl3 .

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

(Nu )

+ Nu c cht kt ta trng khng tan l MgSO4

MgSO4 + NaOH ( Na2SO4 + Mg(OH)2

trng

+ Nu c cht kt ta trng to thnh sau tan trong dung dch NaOH d l Al2(SO4)3 Al2(SO4)3 + 6NaOH ( 3 Na2SO4 + 2Al(OH)3Al(OH)3 + NaOH ( NaAlO2 + 2H2O1.5

2a, Al ( A( B( C( A( NaAlO2- A To ra t Al - A To NaAlO2 A l Al2O3-A l Al2O3 C l Al(OH)3 B L mui tan ca nhm.

Ta c dy bin ho l :

Al ( Al2O3 ( AlCl3 ( Al(OH)3 ( Al2O3(NaA1O2Phng trnh ho hc

(1) 4Al + 3O2 ( 2Al2O3 (2) Al2O3 + 6HCl ( 2AlCl3 + 3H2O

(3) Al3 + 3NaOH ( Al(OH)3 +3NaCl (4) 2Al(OH)3 ( Al2O3 +3H2O

(5) Al2O3 +2NaOH ( 2NaAlO2 +H2O

b, Fe ( D Fe2O3 (D D l mui st III. VD: FeCl3E( Fe2O3 E l Fe(OH)3 G( FeO G l Fe(OH)2

F l mui (II) VD: FeCl2 Ta c dy bin ho :

Fe(FeCl3 (Fe(OH)3(Fe2O3 (FeCl3 (FeCl2 (Fe(OH)2 (FeO

Phng trnh ho hc

(1) 2Fe + 3Cl2 ( 2FeCl3 (2) FeCl3 + 3 NaOH (Fe(OH)3 + 3NaCl

(3) 2 Fe(OH)3 ( Fe2O3 +3H2O (4) FeO3 + 6HCl (2FeCl3 + 3H2O

(5)2FeCl3 +Fe ( 3FeCl2(6) FeCl2 + 2NaOH (Fe(OH)2 + 2NaCl

(7) Fe(OH)2 ( FeO + H2O4.5

31a, Cho t t dd Ba(OH)2 vo dd NH4Cl

Hin tng: Kt ta trng xut hin v tng dn ng thi c kh mi khai thot ra.

Phng trnh ho hc: Ba(OH)2 + (NH4)2SO4 (BaSO4 + 2NH3 + 2H2O

b, Cho mu Na vo dd Al(NO3)3 Hin tng : Ban u mu Na nng chy tn dn, thot ra kh khng mu, ng thi thy xut hin kt ta trng 2 Na +2H2O ( 2 NaOH + H2

3NaOH + Al(NO3)3 (3NaNO3 + Al(OH)

- Kt ta trng c th tan ra 1 phn hoc tan ht to dung dch khng mu nu NaOH d. NaOH + Al(OH)3( NaAlO2 +2H2O.

c, Nh t t dung dch H2SO4 c vo ng Glucoz (C6H12O6)

Hin tng : ng Glucoz mu trng chuyn dn sang mu vng ri thnh mu en, ng thi c kh khng mu thot ra .

C6H12O6 6C + 6H2O C + H2SO4 c nng ( CO2 + SO2 + H2O 2,5

2

a, 1 gi nh my thi ra 250m3 nc thi = 250.000lt

trung ho 5 lt nc thi cn 1 gam Ca(OH)2 trung ho 250.000 lt nc thi cn x (g) Ca(OH)2

X = = 50.000 (g) = 50 kg

Vy trung ho H2SO4 trong nc thi, mi gi cn dng 50 Ca(OH)2b, 1 g(CaOH)2 trung ho c 5 lt nc thi

1 mol (Ca(OH)2 = 74g trung ho c 74x5= 370 lt nc thi

1mol Ca(OH)2 1 mol CaO

- 1mol CaO trung ho 370 lt nc thi hay 56g CaO trung ho 370 lt nc thi

1 ngy nh my thi ra 24 x 250.000 = 600.000 lt nc thi

Lng CaO cn dng cho 1 ngy l = 908.108 (g) = 908,108kg1

4a, nHNO3 = = 0,94 mol nh2(N2O+NO)= = 0,12 mol

t a,b ln lt l s mol ca N2O v NO trong hn hp kh

Ta c = 18,5

a+b = 0,12

=> a = 0,06

b = 0,06

Gi kim loi ho tr (III) l R

PTHH:

11R+ 42HNO3 ( 11R(NO3)3 + 3N2O +3NO +21H2O

11R(g) - 42mol 11mol - 6 mol

5,94(g) x(mol) y(mol) 0,12 mol

Ta c: 11R.0,12 = 6 x 5,94

R=27 R l kim loi nhm : Al

- S mol HNO3 phn ng l : x = 0,84 (mol)

- HNO3 d l: 0,94 - 0,84 = 0,1 mol

- S mol Al(NO3)3 to thnh l y = = 0,22 (mol)

Vy trong dung dch A c HNO3 v Al(NO3)3

mHNO3 d = 0,1.63 = 6,3g m Al(NO3)3 = 0,22 . 213 = 46,86 (g)

mddA= mAl + mdd axt - m kh mddA = 5,94 + 564.1,05 - 0,12.18,5.2

- mddA= 593,7 (g)

C% Al(NO3)3 = . 100% = 7,89% C%HNO3 = . 100% =1,06

b, nKOH = 1.0,8 = 0,08 mol.

