Boi duong hoc sinh gioi hoa hoc 8+9 dao huu vinh p2

Co the lay diTcJng tif dung dich dirdng trong riTdu bang each chifng ca't de lay rifcJu, diTcJng khong bay hdi do do ta thu diTdc nhffng tinh the dirdng tr^ng.. Nhiing mot thanh kim loai

V i theo cdc phan iJng ( 1 , 2, 3, 4) so mol HCl = so mol K O H nen khi cho het

D vao E dung dich trd thanh trung tinh va quy chuyen trd l a i mau tim

B a i 2 Cho cac k i m loai M g , A l , Fe Ian liTdt tdc dung v d i cac dung dich HCl,

N a O H , CUSO4 A g N O j Viet cac phiTdng trinh phan ufng xay ra

-> MgS04 + Cui

-> Mg(N03)2 + 2 A g i

> 2AICI3 + 3H2T 2A1 + 2NaOH + 2H2O > 2NaA102 + S H j t

FeCl2 + H2t -> khong phan ufng

-> A1(N03)3 + N 2 0 t +H2O

4FeS2+1102

2 y A l + 3Fe,0y Fe^Oy + (y - x ) C O 8A1 + 3OHNO3

HNO3 loang (dir CO k h i duy nhat N O bay ra) Viet cdc PTPU xay ra

B a i giai

xFeO + (y - x)C02 ;3nt! = i i.nq -; ^

8A1(N03)3 + 3 N 2 0 t + I5H2O -> 4ZnS04 + H2ST +4H2O

Cdc phan tfug:

Tdc dung vdi H2:

Fe304 + 4H2 CuO + H2

'UO •ad:

3Fe + 4H2O -> Cu + H2O Chat ran B gom Fe, Cu, AI2O3, MgO:

- T^c dung cua B vdi dung dich NaOH va vcli CO2

AI2O3 + 2NaOH > 2NaA102 + H2O CO2 + 2H2O + NaA102

Tic dung vdi dung dich HNO3:

Tim cong thuTc cua cac cha't vCng vdi cac chat A,, B, Vie't cdc phifdng trin},

phan tfng theo sd do do

2 Cho sd do bien hoa:

CaC03

^ A, +x ^ A2 +Y

B, +z +T > B

CaCO.,

T i m cong ihtfc cua cac cha't uTng v d i cac chaft A i , B, Vie't cac phiTdng trinh

phan ufng theo sd do

B a C O j i + 2 N a O H -> F e ( 0 H ) 3 ; + 3 N a N 0 3 (6)

(1) (2) (3)

Co the silr dung cac phan iJng khac, chat khac

B a i 6 Tuf cac nguyen lieu ban dau la khoang vat pirit, muoi an, nU'dc, cac chai

xuc tac va thiet bi can thiet, vie't cac phU'dng trinh phan tfng dieu che' Fe

FeClj, F e ( 0 H ) 3 , NaHS04

B a i giai Cac phan iJng can thiet

4FeS, + 1 1 0 ,

2SO2 + O2 V2O5

2Fc203 + 8 S 0 2 t -> 2SO3

(1) (2)

Fe + 2 H C I > FeClj + H21 (7)

2Fe + 3CI2 > 2FeCl3 (8)

(hoac 2FeCl2 + CI2 > IPeCh)

FeCl3 + 3 N a O H > F e ( 0 H ) 3 i + 3 N a C l (9)

NaCl,,i„, + H2S04dac — ^ NaHS04 + H C l t (10)

Co the siir dung cac sd do khac

Bai 7 T i m cac chat X , , X2, X 3 ihich hdp va hoan thanh cac PTPlJ sau:

XFC2O3 + ( 3 x - 2 y ) C 0 > 2Fc,0y + ( 3 x - 2y)C02

BaS04i + Na2S04 + 2C02t + 2H2O

—> BaS04i +Na2S04 + C02t + H2O

—> AgzOJ' + 2KNO3 + H2O

3Ca(H2P04)2

-> Fe2(S04)3 + 3 S 0 2 t +6H2O ->Fe2(S04)3 + S 0 2 t +4H2O 3Fe2(S04)3 + S 0 2 t + IOH2O

-> Fe2(S04)3 + S02T +6H2O

Fe2(S04)3 + 9S02T + IOH2O Fe2(S04)3 + S 0 2 t +4H2O -> Ag2S04 + SO2T + 2H2O

-> 2BaC03i +2H2O BaCOai +CaC03i + 2H2O

-> Fe(N03)2 + Cu4'

Chu de 3 NhSn bi^'t - T a c h hon h<jrp - Tinh chc cac chlft

Bai 1. Chi diTdc dung kim loai c6 the phan biet diTdc cac dung dich sau day hay

khong: NaCl, HCl, NaN03

Bai giai

Tri/dc het cho vao moi dung dich mot mieng kem (hoSc s^t, magie), ndi nao

CO khi thoat ra la HCl:

Zn + 2HC1 > ZnCl2 + H2t

Sau do trpn dung djch HCl vdi dung dich NaCl dtfdc dung dich A, va trgn

dung dich HCl vdi dung dich NaN03 dUdc dung djch B

Cho hai mieng Cu vao 2 dung dich A va B, ndi nao c6 khi khong mau thoai

ra, ho^ nau trong khong khi la hon hdp HCl va NaNOs Cac phan iJng xay ra:

3Cu + 8HC1 + 8NaN03 > 3Cu(N03)2 + 2N0 T + 8NaCl + 4H2O

Hoac 3Cu + 8HC1 + 2NaN03 > SCuCls + 2N0 T + 2NaCl + 4H2O

2NO + O2 > 2NO2

Bai 2 Co 5 dung dich CUSO4, BaClz, NaHS04, KOH, NaCl Cho cac dung djcli

d6 tac dung vdi nhau tirng doi mot Viet cac PTPLf xay ra Tren cd sd ca^

ph^n i?ng d6 c6 nh$n biS't drfdc cdc dung dich khong, ne'u cdc dung dich bi

NaCl khong c6 phan iJng vdi 4 chat c6n lai

Tren cd sd cac phan tfng ta c6 the nhan biet:

Chat khong c6 phan iJng gi la NaCl

- Cha 't tao 2 ket tua trang la BaCl2

- Chat tao I ket tua trang la NaHS04

- Chat tao 1 ket tua trang, 1 ket tua xanh la CUSO4

- Cha't tao 1 ket tua xanh la KOH

Bai 3 Cho 4 mau kim loai A, B, C, D mau sang bac giong nhau Ian Itfdt tac dung vdi cac dung dich HNO3 dac, ngupi, dung dich HCl va dung dich NaOH Ke't qua thu diTdc nhi/ sau:

-Hoi A, B, C, D la nhffng kim loai nao trong so ci.c kim loai sau: nhom, sat,

kem, magie, bac?

B a i giai

Cho cac kim loai Ian liTdt tac dung vdi HCl, HNO3, NaOH ta c6 cac ket qua sau:

Al + HNO3 dac nguoi > khong phan iJng

2A1 + 6HC1 > 2AICI3 + 3H21

2A1 + 2NaOH + 2H2O > 2NaA102 + 3H21

V$y Al khong ihuoc A, B, C, D

Fe + HNO3 dac, nguoi > khong phan uTng

Fe + 2HC1 > FcCh + H2 T

107

-)• khong phan tfng

^ Zn (N03)2 + 2NO2T + 2H2O ZnCl2 + H 2 t

NazZnOj + H z t

-> Mg(N03)2 + 2NO2T + 2H2O MgCl2 + H 2 t

-> khong phan ufng

AgN03 + N 0 2 t +H2O khong phan uTng

-> khong phan uTng

Vay Ag la kim loai C

B a i 4 Trinh bay phi/dng phap hoa hoc ngan gpn nhat de nhan bie't tiifng muoi

natri trong mot dung dich gom: Na2C03, NaHC03, Na2S03, Na2S04, NaCl,

NaNOj

Bai giai Tru-dc het cho lu-png du" dung dich Ba(CH3COO)2 vao dung dich ban dau thu

