Lời giải đề thi học sinh giỏi Hóa học lớp 9 phần 1

Chi dugc diing thuoc thii de nhgn biet cdc muoi sau: NH4CI, FeClo, FeCls, MgCl2, NaCl, AICI3 Gidi thich vd viet phuang trinh phdn ling neu c6.. Xdc dinh nong do phdn tram cua NosSOj vd

D E SO 1

DE THI HOC S1I\IH GiOl HOA HOC LOfP 9, TP HQ CHI MINH NAM HOC 1998 - 1999

Cdu I Viet 3 phuang trlnli khdc nhau de dicu die muoi ZnCl2

Cdu I I Viet phuang trinh phdn ling de bieu dlen chiiSi bien hoa sau:

FeCls > Fe(0H)2 ^ F e S O , > Fe(N0s)2

FeCh > Fe(0H )3 > FesOs > Fe

Cdu III Co 6 ong nghiem dugc ddnh so tii 1 den 6 chica cdc dung dicli:

NaOH; (NHJ.SO,; NaoCOs; Ba(N03)2; PbiNO^)^; CaCl Hay cho bict 6'ng mang so ndo dUng chat ndo? Viet phdn icng minh hga Biet rdng:

a) Dung dich (2) cho ket tila trdng vai cdc dung dich (1), (3), (4) ''•

b) Dung dich (5) cho ket tua trdng vai cdc dung dich (1), (3), (4)

c) Dung dich (2) khong tgo ket tila vai dung dich (5)

d) Dung dich (1) khong tgo ket tua vai cdc dung dich (3), (4)

e) Dung dich (G) khong phdn i(ng vai dung dich (5)

f) Dung dich (5) bi trung hoa bdi dung dich HCl

g) Dung dich (3) tgo ket tua trdng vai HCl, khi dun nong ket tua nay se tan CduIV

a) Nong do dung dich bdo hoa KCl d 40°C la 28,57%

Tinh do tan cua dung dich KCl a ciing nhiet do

» ) Xdc dinh lugng AgNOs tdch ra khi Idm Ignh 2500 gam dung dich

AgNOs bdo hoa a 60°C xuong 10°C Cho biet do tan cua AgNOs o 60°C Id 525 gam, a 10°C Id 170 gam

Cau V (A) la dung dich H2SO4; (B) la dung dich NaOH

• Trgn 0,3 lit (B) vai 0,2 lit (A) dugc 0,5 lit (C)

Lay 20 nd (C), them mot it quy tim vdo thdy c6 mdu xanli Sau do them tic tit dung dich HCl 0,05M tai khi quy tim ddi thdnh mdu tim thdy het 40 ??il axit

Trgn 0,2 lit (B) vai 0,3 lit (A) dugc 0,5 lit (D)

Lay 20 ml dung dich (D), them mot it quy tim vdo thdy c6 mdu do Sau do them tii tii: dung dich NaOH 0,1M tai khi quy ddi thdnh mdu tim thdy het 80 ml dung dich NaOH

Tim nong do mol/l dung dich (A) vd (B)

Cdu VI Xdc dinh cong thiJtc ciia hai oxit sdt A vd B, biet r&iig:

• 23,2 gam (A) tan vica dil trong 0,8 lit HCl IM

• 32 gam (B) khi k/iii bdng Ho tqo thdnh sdt vd 10,8 gam IhO

L6\I Cdu I * Zn + CI2 > ZnCla

FeS04 + Ba(N03)2 > BaS04^ + Fe(N03)2

2Fe + 3CI2 > 2FeCl3

FeCla + 3NaOH > FeCOIDai + 3NaCl

(ndu do) •

2Fe(OH)3 — > Fe203 + 3H2O

Fe203 + 3 C 0 — > 2Fe + 3CO2T

Cdu I I I Theo cac duf k i e n de b a i neu r a , cac lo d i i n g cac hoa cha t sau:

Cdu rv

a ) G o i S l a do t a n cua K C l d 40°C

K h o i lUcfng dung dich t h u dUgfc: (S + 100) gam

N o n g do p h a n t r a m cua chat t a n t r o n g dung dich bao hoa:

S + 100 S + 100

S + 100 - 3,5S » S = 40 (gam)

Hni Hnr a

' ^ ) ^ 60"C, t r o n g 525 + 100 b a n g 625 gam dung dich c6 525 gam AgNOg

va 100 g a m H2O T r o n g 2500 g a m dung dich c6 x g a m AgNOa va

CJ 10°C, CLf 100 gam ntfcJc h6a t a n 170 gam AgNOa

400 gam niJcJc hoa t a n z gam AgNOa

L a n 2 l a m quy t i m c6 m a u do chufng t o H2SO4 dir

T h e m N a O H de t r u n g h5a H2SO4 di/

2 N a O H + H2SO4 > Na2S04 + 2H2O

Goi X , y l a n l u g t l a n o n g do cua H2SO4 va N a O H

Theo cac p h a n ufng (1), (2), (3) t a c6 phiiofng t r i n h :

Vay oxit sSt (A): Fe304 (sdt tii oxit)

• FexOy + yH2 > xFe + yH20

DE THI CHOIV flOl TUYEN KQC SINH GIOIHOA HOC 9, qUAAl TAN BINH

TP HO CHJ MINH NAM HOC 1998 - 1S99

PHAN A Li thuyet

Cdu I Chi dugc diing thuoc thii de nhgn biet cdc muoi sau: NH4CI, FeClo,

FeCls, MgCl2, NaCl, AICI3 (Gidi thich vd viet phuang trinh phdn ling

neu c6)

Cdu II Viet phuang trinh phdn ling bieu dien cdc bien hoa sau:

a) Fe > Fe2(S04)3 > Fe(0H)3 > FcsOa >Fe

> FeClz > FeCls >FeCl2 > AgCl

b) MgCOs > MgS04 > MgCl2 > Mg(0H)2 > MgO

> MgCOs > CO2 > Ca(HC03)2 > CaCOs

Cdu III Chi dugc dung guy tini vd dung dich AgN03 c6 sdn, neu each

phdn biet cdc dung dich: NaOH, NaCl, HCl, H2S, H2SO4

Cdu IV Sdt nguyen chdt trong khong khi thi khong bi han gl, nhUng sdl

CO tap chdt de Idu ngdy trong khong khi Igi bi han gl Hay gidi thich

hien tugng nay

a Lfll GIAI Oi THI HOC SINH GIOI HOA HOC 9

PHAN B Bai toan

Bai 1- Co mot hon hgp gom Na2S04 vd K2SO4 dugc trgn Idn theo tl Ic

I : 2 ve so mol (1 mol Na2S04 yd.2 mol K2SO4) Hoa tan hon hgp vdo

102 gam nUac thi thu dugc dung dich A Cho 1664 gam dung dich BaCL 10% vdo dung dich A Lgc ket tua, them H2SO4 du vdo tiudc vUa Igc tin thdy tgo ra 46,6 gam ket tua Xdc dinh nong do phdn tram cua NosSOj

vd K2SO4 trong dung dich ddu •

Bai 2 Tinh CM cua dung dich H2SO4 vd NaOH, biet rdng 10 nd dung

dich H2SO4 tdc dung vUa du vdi 30 ml dung dich NaOH Neu lay 20 ud dung dich H2SO4 cho tdc dung vdi 2,5 gam CaCOs thi axit con du vd lugng du ndy tdc dung vita du vdi 10 nd NaOH

