Lời giải đề thi học sinh giỏi hóa học 9

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.. Hay xdc dinh kim loai nghien ciiu, viet phuang trinh phd

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

BaCl2 + K2SO4 > B a S 0 4 i + 2KC1 (2) (mol) 2x 2x ^ 2x

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

-> CaO + C O a t

> ZnS04 + H 2 t

— > ZnS04 + S 0 2 t + 2H2O -> FeS04 + H 2 S t

hoac

N a O H + NH4NO3 Na2S03 + H2SO4

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:

FeCl,, [Cu(NH3)J(OH)2 +HC1 ->CuCL +NaOH du

+ N H O H dil

-^Cu(OH),

Fe(OH), cliaii kliong -^FeO

P h a n ufng:

• 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:

=> 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

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

0,015

V 2 V ' • Theo cau (a): - n^^^.,^ = n^^^^^ = Cx-V = 0 , 2 5 ( m o l )

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

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

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

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)

"2 22,4

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)

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

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

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

Cdu 2 Phan ijfng:

2Fe + 3CI2 FeCl, + 3NaOH

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)

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:

2FeCl3 + F e > SFeClg

Cdu 2: Viet phUcfng trinh

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

- 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

Hoc sinh A lam bay hoi phan nijfa luong niidc

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

Dinh luat bao to^n khoi liicfng cho:

•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

3 ) 2Fe(OH)3 -> FeaOa + SHgO

4) FeaOg + SCO

5 ) Fe + 2HC1

-> 2Fe + 3CO2T FeCl2 + H2T

6) FeCl2 + 2NaOH > Fe(0H)2i + 2NaCl

tvang xauh

7) Fe(0H)2 + H2SO4 > FeS04 + 2H2O

Cac p h a n ufng dieu che:

18

(mol)

P h a n ijfng: CaO + H2O > Ca(0H)2

Ca(0H)2 + CUSO4 > C u ( 0 H ) 2 i + CaS04 (1)

K h o i iLTcfng d u n g d i c h Boocdo t h u dirge l a : 100 + 1 + 10 = 111 (kg)

_ 10^

•^4 250 (mol)

m C"(o")2 250 X 98 = 0,392.10^ (gam)

= 46 (gam)

p h a n ttjf k h o i cua ( Q ) : = 3 0 (gam)

0, 05

M a t k h a c , ( N ) tac d u n g vdi N a nen ( N ) la: C2H5OH

( Q ) tac d u n g v d i CI2 chieu sang n e n ( Q ) la: C2H6

• Theo de: p h a n ttf k h o i cua ( N ) = - k h o i liJgng p h a n t i i cua M

V a p h a n tuf k h o i cua (X) l a : M x = 3M R = 3 X 60 = 180 (gam)

va (X) dieu che r a C2H5OH va CH3COOH => (X): C6H12OC (glucozcf)

Cdu rv

1. P h a n ling: 3M(N03)2 + 2K3PO4 > M 3 ( P 0 4 ) 2 i + 6KNO3 ( l i

(Vdi M la k i m loai hoa t r i I I )

TU p h a n ufng (1): Su khac nhau ve khói lijgng la do thay 6N()

CkH2k;-2-Ta c6: n hon hep 3 khi = " ^'^^

1. T i n h p h a n t r a m the" tich m o i hidrocacbon

(mol) 3a 3ak 3ăk - 1)

% V B =

0,06 0,02

Vi m , n , k la nguyen va chSn nen n g h i e m hgp l i la: n = 2; m = 4; k = 2

^ 3 hidrocacbon can t i m la: (A): C2H6; (B): C4H8 va (C): C2H2

(*)

DE SO 10

flE THI CHQN HOC SINH GIQ! HOA HOC 9, TJNH BJNH OjNH NAM HOC 2001 - 2002

1- Chi dugc dung CO2 vd HoO, hay trinh bay each phdn biet 4 Ig chila

^ chdt rdn: K2CO3, BaCOj, KNO3, BaSỘ Viet phdn iCng de niinh hgạ

2 Sue a (mol) COo vdo dung dich chíia 1 mol CafOWs Tinh so mol CaCOs tgo thdnh ting vai gid tri a = 0; a = 1; a ^ 2

-Ve duang bleu dien so mol CaCOs tgo thdnh theo so mol CO2 da chọ

® " 3 iV/uê phdn m 1 gam hSn hgp Mg, MgCOs ngodi khong khi den khi phdn ling xong ta thu dugc 7712 ga77i mot chat rán Biet 7711 = 77x2. Tinh %

lugng Mg t7-o7ig hdn hap đụ

: THI

Cdu 4 Hoa tan hon hap Na20, NaHCOs, BaCl2, NH4CI c6 cung so mol vaa

nuac du, dun nong nlie thu dugc dung dich A vd ket tiia BaCOs- Hoi

dung dich A chiia gi? Viet phdn iCng ininh hga

Cdu 5 Trgn 11,2 gam bgt Fe vd 4 gam bgt S trong chen sii dem ?iuiig

khong CO khong khi de phdn iCng xdy ra tao FeS vai hieu sudt 80% Lay'

chat rdn tim dugc trong chen sii cho tdc dung viia du vai V lit dung dich

HCl IM, thodt ra a (mol) hon hap khi vd m (gam) chat rdn khong tan

a) Viet tat cd phdn ling xdy ra

b) Ttnh gid tri V, a, m

Cdu 6. Dem hSn hgp gom 0,1 mol Mg vd 0,2 mol Al tdc dung vdi mot

lugng H2SO4 d4c, nong viia du thu dugc hon hgp mudi, 0,075 mol S va

0,175 mol SO2

a) Tinh khdi lugng hdn hgp mudi tgo thdnh

h) Tinh sd mol H2SO4 phdn iJCng viia du

LCil GIAI

Cdu 1

- Lay moi lo mot it cho vao 4 co'c

- Che niidfc vao 4 co'c, phan diioc 2 nhom: nhom (I) tan: dung dich

K2CO3, KNO3; nhom (II) khong tan; BaCOg, BaS04

- Sue k h i CO2 vao 2 coc nhom (II):

Neu tan la coc chtTa BaCOs, phan tfng xay ra tao ra dung dich BadlCOa)^:

CO2 + H2O + BaCOg > Ba(HC03)2 Coc khong tan chufa BaS04

- Lay it dung dich Ba(HC03)2 nho vao 2 coc cua nhom (I)

Neu coc nao tao ra ket tua trSng do la coc chufa K2CO3, phan ufng tao

raBaCOg

K2CO3 + Ba(HC0g)2 > BaCOg + 2KHCO3 Coc khong ket tiia chufa KNOg

Cdu 2

a) Tinh so mol CaCOg: K h i a = 0 khong c6 phan ufng, so mol CaCOg = 0

K h i a = 1 phan ufng xay ra:

CO2 + Ca(0H)2 > CaCOg + H2O (D

=> So mol CaCOg = 1

Khi a = 2 phan ufng xay ra:

2CO2 + Ca(0H)2 > Ca(HC03)2 (2'

=> So mol CaCOg = 0

Triicfng hop 2 c6 the viet phan ufng hoa tan het CaCOg

If) Chon 3 diem:

n

^' "^CaCO.j = 0 C02 " •' '•'CaCO;, = 1 C02 ~ ^ ^CaCOg = 0

C&u3

Phan ufng xay ra:

M g + ^02 MgCOg —

MgO

MgO + C02t

>oi a, b Ian iLTcft la so mol cua Mg va MgCOg trong hSn hgp dau

Vi m i = m2 nen khoi lirgng O2 (1) = khoi liigng CO2 (2) =^ 16a = 44b

(1) (2)

2NaOH

> NagCOg + H2O

(1) (2) (3) (4)

Vi so mol cua 4 chat: NagO, BaClg, NaHCOg, NH4CI bSng nhau, nen theo (1), (2), (3), (4) dung dich A chi chila NaCl

Cdu 5

-> BaCOgi + 2NaCl NaCl + H2O + NHgt

Vi CO hieu suat 80% nen chat ran gom: FeS, Fe di/, S diT

FeS + 2HC1 > FeCl2 + HaSt

Fe + 2HC1 > FeCl2 + H2t

(2) (3)