diTdc ke't tua A va dung djch B

Ket tua A gom BaS04, BaS03, B a C O j

Na2S04 + Ba(CH3COO)2 > BaS04 i + 2CH3COONa

NasCOj + Ba(CH3COO)2 > B a C 0 3 i + 2CH3COONa

Na2S03 + Ba(CH3COO)2 )• BaS03 i + 2CH3COONa

Hoa tan ke't tua A bkng dung dich H C l diT, mot phan khong tan la BaS04:

nhan bie't diTdc Na2S04

Na2C03 + 2HC1 > 2NaCI + H2O + CO2 T

Na2S03 + 2HC1 > 2NaCl + H2O + S O 2 1

Nhan bie't hon hdp CO2, SO2 nhiT cac bai tren Nhif vay ta bie't dufdc Na2C03,

Na2S03

Phan dung djch B c6 NaHC03, NaCl, NaNOj, C H 3 C 0 0 N a , Ba(CH3COO)2

La'y mot phan dung dich B cho tac dung vdi dung dich H C l diT, thay khi bay

ra: nhan bie't N a H C O j

NaHC03 + H C l > NaCl + H2O + C O 2 1

108

Sau do them tie'p Cu vao thay dung dich c6 mau xanh, c6 k h i khong mau

thoat ra hoa nau trong khong khi: nhan bie't dtfdc NaNOs:

3Cu + 8HC1 + 2NaN03 > 3CuCl2 + 2 N O T + 2NaCl + 4H2O

Lay mot phan dung dich B, them mot liTdng d\i HNO3 va sau 66 AgN03 tha'y

j{6't tua trang: nhan bie't dtfdc NaCl

A g N 0 3 + NaCl > A g C U + N a N 0 3 pai 5 M u o i NaCl bi Ian cac tap chat Na2S04, MgCl2, CaCl2, CaS04, NaBr

Trinh bay phu'dng phap hoa hoc de thu dtfdc NaCl nguyen cha't

Bai giai Hoa tan hoan toan NaCl vao niTdc (CaS04 v i liTdng it nen tan het) sau do th6m luTdng du" BaCl2 de ke't tua het muoi sunfat:

Na2S04 + BaCl2 > B a S 0 4 i + 2NaCl ' Na2C03 + MgCl2 > MgC03 i + 2NaCl

Na2C03 + CaCl2 > C a C O j i + 2NaCl Phan nirdc loc con lai (gom NaCl, NaBr, Na2C03) cho tac dung vdi dung dich HCl d i / , l u c d 6 :

Na2C03 + 2HC1 > 2NaCl + H2O + CO2 T ' Trong dung dich chi con lai NaCl, NaBr Sue k h i CI2 tdi du" va c6 can dung

dich ta se c6 NaCl nguyen cha't

2NaBr + Cl2 )• 2NaCl + Br2 Bai 6 Co hon hdp M chiJa cdc chat CaC03, Fe203, AI2O3, Si02 Hay tnnh bay

phiTdng phap hoa hoc de lay tiifng cha't rieng le nguyen cha't, khoi liTdng khong ddi

B a i giai ' Cho hon hdp vao niTdc r o i sue khi CO2 drf vao Luc do xay ra phan iJng hoa

tan C a C O j , con lai chat r^n R

CO2 + H2O + CaC03 > Ca(HC03)2 (1) Lay dung dich thu du'dc dem dun nong ta c6 CaCOs

Ca(HC03)2 — ^ CaC03 + H20 + C02T (2) I Hoa tan R bkng dung dich H C l diT, thu dUdc SiOj ^