LdlGIAI

PHAN A Li thuyet Cdu I - Trich m6i lo mot it lam mt\ thuf

- Cho dung dich NaOH dii Ian liicft vao cac mau thif tren va dun nhe;

Mau thuf nao c6 k h i miii khai bay ra la NH4CI:

NaOH + NH4CI — > NaCl + NHgt + H^O

Mau tao ket tiia trSng xanh, hoa nau do trong khong k h i la FeC]2:

A1(0H)3 + NaOH > NaAlOa + 2H2O

Mau tao ket tua trSng la MgCl2:

2 N a O H + MgCl2 ^ Mg(0H)2 + 2 N a C l

Mau khong c6 hien tJong gi la NaCl

Lfll GIAI T H I unr emu ni/\i un» unr a ^

Cdu IỊ

a) 2Fe + 6H2SO4 damdac > 1^02(804)3 + 3SO2T + 6II2O

Fe2(S04)3 + 6 N a O H > 2 F e ( O I I ) 3 i + 3Na2S04

2Fe(OH)3 — > FeaOs + 3H2O

FeaOg + SCO — > 2Fe + SCOat

hoac CăHC03)2 + 2 N a O I I > C a C O a i + NaaCOg + 2H2O

hoac CăHC03)2 + Că0H)2 2 C a C 0 3 i + 2H2O

Cau HỊ

T r i c h m 6 i lo m o t i t l a m m a u thtf

- Cho quy tim tam ni/dc I a n \\iat vao cac mau thijf t r e n :

• M a u thuf nao l ^ m quy t i m hoa x a n h l a N a O H

• M a u k h o n g c6 h i e n t i i o n g la N a C l

• Nhufng m a u l ^ m quy t i m hda do: I I C l , H2S, H2SO4

- Sau do cho dung dich AgNOa Ian li/gt vao cac mau l a m quy t i m hoa dọ

- K h i t r o n g sat b i I a n t a p chat, de l a u ngay b i b a n g i do x a y r a su

a n m o n k i m loai tufc l a bién sat t h a n h hop chat cua sat

Gidi thidi: T r e n be m a t k i m loai c6 Idp nUdrc a m , da hoa t a n m o t

l u o n g oxi n e n chuyen Fe -> Fê\

Va oxi hoa t a n t r o n g niidc theo qua t r i n h : O2 + 2H2O >• 4011"

Sau do Fế k e t h o p vdri O H " > F e ( 0 H )2 (trdng xanh)

M o t p h a n Fe(0H)2 b i oxi hoa tao F e ( 0 H ) 3 (nau do)

4Fe(0H)2 + O2 + 2H2O - > 4Fe(OH)3

V i k h i t h e m dung dich H2SO4 vao n L f d c loc l a i tao k e t t u a n e n

t r o n g niidc loc con dtf BaClạ

K h o i l a g n g dung dich A l a : nidungdich = mNạsô + " ^ K , S O , + "^H^O

« mdung dich = 0,2 X 142 + 0,4 x 174 + 102 = 200 (gam)

5 Mot hon hap NaCl vd MgCl2 them nuac vao hon hap ta c6 dung dich A

Them dung dich AgNOs vao dung dich A, khi phdn iing ket thuc logi

bo chat ket tua trdng, phdn dung dich con Igi la dung dich D

Chia B lam 2 phdn: a vd b

Phdn a: Sau khi c6 can tiep tuc dun nong thl diigc mot hon Jigp khi

C Cho hon hap khi nay qua binh dUng KOH

Phdn b: Cho vdo lugng dU dung dich HCl thi thu dugc ket tua trdng D

Viet cdc phuang trinh phdn dng

6 Lam the ndo de phdn biet cdc Ig hoa chat diidi day md khong dugc

dung theni hoa chat ndo khdc: MgCl2, H2SO4, NaCl, CuSO^, NaOH

7 Mot hon hap gom: CuO, FeO, AI2O3 Lam cdch ndo de tdch chiing ra

khoi nhau

Cdu II Bai todn

Cho 26,Ig MnOo tdc dung vdi dung dich HCl c6 20 gam HCl Cho hct khi do qua mot lit dung dich NaOH lodng dU

a) Lugng HCl nay c6 du de phdn dng het vdi Mn02 khong?

b) Tinh nong do mol 11 cua muoi thu dUgc trong phdn I'ing giOa do vd NaOH

c) Nung qugng pyrit sdt 'de tgo ra SO2 Cho khi SO2 sue vdo dung dich

chda 2 muoi tren

Sau do them vdo mot lugng du Ba(N03)2 Tim khoi lugng ket tua va khoi lugng pyrit can dung Biet rdng lugng SO2 tdc dung vUa du

dung dich muoi

T r o n g 20 m l dung dich H2SO4 chufa 0,025 m o l H0SO4

10 m l dung dich H2SO4 chtfa x mol H2SO4

T r o n g 10 m l dung dich N a O H chijfa 2a mol N a O H

30 m l dung dich N a O H chufa y m o l N a O H 2a X 30

1 Tit 7 Ig hoa chat sau, em c6 the dieu chc nhUng chat khi nuo?

Axit sunfuric; natri hidroxit; amoni nitrat; canxi cacbonat; natri

sunfit; sdt sunfua vd kim logi keni

2 Ta H2SO4 CO may cdch de dieu che CaSOJ

3 Viet cong thUc vd ten ggi 3 muoi diing trong nong nghiep (phdn dam,

phdn Idn vd plidn kali) Hay gidi thich tgi sao ngUdi ta khong trgn tro

bep vdi phdn dam de ban rugng?