PTHH: HNO3 +KOH (KNO3 +H2O

1mol 1mol

0,1mol 0,1mol

S mol KOH cn li l 0,8- 0,1 = 0,7 mol

Al(NO3)3 + 3KOH ( Al(OH)3 +3KNO3 1mol 3mol

o,22mol 0,66mol 0,22mol

nKOH cn d l : 0,7 - 0,66 = 0,04 mol Al(OH)3 + KOH ( KAlO2 + 2H2O

Trc phn ng : 0,22 0,04

Phn ng: 0,04 0,04

Sau phn ng : 0,18 0

Vy : nAl(OH)3 thu c l 0,18 mol

- mAl(OH)3 = 0,18.78 = 14,04 (g) 5

5a, Phng trnh ho hc

2NaHCO3 Na2CO3 +CO2 + H2O (1)

Na2CO3 + 2HCl ( 2NaCl +H2O+CO2 (2)

b, Gi a, b, c ln lt l s mol ca NaHCO3Na2CO3 v Na2SO4 trong 17,8 g hn hp

Theo (1) Ta c nNaHCO3 = 2nCO2 = 2 = 2 x 0,25 mol = 0,5mol

- mNaHCO3 = 0,5 x 84 = 42 (g)

% NaHCO3= . 100% 23,6%

- Theo (1) nNa2CO3 = nCO2 = 0,25 mol

Vy trong A c b+ 0,25 mol Na2CO3Theo (2) nNa2CO3=nCO2 = = 0,55(mol)

b+ 0,25 = 0,55 b = 0,3 (mol)

Khi lng Na2CO3 l 0,3x106 = 31,8(g)

% Na2CO3 = .100% = 17,8% % Na2SO4 = 100% - (23,6% + 17,8%) = 58,4

P N 6Cup nim

11Fe3O4 + 8HCl FeCl2 + 2FeCl3 + 4H2O

(A) (B) (D) (G)

Fe3O4 + 4CO 3Fe + 4C

X)

Fe3O4 + 4H2 3Fe + 4H2O

(Y)

Fe3O4 + 2C 3Fe + 2CO2 (Z)

Fe + 2HCl FeCl2 + H2

FeCl2 + 2Cl2 2FeCl3

(E)1.5

Gi hn hp cc cht cn tch l A.

S tch cht:

dd X(NaCl, CaCl2)

A dd B(NaCl, CaCl2, Ca(OH)2)

CaCO3 CaO

dd Y(NaCl, Na2CO3)

dd X

CaCO3 dd CaCl2 CaCl2 khan

Dd Y dd NaCl NaCl khan

Cc PTHH minh ha:

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

CaCO3 CaO + CO2CaCl2 + Na2CO3 CaCO3 + 2NaCl

CaCO3 + 2HCl CaCl2 + H2O + CO2

Na2CO3 + 2HCl 2NaCl + H2O + CO2

1,5

21a. Kim loi Natri tan dn, c kh khng mu bay ra, xut hin cht kt ta mu xanh.

2Na + 2H2O 2NaOH + H2

2NaOH + CuCl2 2NaCl + Cu(OH)2

b. Ban u thy nc vi trong vn c, sau dd li tr nn trong sut.

Ca(OH)2 + CO2 CaCO3

CaCO3 + CO2 + H2O Ca(HCO3)2 (tan)

c. Thuc tm mt mu, xut hin kh mu vng lc.

2KMnO4 + 16HCl 2KCl + 2MnCl2 + 5Cl2 + 8H2O

d. Dung dch st(III)sunfat mu vng nu nht mu dn ri chuyn dn thnh dd mu xanh nht.

Cu + Fe2(SO4)3 2FeSO4 + CuSO4

2

nh s th t cc l ho cht. Ly mu th vo cc ng nghim c nh s tng ng.

Ln lt nh mt dd vo cc dd cn li. Sau 5 ln th nghim ta c kt qu sau:

NaHCO3Na2CO3BaCl2

Na3PO4H2SO4NaHCO3CO2

Na2CO3BaCO3

CO2

BaCl2BaCO3

Ba3(PO4)2

BaSO4

Na3PO4Ba3(PO4)2

H2SO4CO2

CO2

BaSO4

Kt qu

1

1, 1

3

1

2, 1

Nhn xt: Khi nh 1 dd vo 4 dd cn li:

Nu ch si bt kh mt mu th dd em nh l NaHCO3, mu to kh l H2SO4.

Nu ch xut hin mt kt ta th dd em nh l Na3PO4, mu to kt ta l BaCl2.

Mu cn li l Na2CO3.2.5

311. Phng trnh phn ng :

Zn + CuSO4 ZnSO4 + Cu (1)

0,1 0,1 0,1

Zn + FeSO4 ZnSO4 + Fe (2)

0,1 0,1 0,1

Theo (1), nCu = nZnSO= nZn tgp= n= 0,1 (mol)

Sau phn ng (1), CuSO4 phn ng ht, Zn cn d 0,2 0,1 = 0,1 (mol) v tgp (2).

Theo (2), nFe = nZnSO= nFeSOtgp = nZn =o,1 (mol).

Sau phn ng (2), Zn phn ng ht, FeSO4 cn d 0,2 0,1 = 0,1 (mol).

Tng s mol ZnSO4 c to ra l: 0,1 + 0,1 = 0,2 (mol)

Vy dung dch sau phn ng cha 0,1 mol FeSO4 v 0,2 mol ZnSO4.

Khi lng dung dch sau phn ng l:

mdd sau p = mZn + mX mCu mFe = 13 + 100 0,1(64 + 56) = 101 (gam)

Nng phn trm ca dd FeSO4 l:

Nng phn trm ca dd ZnSO4 l: (0,5 im) (0,25 im)(0,25 im) (0,5 im) (0,5 im) (0,5 im) (0,5 im)

12.Khi lng Fe2O3 trong 200 tn qung l: (tn)

V H = 96% nn lng Fe2O3 thc t tham gia phn ng l:

(tn) Phn ng luyn gang:

Fe2O3 + 3CO 2Fe + 3CO2

Theo ptp, nu c 160 tn Fe2O3 tgp s to ra 112 tn Fe.

Vy, c 57,6 tn Fe2O3 tgp s to ra x tn Fe.

x = (tn)

Lng Fe ny ho tan mt s ph gia khc (C, Si, P, S) to ra gang. Lng Fe chim 80% gang. Vy khi lng gang thu c l:

(tn) (0,5 im)(0,5 im)(0,5 im)(0,5 im) (1 im)

Gi M l k hiu ca kim loi v l nguyn t khi ca kim loi. Cng thc ca oxit v mui sunfat kim loi ln lt l MO v MSO4.