^) Theo (1), neu hieu suat bkng 100% thi S het, Fe dif

Khi hieu suat 80% t h i : S dii = 0,8 gam = a; Sd mol S phan ufng = Sd oiol Fe phan ufng = So mol FeS sinh ra = 0,1 mol

Vay sdmol Fe = 0,1 mol

(2) va (3) ^ Sd mol HgS = Sd mol FeS = 0,1 mol

Sd mol H2 = Sd mol Fe = 0,1 mol =:> a = 0,2 mol, sd mol HCI dung la 0.4 mol V = 0,4 l i t

Cdu 6

a) Tinh khS'i Itfgng hon hcfp muol:

CiJ 0,1 mol M g tao ra 0,1 mol MgS04 (bao toan khoi lirgng)

Cur 0,2 mol A l tao ra 0,1 mol Al2(S04)3 (bao toan khoi lifgng)

Vay khoi li/crng hon hgp muoi = (120.0,1 + 342.0,1) = 46,2 gam

b) T i n h so mol H2SO4 dac n6ng da dung vCra du

Phan ufng xay ra:

3Mg + 4H2SO4 > 3MgS04 + S i + 4 H 2 O (1)

2A1 + 4H2SO4 > Al2(S04)3 + S i + 4H2O (3)

2A1 + 6H2SO4 > Al2(S04)3 + 3 S 0 2 t + 6 H 2 O (4)

Theo (1) va (3) ^ ^H^SO, dung = ^.ng = (0,075 x 4 ) mol = 0,3 (mol)

Theo (2) va (4) => n^^^gg^ ^^^^ = 2n^^^ = (0,175 x 2) mol = 0,35 (mol)

Vay so' mol H2SO4 da dung viTa du 1^ 0,65 (mol)

a) Tinh so mol do trong 7,19 gam do;

b) Tinh so mol O vd O2 trong 8 gam oxi;

c) Tinh khoi lugng cua 0,05 mol kem;

d) Tinh khoi lugng cua 0,75 mol nudc;

d) Tinh so nguyen tii C trong 0,02 gam C;

g) Tinh so phdn tvC CO2 trong 1,1 gam khi CO2;

h) Tinh so g cua 1 nguyen tii Na;

i) Tinh so g cua 1 phdn tii SO2

2 Mot hdn hgp X gom FeO vd Fe203 cd khoi lugng Id 30,4 gam Nun.^

hon hgp nay trong mot binh kin cd chica 22,4 lit CO (dktc) Kho'

lugng hon hgp khi thu dugc sau khi nung la 36 gam

a) Hay xdc dinh thdnh phdn hon hgp khi Biet rdng hdn hgp X b\

khv[ hodn toan thdnh Fe

b) Tinh khoi lugng Fe thu dugc vd khoi lugng cua mSi oxit sdt trong X

Co 6 to mat nhan diCng cdc dung dich khong mdu Id: iVa^SOj (1); MasCOs (2); BaCls (3); BafNOaJs (4); AgNOa (5); MgCls (6) Bdng phuang plidp hoa hoc vd khong dung them cdc hoa chat khdc hay trinh bay each nhan biet cdc dung dich tren, biet rdng chung deu c6 nong do du Ian de cdc ket tua it tan cUng c6 the tao thdnh (Khong can viet phuang trinh phdn ling)

Cdu 3- ^°<^ dinh nong do cua cdc muoi NaHCOa vd Na2C03 trong mot

dung dich hdn hgp cua chung (dung dich A), nguai ta lam cdc thi nghieni nhu sau:

Thi nghiem 1: Lay 25 ml dung dich A cho tdc dung vai 100 nd dung dich HCl IM (du) dun nong hdn hgp, sau dd trung hoa lugng axit du bdng lugng vita du la 14 ml dung dich NaOH 2M

Thi nghiem 2: Lai lay 25 ml dung dich A, cho tdc dung vai lilang du dung

dich BaCl2 Loc bo ket tua mdi too thdnh, thu lay nUdc Igc vd nuac nla gap

Igi roi cho tdc dung vai lugng vica du la 26 ml dung dich HCl IM

1 Viet cdc phuang trinh phdn ling xay ra vd gidi thich vdn tdt

2 Tinh nong do mol cua moi muoi trong dung dich A

Cdu 4 Mot dung dich axit axetic CH3COOH cd C% = 10% Lay 300 gam

dung dich axit nay cho tdc dung vai 300 ml dung dich NaOH 2M tao ra dung dich A Dung dich A cd tinh axit hay baza?

Tinh nong dd % cdc chat tan trong dung dich A, biet rdng dung dich NaOH2M cd d = 1,2 giml

Ghi chu: Hoc sink dugc phep svt dung bdng tudn hodn, bdng tinh tan, gido vien coi tin khong gidi thich gi them

c ) mzn = 0,05 X 65,38 = 3,27 gam

d ) mj^^Q = 0,75 mol x 18 g/mol = 13,5 gam

iSlSi^li^ THI HOC SINH G|6'HOA HQC 9

2, a) T a c6: nco (ban ddu) = 1 mol => mco (ban dsuj = 28 gam

Do t a n g k h d i liiOng: 36 - 28 = 8 gam = mo => n^^ = 0,5 mol

V a y CO 0,5 mol C O ket hop v d i 0,5 mol O cho r a 0,5 mol CO2

T h ^ n h p h a n h o n hap k h i la: 0,5 mol CO v a 0,5 mol CO2

b ) T a c6: n^^ (c6 t r o n g FeO v a Fe203) = n^^ lay r a (d t r e n ) = 0,5 mol

Theo d i n h l u a t bao t o a n nguyen to t h i :

npe ( t h u dLToc) = npe ( t r o n g FeO) + npe ( t r o n g Fe203)

= a + 2b = 0,4 mol

Vay: nipe ( t h u duoc) = 0,4 x 56 = 22,4 gam

Cdu 2

1. L a y m o t dung dich b a t k i cho vao 5 dung dich con l a i , ta c6 b a n g sau:

Na2S04 Na2C03 BaCl2 Ba(N03)2 AgNOa MgCl2

-50 LOI GIAI at THI HOC SINH GIQI HOA HOC 9

Tii bang tren t a thay:

D u n g dich n a o cho vao tao ra 4 Ian ket t u a l a dung dich Na2C03 va

• AgNOa (cap dung dich 1) D u n g dich nao cho vao tao r a 3 I a n k e t tua

la dung dich Na2S04 va BaCla (cap dung dich 2) D u n g dich nao cho vao tao r a 2 I a n k e t tiia l a dung dich MgCls v a Ba(N03)2 (cap dung dich 3)

+) L a y m o t trong h a i chat a c&p dung dich 3 I a n lirgt cho vao 2 dung

dich 6 cap 2, n e u c6 tao r a ket tiia: thi chat cho vao l a Ba(N03)2, con

l a i l a MgCl2

Chat tao r a ket t u a a cap 2 l a Na2S04, con l a i l a BaCl2

+) L a y Ba(N03)2 da tim diTOc a cap 3 cho vao h a i dung dich d cap 1,

neu CO ket t u a thi: Chat tao r a ket tiia vcfi Ba(N03)2 l a Na2C03 con lai la AgNOa

Cdu 3 Bat so mol Na2C03 v a NaHCOa trong 25 m l dung dich A I a n liTot l a x, y

Doi v(5i thi nghiem 1, t a c6:

Na2C0a + 2HC1 > 2 N a C l + COgt + H2O (1)

I NaHCOa + H C l > N a C l + CO.T + H2O (2)

H | | H C l (dii) + N a O H > N a C l + H2O (3)

So mol H C l trong 100 ml dung dich l a : 0,1 x 1 = 0,1 mol

So mol H C l dir sau phan ufng (1) va (2) l a : 0,014 x 2 = 0,028 mol

k So mol H C l da tac dung vdri dung dich A la:

nong do mol cua NaHCOg l a : 0,026 : 0,025 = 1,04M

i J ^ i i l i l n f

Cdu 4

300 g a m C H 3 C O O H 10% chOra 30 gam C H 3 C O O H

^ So m o l C H 3 C O O H = 0,5mol

300 m l dung dich N a O H 2 M chufa nNaOH = 0,3 x 2 = 0,6 m o l

P h a n ling: CH3COOH + N a O H > CHaCOONa + H2O

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

T i le p h a n lifng l a 1 : 1

V a y sau p h a n ufng con N a O H = 0,6 - 0,5 = 0,1 ( m o l )

dung d i c h A c6 t i n h bazcf

T i n h k h o i lufOng dung dich A :

m (dung d i c h A ) = m (dung dich CH3COOH) + m (dung dich N a O l I

a) Cho biet dp tan mot loai muoi bii'n doi theo nhiet dp theo bang sau

neu lam Iqnh dung dich c6 nong dp % ciia muoi do Id 22% tie 50 °Cl

thi phqm vi de bdt ddu xudt Men ket tinh cua muoi do trong khodni^

nhiet do ndo?