1 9

2. D i e u che CaS04 t i f H2SO4:

Cdch 1: Tac d u n g vdri canxi: H2SO4 + Ca

Cdch 2: Tac d u n g vdri CaO: H2SO4 + CaO

Cdch 3: Tac d u n g v6i bazcJ Ca(0H ) 2 :

H 2 S O 4 + Ca(0H ) 2 > CaS04 + 2H2O Cdch 4: Tac d u n g vcri muo'i cija canxi:

H2SO4 + CaC03 > CaS04 + COat + H2O

Chu y: Con nliieu cdch klidc, xin nJiuang ban doc!

m a n g t i n h k i e m n e n ngi/di t a k h o n g t r o n p h a n d a m vcfi t r o bep

4 Ong 1: BaCl2 + K2CO3 > B a C O g i + 2KC1

2 H N O 3 + BaCOg > B a ( N 0 3 ) 2 + C 0 2 t + H2O

Ong 2: BaCIa + Na^CO^ > B a C O g i + 2 N a C l

2 H N O 3 + BaCOa -> Ba(N03)2 + C02t + H2O

dng 3: BaCla + 2AgN03 > Ba(N03)2 + 2 A g C l i

A g C l + HNO3

T r o n g 3 ong n g h i e m t h i t a t h a y k e t t i j a A g C l k h o n g t a n t r o n g a x i t mac dti l a a x i t HNO3 v i t h e t r o n g ong 3 con chat k e t t u a

5 P h a n ufng: N a C l + A g N 0 3 > A g C U + NaNO g

MgCl2 + 2 A g N 0 3 > 2 A g C U + Mg(N03)2 Sau k h i Ipc bo k e t tua t h i dung d i c h (B) gom: NaNOs, Mg(N03)2

2NO2 + 2 K 0 H — > KNO3 + K N O 2 + H2O

Phdn b: K h i cho H C l d u vao dung d i c h B t h u diJOc k e t tua D i e u

n a y cho t a t h a y r a n g t r o n g dung dich con AgNOs dif

H C l + AgNOs > A g C l i + HNO3

(D)

6, T a n h a n t h a y : t r o n g t a t ca cac dun g d i c h t r e n t h i ch i c6 m o t dung

d i c h CO m a u x a n h la: CUSO4, cac dung dich con l a i la k h o n g mau

T r i c h m o i lo m o t i t l a m m a u thijf

- Cho d u n g d i c h CUSO4 I a n lifot vao cac m a u thuf t r e n :

• M a u thCf cho ke't t i i a m a u x a n h la N a O H :

CUSO4 + 2 N a O H > C u ( 0 H ) 2 i + Na2S04

• Cac m l u con l a i k h o n g c6 h i e n tu'ong

- D u n g N a O H l a m thuoc thiif, cho I a n liiot vao cac m a u ihxi con l a i :

• M a u t h L f C O k e t t u a m a u t r S n g la MgClg:

2 N a O H + MgCla > M g ( 0 H ) 2 + 2 N a C l

• M a u t h i i tao dung dich t r o n g suot va toa n h i e t m a n h l a H2SO4

2 N a O H + H2SO4 > Na2S04 + 2H2O

• M a u k h o n g c6 h i e n t u g n g l a N a C l

7 T a c h CuO, F e O , AI2O3 r a khoi nhau:

• AI2O3 + 2 N a O H > 2NaA102 + H2O

NaAlOz + CO2 + 2H2O > A l ( 0 H ) 3 i + NaHCOg

36,5

> MnCla + Cl2t + 2H2O

0,1375 -> N a C l + N a C l O + H2O 0,1375

a) L i f g n g H C l n a y c6 du de p h a n lifng h e t vcfi M n 0 2 khong?

V a y lirang H C l n a y k h o n g dii de p h a n ufng h e t liicfng M n 0 2 da cho

hay I I C l p h a n ijfng h e t va M n 0 2 con dii

b) The tich dung dich thu dUcfc chinh 1^ the tich cua N a O H tufc la:

Vduiigdich = 1 (lit) So' mol cua N a C l trong phan ufng (2): 0,1375 (mol)

=> mpes, = - ^ - r — X 120 = 8,25 (gam)

DE SO 4

OE THI HOC SiNH GIOI HOA HOC 9 , CAP TP HO CHI MINH N A M HOC 1999 - ZOOO

Cdu I Khi cho kirn loai vdo dung dich mud'i c6 the xdy ra nhitng phan

ling gi? Cho vi du minh hoa

Cdu I I Viet cdc phuang trinh phdn vCng theo chuSi bien hoa sau:

FeCh > Fe(0H )3 > FezOs

Thuoc thii Kim loai I Kim loai II Kim loai III

Trong do ddu (+) de chi trudng hap kim loai hoa tan, ddu (-) chi triiang

hap kim loai khong tdc dung vai dung dich kicm hay axit

Hay xdc dinh kim loai nghien ciiu, viet phuang trinh phdn ling vd gidi

thich vi sao kim loai khong tdc dung vai cdc chat dd cho

Cdu TV Chi diing kim loai, hay nhgn biet cdc dung dich sau day:

HCl, HNO3 ddc, AgNOa, KCl, KOH

Vii't cdc phuang trinh phdn dug xdy ra trong qua trinh nhgn biet

Cdu V Hoa tan oxit cua kim logi hoa tri II trong mot lilgng vita dil dung

dich H2SO4 20% thi thu dicgc dung dich mudi c6 nSng do 22,69c

Xdc dinh kim logi do

Cdu VI Co mot hon hap gom Na2S04 vd K2SO4 dilgc trgn Idn iheo ti le

I : 2 ve so mol Hoa tan hon hap vdo 102 gam nUac thi thu dilgc dung

dich A Cho 1664 gam dung dich BaClz 10% vao dung dich A, xudt hien

ket tua Loc bo ket tua, them H2SO4 dU vdo nilac Igc thi tlidy tgo ra

46,6 gam ket tua Xdc dinh nong do phdn tram cua A^a^SOj vd K2SO4

trong dung dich A ban ddu

(Cho Na = 23; S = 32; K = 39; Ba = 137; CI = 35,5)

Ld\I

Cdu I K h i cho k i m loai vao dung dich muoi c6 t h e xay r a :

- N e u k i m l o a i cho vao khac k i m loai t r o n g m u o i t h i xay r a p h a n uTng

the - oxi hoa khijf

• A l k h o n g tac d u n g v d i HNO3 v i A l b i t h u dong t r o n g m o i t r U c f n g

HNO3 dac nguoi

Cdu IV T r i c h m o i lo m o t I t l a m m a u thijf

- Cho bot k i m loai Cu dti I a n luat vao cac m a u thtf t r e n :