Gi x, y, z ln lt l s mol ca M, MO v MSO4.

Theo bi ra, khi lng ca hn hp l 14,8 gam.

Ta c: x.M + (M + 16)y + (M + 96)z = 14,8 (I)

- Phn ng ca hn hp vi dd H2SO4: M + H2SO4 MSO4 + H2 (1)

x mol x mol x mol

MO + H2SO4 MSO4 + H2O (2)

y mol y mol

MSO4 + H2SO4 khng phn ng

z mol

Theo bi ra, nH= x = (mol)

Theo (1), nM = nH= x = 0,2 (mol) (*)

Dung dch A cha (x + y + z) mol MSO4 v H2SO4 d sau cc p. (1) v (2).

- Dung dch A tc dng vi NaOH:

MSO4 + 2NaOH Na2SO4 + M(OH)2 (3)

(x + y + z) mol (x + y + z) mol

NaOH + H2SO4 Na2SO4 + H2O (4)

- Nung kt ta B: M(OH)2 MO + H2O (5)

(x + y + z) mol (x + y + z) mol

Theo bi ra, khi lng cht rn thu c sau khi nung kt ta B l 14 gam.

Ta c: (M + 16) (x + y + z) = 14 (II)

- Phn ng ca hn hp vi CuSO4: Ch c M phn ng.

Theo bi ra, nCuSO= 0,2.2 = 0,4 (mol)

M + CuSO4 MSO4 + Cu (6)

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

Theo (*), nM = 0,2 mol.

T (6) suy ra nCuSOtgp = nM = 0,2 (mol)

Sau p. (6), CuSO4 cn d 0,4 0,2 = 0,2 (mol)

Vy cht rn thu c sau khi chng kh dung dch gm (z + 0,2) mol MSO4 v 0,2 mol CuSO4.

Ta c: (M + 96) (z + 0,2) + (0,2.160) = 62 (III)

T (I), (II) v (III) ta c h phng trnh sau:

x.M + (M + 16)y + (M + 96)z = 14,8 (I)

(M + 16) (x + y + z) = 14 (II)

(M + 96) (z + 0,2) + (0,2.160) = 62 (III)

xM + My + 16y + Mz + 96z = 14,8 (a)

Mx + My + Mz + 16x + 16y + 16z = 14 (b)

Mz + 0,2M + 96z + 19,2 + 32 = 62 (c)

Ly (a) tr (b) ta c: 80z 16x = 0,8 (d)

Thay x = 0,2 (*) vo (d) ta c: 80z = 4 z = 0,05

Thay z = 0,05 vo (c) ta tm c M = 24. Vy M l kim loi Magie: Mg.(0,5 im)(0,5 im)(0,5 im)(0,5 im)(0,5 im)(0,5 im)(0,5 im)(0,25 im)(0,5 im)(0,5 im)(0,5 im)(0,5 im)(0,5 im)

5Cho m gam nhm phn ng va vi 2 lt dung dch HNO3 thu c 5,6 lt hn hp hai kh: N2O v kh X. T khi hi ca hn hp kh so vi H2 l 22,5.

a. Tm kh X v tnh m.

Tnh CM ca HNO3.

a. = 22,5 x 2 = 45

Trong hn hp kh phi c mt kh c M < 45, kh cn li c M > 45. V N2O c

M = 44 ( 45.

Trong cc kh c th c sinh ra do kim loi tc dng vi HNO3, ch c kh NO2 c M = 46 l tho mn iu kin trn.

Vy X l NO2.

Gi x, y ln lt l s mol ca NO2 v N2O.

Theo bi ra ta c: nhh kh = x + y = (mol) (I) Phng trnh phn ng:

Al + 6HNO3 Al(NO3)3 + 3NO2 + 3H2O (1)

mol 2x mol x mol

8Al + 30HNO3 8Al(NO3)3 + 3N2O + 15H2O (2)

mol 10y mol y mol

Theo bi ra ta c:2

P N 7Cup nim

1a- V (A) l kim loi thng dng c 2 ho tr thng gp l (II) v (III) kh bn, ng thi theo chui bin i (M) ch c th l Fe . 2Fe + 3Cl2 2FeCl3

2FeCl3 + Fe 3FeCl2

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

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

2Fe(OH)3 Fe2O3 + 3H2O


b- Cc phn ng xy ra trong qu trnh iu ch H2SO4 :

4FeS2 + 11O2 2Fe2O3 + 8SO2

2SO2 + O2 2SO3 + Q (kJ )

SO3+ H2O H2SO4 (0,5 im)(0,5 im) (0,5 im)(0,5 im) (0,5 im) (0,5 im) (0,5 im) (0,5 im) (0,5 im)

2aa) MgCO3 + H2SO4 MgSO4 + CO2 + H2O (1)

x x x x

RCO3 + H2SO4 RSO4 + CO2 + H2O (2)

y y y y

Nung B to CO2 B cn , X d. Vy H2SO4 ht.

T (1) v (2) : nH2SO4 =nCO2 = = 0,2 mol.

CMH2SO4 = = 0,4(M) .