Nhiet dp 10° C 20°C 30°C 40°C

Dp tan (trong lOOg nicac) n,3g 15,2g 18,9g 24,8g 36,7g

b) Co 6 dung dich dugc ddnh so ngdu nhien tic 1 den 6, moi dung c/?'

chiia mpt chdt gSm: BaCh, H2SO4, NaOH, HCl, MgCl., NazCOa

La>-luat thuc hien cdc thi nghiem ud dugc ket qua sau:

Thi nghiem 1: Dung dich (2) cho ket tua vai cdc dung dich (3) vd (4'i'

Thi nghiem 2: Dung dich (6) cho ket tua vai cdc dung dich (1) vd ^4'

Thi nghiem 3: Dung dich (4) cho khi bay len khi tdc dung vai

dung dich (3) vd (5)

Hay xdc dinh s6 cua cdc dung dich

IL

Viet 5 phuang tri/ih phan icng dieu che CaCOs

Cho 4 kim loai sau: Be, Na, Al, Ca Hoi kim loai ndo khi tqo hap kim vdi Mg thi khi dot chdy hap kim do, oxlt tqo nen c6 khoi lugng gap doi khoi lugng hgp kim ban ddu

c) Dot chdy hodn todn 5,6 lit (dktc) hdn hgp khi gom H2 vd H2S sdn

phdm tqo thdnh la nuac vd mpt chdt khi Lugng khi nay tdc dung viCa

du vdi 100 ml dung dich NaOH 2M de tqo thdnh muoi trung hda

Tinh khoi lugng inol trung binh ciia hon hgp khi ban ddu

Cdu HI Dot chdy 1,76 gam sunfua cua mpt kim loqi MeS (kim loqi Me

trong cdc hgp chat chi cd hai hda' tri 2 vd 3) trong oxi du Hda tan chdt rdn sau phan ilng bdng mpt lugng viia du dung dich H2SO4 29,4% Dung dich tqo nen c6 nong dp 34,5% Lam Iqnh dung dich thi thu dugc 2,9 gam tinh the hidrat Idng xud'ng vd dung dich con Iqi cd nong dp muoi Id 23% Xdc dinh cong thUc cua tinh the hidrat

Cdu rV Cho m gam hon hgp CaCOs vd FeS tdc dung vUa du vdi V ml

dung dich HCl (D ^ 1,1 g/ml) thi thu dugc a lit hon hgp klii X (dktc) c6 khoi lugng mol trung binh Id 40,67 (gam/mol) vd dung dich Y cd khoi ugng la b gam

<.) Tinh a theo m, V, b

b) Ap dung: Cho m = 1,44 gam; V = 400 ml; b = 440,83 gam Tinh a

c) Neu chi sii dung mpt dU kien khoi lugng mol trung binh ciia X la

^^0,67 (gam/mol), hay tinh % theo khoi lugng cua hdn hgp ban ddu

Cdu V Mpt hon hgp A gom Na2C03 vd NaoSOs cho tdc dung vdi dung

(iich HCl du thu dugc hon hgp khi X cd klidi lugng mol trung binh Id

56 (gam/mol) C/io 0,224 tit (dktc) khi X di qua 1 lit dung dich

Ba(0H)2-•Sau thi nghiem phdi diing 50 ml dung dich HCl 0,2M de trung hda

L d l GIAI

Cdu 1

a) Dap so:

- K h o a n g n h i e t do x u a t h i e n k e t t i n h tit 30°C -> 40°C

- Kho'i lijgng muo'i k e t t i n h k h o a n g [24,8 < m„,u6i < 36,7]

b ) Thi nghiem 1: D u n g d i c h (2) cho k e t t i i a v d i dung dich (3) va (4)

=> d u n g d i c h (3) l a H2SO4 va dung d i c h (4) la Na2C03 hoac ngagc l a i

Thi nghiem 2: D u n g dich (6) cho k e t t u a v 6 i dung dich (1) va (4)

=> dung d i c h (1) l a N a O H va dung d i c h (4) l a Na2C03 hoac ngu'cfc l a i

Thi nghiem 3: D u n g dich (4) cho k h i bay l e n v d i dung dich (3) va (5),

Do do d u n g d i c h (4) l a NaaCOs, dung dich (3) la H2SO4 => d u n g dich

(5) p h a i l a H C l , dung d i c h (1) la N a O H , dung d i c h (6) la MgCla, coii

l a i d u n g d i c h (2) la BaCla

Vay: d u n g d i c h (1) l a N a O H ; dung d i c h (2) la BaCl2

d u n g d i c h (3) l a H2SO4, dung d i c h (4) la NaaCOg

(mol)

I ^ h o n hap — = 0,25 (mol)

Vay: M h h =

22,4 nj^^ = 0,25 - 0,1 = 0,15 (mol)

™ 0,1 X 34 + 0,15 X 2

0,25 = 14,8 (g/mol) /// G o i a l a so m o l ctia MeS dem do't chay

2 M e S + - O 2

2 MeaOs + 2 S 0 2 t

2 Me203 + 3H2SO4 Me2(S04)3 + 3H2O

3678

= 56 (Fe) 65,5

Khoi lUctng dung dich luc sau: mdung dkh sau = 11,6 - 2,9 = 8,7 (gam)

Trong (400 + 18x) gam Fe2(S04)3.xH20 thi chufa 400 gam Fe2(S04)3

=> 2,9 gam Fe2(S04)3.xH20 thi chura a gam Fe2(S04)3

Vay khoi lifong Fe2(S04)3 bi t^ch ra:

Khoi lifgng dung dich cua HCl: 1,1.V (gam)

Ap dung dinh luat bao to^n khoi lugrng cho phan ufng (1), (2):

m + 1,1 X V = b + 44x + 34y

<=> 44x + 34y = m + 1,1.V ~ b c:> 34(x + y) + lOx = m + 1,1V - b

a „ , a

22,4 22,4 Theo de bai, ta c6 he phiiOng trinh:

+ lOx = m + 1,1V - b

lOOx + 88y = m 44x + 34y

X + y

lOOx + 88y - m 44x + 34y = 40,67x + 40,67y

= 40,67 100x +

3y ~ m

X = 2y

22,4 34a

144

134 m 22,4 144 '67m

+ 1 , 1 V - b

a = 72

b ) K h i m = 1,44 gam; V = 400 ml; b = 440,83 gam

67 Vay a =

72 X 1,44 + 1,1 X 400 - 440,83

22 4

X ^ =^a = 0,336 (ht)

34 c) Theo cSu a,

88y 200y + 88y xlOO = 30,56%

V

Na2C03 + 2HC1 (mol) a

Vi Ba(0H)2 dif nen tao mudi trung hoa

CO2 + Ba(0H)2 > BaCOai + H2O (mol) a ^ a

SO2 + Ba(0H)2 > BaSOgi + H2O (mol) b -> b

naci = 0,05 x 0,2 = 0,01 (mol)

Ba(0H)2 + 2HC1 > BaCh + 2H2O

(mol) 0,005 <- 0,01

(1) (2)

(3) (4)

a l 0 6

Na2C03 106a + 126b

106a 106a + 126 X I a

So jnol Bă0H)2 dH = ]- só mol cua H C l = 0,005 (mol)

1. Nung nong Cu trong khong khi, sau mot thai gian dugc clidt rdn Ạ

Hda tan A trong H2SO4 dgc, nong dugc dung dich B vd klii C Khi C

tdc dung vai dung dich KOH dugc dung dich D D vica tdc dung dugc

vai BaCl2 vila tdc dung vai NaOH B tdc dung vai (lung dich KOU

Viet cdc phuang trinh phdn iCng xdy rạ

2. Hodn thdnh cdc phuang trinh phdn iCng sau:

a) Cu(N03)2 + ? > CuS + •?

b) Cu + ? > CuCls

3 Cho tU tU tUng mdu natri kirn logi den du vdo dung dicli AICI3 vc

dung dich CuSỘ Hien tugng xdy ra c6 giong nhau khong?