+) M a u t h i i nao dung dich id k h o n g m a u chuyen sang x a n h la AgNOs

Cu + 2 A g N 0 3 > Cu(N03)2 + 2 A g

+) M a u thuf nao vifa tao dung dich mau x a n h va c6 k h i nau do bay r u

la HNO3

Cu + 4HNO3 dac > Cu(N03)2 + 2 N 0 2 t + 2II2O

- Sau do cho dung dich viia t h u di/oc a t r e n I a n liicft vao cac m a u con l a i

+) M a u CO k e t tua m a u x a n h la K O H

Cu(N03)2 + 2 K 0 H > 2KNO3 + C u ( 0 H ) 2 i •

+) Cac m a u con l a i k h o n g c6 h i e n tugng

- Cho dung dich K O H vao 2 mau con l a i , m a u nao c6 p h a n ufng toa

n h i e t la H C l

K O H + H C l > K C l + H2O

M a u con l a i la K C l

Cdu V Goi k i m loai hoa t r i I I la A va c6 a m o l => oxit l a : A O

A O + H2SO4 > ASO4 + H2O

K l i i them dung dich H2SO4 vao lo niidfc loc thi tao ket tua nCfa nen

trong nude loc con dii B a C l 2

OE THI HOC SINH GlDl HOA HOC, QUAN 9 (VONG 2) TP HCM NAM HOC 19S9 - 2000

Cdu I Cho chuoi phuang trinh plidn ling sau:

+A, I , +B,

Tim cdc chat A, B, C, X, Y, Z (Id cdc chat khdc nhau vd khdc

CaCOs)-Viet cdc phuang trinh phdn ting

Cdu II Dung mot kim loai de nhdn biet cdc lo dung dich sau:

FeCh; FeCh; HCl; BaCh; (NH^2S04; AlCh; NH4CI

Cdu I I I Cho Na vdo 2 dung dich mud'i Al2(S04)3 vd CuSOj thi thu dagc khi A, dung dich B vd ket tua C Nung ket tila dugc chat rdn D Cho H2

di qua D nung nong dUgc chat rdn E Hda tan E vdo dung dich HCl thi thdy E tan mot phdn

Gidi thich Viet phuang trinh phdn iCng

Cdu IV

a) Cho 0,25 mol CuO tan het trong dung dich II9SO4 20% dcm nung

nong lugng vila dii, sau do Idm ngugi dung dich den 10°C Tinh khoi

lugng tinh the CUSO4.5H2O tdch ra khoi dung dich Biet do tan CUSO4 a 10°C Id 17,4g

b) Clio biet do tan CaSO^ Id 0,2g (a 20°C) vd khoi lugng rieng cua dung

dich bdo hda coi bdng Iglml

• Tinh do tan cua CaSO^ theo nong do mol

• Khi trgn 50 ml dung dich CaCl2 0,012M vai 150 ml dung dich

NaoSO^ 0,004M a 20°C thi c6 ket tila xudt hien khong?

Cdu V

a) Hda tan hodn todn 1 hidroxit cua kim logi M bdng mot lugng vita dii

dung dich HCl 10% Sau phdn ling thu dugc dung dich A Them vdo dung dich A mot lugng vUa dii dung dich AgNOs 20% thu dugc dung dich muoi c6 nong do 8,965% Xdc dinli cong thvCc hidroxit tren

b) Khi phdn tich 2 oxit vd 2 hidroxit tUong iCng cila ciuig mot nguyen to

hda hoc dugc so lieu sau: ti so thdnh phdn % ve khoi lugng ciia 0x1

a) Khi trgn V (lit) dung dich X vol V (lit) dung dich Y thu dUgc 2 (lit^

dung dich Z Tinh CM dung dich Z

b) Neu lay 100 nd dung dich X vd lay 100 nd dung dich Y cho tdc dung

het vdi kim logi Fe thi lugng hidro thodt ra trong hai trUdng hgp lech iihau 0,448 (lit) (dktc) Tinh CM dung dich X, Y

Riai n c T u , „ „ o unA u n r a 2 1

Cdu VII Trgn Idn 10 ml dung dich HCl vai 20 ml dung dich HNO3 vd

20 nd dung dich H2SO4 thu dugc dung dich A, pha them II2O vdo dung

dich A de diCgc dung dich B c6 the tich gap doi Trung hoa 25 nd dung

dich B can 8 ml dung dich NaOH 8% (D = l,25g/nd) Bern c6 can dung

dich tqo thdnh dugc l,365g muoi khan Neu cho 40 tnl dung dich B tdc

dung vai mot liCgng dU dung dich BaCl-z thi thu dugc 0,932 ket tiia

a) Tinli CM dung dich axit ban ddu

b) Dung dich C chi'ia hon hap NaOH 0,8M vd Ba(0H)2 0,2M Can bao

nhieu ml dung dich C de trung hoa het 50 ml dung dich B?

(B)

> CaCl2 + 2H2O

C a ( 0 H )2 + 2HC1 —

CaCl2 + NaaCOa > CaCO^i + 2 N a C l

CO2 + 2 N a O H > Na2C03 + H2O

(Y)

NaaCOs + CO2 + II2O > 2 N a H C 0 3

(Z) -> CaCOyi + 2 N a N 0 3 + CO.T + II2O

2 N a I i C 0 3 + Ca(N03)2 —

Cdu II T r l c h m o i dung dich m o t i t l a m m a u thuf

Cho k i r n loai Ba vao cac m a u ihii t r e n , dau t i e a c6 p h a n ufng:

Ba + 2H2O Ba(0H)2 + H2T

M a u nao cho k e t tiia t r f i n g x a n h la FeCl2:

Ba(01i)2 + F e C L — > FeCOIDai + 2BaCl2

22 I r i i p i A i n c T u i u n r Qtwu n n i u n A u n r Q

M & u nao cho k e t t u a nau do la FeCla:

3Ba(OH)2 + 2FeCl3 > 2 F e ( O H ) 3 i + 3BaCl2

M a u CO k e t t i i a va k h i m i i i k h a i bay ra la (NH4)2S04:

B a ( 0 H ) 2 + (NH4)2S04 > BaS04^ + 2NH3T + 2H2O

M S U cho k e t tiia keo t r a n g la AICI3:

3Ba(OH)2 + 2AICI3 > 2 A l ( O H ) 3 i + SBaCla

M a u c h i CO k h i m t i i k h a i bay r a la NH4CI:

B a ( 0 H ) 2 + 2NH4CI > BaCl2 + 2NH3t + 2H2O

M a u cho dung dich t r o n g suo't va toa n h i e t la H C l :

K h i h a n h i e t do:

• C U S O 4 + 5H2O — > CUSO4.5H2O

(mol) a - > a

Kho'i liicrng C U S O 4 c6n l a i t r o n g dung dich: 40 - 160a (gam)