Theo nh lut BTKL: mx + mH2SO4 = mA + mB + mH2O + mCO2

mB = 115,3 + 0,2.98 12 0,2(18+44) = 110,5 (g)

Nung B thu 11,2 lt CO2 v rn C

mC=mB-mCO2 = 110,5-0,5.44=88,5 (g) (0,25 im)(0,25 im)(0,25 im)(0,25 im)(0,25 im)(0,25 )

bb. T (1) v (2): x+y= 0,2 mol

nCO2 = 0,2 mol mSO4 = 0,2 . 96 = 19,2g > 12g

c mt mui tan MgSO4 v RSO4 khng tan

nMgCO3 = nMgSO4 = = 0,1 mol nRCO3 = nRSO4 = 0,2-0,1 =0,1 mol

Nung B, RSO4 khng phn hy, ch c X d b nhit phn

t a = nMgCO3 RCO3 = 2,5a (trong X)

MgCO3 MgO + CO2 (3)

a- 0,1 a-0,1

RCO3 RO + CO2 (4)

2,5a 0,1 2,5a 0,1

T (3) v (4) : nCO2 = 3,5a 0,2 = 0,5 a = 0,2

mX = 84.0,2 + 2,5.0,2(R + 60) = 115,3 R = 137 (Ba)

(0,25 )(0,25 )(0,25 )(0,25 )(0,25 )(0,25 )

3Cc PTP: Mg + H2SO4 MgSO4 + H2 (1)

Zn + H2SO4 ZnSO4 + H2 (2)

nH2 TNI = = 0,4 mol

nH2 TNII = = 0,5 mol

a. Vi hh kim loi X khng i , th tch dd axit Y tng gp 3:2 = 1.5 ln m khi lng H2 gii phng tng 0,5 : 0,4 < 1,5 ln. Chng t trong TNI cn d kim loi, trong TNII kim loi phn ng ht, axit cn d.

T (1) v (2) : nH2SO4 = nH2 = 0,4 mol ( TNI)

b. Gi x l s mol Mg, th 0,5 x l s mol ca Zn, ta c:

24x + (0,5 x)65 = 24,3

Suy ra : x = 0,2 mol Mg

Vy : mMg = 0,2 . 24 = 4,8 g.

mZn = 24,3 4,8 = 19,5 g.

CMH2SO4 = 0,4 : 2 = 0,2M

(0,25 )(0,25 )(0,25 )(0,25 )(0,25 )(0,25 )(0,25 )(0,25 )(0,25 )(0,25 )(0,25 )(0,25 )

Gi 2a v 2b l s mol Fe v M trong 5,6g A.

Khi lng mi phn ca A l:

= 56a + Mb = = 2,78g.

Phn tc dng vi HCl:

Fe + 2HCl FeCl2 + H2 (1)

a a

M + nHCl FeCln + n/2 H2 (2)

b b

Theo (1) v (2) :

nH2 = a + b = = 0,07 mol ; hay 2a + nB = 0,14 (I)

Phn tc dng vi HNO3:

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

a a

3M + 4nHNO3 3M(NO3)n+ NO + 2nH2O (4)

b b

Theo (3) va (4) :

nNO = a + b = = 0,06 mol.

Hay 3a + nb = 0,18 (II)

Gii h PT (I,II) ta c : a = 0,04 mol Fe.

Thay vo biu thc trn : 56 . 0,04 + Mb = 2,78

Mb = 2,78 2,24 = 0,54

Thay vo (I) : nb = 0,14 0,08 = 0,06

=

EMBED Equation.3 = = 9 . Hay M = 9n

Lp bng :

n 1 2

3

4

M 9

18

27 36

Cp nghim thch hp : n = 3 ; M = 27 . Vy M l Al

Thay n = 3 vo (I) v (II) c b = 0,02

Thnh phn % khi lng mi cht :

%mAl = . 100 = 19,42%

%mFe = . 100 = 80,58%

(0,25 )(0,25 )(0,25 )(0,25 )(0,25 )(0,25 )(0,25 )(0,25 )(0,25 )(0,25 )(0,25 )(0,25 )(0,25 )(0,25 )(0,25 )(0,25 )(0,25 )

--------------------------------------------------------------------------------------------

8 p n - Biu im mn Ho hc

CuNi Dung

11(1 im) p n C v :

Khi lng 1 nguyn t C = 1,9926.10-23gVy 1 nguyn t Na = 1,9926.10-23g x

EMBED Equation.3 = 3,81915.10-23g 3,82.10-23g.

21(1 im) 1 mol CO2 c KL mol = 44g -> 1g CO2 c lng cht = mol

1 mol CO c KL mol = 28g -> 1g CO c lng cht = mol

1 mol NO2 c KL mol = 46g -> 1g NO2 c lng cht = mol

1 mol N2O c KL mol = 44g -> 1g N2O c lng cht = mol

Vy : (a) v (d) ng

2(1 im) Nc vi ban u c v :

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

rn, trng

tip tc thi th lng CO2 tng ln v d nn xy ra PTHH

CO2 + H2O + CaCO3 -> Ca(HCO3)2 tan

v tng CO2 nn phn ng song song xy ra

2CO2 + Ca(OH)2 -> Ca(HCO3)2

2 cht v c tho mn l NaCl v CaCO3 CaO Ca(OH)2 CaCl2

(1 im) CaCO3 CaCO3 CaCO3 CaCO3

CO2 NaHCO3 Na2CO3 Na NaOH Na2SO4 (1 im) NaCl NaCl NaCl NaCl

Cl2 HCl BaCl2

4(1 im ) Xt 2 trng hp v CO2 + Ca(OH)2 -> CaCO3 + H2O (1)

CO2 + Ca(OH)2 -> Ca(HCO3)2 (2)

= 0,02 . 4 = 0,08 mol

= < 1

(0,5 im ): TH 1 - Ch sinh ra 5g kt ta v Ca(OH)2 d ta c :

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

44 100

2,2g = = 0,05 mol

Vy = 22,4. 0,05 = 0,112 (lt)

(0,5 im) :TH2 - Nu sinh 2 mui th phn ng (1) ht 0,05mol

Vy mol Ca(OH)2 tham gia phn ng (1) l 0,08 - 0,05 = 0,03 mol

2CO2 + Ca(OH)2 -> Ca(HCO3)2 2 mol 1 mol

0,06 mol = 0,05 + 0,06 = 0,11 mol

= 0,01. 22,4 = 2,464 (lt)

P N 9

Ta c: p + n + e = 40. Trong mt nguyn t s p = s e

2p + n = 40

n = 40 - 2p

(1)

Mt khc: p + e - n = 12

n = 2p - 12(2)

T (1) v (2)

40 - 2p = 2p - 12Gii ra: p = 13

Vy R l Nhm Al.