Dan 8 lit hSn hgp khi A d dieu kien tieu chudn gom hidro, etan vd axctilen di qua bgt Ni nung nong thi thu dugc 5 lit chat khi day nhdt

Hoi hdn hgp khi A ngng han hay nhe han khong khi boo nhieu Idn?

2. Dung dich A chUa hdn hgp KOH 0,02M vd BafOHJs 0,005M; dung

' dich B chUa hon hgp HCl 0,05M vd H2SO4 0,05M

a) Tinh the tich dung dich B can de trung hda 1 lit dung dich Ạ

b) Tinh nong do mol cua cdc mud'i trong dung dich thu dugc sau phdn ling, cho rdng the tich dung dich khong thay đị

Cdu 4 Hda tan l,18g hon hgp A gom bgt luu huynh vd bgt nhom trong

375 ml dung dich HCl 0,2M thu dugc 0,672 lit khi do d dieu kien tieu chudn vd dung dich B

a) Xdc dinh nong do mol cdc chdt trong dung dich B

b) Nung nong 3,54 gam cung hon hgp A noi tren d nhiet do cao thich hgp trong binh kin khong c6 oxi cho den khi phdn Ung xong tlii thu dugc chdt rdn C Xdc dinh phdn tram khoi lugng cdc chdt trong chdt rdn C

Cdu 5. A Id hgp chdt hUu ca chUa 2 hogc 3 nguyen to C, H, Ọ Trgn

1,344 lit CH4 vai 2,688 lit khi A dcu d dieu kien tieu chudn, thu dugc

4,56 gam hon hgp khi B Tinh khoi lugng mol cua Ạ

Dot chdy hodn todn hon hgp B, cho sdn phdm chdy hap thu hodn todn

vdo dung dich Bă0H)2 dU thdy tgo thdnh 35,46 gam kct tuạ Xdc dinh

cdng thícc phdn tie vd cong thvtc cdu tgo cua Ạ

(Biet: C = 12, H = 1, O = 16, N = 14, CI = 35,5, Ba = 137, S = 32,

Al =27,K = 39, Ca = 40)

9 LCJl G I A I Cdu 1

1 Nung nong C u trong khong k h i thu diicrc chat r ^ n A la CuO va Cụ

Hoa tan A trong H2SO4 dac, nong dugc dung dich B va khi C

C u O + H2SO4 > CUSO4 + H2O (1)

C u + 2H2SO4 > CUSO4 + SOat + 2H2O (2)

vu

phuang trinh phdn dng vd gidi thich

Cdu 2

a) Cho V lit COo a diSu kien tieu chudn hdp thu hodn todn vdo 200 //^

dung dich chica hdn hgp KOH IM vd CăOHh 0,75M thu dUgc 12 ga»^

ket tuạ Tinh V?

Dung dich B la CUSO4; k h i C la SO2; dung dich D la KHSO3:

SO2 + K O H > K H S O 3

2 K H S O 3 + BaCl2

2 r a i S 0 3 + 2 N a O H CUSO4 + 2 K 0 H —

K2SO3 + BaSOs + 2IIC1 Na2S03 + K2SO3 + 2H2O -> C u ( 0 H )2i + K2SO4

2, H o ^ n t h a n h cac p h u a n g t r i n h p h a n ufng:

a) Cu(N03)2 + NazS > C u S i + 2 N a N 0 3

b) Cu + CI2 > CuCh

3 K h i cho tijt tii tiing m a u n a t r i k i m l o a i d e n dii vko d u n g d i c h AICI3 tlii

t h a y CO k l i i bay l e n , r o i c6 k e t tiia, sau cung k e t t u a t a n (Hoc s i n h t u

v i e t 3 p h i i o n g t r i n h p h a n lifng). C 5 n k h i cho v^o d u n g d i c h CuSO;

t h a y CO k h i bay l e n , r o i c6 k e t tija, k e t t i i a k h o n g t a n (Hoc s i n h t i ;

22,4 g a m n ^ y l a i nho h o n 24 gam chat r a n tao t h a n h sau p h a n ufng la

v6 l i , CO n g h i a l a p h a n ufng xay r a chaa h o a n t o a n , con duf CuO

G o i so' m o l CuO da p h a n ufng vori H2 l a x, t h i s o m o l H2 t h a m g i a

p h a n ufng v a so' m o l Cu, H2O tao t h a n h sau p h a n ufng cung la x; 2-^

g a m c h a t r S n t h u di/gc sau p h a n ufng g o m CuO diJ v a Cu tao t h a n h

1 duy n h a t con l a i la C2H6 Cac p h a n ufng (1) va (2) xay ra hoan

^ n , H2 va C2H2 d i u he't

h o n h o p (A) g i a m la do V^j^ p h a n ufng = 8 - 5 = 3 l i t

„ b a n dau = ,rV„ = 1 , 5 l i t V„ „ b a n dau = 8 - 1,5 - 3 = 3,5 l i t

Do do t i le kho'i l i j g n g h o n h o p A so v d i k h o n g k h i b a n g :

3 , 5 3 0 + 1,5.26 + 3.2 ^ ^ c i - - u i dh6n h?p AJkk = = 0 , 6 5 I a n nhe h o n

2 D u n g d i c h A chufa h'on h o p K O H 0 , 0 2 M va B a ( 0 H )2 0 , 0 0 5 M ; dung dich

B chtia h o n h o p H C l 0 , 0 5 M va H2SO4 0 , 0 5 M a) Trong 1 l i t dung dich A ta c6 tong so mol O H ' = 0,02 + 0,005 x 2 = 0,03

so'

=> So m o l i o n SO^" c5n diT = 0,01 - 0,005 = 0,005 m o l

V i t h e t i c h d u n g d i c h k h o n g t h a y d o i n g h i a la t h e t i c h t o n g cong 1 1,2 l i t m a n , = 0,02 m o l ; n^,_ = 0,01 m o l v a n „_ = 0,005 mo

K CI

N e n n o n g do mol/1 cua m u o i clorua:

[ K C l ] = M l = 0,0083M ; [K2SO4] = = 0,0042M

1,2 1,2

Cdu 4

a) Hoa tan hon hgp A vao dung dich HCl chi c6 Al tan:

2A1 + 6HC1 2AICI3 + SHat 0,672

(1)

va n A i = - n „ = - x 0,03 = 0,02 mol

3 "2 3

So mol HCl tham gia phan ufng (1) = 2nj^ = 0,03 x 2 = 0,06 mol

iiHci ban diu = 0,375 x 0,2 = 0,075 mol

Sau phan ufng con dtr HCl = 0,075 - 0,06 = 0,015 mol; do vay A l taii

het, dung dich B trong binh phan ufng gom HCl dii 0,015 mol va A 1C1;J,

(S khong tan)

Coi V dung dich B thay doi khong dang ke, gan bang the tich dung

dich B ban dau = 0,375 ht

Trong 1,18 gam hon hgp A c6 0,02 mol Al hay 0,54 gam A l :