K h o i l i i o n g dung d i c h con l a i : 142,5 - 250a (gam)

^Na^so^ = O'^'^ ^ 0'004 = 0,0006 (mol)

d 2Q°C t r o n g 1 l i t dung dich CaS04 bao hoa c6 0,015 m o l CaS04

t r o n g 0,2 l i t dung dich CaS04 bao hoa c6 0,003 m o l CaS04

V i 0,6.10"^ < 3.10"^ n e n k h o n g c6 h i e n t i / g n g k e t t i i a

Cdu V

a ) G o i cong thufc cua h i d r o x i t k i m loai M : M ( O H ) a

M ( O H ) a + a H C l > MCla + aHaO ( I ) MCla + aAgNOg > a A g C l l + M ( N 0 3 ) a (2) Klio'i l u g n g dung d i c h t r o n g (1)

= M + 17a + 36,5a x 10 = M + 382a (gam)

K h o i lucfng dung d i c h t r o n g (2):

m d u „ g d i c h 2 = ra ^i^ + ra ^^j^^o, = + 382a + 170a : 2 0 % - 143,5a

= M + 1232a - 143,5a (gam) Theo de, n o n g do cua dung dich m u d i t h u dirge sau p h a n ufng (2) l a :

M + 62a C% M ( N 0 , ) ,

Cdu VI

a) H C l + AgNOa

(mol) 0,25

H C l + N a O H (mol) 0,15 <- 0,15

V 2HC1 + Fe 0,015

FeCl2 + H a t

0,025 2V -> FeCl2 + H 2 t

3V - 5(2 - V) = 8V(2 - V) 3V - 10 + 5V 16V 8V^

V, =1,7 V2 = -0,72 (loai)

2 N a O H + H2SO4 —

H C l > N a C l + PI2O

0,0025CA

— > NaNOs + H2O 0,005CB

0,0025CA HNO3 -

> Na2S04 + 2H2O 0,01Cc <- 0,005Cc -> 0,005Cc H2SO4 + BaCla > B a S 0 4 i + 2H2O 0,004 <- 0,004

The t i c h cua dung dich (B): 100 m l = 0,1 l i t

Goi CA l a n o n g do m o l cua dung dich H C l

So m o l cua H C l t r o n g 25 m l dung dich B :

0 ^ 0 , 0 1 x 0 , 0 2 5 0,1 = 0,0025CA Goi CB l a n o n g do m o l cua dung dich HNO3

=> So m o l ciia HNO3 t r o n g 25 m l dung dich B :

CB.0,02 x b , 025 0,1 = 0,005CB Goi Cc l a n o n g do m o l cua dung dich H2SO4

=> So m o l ciia H2SO4 t r o n g 25 m l dung dich B :

Khoi i L f o r n g dung dich N a O H = 8 x 1,25 = 10 (gam)

Suy ra: C A = 6 - 2 C B = 4 (mol)

b ) Cac p h a n ufng t r u n g hoa:

Ba(0H)2 + 2HC1 > BaCla + 2H2O (4

Goi a, b, c I a n Ixxat Ik so m o l ciia H C l , HNO3, H2SO4 t h a m gia p h a n

L^ng (1), (2), (3)

a', b', c' Ian liiat la so m o l cua H C l , HNO3, H2SO4 t h a m gia p h a n

ang (4), (5), (6)

, C A X 0 , 0 1 x 0 , 0 5 , iiHci = a + a' = = 0,02 m o l

, , , C B X 0 , 0 2 x 0 , 0 5 • ,

nHN03 = b + b' = — = 0,01 mol

, C c x O , 0 2 x O , 0 5 , nH2S04 = c + c = - ^ - ^ 1 ^ = 0,005 m o l

a + b + 2c = 0,8V (vdi V la the tich ciia dung dich C)

^ + K +c' = 0,2V

2 2

2]n,^^ = a + a' + b + b' + c + c' = 0,035 (0 ,8V - 2c - b) + a' + b + b' + c' + c = 0,035

b) Cho cac hoa chat: Na, MgCh, FeCls, FeCk, AICI3 Chi dung them H2O

hay nhdn biet chung

Cdu in Cho phuang trinh phan ling c6 dang sau: BaCh + ? = NaCl + ?

Hay viet 4 phuang trinh phan ling xdy ra Biet rdng cac phan i2ng deu xdy ra hodn todn

.au rV.d 25°C nguai ta da hda tan 450 gam kali nitrat vao trong 500 gain

niiac cat (dung dich A) Biet rdng dp tan ciia nitrat kali la 32 gam „

20°C Hay xdc dinh khoi lugng kali nitrat tdch ra khoi dung dich khj

Idm lanh dung dich A den 20°C

~lau y Cho 3 gam hdn hap hai kim loqi vun nguycn chat Id nhom v,i

magie tdc dung hct vai H2SO4 lodng thi thu duqc 3,36 lit mot chat khi a

dieu kien ticu cliudn

Xdc dinh thdnh phdn phdn tram ve khoi lugng ciia nlwm vd magic

trong hdn hap

^du VL Khi cho a gam Fe vdo trong 400 ml dung dich HCl, aau khi phun

ling ket thuc dem c6 can dung dich thu dicgc 6,2 gam chat rdn X

Ncu cho hon hap gom a gam Fe vd b gam Mg vdo trong 400 ml

dun/-dich HCl thi sau kJii phdn ling ket thiic thu diigc 896 ml khi H2 (dicii

kien chuan) vd c6 can dung dich thi dugc 6,68 gam clidt rdn Y

Tilth a, b, nong do phdn td gam ciia dung dich HCl vd thdnh phdn kiwi

lugng cdc chat trong X, Y

(Gid sii Mg khong phdn ling vai nitdc vd khi phdn I'ing vol axit thi Mij'

pJidn ling trade, het Mg mai den Fe Cho biet cdc phdn ling dcu xdy ru

hodn todn)

Biet: H=1;N= 14; O ^16; Mg = 24, Al = 27; CI = 35,5; K = 39; Ca = 40; Fe = 56

LCilGIAI

du I Chuoi bien hoa:

CaCOs ^ - ^ ^ — > CaO + CO.T

CaCOa + 2HC1 > CaCh + COgt + H2O

CaO + 2HC1 > CaCl2 + H2O

CaO + H2O > Ca(0H)2

Ca(0H)2 + CO2 -> CaCOsi + IlaO

CaCl2 + 2AgN03 > Ca(N03)2 + 2AgCU

Ca(0H)2 + 2HNO3 ^ Ca(N03)2 + 2H2O

Ca(N03)2 + Na2C03 ^ CaC03>l + 2NaN03

Cau

II-a) Scf do tach:

BaCOy BaSO

+HC1

I d C

BaCl - ^ M i _ ^ B a C O i loc

BaSO^ i BaCO,,

BaSO^ +H;,o KCl

MgCl^

KCl MgCl^

+K011

vL/a du

KCl

'Mg(OH)^ i + HC1 ->MgCl,

b) Trich mSi chat 1 i t lam mau thuf

+) Cho ni/orc Ian liJOt vao cac mau thiJf tren Mau nao c6 k h i bay ra la natri

Na + H2O -> NaOH + - H a t

2 +) Cho dung dich N a O H Ian liigft vao cac mau thtf tren thi:

- M S U CO ket tua mau trSng la MgCl2

- Mau cho ket tua mau nau do la FeCls

FeCla + 3NaOH > F e ( 0 H ) 3 i + 3NaCl

- Mau cho ket tua keo trang la AICI3

AICI3 + 3NaOH -> A1(0H)3^ + 3NaCl

Neu N a O H dii thi ket tiaa tan dan:

A1(0H)3 + N a O H > NaAlOa + 2H2O

BaCl2 + Na2C03 > B a C 0 3 i + 2NaCl

BaCl2 + Na2S04 — > B a S 0 4 i + 2NaCl

3BaCl2 + 2Na3P04 > Ba3(P04)2i + 6NaCl

BaCla + Na2S03 > B a S O g i + 2NaCl

(hoac BaCl2 + NaaSiOa > BaSiOgi + 2NaCl

LOi r i A i ,

Cdu IV Bap so: Kho'i Itfang kali nitrat t^ch ra khoi dung dich:

290 (gam) KNO3

Cdu V. Goi a va b Ian liigt la so' mol cua Mg A l

Phan ijfng: Mg + H2SO4 > MgS04 + Hgt (1)

(mol) a ->• a

2A1 + 3H2SO4 > Al2(S04)3 + 3H2t (2)

3b (mol) b ->

Theo de bai, ta c6 he phuong trinh:

3

Cdu VL

+) Tinh a va thanh phan cua (X)

Thi nghiem 1: K h i cho a gam Fe + 400 m l dung dich H C l , c6 can dung

dich t h i thu difgc 6,2 gam chat r^n Neu Fe phan ijfng het tufc chat rSn

Thi nghiem 2: Cho a gam Fe va b gam Mg vao 400 m l dung dich HCl, c6

can dung dich t h i thu diigc 6,68 gam chat r ^ n va 0,896 l i t H2 (dktc)

Fe + 2HC1 > FeCla + Hat (2) (mol) y y

(Vdi X, y Ian liTOt 1^ so' mol cua M g va Fe tham gia phan iJtng tren)

So sanh (*) va (**) ta thay: Khoi lugng k i m loai d t h i nghiem 2 nhieu

han d t h i nghiem 1 nhifng so mol H2 thu diioc l a i i t hon Do do trong

thi nghiem (1) t h i Fe di/ va HCl het so mol H2 cf t h i nghiem 1 la:

0,04 (mol)

Fe + 2HC1 > FeCla + Hgt (3) (mol) 0,04 0,08 0,04 ^ 0 , 0 4 i

Tif (3) ra^^cA., = 0,04 x 127 = 5,08 (gam)

Khoi lugng Fe phan uTng: 0,04 x 56 = 2,24 (gam)

va khoi lircfng Fe diT: 6,2 - 5,08 = 1,12 (gam) Vay: a = 2,24 + 1,12 = 3,36 (gam)

+) Tinh b va thanh phan cua (Y)

Neu khoi lircfng M g dung du t h i :

mcLatrdn = 0,04 X 95 + 3,36 = 7,16 (gam) Theo de bai: nichat rin = 6,68 < 7,16 => Mg khong du nen Fe da phan ijrng

Phan Lfng: M g + 2HC1 > MgCh + H2t (4)

Fe + 2HC1 > FeCla + Hat (mol) y -> y y

Vay: niMg = 0,02 x 24 = 0,48 (gam)

^) Chi dung nuac hay nhdn biet 3 bgt kim loai: Ba, Al vd Ag

^) Til cdc chat sau: Na^O, HCl, H2O, Al c6 the dieu che dugc n/ulng clid't fnai ndo md kJiong dung them phiCang tien ndo khdc Viet phan ling

Cdu 3 Bern nij gam hon hap ZnCOs, Zn dun nong ngodi khong ' i dc

plidn ling xdy ra hodn todn, thu dugc nig gam chat rdn

Biet nil = mo Tinh % khoi lugng ZnCOs trong hSn hap ddu

Cuu 4 Bem dung dich chiia 0,1 mol sdt clorua tdc dung vai dung dich

NaOH du thu dugc 9,05 gam ket tua

Xdc dinh cong thiCc sdt clorua vd tinh hieu sudt phdn ling

Cdu 5 Bem 46,4 gain Fe^Oy tdc dung vdi Ho dun nong thu dugc chat rdn B

gom Fe vd Fe^Oy du Bem chat rdn B tdc dung het vai dung dich HNO

lodng du thu dUgc dung dich C c6 chUa 145,2 gam mud'i Fe(N03)3 va

a mol NO thodt ra Tat cd phdn Ung xdy ra hodn todn

a) Xdc dinh cong thUc Fe^Oy

h) Biet a = 0,52, tinh khoi lUgng tiCng chat trong B

L d i G I A I

Cdu 1

a ) Cho 3 k i m loai vao 3 coc niicfc

- Tan C O bot k h i bay len la Ba:

Ba + 2H2O > Ba(0H)2 + Hat '

- Kliong tan la A l va Ag

- Cho 2 k i m loai A l va Ag Ian liicft vao hai coc chufa dung dich

Ba(0H)2: K i m loai nao tan c6 bot k h i bay len la Al:

2A1 + Ba(0H)2 + 2H2O > Ba(A102)2 + Sllat

- Khong tan la Ag

3 ZnCOo -> ZnO + CO2

2Zn + O2 2ZnO

I

(1) (2)

Goi X, y Ian lircft la so mol ZnCOg va Zn trong h6n hop dau Vi mj = m2

Khoi liiong CO2 thoat ra 0 (1) = khoi lugng oxi tham gia a (2)

=> 44x = 16y ^ - = ~

y 11 Vay % khoi iLforng ZnCOg = 41,15%

Cdu 4 Neu la FeCl2: FeCl2 + 2NaOH > Fe(0H)2i + 2NaCl

Khi phan ufng xay ra 100% thi khoi lugng ket tiia Fe(0H)2 = 9g < 9,05g

=i> V6 11, vay do la FeCls

FeCla + 3NaOH > Fe(0H)3i + SNaCl

=> Hieu sua't (H) = = 84,58%

Cdu 5

a ) Khoi liiang Fe c6 trong 145,2 gam FeCNOsJa:

145,2 X 56

b ) NasO + H2O > 2NaOH

2A1 + 2NaOH + 2H2O > 2NaA102 + 3 H 2 t

NaAlOa + HCl + H2O > A l ( 0 H ) 3 i + NaCl

NaOH + HCl > NaCl + H2O

Al + 6HC1 > 2AICI3 + 3H2T

242 = 33,6 gam

Cdu 2 Phan ijfng:

2Fe + 3CI2 FeCl, + 3NaOH

-> 2FeCl3

> Fe(0H)3i + 3NaCl 2Fe(OH)3 FeaOa + 3H2O

A: Fe;

X: CI2;

Fe203 + 3H2 B: FeCla;

b ) Fe + 4HNO3 > Fe(N03)3 + N O t + 2H2O

3Fe304 + 28HNO3 > 9Fe(N03)3 + NOT + UlUO

Goi X , y Ian liigt la so mol ciia Fe va Fe304 trong hon hop

So mol NO: x + ^ = 0,52

3 Bao toan Fe: 56x + 168y = 33,6

=> X = 0,51 mol ^ m p e = 28,56 (gam)

V',-A, B, D, X, Y, Z CO the khac, nhUng C phai la Fe(0H)3 y = 0,03 mol ^ m^^ ^ = 6,96 (gam)

Wl Rli, „ S

DE SO 8

flE THI HOC S I N H GlDl HOA HOC 9, CAP TP HQ CHI IVIINH N A M HOC 2 0 0 1 - ZOOZ

Cdu 1: Bo tiic ud can bdng cdc phitang trinh phdn ling sau:

Cdu

CaCh + ? Ba(HC03)2 + ? CaS03 + ?

HCl FeClo FeCh

Ca3(P04)2^ + ?

^ BaCOaJ- + ? -> SO f + ? + ?

+ ?

+ ?

+ ?

-> NallCOs + ? -> FeCh

FeCU

Cdu 2: Vict cdc pliUang trinh phdn I'tng de bleu dien chuoi bicn lioa sau:

FcS2 ^ SO2 -> SO3 ^ H2SO4 -> SO2 Na2S03 ^ BaSOa

Cdu 3:

a) Cho 6 dung dich gom: NaCl, BaCU, CUSO4, NaOH, MgCh, AgNO,

KJioiig dung them hoa chat ndo khdc, hay nhan biet chung

b) Cho 3 dung dich: BaCU, BafNOi)., Ba(HC03)2 Chi diCgc si'i dung

them mot hoa chat, hay nlidn bict cluing

c) Cho hdn hap khi gom CO2, SOo Bdng pliuang phdp hoa hoc, hay tdch

rieng CO2

Cdu 4: A vd B Id 2 loai clidt chi cliiia cdc nguyen to X, Y Thdn.Ii pJidn

phdn tram cua nguyen to X trong A vd B Idn liigt Id 30,4% vd 25,97/

Neu cong thi'ic phdn til cua A la XY2, thi cong thiic phdn tii cua B Id gi?

Cdu 5: De gia tang nong do cila 50 gam dung dich CuSO^ 57c len gap luu

Idn, CO boil hoc sinh da thUc Jiien bdng bdn cdch khdc nJiau:

Hoc sinh A: dun nong dung dich de Idm bay hai phdn ni'ca lugng nitac

Hoc sinh B: them 2,78 gam CuSO^ khan vdo dung dich

Hoc sinh C: them 4,63 gam tinh the CUSO4.5H2O vdo dung dich

Hoc sinh D: them 50 gam dung dich CuSO^ 157c vdo dung dich

Hoi hoc sinJi ndo da lam diing, gidi thich

Cdu 6: Hop chat A bi phdn hiiy a nhiet do coo thco phaang trinh plidn dug:

L6\I 1: Bo tuc v a can bkng:

3CaCl2 + 2Na3P04 > Ca3(P04),i + 6NaCI

Ba(HC03)2 + Ba(0H)2 > 2 B a C 0 3 i + 2H2O

CaSOa + 2HC1 > SOat + CaCl2 + H2O

H C l + NaaCOa > N a H C 0 3 + N a C l 2FeCl2 + CI2 > 2FeCl3

2FeCl3 + F e > SFeClg

Cdu 2: Viet phUcfng trinh

4FeS2 + IIO2 2Fe203 + 8SO2 2SO2 + O2 ^

450*^ C : i 2SO3

SO3 + H2O

a )

San phdm tgo thdnh deu a the khi, khoi lugng mol trung blnh cua hon

hgp khi sau phdn ting Id 22,86 (gimol) Tinh khoi lugng mol cua A

Cho so lieu: H = 1; O = 16; S = 32; Cu = 6"-^

Hoc sinh c6 the sii dung bdng do tan vd bdng he thong tudn hodn cdc nguyen to

hoa hoc

> H2SO4

2H2SO4 + Cu > CUSO4 + S02t + 2H2O

SO2 + 2 N a O H > Na2S03 + H2O

Na2S03 + BaCl2 > B a S O s i + 2 N a C l

N h a n biet CUSO4: mau xanh Cho mSu CUSO4 tac dung vdi cac mau c5n lai c6 1 mau cho ket tua xanh lam l a N a O H , mot mau cho ket tiia trSng la BaCl2

L a y mSu N a O H cho tac dung vdi cac mau con l a i : mau cho ket tua

trSng la MgCla, mSu cho ket tua den 1^ AgNOa, mau khong tao ket tua l a N a C l

CUSO4 + 2 N a O H

CUSO4 + BaCl2

> C u ( 0 H ) 2 i + Na2S04

MgCl2 + 2 N a 0 H — AgNOg + N a O H -

2 A g O H > Agp + up

•> B a S 0 4 i + CuCla

- > M g ( 0 H )2i + 2 N a C l

^ A g O H + N a N O ,

b) - Dun nong mau cho ket tua la Ba(HC03)2

Ba(HC03)2 ^ B a C O g i + CO2T + H2O

- Hai mau con lai cho tac dung vdri dung dich AgNOs Mau cho i

BaCl2 mSu con la i la Ba(N03)2

BaCl2 + 2AgN03 > Ba(N03)2 + 2AgCU

c) - Cho qua dung dich thuoc t i m SO2 bi hap thu, con CO2 bay ra (cho

qua dung dich Br2 chi c6 SO2 bi hap thu)