S e = s p = 13 (ht)

S n = 40 - 2.13 = 14 (ht)

Nhn bit c mi oxit

0,25 x 4 = 1,0

Vit ng mi phng trnh

0,25 x 3 = 0,75

* Hai thuc th nhn bit Nc v Qu tm

- Cho 4 mu oxit vo nc:

Hai mu tan hon ton:

Na2O+H2O

2NaOH

P2O5+3H2O

2H3PO4

- Cho qu tm vo 2 dung dch thu c:

Qu tm xanh dd NaOH, nhn bit Na2O

Qu tm dd H3PO4, nhn bit P2O5

- Cho dd NaOH trn vo hai mu cn li:

Mu tan l ZnO do ZnO + 2NaOH Na2ZnO2 + H2O

Mu khng tan l MgO.

(1im) 4 phn ng iu ch O2 (Vit c mi phn ng c 0,25 im)

5t cng thc ca oxit l RxOy, ha tr kim loi bng 2y/x.

Phn ng ha tan:

(1)

Ta c nHCl = 0,3 . 1 = 0,3 mol.

Gi M l khi lng nguyn t ca R ta c t l:

Khi n = 1 : loi

n = 2 : loi

n = 3 l Fe, oxit l Fe2O3

Gi Z, N, E v Z', N', E' ln lt l s ht proton, ntron, electron ca hai nguyn t A, B. Ta c cc phng trnh :

Z + N + E + Z' + N' + E' = 78 .

hay : (2Z + 2Z' ) + (N + N') = 78 (1)

(2Z + 2Z' ) - (N + N') = 26 (2) (2Z - 2Z' ) = 28

hay : (Z - Z' ) = 14 (3)

Ly (1) + (2) sau kt hp vi (3) ta c : Z = 20 v Z' = 6 Vy cc nguyn t l : A l Ca ; B l C .

Cu 6. (3,0 im)a. Tnh V

Theo bi ra ta c h:

T (1) v (2):

b. Tnh khi lng hn hp E (Fe2O3 v CuO)

Gi khi lng hn hp E l m gam

Theo ra:

v

T (1), (2), (3), (4) suy ra: 0,009m + 0,004m = 0,13

Vy m = 10 (gam).

0,250,250,50,250,250,250,250,250,250,250,25

10 Hng dn chmCuNi dungim

1aa) Cc phng php iu ch Baz

- Kim loi tc dng vi nc

2Na + H2O -> 2NaOH + H2

- Oxit ba z tc dng vi nc

CaO + H2O -> Ca(OH)2

- Kim tc dng vi mui tan

KOH + FeCl2 -> Fe(OH)2 + 2KCl

- in phn mui c mng ngn:

2KCl + 2H2O 2KOH + H2 + Cl2

- iu ch Hr lng tnh cho mui ca nguyn t lng tnh tc dng vi NH4OH (hoc kim va ).

AlCl3 + 3NH4OH -> Al(OH)3 + 3NH4Cl

ZnSO4 + 2NaOH -> Zn(OH)2 + Na2SO40,250,250,250,250,250,25

bb) Cc phng php trn ch c phng php kim tc dng vi mui tan l ph hp

- Dung dch kim nh: NaOH, KOH, Ca(OH)2, Ba(OH)2

- Mui tan:CuCl2, Cu(NO3)2 ; CuSO4

Phng trnh phn ng:

2NaOH + CuCl2 -> Cu(OH)2 + 2NaCl

2NaOH + Cu(NO3)2 -> Cu(OH)2 + 2NaNO3

2NaOH + CuSO4 -> Cu(OH)2 + Na2SO4

2KOH + CuCl2 -> Cu(OH)2 + 2KCl

2KOH + Cu(NO3)2 -> Cu(OH)2 + 2KNO3

2KOH + CuSO4 -> Cu(OH)2 + K2SO4

Ca(OH)2 + CuCl2 -> Cu(OH)2 + CaCl2

Ca(OH)2 + Cu(NO3)2 -> Cu(OH)2 + Ca(NO3)2

Ca(OH)2 + CuSO4 -> Cu(OH)2 + CaSO4

Ba(OH)2 + CuCl2 -> Cu(OH)2 + BaCl2

Ba(OH)2 + Cu(NO3)2 -> Cu(OH)2 + Ba(NO3)2

Ba(OH)2 + CuSO4 -> Cu(OH)2 + BaSO4

0,5

0,5 0,5 0,25 0,25 0,25 0,25 0,25 0,25 0,25 0,25 0,25 0,25 0,25 0,25

2a) Magi chy trong khng kh, n tc dng vi oxi dng t do. Magi cn c th chy trong kh SO2, CO2 n tc dng vi xi dng hp cht to ra oxit.

- Cht bt A mu trng l Magi oxt

- Cht bt B mu vng, khng tc dng vi H2SO4 long l lu hunh. Lu hunh chy trong khng kh to ra lu hunh ioxit.

- Cht C l sn phm ca MgO vi dng dch H2SO4 long. Vy C l Magi Sunfat MgSO4.

0,5 0,5 0,5 0,5

bb) Phn ng ca Magi chy trong SO2

2Mg + O2 ( 2MgO

+ Phn ng trn thuc loi phn ng xi ho _ kh

+ Mg l cht kh (cht b oxi ha)

+ SO2 l cht xi ha (cn gi l cht b kh)

- Phn ng ca A vi H2SO4 long:

MgO + H2SO4 ( MgSO4 + H2O

- Phn ng ca B chy trong khng kh:

S + O2 ( SO2

0,5 0,5 0,5 0,5 0,5 0,5

a) Ly mi dung dch mt t lm mu th sau vo nhau tng cp mt. Cp no c bt kh thot ra l Na2CO3 v HCl, cn cp kia l NaCl v H2O

Na2CO3 + 2HCL -> 2NaCl + H2O + CO2

Nhm 1 l Na2CO3 v HCl

Nhm 2 l NaCl v H2O

- un n cn nhm 1:

+ Khng c cn l HCl

+ C cn l Na2CO3

- un n cn nhm 2:

+ Khng c cn l H2O

+ C cn l NaCl

0,5 0,5 0,5 0,5 0,5 0,5 0,5 0,5

b) p n B l ph hp

Gii thch:

0,5 0,5

Trng hp 1:Gi M l kim loi ha tr II.