So g S con lai la: 1,18 - 0,54 = 0,64 gam hay ng = 0,02 mol

Nhii vay, t i le so mol Al va S la ^' = 1-: 1; ma 3,54 gam h6n hgp A

Nhu vay S phan (ing het, con Al d\J 0,02 mol;

phan ufng tao thanh 0,02 mol AI2S3

5 Khoi lugng 1,344 lit CH4 (dktc): ^^22 4^^^ "

Khoi lugng cua 2,688 lit khi A: 4,56 - 0,96 = 3,6 gam

3,6.22,4 Vay khoi lugng mol ciia A:

Dung dich Ba(0H)2 hap thu CO2 c6 the xay ra:

Ba(0H)2 + 2 C O 2 > Ba(HC03)2

Ba(0H)2 + CO2 - > B a C 0 3 i + IhO Theo (4) CO = 0,18 mol BaCOgi difgc tao ra do 0,18 mol CO2;

197

f so mol Ba(0H)2 con dii la 0,24 - 0,18 = 0,06 mol, dieu do chufng to

phan ufng (3) khong xay ra 0,18 mol CO2 la do (1) va (2) tao ra; (1)

1,344 chi tao

2 688 0,18 - 0,06 = 0,12 la do = 0,12 mol C J i y O , chay tao ra

Do vay, ta c6 the ket luan phan ttf C^HyO^ chi c6 1 nguyen tuf C

Mat Uiac, phan tuf khoi ciia CJIyO, = (4,56 - 0,96) 22,4

2,688 = 30 (gam) Nen cong thufc phan ttf CxHyO^ la CH2O, c6 cong thufc cau tao la: HCHO

OE SO 14

T H I C H Q N H O C S I N H G I O I H O A H O C 9 , Q O A N 1 , T P H C M N A M H O C 2 0 0 2 - 2 0 0 3

o>) Hay chgii 6 chat rdn khdc iihau de khi clio moi chat do tdc dung uoi

dung dich HCl ta thu duac 6 chat khi khdc nhau Viet cdc pliUang trinh phan ling xay ra

^) Co 4 dng nghiem, moi dug chiia mot dung dich mudi (khong triing kini loai cung nhu gdc axit) clorua, sunfat, nitrat, cacbonat cua cdc kini loai Ba, Mg, K, Pb

^oi dung dich mudi ndo da chiia trong 4 dng nghiem trcn

" phuang phdp de phan biet 4 dng nghiem do

« 3

Cdu II

a) Viet pliUang trliih phdn iciig de bieu dien chudi phdn I'cng sau:

^ SO3 > H2SO4 \

FeSs > SO2 ^ > SO,,

b) Neu hien tiigng xdy ra vd uie't phuang trinh phdn ling

- Sue khi CO2 vdo dung dich nUac voi trong

- Cho til tit dung dich HCl vdo dung dich Na2C03

Cdu HI Trgn 100 ml dung dich Fe2(S04)3 1,5M vai 150 ml dung die],

Ba(0H)2 2M thu duge ket tila A vd dung dich B Nung ket tua A troniy

khong khi den khoi lugng khong doi thu dUgc chat rdn D Them BaCl,

du vdo dung dich B thi tdch ra ket tua E

a) Viet phuang trinh phdn ling Tinh khoi lugng D vd E

b) Tinh khoi lugng mol chat tan trong dung dich B (coi the tich thay dd'l

khong dang ke)

Cdu IV

a) Hoa tan mot lugng muoi cacbonat cua mot kim logi hoa tri II bdng

dung dich H2SO4 14,7% Sau khi phdn iCng ket thuc khi khong con

thodt ra nUa, thi con Igi dung dich 17% muoi sunfat tan Xdc dink

khoi lugng nguyen tii cua kim logi

b) Lay 40 gam dung dich bdo hoa cua FeCl2 them vdo 10 gam muoi

FeCl2 khan Dun nong de hba tan het Khi de ngugi den nhiet do ban

ddu thi Idng xuong 24,3 gam tinh thi muoi hidrat Xdc dinh cdiig

thiic cua tinh the hidrat, biet rdng dung dich bdo hoa chiia 38.''

muoi khan

Cdu V

1 Cho 100 gam dung dich Na2C03 16,96%, tdc dung vai 200 gam dui'S

dich BaCh 10,4% Sau phdn ling, Igc bo ket tua dugc dung dich ^

Tinh nong do % cac chat tan trong dung dich A

2 Tii 9,8 gam H2SO4 c6 the dieu che dUgc:

a) 1,12 lit SO2 (dktc) khi cho tdc dung vai kim logi

b) 2,24 lit SO2 (dktc) khi cho tdc dung vai muoi

Viet vd can bdng cdc phUang trinh phdn iCng xdy ra

BaS + 2HC1 —

> 2 N a C l + COat + H2O -> BaCla + H2ST

Mn02 + 4HC1 CaSOs + 2HC1 CaC2 + 2HC1 -

— - > MnCl2 + CI2 + 2H2O

— > CaCl2 + S02t + H2O CaCl2 + C2H2t

b) Cac dung dich muoi chufa trong 4 ong nghiem phai la muoi tan nen

ta c6: BaCl2, MgS04, Pb(N03)2, K2CO3

• Nhan bie't 4 ong nghiem tren:

Trich moi dung dich mot i t lam mau thuf

Cho dung dich HCl Ian li/cft vao 4 mau thi^ tren:

- Mku thijf nao cho ket tiia trSng la Pb(N03)2:

Pb(N03)2 + 2HC1 > PbClsi + 2HNO3

- Mau thtf nao c6 hien tLTomg sui bpt k h i la K2CO3:

K2CO3 + 2HCI > 2KC1 + CO2T + H2O

f Cho dung dich NaOH vao hai mSu con lai:

- Mau iiao cho ket tua trSng la MgS04:

MgS04 + 2NaOH > MgCOIDai + Na2S04 C6n la i la BaClz

SO2 + NaOH 2NaHS03 •

> C U S O 4 + SO2T + 2H2O) -> NaHSOg

b ) Hien tiJcrng

Khi sue khi CO2 vao dung dich nUdc voi trong thi lam due nUdc voi troiijj

CO2 + Ca(OH)2 > CaCOs^ + H2O

Neu sue tiep CO2 t h i ket tua tan dan v^ dung dich trong suot:

CO2 + H2O + CaCOa > C a ( H C 0 3 ) 2

K h i nho tCr ti^ dung dich HCl vao dung dich Na2C03, luc dau chu;,

CO h i ^ n tUcmg gi v i :

HCl + NaaCOa > NaHCOg + NaCl Khi het Na2C03 ma vSn nho tiep H C l thi c6 hien tLfcfng siii hot khi \

H C l + NaHCOg > NaCl + C 0 2 t + H2O

Liiu y: Neu cdc em cho ngUgc lai, ticc Id cho Na^COs vao HCl tin c6 hi 1,

tugng sui bgt khi ngay tiic khdc

Theo de: dung dich (B) chufa Fe2(S04)3 di/

Fe2(S04)3 + SBaCla > 2FeCl3 + 3BaS04i Ci'

(mol) 0,05 -> 0,15

So mol cua Fe2(S04)3 = 0,15 - 0,1 = 0,05 (mol)

T i n h khoi lifcfng D: Theo (1) va (2), ta c6:

mo = mg^so^ + mp^^o^ = 0,3 x 233 + 0,1 x 160 = 85,9 (gam'

T i n h khoi liXcfng E: Theo (3), ta c6:

niE = me^so^ = 0,15 x 233 = 34,95 (gam)

b ) Tinh nong do mol/1 cua Fe2(S04)3

The tich dung dich thu dugc la: 0,1 + 0,15 = 0,25 (lit)

Nong do mol cua Fe2(S04)3 d^r: ~ J = 0,2M

" ifli m i l O F THI Hnn SINH GIQI HOA HOC 8

IV

- a) Goi k i m loai hoa t r i 2 la: M Phan ufng: MCO3 + H2SO4 (mol) 0,15 <- 0,15 ^

Cong thile muo'i caebonat: MCO3

MSO4 + CO2T + H2O

b ) Goi cong thufc ciia muo'i tinh the hidrat: FeCl2.xH20 Khoi liJOng FeCl2 k h a n trong dung dich bao hoa:

38,5 X 40

Sau k h i them lOg FeCl2 vao dung dich bao hoa t h i :

m d u n g d i c h = 10 + 15,4 = 25,4 (gam)

trong do m^.o = 50 - 25,4 = 24,6 (gam)

Do t a n ciia FeCl2 of n h i e t dp b a n dau: 15,4

Vay cong thufc tinh the hidrat: FeCl2.4H20

V i _ J _ J L = 0,16 > = 0,1 =i> sau p h a n ufng NaaCOa con di^

Do do, dung dich A gom: N a C l tao t h a n h va Na2C03 du

Na.COsdu 280,3

2 T a c6: n„^,„^ = H = ^'^ ^""^^^

a) De t h u diicfc 1,12 l i t SO2 t h i n^.^^ = ^ n „ => K i m loai do l a Cu

Cu + 2H2SO4 dung dich CUSO4 + SO2T + 2H2O (mol) 0,1 0,05

^ Vyo, = 0,05 X 22,4 = 1,12 ( l i t )

b) De t h u di/gc 22,4 l i t SO2 t h i ngg^ = n^^g^^ => muoi do 1&: NaaSOg

Na2S03 + H2SO4 > Na2S04 + SOgt + H2O (mol) 0 , 1 - > 0 , 1

^ Vyo,= 0,1 X 22,4 = 2,24 ( l i t )

Luu y: Cdc em c6 the dung chat klidc nliU cacboii (C)

Ldl GIAI BE THI HOC SINH GIOI HOA HnC 9

DE SO 15

OE THI HOC SINH GIOI HOA HOC LifP 9 (VONG 2), QUAN THU flUfC, TP HCM

NAM HOC 2002 - 2003

C&u I- Hodii thien chuoi bien hoa sau day:

Fe < = ± Fe2(S04)3 > FeS04 > FeCNOsJs > Fe(0H)2

> Fe(0H )3 > FesOs > FC3O4 , Fc Cdu 11

a) Co 7 lo mat nhdn chi'ia rieng biet tiing dung dich sau: BaClz, NaOH, NaCl, NaoCOa, H2SO4, NH4CI vd AySOJs- Chi dugc dung quy tim,

hay nhdn biet tiing Ig

b) Phan biet hai dung dich KOH vd Al2(S04)3 md khong diing them hoa

chat ndo khdc

Cdu I I I Hoa tan 15 gam tinh the sdt sunfat nggm nUac FeS04.7H20 vdo

nUac, them dan dung dich NaOH cho den diC, roi dun nong trong khong khi, Igc ket tila tgo thdnh, rvCa sgch, say kho vd nung a nhiet do cao San pJidm thu dugc can ngng 4 gam Hoi muoi sunfat sdt nay c6 tinli khiet khong'?

Cdu IV

a) De sdn xudt 1 tan voi chiia 85% CaO, nguai ta phdi tieu thu bao nhieu

kg dd voi clii'ia 94% canxi cacbonat Biet hieu sudt phan ling Id 85%

b) Diet oleum c6 cong thilc la H2S04.nS03 Hoa tan 6,76 gam oleum ndy vdo nude thdnh 200 ml dung dich H2SO4 Biet rdng 10 ml dung dich

ndy trung hoa vica het 16 ml dung dich NaOH 0,5M Xdc dinh n?

Cdu V Cho 5,4 gam hdn hap gom hai kim loai (diing trUac H trong ddy

Beketop) Idn lugt c6 hoa tri II vd III tdc dung vai dung dich H2SO4

lodng sinh ra 0,45 mol khi Biet rdng nguyen tH khdi cila kim loai dan nho han ba Idn so vai nguyen tH khdi cua kim loai sau Ti le so mol trong hon hgp Id 3 : 1

^dc dinh ten cdc kim loai trong hdn hgp

Cdu VI Nu?ig nil gam Cu trong m2 gam O2 thu dugc sdn phdm Aj Dun nong Aj trong ms gam dung dich H2SO4 98% khi tan h^t dugc dung dich

vd khi A3 Todn bg khi A3 dugc hap thu bdi 200 ml dung dich NaOH

0,15M tgo ra 2,3 gam muoi Dem c6 can dung dich Ao thu dugc 30 gam

USO4.5H2O

Neu cho Ao tdc dung vai NaOH, phdi dung it nhdt 300 ml dung dicli

NaOH IM mai tao dugc lugng ket tua toi da Cho lugng ket tua nay tan

trong dung dich HCl viCa du Sau do nhung 1 thanh sdt vdo dung dich

thu dugc, sau phdn ling thdy khoi lugng thanh sdt tang 0,8 gam

a) Tinh mj, m2, ni3?

b) Tinh khoi lugng sdt dd tan vdo dung dich?

c) Tinh khoi lugng muoi c6 trong dung dich sau khi nhung thanh sdt?

Cdu VII Trgn 120 ml dung dich H2SO4 vai 40 ml dung dich NaOH Dung

dich sau khi trgn chUa mot muoi axit vd con du H2SO4 c6 nong do 0,1M

Mat khdc, neu trgn 40 ml dung dich H2SO4 vai 60 ml dung dicli NaOll

ndy thi trong dung dich sau khi trgn con dU NaOH c6 nong do 0,16M

Xdc dinh nong do moll I cua hai dung dich H2SO4 vd NaOH ban ddu

Biet: Fe = 56, S = 32, O = 16, H = 1, Na = 23, C = 12, Ca = 40, Mg = 24,

Cu = 64, CI = 35,5

LCil GIAI

Cdu 1 2 F e + 6H2SO4 dac — > Fe2(S04)3 + SSOgt + 6H2O

Fe2(S04)3 + 2A1 > Al2(S04)3 + 2F e

SFeaOg + C O —> ^Fe^Oi + C O a t Fe304 + 4 C 0 —> 3 F e + 4 C 0 2 t Cdu II

a ) T r i c h m6i lo mot it l a m mau thtf

Cho quy t i m Ian liiot vao cac mau thuf tren:

- Mau thijf l a m quy tim hoa xanh l a : N a O H

- Mau thtf l a m quy tim hoa do l a : H2SO4

Dung N a O H l ^ m thuoc thuf cho Ian lifot v^o cac mau thif con

lai, dun nong nhe

- M a u thijf c6 k h i mui khai bay r a \k: N H 4 C I

N H 4 C I + N a O H — > N a C l + N H 3 T + H2O

- Mau thCf cho ket tua keo trSng, sau d6 bi tan dan den het Ik Al2(S04)3:

Al2 (S04)3 + 6 N a O H > 2 A l ( O H ) 3 i + 3Na2S04 A1(0H)3 + N a O H > NaAlOa + 2H2O

Dung H2SO4 lam thuoc thur va Ian liigt cho vao cac mSu thuf cbn lai

- Mau thijf CO ket tiia trang khong tan l a BaCl2:

H2SO4 + BaCla > B a S 0 4 i + 2HC1

- M a u thtf CO h i e n tLfcfng siii bot k h i l a Na2C03:

H2SO4 + NaaCOg > Na2S04 + C 0 2 t + H2O

- Mau thtf khong c6 hien tiXOng l a N a C l

15

278

(mol)

278 -> F e ( 0 H ) 2 i + Na2S04

TU (1), (2), (3): So mol Fe203thu di/oc: — ( m o l )

K h o i liiong Fe203 thu diicfc: 15

T r o n g 10 m l dung dich H2SO4 chuTa 0,004 m o l H2SO4

200 m l dung dich H2SO4 chufa x m o l H2SO4

Cdu V Goi hai k i m loai can t i m la X, Y c6 so' mol Ian I j o t la 3a va a;

Kim loai X c6 hoa t r i I I

Cdu VI

a) Phan ufng:

2Cu +O2 > 2CuO

Cudu + 2H2SO4 dac CuO + H2SO4 dac -

SO2 + 2NaOH — (mol) a 2a

SO2 + NaOH (mol) b -> b

-> CUSO4 + SOot + 2H2O

> CUSO4 + H2O -> NaaSOa + H2O

(1) (2) (3) (4) (5) (6)