- Co the dot chay SO2 SO3 hap thu vao Ba(0H)2 sau do cho HCl

vao de giai phong CO2

Cdu4: - Cong thufc A: XY2

- Cong thufc B: XaYb

- Trong A: % X = 30,4% % Y = 100% - 30,4% = 69,6%

- Trong B:

TCrd), (2)

X _ 30,4 _ X 60,8 2Y 69,6 Y 69,6

%X = 25,9% %Y = 74,1%

aX _ 25,9 _ X 25,9b

bY 74,1 Y 74,1a 60,8 25,9b a _ 2

Hoc sinh A lam bay hoi phan nijfa luong niidc

m^^Q b i i o a i = = 23,75 = m^^o cbn

m d u n g d c h sau = 50 - 23,75 = 26,25 gam [CuS04]% = I T^M^rJ X 100 = 9,52% ^ 10%

Neu lay 2a mol A nhiet phan se tao thanh 7a mol k h i

Dinh luat bao to^n khoi liicfng cho:

Hay cho blet vi tri (1), (2), (3), (4)., (5), (6), (7), (8) la cdc ti( gi

•2 Neu hien tUgng, viet phUpng trinh phdn ling cho cdc thi nghieni sau:

a ) Nhung dinh sdt da cqo sach gi vao dung dich CuSOj

Sue khi SO2 vao dung dich Ca(HC0a)2

^) Dan khi etilen qua dung dich nUac brom

^' Cho day chuyen hoa sau:

Fe ->A -^B ->C-^Fe ->E->-F-^D dinh A, B, C, D, E, F Viet cdc phiiang trinh phdn ring

Cau II

1 Dung dich Boocdo dung chong nam cho cay dugc pha theo tl le:

1kg CUSO4.5H2O + 10kg voi song (CaO) + 100 lit nuac

Hay tinh thdnh phdn % theo khoi lUgng cdc chat c6 trong dung die/,

Boocdo Viet cdc phiiang trinh phdn ling

2 Tie glucoza va cdc chat v6 ca can thiet, viet cdc phuang trinh phcn,

ling dieu che: etylaxetat

Cdu III

1 Ba khi A, B, C c6 phdn ti2 khoi bdng nhau vd bdng 28 dvC A, B cu

the hi dot chdy trong khong khi, sdn phdm sinh ra deu c6 khi COo, li

CO the kill} dugc CuO a nhiet do coo, C la thdnh phdn quan trgiii:

trong phdn bon hoa hoc Xdc dinh cong thilc phdn ti2 cua A, B, C, vii')

cdc phuang trinh phdn iCng

2 (M), (N), (P), (Q), (R), (X) Id nhUng hap chat hUu ca dugc biet den

trong chuang trinh hoa hoc pho thong cap trung hoc ca sd (P) vd (Ni

CO cung cong thvCc phdn tit

- Ve khoi lugng phdn tit (klpt):

klpt (N) = -klpt (M); klpt (X) = 3klpt (R) - Gklpt (Q)

2

2,3 gam (N) hay 1,5 gam (Q) c6 the tinh bdng the tich cua 1,6 gam 0^

cung dieu kien

- Ve tinh chat: (M), (N), (R) cd phdn itng vai Na, tic 0,1 mol (M) c6 the

cJio the tich H2 Ian nhdt la 3,36 lit (dktc), chi c6 (R) phdn itng dugc

vai dung dich NaOH Tit (X) cd the dieu che ra (N), (R), (Q) c6 phdn

ling vai CI2 (chieu sang) Xdc dinh cong thitc cdu tgo cua (M), (N), (P'

(Q), (R) vd cong thUc phdn tii cua (X) Gidi thich

Cdu IV Hoa tan mudi nitrat cHa mot kim logi hoa tri 2 vao nudc dugc 200 lu'i

dung dich (A) Cho vao dung dich (A) 200 ml dung dich K3PO4, phan

iint-xay ra vita du, thu dugc ket tua (B) vd dung dich (C) Klioi lugng ket tua (I'

vd khoi lugng mudi nitrat trong dung dich (A) khdc nhau 3,64 gam

1 Tim nong do mol/lit cua dung dich (A) vd (C), gid thiet the tich dwni

dich khong thay doi do pha trgn vd the tich ket tua khong ddng ke

2 Cho dung dich NaOH (lay du) vao 100 ml dung dich (A) thu dugc kH

tua (D), Igc lay ket tua (D) roi dem nung den khoi lugng khong do'

can dugc 2,4 gam chat rdn Xdc dinli kim logi trong mudi nitrat

4 0 1 ^ 1 niAi np TMi unr ciuu mni un/t unr Q

V Doi c/'idy hodn todn 1,344 lit (dktc) hon hgp 3 hidrocacbon the khi:

C H'>n + 2! CinH2m', CkH2k - 2- Sau phdn Ung, dan lion hgp sdn piidm Idn

lugt Q'^'^ H2SO4 fdgc), dung dich NaOH (du) thdy khoi lugng H2SO4 (dd^^) ^'^"^ ^'^^ khoi lugng dung dich NaOH tang 7,04 gam

1 Tinh thdnh phdn % theo the tich hSn hgp 3 hidrocacbon, biet the tich

hidrocacbon CkH2k-2 trong hon hgp gap 3 Idn the tich

CJl2„+2-2 Xdc dinh cong thiic phdn tiCJl2„+2-2 3 hidrocacbon, biet rhng cd CJl2„+2-2 hidrocacbon

CO so nguyen tit cacbon bdng nhau vd bdng so nguyen tit cacbon

cua hidrocacon con Igi

L d l GIAI

Cdu I

1 Ten goi cac v i t r i : (1): axit; (2): oxit; (3): oxit khong tao muoi; (4); chat k h i ; (5): dcfn chat; (6): baza; (7): chat hau ccf; (8): bazcf kiem

2. Viet phan ijfng:

a) Fe + C U S O 4 > FeS04 + C u i

- Dung dich mau xanh bi nhat dan

- Co ket tiia cua dong xuat hien

b ) SO2 + Ca(HC03)2 > CaSOgl- + 2C02t + H2O

(cd ket tua va c6 khi)

hay 2SO2 + Ca(HC03)2 > Ca(HS03)2 + 2CO2T

(c6 khi bay ra) '

c) CH2 = CH2 + Br2 > CH2Br-CH2Br

(mat mau ndu do dung dich nifdc brom)

3. Thiic hien day chuyen hoa:

Pe y i _ > FeClg : > Fe(0H)3 Fe203 Fe

FeCl2 Fe(0H)2 > FeS04 > FeCl2

Phan ling:

1) 2Fe + 3CI2 2FeCl3

2) FeClg + 3NaOH > Fe(OH)34 + 3NaCl

nau do