Ta c cng thc MCO3

Phng trnh phn ng:

MCO3 -> MO + CO2

(1)

CO2 + Ba(OH)2 -> BaCO3 + H2O(2)

S mol BaCO3 l:

Ta c s : MCO3 -> CO2 -> BaCO3

1 mol

1 mol

0,1 mol MO + CO2

CO2 + Ba(OH)2 -> BaCO3 + H2O

0,15 mol 0,15 mol

d: 0,15 0,1 mol = 0,05 mol

BaCO3 + CO2 + H2O -> Ba(HCO3)2

S mol CO2 phn ng l:

= 0,15 + 0,05 = 0,2 mol

a) Khi lng cht rn A l:

MMO = 20 44.0,2 = 11,2 gam

b) Khi lng mol ca MCO3 l:

Nguyn t khi kim loi M = 100 60 = 40 VC.

l Ca v cng thc l CaCO3

Hc sinh gii cch khc ng cho im ti a.0,25 0,25 0,25 0,25 0,5 0,25 0,25 0,25 0,25 0,25 0,25 0,25 0,25 0,25 0,25

11

Ni dungim

V (A) l kim loi thng dng c 2 ho tr thng gp l (II) v (III) kh bn, ng thi theo chui bin i (M) ch c th l Fe 2Fe + 3Cl2 2FeCl3

2FeCl3 + Fe 3FeCl2

FeCl2 + 2NaOH Fe(OH)2 + 2NaCl4Fe(OH)2 + O2 + 2 H2O 4 Fe(OH)3 2Fe(OH)3 Fe2O3 + 3H2O


(0,5 )(0,5 )(0,5 )(0,5 )(0,5 )(0,5 )(0,5 )

b- Cc phn ng xy ra trong qu trnh iu ch H2SO4 :

4FeS2 + 11O2 2Fe2O3 + 8SO2

2SO2 + O2 2SO3 + Q (kJ ) SO3+ H2O H2SO4 (0,5 im)(0,25im)(0,25im)

Cu 2:(5,0 im )

a) MgCO3 + H2SO4 MgSO4 + CO2 + H2O (1)

(0,25 im) x x x x

RCO3 + H2SO4 RSO4 + CO2 + H2O (2)

(0,25 im)

y y y y

Nung B to CO2 B cn , X d. Vy H2SO4 ht.

T (1) v (2) : nH2SO4 =nCO2 = = 0,2 mol.

(0,25 im) CMH2SO4 = = 0,4(M) .

(0,25 im)Theo nh lut BTKL: mx + mH2SO4 = mA + mB + mH2O + mCO2

mB = 115,3 + 0,2.98 12 0,2(18+44) = 110,5 (g)

(0,25 im) Nung B thu 11,2 lt CO2 v rn C

mC=mB-mCO2 = 110,5-0,5.44=88,5 (g) (0,25 im)b. T (1) v (2): x+y= 0,2 mol

nCO2 = 0,2 mol mSO4 = 0,2 . 96 = 19,2g > 12g

(0,25 im)

c mt mui tan MgSO4 v RSO4 khng tan

nMgCO3 = nMgSO4 = = 0,1 mol nRCO3 = nRSO4 = 0,2-0,1 =0,1 mol(0,25 im)Nung B, RSO4 khng phn hy, ch c X d b nhit phn

t a = nMgCO3 RCO3 = 2,5a (trong X)

MgCO3 MgO + CO2 (3)

(0,25 im) a- 0,1 a-0,1

RCO3 RO + CO2 (4)

(0,25 im) 2,5a 0,1 2,5a 0,1

T (3) v (4) : nCO2 = 3,5a 0,2 = 0,5 a = 0,2

(0,25 im)mX = 84.0,2 + 2,5.0,2(R + 60) = 115,3 R = 137 (Ba)

(0,25 im)

Cu 3:(5,0 im ) .

Cc PTP: Mg + H2SO4 MgSO4 + H2 (1)

(0,25 im)

Zn + H2SO4 ZnSO4 + H2 (2)

(0,25 im)nH2 TNI = = 0,4 mol

(0,25 im)nH2 TNII = = 0,5 mol

(0,25 im)a. Vi hh kim loi X khng i , th tch dd axit Y tng gp 3:2 = 1.5 ln m khi lng H2 gii phng tng 0,5 : 0,4 < 1,5 ln. Chng t trong TNI cn d kim loi, trong TNII kim

loi phn ng ht, axit cn d. (0,25 im) T (1) v (2) : nH2SO4 = nH2 = 0,4 mol ( TNI)

(0,25 im)b. Gi x l s mol Mg, th 0,5 x l s mol ca Zn, ta c:

24x + (0,5 x)65 = 24,3

(0,25 im)Suy ra : x = 0,2 mol Mg

(0,25 im)Vy : mMg = 0,2 . 24 = 4,8 g.

(0,25 im)

mZn = 24,3 4,8 = 19,5 g.

(0,25 im) CMH2SO4 = 0,4 : 2 = 0,2M

(0,25 im)

Cu 4:(5,0 im )

Gi 2a v 2b l s mol Fe v M trong 5,6g A.

(0,25 im)Khi lng mi phn ca A l:

= 56a + Mb = = 2,78g.

(0,25 im)Phn tc dng vi HCl:

Fe + 2HCl FeCl2 + H2 (1)

(0,25 im) a a

M + nHCl FeCln + n/2 H2 (2)

(0,25 im) b b

Theo (1) v (2) :

nH2 = a + b = = 0,07 mol ; hay 2a + nB = 0,14 (I)

(0,25 im)Phn tc dng vi HNO3:

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

(0,25 im) a a

3M + 4nHNO3 3M(NO3)n+ NO + 2nH2O (4)

(0,25 im) b b

Theo (3) va (4) :

nNO = a + b = = 0,06 mol.