H2SO4 + 2 N a O H > Na2S04 + 2H2O (7)

Goi a la so mol SO2 tham gia phan ilng (4)

b la so' mol SO2 tham gia phan t f n g (5)

So mol SO2 tao thanh trong phan Ofng (2): 0,02 (mol)

=> So mol CUSO4 tao thanh trong phan ufng (2): 0,02 (mol)

=> So mol CUSO4 tao thanh trong phan ung (3): 0,12 - 0,02 = 0,1 (mol)

=> So mol CuO tao thanh trong phan ufng (1): 0,1 (mol)

=> So mol dong dif tham gia phan ufng (2): 0,02 (mol)

=> So mol dong tham gia phan ufng (1): 0,1 (mol)

Vay so mol Cu ban dau: 0,1 + 0,02 = 0,12 (mol)

=> mi = mcu = 0,12 X 64 = 7,68 (gam) So' mol O2 tham gia phan ufng (1):

" 0 , = ^ n ^ u 0 = ^ x 0.1 = 0.05 (mol)

Vay: nig = m^^ = 0,05 x 32 = 1,6 (gam)

So mol H2SO4 tham gia trong phan ufng (2) v^ (3):

" =i> c = 0,03 (mol)

Vay so mol H2S04ban dau la: 0,14 + 0,03 = 0,17 (mol)

Khoi l i / g n g H2SO4: 0,17 x 98 - 16,66 (gam)

Ap dung: C% = x 100

m ^ j X l O O 16,66 x100 , ,

^ m , , = m„ = = — = 17 (gam)

ddHjSO^ 3 C% 98

T a c6: n^^^^H),^,,,^, = "cuso, = 0.12 (mol)

So mol CuCl2 tham gia trong phan ijfng (9): 0,1 (mol)

V a y so mol CuCl2 con diT sau phan ufng (9): 0,12 - 0,1 = 0,02 (mol)

==> m^^ci, = 0.02 X 135 = 2,7 (gam)

So mol FeCl2 tao thanh trong phan ufng (9): 0,1 (mol)

mp ^ci, = 127 X 0,1 = 12,7 ( g a m ) Suy r a : m„u,6, = 12,7 + 2,7 = 15,4 (gam)

Cdu VII

Goi C i la nong do cua H2SO4

C2 la nong do ciia NaOH

T r o n I a n 1: H2SO4 + N a O H > N a H S 0 4 + H2O ('

(mol) 0,04C2 <- 0,04C2

Theo de bai, ta c6 phifang trinh:

= 0,1 « 0,12c, - 0,04C = 0,016 0,16

(;:

T r 6 n l a n 2 : H2SO4 + 2NaOH >• Na2S04 + 2H2O (^

(mol) 0,04Ci ^ 0,08Ci

Theo de ta c6 phiiong trinh:

0,06C^ - 0 , 0 8 C j

- 0,16 0,1

Cdul-trinh lap 8, ghi du dieu kien phan ling (neu cd)

b) Bdng phuang phdp hoa hoc, em hay trinh bay each tdch rieng Fe203

trong hon hap c6 chiia Fe203 vd CaO

Cdu II- L^y "^9^ ^'^n hap gom 6,9 gam natri vd 6,2 gam natrioxit vdo 500

ml niCac tao thanh dung dich A Can lay bao nhieu gam NaOlI (c6 lun 20% tap chat) cho vdo dung dich A de dugc dung dich B cd nong do Id 2M Biet the tich thay doi khong dang ke trong qua trinh thuc hien tren

Cdu III

a) Can lay bao nhieu gam CUSO4 hoa tan vdo 400 ml dung dich CuSO^

10% (D - 1,1 g/ml) de tao thdnh dung dich C cd nong do Id 20,8%

b) Khi ha nhiet do dung dich C xudng 12°C thi thdy c6 60 gam muoi CUSO4.5H2O ket tinh, tdch ra khoi dung dich Tinh do tan cua CuSOj

d 12°C (Dugc phep sai so nhd han 0,1%)

Cdu IV Cho 7,73 gam hon hap gom kem vd sdt cd ti Ic nzn • np^ -5:8 vdo

dung dich HCl du ta thu dUgc V lit khi H2 (dktc) Dan todn bg lugng khi

H2 ndy qua hon hgp E (gom Fe203 chiem 48%; CuO chiem 32%, tap chat

chiem 20%) c6 nung nong

a) Tinh V

b) Tinh khoi lugng hon hgp E vUa du de phan Ung hodn todn vdi V lit

khi H2 ndi tren Biet rdng tap chat khong tham gia phan I'Cng

Cdu V KhvC hodn todn 4,64 gam mot oxit kim logi thi can 1,792 lit khi CO

(dktc) Neu lay todn bg lUgng kim logi thu diigc d tren cho vdo dung dich

HCl du thi thu dugc 1,344 lit khi H2 (dktc) Xdc dinh cong thi'Cc hoa hoc

cua oxit ndi tren

Cdu VI " •

a> Tinh nong do phan trdm cua dung dich H2SO4 6,95M (D = 1,39 g/ml)

^) Trong mot binh kin, ngUdi ta thUc Men mot phan vCng hoa hoc theo

phuang trinh sau: 3A + 2B — > C + 2D

Trong do A, B, C, D Id cdc hgp chat hoa hoc Tong so mol cac chat ban

ddu la 1,5 mol ThUc hien phan Ung dUgc 5 phut ngUdi ta dUng Igi thi

luc do tong so mol cdc chat trong binh la 1 mol Hoi luc diing phan iing thi so phan tii cua moi logi hgp chat C vd D thu dugc Id bao nhieu?

Cho: Na = 23; O = 16; H = 1; Cu = 64; S = 32; Ca = 40; Fe = 56; Zn = 65; CI = 35,5

H2O dien phan H a t + ^ Oat

b) Cho t o a n bo h o n hop vao niJdc d i i va k h u a y deu, CaO t a n h e t trong

H2O, FeaOs k h o n g t a n t a dun g pheu de loc FeaOs

CaO + H2O > Ca(0H)2 Cau 11 T a c6: n 6,9

NaaO + H2O

-> N a O H + i H a t 0,3

> 2 N a O H 0,2

(1)

(2) (mol) 0,1

So' m o l cua dung dich (B): 0,5 x 2 = 1 (mol)

TO p h a n ufng (1), (2): n a ^ n g d i c h A = 0,3 + 0,2 = 0,5 (mol)

Suy r a , so' m o l ciaa N a O H cho vao dung d i c h (A) l a :

1 - 0,5 = 0,5 (mol)

V a y k h o i liicfng N a O H (chila 2 0 % tap chat) can t h e m vao dung dich

(A) la: 0,5 X 40 X 1^ = 25 (gam)

80

Cau III

a ) Goi a l a kho'i l i i g n g CUSO4 can t i m

K h o i iLTgng dung dich CUSO4 10%: 400 x 1,1 = 440 (gam)

K h o i li/grng dung dich CUSO4 20,8%: a + 440 (gam)

K h i pha t r o n t a c6 phifong t r i n h ciia chat t a n :

440 x 10 ( a + 440) 20,8

a +

100 100 100a + 4400 = 20,8a + 9152 79,2a = 4752 => a = 60 (gam)

b) K h i ha n h i i i do dung dich C xuong 12°C t h i :

Z n + 2HC1 (mol) 0,05

Fe + 2HC1 (mol) 0,08

-> ZnCl2 + H 2 t

0,05 FeCl2 + H a t

0,08 Ti^ (1), (2): V „ = (0,05 + 0,08) x 22,4 = 2,912 ( l i t )

"2 (dktc)

b) T i n h k h o i li/cfng h o n hop E (FezOg va CuO)