(0,25 im)Hay 3a + nb = 0,18 (II)

Gii h PT (I,II) ta c : a = 0,04 mol Fe.

(0,25 im)Thay vo biu thc trn : 56 . 0,04 + Mb = 2,78

(0,25 im) Mb = 2,78 2,24 = 0,54

(0,25 im)Thay vo (I) : nb = 0,14 0,08 = 0,06

(0,25 im)

=

EMBED Equation.3 = = 9 . Hay M = 9n

(0,25 im)Lp bng :

n 1 2

3

4

M 9

18

27 36

Cp nghim thch hp : n = 3 ; M = 27 . Vy M l Al

(0,25 im)Thay n = 3 vo (I) v (II) c b = 0,02

(0,25 im)Thnh phn % khi lng mi cht :

%mAl = . 100 = 19,42%

(0,25 im)%mFe = . 100 = 80,58%

(0,25 im) .

12P N V BIU IMCu 1 (1 im) : n H2SO4 = 0,1.1 = 0,1mol; nNaOH = 0,3.1 = 0,3mol (0.25)

Theo bi ra ta c phng trnh : H2SO4 +2NaOH Na2SO4 + 2H2O (0.25)

Theo phng trnh c 1mol H2SO4 tc dng ht vi 2mol NaOH (0.25)

Theo bi ra c 0,1mol H2SO4 tc dng vi 0,3mol NaOH do lng NaOH d v vy lm cho qu tm chuyn sang mu xanh

(0.25)

Cu 2 (2 im) : Mun iu ch Canxi sunfat t Lu hynh v Canxi cn thm t nht 2 cht l nc v Oxi (1)

S + O2 SO2 (0.25)

SO2 + O2 2SO3 (0.25)

SO3 + H2O H2SO4 (0.25)

Ca + H2SO4 CaSO4 + H2 (0.25)

Cu 3 (2im) :

a/ Mu xanh ca dung dch CuSO4 nht dn, c kt ta xut hin (0.5)

b/

SO2 + Ca(HCO3)2 CaSO3 + 2CO2 + H2O

c kt ta v c kh that ra (0.5)

c/ Brm mt mu nu (0.5)

phng trnh ha hc : CH2 = CH2 + Br2 CH2Br CH2Br

d/ C kh that ra mi hc

Cu + 2H2SO4 c nng CuSO4 + SO2 + 2H2O (0.5)

Cu 4 (1.5 im) : Cho CO2 li chm qua nc vi trong xy ra phn ng ha hc

CO2 + Ca(OH)2 CaCO3 + H2O (c vn c xut hin) (0.5)

Tip tc cho nc vi trong vo, dung dch vn c s tan ra

CO2 + CaCO3 + H2O Ca(HCO3)2 (0.5)

sau li vn c do

Ca(OH)2 + Ca(HCO3)2 2CaCO3 + 2H2O (0.5)

Cu 5 (2 im) : Trong hp cht H3PO4 c =

EMBED Equation.3 suy ra mH = (0.5)

Trong hgp cht H2SO4 c (0.5)

V Hyr A bng Hyr B nn = (0.5)

Suy ra (0.5)

Cu 6 (2 im) : MCuSO4 = 160

Khi lng CuSO4 trong 8 gam CuSO4.5H2O l : (1 )

Phng trnh ha hc : Zn + CuSO4 ZnSO4 + Cu (0.5)

160g 64g

5.12g xg

x = 5,12 x 64 : 160 = 2,05(g) Cu (0.5)

Cu 7 (2 im) : S bin ha c th l :

- Tm D : do CuSO4 l sn phm sinh ra t D nn D c th l Cu(OH)2 (0.5)

- Tm C : do Cu(OH)2 l sn phm sinh ra t C nn C c th l CuCl2 (0.5)

- Tm B : do CuCl2 l sn phm sinh ra t B nn B c th l CuO (0.5)

- Tm A : do CuO l sn phm sinh ra t A nn a c th l Cu (0.5)

T ta c s bin ha :

Cu CuO CuCl2 Cu(OH)2 CuSO4 ( Hs c th tm cc cht khc trong s bin ha)

Cu 8 (2.5 im): n H2SO4 = 100 x 0,8 : 1000 = 0,08 mol (0.25)

Gi kim loi ha tr II l R ta c phng trnh ha hc l

RO + H2SO4 RSO4 + H2O (0.25)

1 mol 1 mol 1 mol

0.08mol 0.08mol 0.08mol

Theo bi ra ta c R + 16 = 4,48 : 0.08 = 56 suy ra R = 40 (0.25)

Vy kim loi l Ca nn cng thc ha hc ca mui : CaSO4 m CaSO4 = 0,08 x 136 = 10,88 (g) (0.25)

m H2SO4 kt tinh bng : 13,76 10,88 = 2,88 (g) (0.25)

n H2SO4 bng 2,88 : 18 = 0,16 mol (0.25)

t l ca n CaSO4 vi n H2SO4 l 0,08 : 0,16 = 1 : 2 (0.25)

Vy CTHH ca mui ngm nc CaSO4.2H2O (0.25)

Cu 9 (3 im) : n H2 = 8,96 : 22,4 = 0,4 mol (0.25)

Khi lng cht rn khng tan bng 3 gam l khi lng ca Cu nn khi lng ca

Al v Mg l : 10,8 3 = 7,8 g (0.25)

Gi s mol ca Al v Mg ln lt l x; y ta c cc phng trnh ha hc

a/ 2Al + 6HCl AlCl3 + 3H2 (1) (0.25)

2 mol 6 mol 3 mol

x mol 3x mol 1,5 x mol

Mg + 2HCl MgCl2 + H2 (2) (0.25)

1 mol 2 mol 1 mol

y mol 2y mol y mol

t (1) v (2) ta c : 27 x + 27 y = 7,8 0.25)

v 1,5 x + y = 0,4 (0.25)

gii h PT trn ta c : x = 0,2 v y = 0,1 (0.5)

mAl = 0,2 x 27 = 5,4 (g) (0.25)

m Mg = 0,1 x 24 = 2,4 (g) (0.25)

m Cu = 3(g)

b/ Th tch HCl cn dng :

n HCl cn dng cho phn ng (1) v (2)

n HCl = 3x + 2y = 0,8 (mol) (0.25)

V HCl = 0,8 : 0,5 = 1,6 (lt) (0.25)

13

HNG DN CHM HC SINH GII LP 9 CuNi dungim

11) -Dng Cu th 4 dung dch, nhn ra ddAgNO3 nh to ra dung dch mu xanh lam:

Cu + 2AgNO3 ( Cu(NO3)2 + 2Ag (

-Dng dung dch Cu(NO3)2 to ra th cc dung dch cn li, nhn ra ddNaOH nh c kt ta xanh l:

Cu(NO3)2 + 2NaOH ( Cu(OH)2 ( + 2NaNO3 -Cho AgNO3 ( nhn ra trn) vo 2 cht cn li, nhn ra ddHCl nh c kt ta trng. Cht cn li l NaNO3

AgNO3 + HCl ( AgCl ( + HNO3 ( HS c th dng Cu(OH)2 th, nhn ra HCl ho tan c Cu(OH)2 )

2) a.SO3 + H2O ( H2SO4

H2SO4 + BaCl2 ( BaSO4 ( + 2HCl

b. 4Fe(OH)2 + O2 2Fe2O3 + 4H2O (

c. 2NaCl + 2H2O 2NaOH + H2 ( + Cl2 ( 0,1

0,1

0,1

0,1

0,1

0,1

0,1

0,1

0,1

0,1

2

2Na + 2H2O 2NaOH + H2

6NaOH + Fe2(SO4)3 2 Fe(OH)3 + 3Na2SO4

0,12 mol 5,24g ( loi )

Trng hp 2 : NaOH d

S mol NaOH d : =

NaOH + Al(OH)3 NaAlO2 + 2H2O

Ban u : 0,08mol

Phn ng :

SP : 0 mol 0,44 - mol

2 Fe(OH)3 Fe2O3 + 3 H2O

0,04 mol 0,02 mol

2 Al(OH)3 Al2O3 + 3 H2O

(0,44-)mol (0,22-) mol

Thnh phn khi lng cht rn

(0,02x 160) + 102( 0,22 - ) = 5,24 -> a= 9,2 gam

= =>

Khi lng hn hp = 160 x 1,25= 200 gam

= =>

Khi lng ca dung dch = 9,2 +200-(0,04x107)-78(0,44-)-0,4 = 201,4 gam

C%Na2SO4=

C% NaAlO2=

0,1

0,1

0,1

0,1

0,1

0,1

0,1

0,1

0,1

0,1

0,1

0,1

0,1

0,1

0,1

0,1

0,1

0,1

0,1

0,1

0,1

0,1

0,1

0,1

0,1

3Gi CTHH ca ba mui trn l : M2CO3, MHCO3, MCl

Gi x, y, z ln lt l s mol ca ba mui trn dng

M2CO3 +2HCl 2 MCl + CO2 + H2O

x mol 2x mol 2x mol x mol

MHCO3 + HCl MCl + CO2 + H2O

y mol y mol y mol y mol

Gi s dung dch A cn d 2 a mol HCl d nh vy mi phn dung dch A c a mol HCl d

Phn ng phn 1:

HCl + AgNO3 AgCl + HNO3 a mol a mol

MCl + AgNO3 AgCl + MNO3

-->

Phn ng phn 2:

HCl + KOH KCl + H2O

a mol a mol a mol

=> 29,68 gam hn hp mui khan gm a mol KCl

Do ta c h phng trnh

x( 2M + 60) + y(M +61) +z(M +35,5) = 43,71

x+ y=

a +

a = 0,125 x 0,8=0,1

Gii h phng trnh trn ta tm c M = 23. vy M l Na

x= 0,3 mol

y= 0,1 mol

z= 0,6 mol

Vy% Na2CO3=

% NaHCO3=

% NaCl = 100%-(72,75% + 19,25%)=8,03%

S mol HCl ban u dng = 2x+y+2a = 2 . 0,3 + 0,1 + 2. 0,1 =0,9 mol

Th tch dung dch HCl =

0,1

0,1

0,1

0,1

0,1

0,1

0,1

0,1

0,1

0,1

0,1

0,1

0,1

0,1

0,1

0,1

0,1

0,1

0,1

0,1

4Gi s mol Fe, Cu trong hn hp ln lt l a, b ( a, b > 0 )

Gi s Fe, Cu phn ng ht vi AgNO3 theo phn ng

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

a mol a mol 2a mol

Cu + 2 AgNO3 Cu(NO3)2 + 2 Ag

b mol 2b mol

Theo bi ra ta c :

56a + 64b = 10,72

Nhng 64(a+b) > 56a +64b => 64(a+b)> 10,72

=> a+b >

=> mAg=2a + 2b > 2. 0,1675=0,335

=> S gam Ag thu c 108(2a+2b) > 0,335.108=36,18 g> 35,84 g

=> Fe v Cu khng ht m cn trong B

C 2 kh nng i vi B

Gi s trong B cn d Fe, Cu cn nguyn

Gi x l s mol Fe phn ng vi AgNO3 Fe + 2 AgNO3 Fe(NO3)2 + 2 Ag

xmol x mol 2x mol

C 1 mol Fe tham gia gy tng 2.108 56 ( g )

Vy x mol Fe tham gia gy tng (2.108 56 ) x = 160 x gam

M khi lng cht rn tng : 35,84-10,72=25,12 (g)

=> 160x=

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21 de Thi Hsg Co Da Chi Tiet(2)