FeaOs 0,003m CuO •

-3Ha + (mol) 0,009m <

Ha + (mol) 0,004m ^ 0,004m

G o i k h o i liJcfng h o n hop E l a m (gam)

m , Theo de: %m„„ „ =

Cdu V G o i oxit c a n t i m l a M x O y c6 a m o l

M ^ O y + yCO > x M + yCOat (mol) a - » ay ax

2 M + 2 n H C l > 2MC1„ + nH2t

ax 0,5anx (mol)

C o n g thijfc oxit: Fe304

So m o l chat dii sau p h a n ijfng: 1,5 - 5x (mol)

T o n g so m o l chat t a o t h a n h sau p h a n ufng:

OE THI HOC SINK GIQI HQA HOC 9 , TP HO CHI MINH M A M HOC Z 0 0 3 - 2 0 Q 4

^ pHAN BAT BUOC

CduJ-d) Trong phong thi nghiem thuang dieu che CO2 bdng each cho HCl tdc dung vai CaCOs, do do CO2 bi Idn mot it khi hidroelorua vd hai nude

Ldm the ndo de ed CO2 hodn todn tinh khiet?

b) Bdng phuang phdp hda hoe, hay tdeh SO2 ra kiwi hon Iigp gom SO2, SO3, O2

c) Co 5 lo dung dich dugc ddnh so ngdu nhien tie 1 den 5 gom: Na2S0j,

CalNOs)?, Al (N03)3, NaOH, BaCl2 Thue hien cdc thi ngliiem vd dugc

ket qua sau:

Thi nghiem 1: Dung dich (4) tdc dung vai dung dich (3) cho ket tiia trdng

Thi nghiem 2: Dung dich (2) tdc dung vai dung dich (1) cho ket tila trdng vd bi hda tan khi nhd du dung dich (2)

Thi nghiem 3: Dung dich (4) tdc dung vai dung dich (5) khong cd ket tua ngay Hoi Ig ndo chiia chat ndo'?

Cdu 11

a) Viet phuang trlnh phan iing de bieu dien

sa do bien hda sau:

b) Trong phong thi ngliiem ngudi ta thudng diing cdc hda clidt la H2SO4 dgc, CaO de

ldm kho cdc chat khi Hoi plidi diing chat ndo de ldm kho cdc khi dm sau day: SO2, CO2, O) Hay gidi thich sU lua cJign dd

c) Cho 45,9 gam BaO tan hodn todn trong nUac thu dUgc dung dich A

Mat khde nguai ta lay 36,8 gam hSn hgp gom CaCOs, MgCOs tdc

dung vdi mot lugng vila du dung dich HCl thu dugc khi B Hoi neu cho khi B hap thu hodn todn vao dung dicJi A thi cd ket tua khong?

Vi sao?

^ a u / / / Cho 39,09 gam hon hgp X gom K2CO3, KHCO3 vd KCl tdc dung

^oi V ml dung dich HCl (du) 10,52% (d = 1,05 g/nd) thu dugc dung dich

Y vd 6,72 lit khi COo (dktc) Chia Y ldm 2 phan bdng nhau

~ Phan 1: De trung hda dung dich thi can 250 ml dung dich NaOH

0,4M Sau dd c6 can dung dich, thu dUgc m gam muoi khan

~ Phan 2: Cho tdc dung vdi mot lugiig du dung dich AgNOs thu dugc

51,66 gam ket tua

Tinh khd'i lugng cdc chat trong hon hgp X

^> Tinh V vd m

N a O H N a O I I

B P H A N T U C H Q N Thi sinh chgn mgt trong hai cau sau day:

Cdu I (Chuang trliih THCS iiiai, thi diem tai Qudn 3)

a) Hay cho biet gid iri cua pH (> 7, = 7, < 7) trong cdc dung dich sau:

- Nuac tilth khiet de ngodi khong khi (CO2 trong khong khi hoa tun

vdo nuac)

- Nuac tinh khiet

- Nuac v6i

- Gidin

b) Thiic hien 2 thi nghiem sau:

Thi nghiem 1: Cho dinh sdt vdo dung dich CUSO4

Thi nghiem 2: Cho nigt day dong vdo dng nghiem dUng dun:;

dich AgNOs

- Cho biet hien tugng xay ra a hai thi nghiem tren va gidi thich

- Tie ket qud thi nghiem, hay sdp xep theo chieu do hogt dgng hue

hoc tang dan cua cdc kim logi neu tren

c) Cho mgt lion hgp khi thai gom HCl, H2S, COo, SO2, c6 the dung chu:

ndo sau day de logi bo chiing la tot nhdt?

1 Nuac voi trong 2 Dung dich HCl

3 Dung dich NaCl 4 Nuac

Gidi thich vd viet cdc phuang trlnh phan Ung neu c6

b) Xdc dinh lugng MSO4.7H2O tdch ra khi Idm Ignh 800 gam dung dich

MSO4 bao hba a 70°C xuong 20°C Cho biet do tan cua MSO4 d 70''(

A sau do cho qua lo chufa dung dich AgNOs cuoi cung t a t h u difgc CO2

k h i 0 „

dd H2SO4

SO2 t i n h khiet

c) T^ thi nghiem 2: V i dung dich (1) t a n k h i cho d u dung dich (2) n e n

dung dich (1) la A1(N03)3 va dung dich (2) la N a O H

Tit thi nghiem 1: D u n g dich (4) c6 the la BaCl2 hoac Na2S04 va dung

dich (3) cung c6 the la BaCl2 hoac Na2S04

Tit thi nghiem 3: Dung dich (4) la Na2S04 va dung dich (5) la Ca(N03)2

Vay dung dich 1: A1(N03)3; dung dich 2: N a O H ; dung dich 3: BaCla

dung dich 4: Na2S04; dung dich 5: Ca(N03)2

Cdu I I

a) V i e t phirang t r i n h p h a n ufng:

1000°c

(1) CaCOg (2) CO2 + 2 N a O H

—> CaCOai + NagCOg + 2H2O -> CaCOsi + 2 N a N 0 3

(6) 2 N a H C 0 3 hoac NaHCOa + N a O H

> NaaCOg + C02t + H2O

> Na2C03 + H2O (7) Na2C03 + CO2 + H2O -> 2 N a H C 0 3

Lfll

^) - H2SO4 dac CO tac dung hap t h u ni/6c r a t tot va k h o n g c6 p h a n ijfng xay r a v d i cac k h i : SO2, CO2, O2 nen dung H2SO4 l ^ m klio cac k h i do

- CaO cung c6 k h a n a n g hap t h u niidc n h i f n g v i c6 p h a n ijfng v6i

SO2, CO2 n e n k h o n g d u n g CaO de l ^ m kho h a i k h i do m a chi dung

de l a m k h o k h i O2 t h o i

c) T a c6: n s a o = 4 5 , 9

153

BaO + H2O (mol) 0,3

= 0,3 (mol)

Ba(0H)2

0,3

V a y t r o n g d u n g d i c h A chuTa 0,3 m o l Ba^* v a 0,6 m o l O H "

V i p h a n tfng h o a t a n 36,8 g a m h o n h g p CaCOa, MgCOg troug

H C l vLTa d u n e n t a goi cong thufc tiicfng (ing c u a h a i m u o i : MCO^ vui

tija, tufc la k h i sue k h i CO2 vao dung dich A t h i chi tao muoi axit

Cdc phan ling: Hoc sink tU uiet

Cdu I I I G o i X , y, z I a n l u a t l a so m o l cua K2CO3, KHCO3 v a K C l

K C l + A g N 0 3 2x + y + z

b ) Thi nghicm 1: D u n g dich iU mau x a n h chuyen sang k h o n g mau

G i a i t h i c h : Fe + CuSp4 > FeS04 + Cu

(mau xanh) ( k h o n g m a u )

Thi nghiom 2: D u n g dich i\i k h o n g m a u chuyen sang m a u x a n h

Cho tiep xuc ciia day dong v6i dung dich AgN03 t h i c6 Idp bac bam vao

G i a i t h i c h : Cu + 2AgN03 > Cu(N03)2 + 2Ag>l

( k h o n g mau) (mau xanh) +) K h a n a n g boat dong cua cac k i m loai du'gc sap x e p theo chieu

t a n g d a n : A g < Cu < Fe