Si02 + H C l > khong phan u-ng ' ' '

Fe203 + 6HC1 > 2FeCl3 + 3H2O (3) AI2O3 + 6HC1 > 2AICI3 + 3H2O J (4)

109

Lay dung djch thu dUcJc (gom FeCh, AICI3, HCl du) cho tac dung vdi duiig

dich NaOH dU' Cac phtfcJng trmh phan i?ng xay ra:

(Iri) tC FeCl3 + 3 N a O H > Fe(OH)3 4. + 3 N a C l ( 6 )

AlCl3 + 3 N a O H > A 1 ( 0 H ) 3 i + 3 N a C l (7)

Al(OH)3 + NaOHdu > NaAlOj + 2H2O (8)

Lay ket tua nung 5 nhiet do cao ta c6 Fe203

Sue khi C O 2 vao phan ni/dc loc (gom N a A 1 0 2 va NaOH) sau do nung d nhiet

do cao ta diTcfc AI2O3 > >

C O 2 + NaOH > NaHC03 (10)

C O 2 + 2H2O + NaAIO2 > A1(0H)3 i + NaHC03 (11)

2A1(0H)3 5 ^ Al203 + 3 H 2 0 t (12)

B a i 7

1 Co the tach s^t kim loai khoi hon hdp kim loai sat, dong, nhom, tach s^t kim

loai khoi s^t sunfua hlng each diing nam cham hay khong?

2 Mot loai dirdng kinh bi Ian mot it cat Lam the nao de c6 dufcJng nguyen chat

Co the lay diTdng tiT dung dich di/dng tan trong riTdu etylic dtfdc khong?

Bai giai

1 Co the dung nam cham de tach sat kim loai ra khoi hon hdp sat, nhom, dong vl

nhom, dong kim loai khong bj nam cham hut, nhiftig khong the tach Fe ra khoi

FeS VI day la hdp chaft, nam cham khong the pha v9 hen ket giiJa Fe va S

2 TriTdc het hoa tan diTdng kinh vao nu'dc, loc (gan) bo phan cat khong tan, sau

d6 c6 can can than ta thu dufdc dtfdng kinh Co the lay diTcJng tif dung dich

dirdng trong riTdu bang each chifng ca't de lay rifcJu, diTcJng khong bay hdi do

do ta thu diTdc nhffng tinh the dirdng tr^ng

Bai 8

1 Tai sao do vat b^ng bac de trong khong khi van giff difdc anh kim, nhi^ng

neu khong khi bi nhiem ban H2S thi do vat bang bac bi nhanh chong doi maU

thanh den

2 Khi v6 tinh lam v3 nhiet ke thuy ngan, thuy ngan kim loai luc 66 rdi va'

kh^p nha thanh nhQ-ng hat nho l i ti khong thu gom het V i thuy ngan rat dpc

nen ngu'di ta dung bien phap r^c bot lu\ huynh vao nhffng cho c6 thuy ng'"'

rdi vai Tai sao?

m

Tai sao nh6m hoat dong hdn s^t, dong nhuhg khi de cic d6 vat hkng nhom,

^ siu dong thi do vat b i n g nhom rat ben khong bi htfhong, trai lai cdc do vat

b^ng s^t, dong thi bi hoen r i

Bai giai

£)5 v$t b^ng bac de trong khong khi van giiir duTdc anh kim vi A g khong tdc

dung vdi O2 cua khong khi thanh A g 2 0 ; nhiftig khi khong khi nhiem ban H2S

fhi bi den xam do phan tfng tao thanh Ag2S (den) sau:

2Ag + 0,502 + H2S > Ag2S + H2O

2 Khi bi rdi vai thuy ngan ra't doc (do bay hdi), khong the thu gom, v l Hg bi

phan tin thanh nhffng hat ra't nho, do 66 ngiTdi ta phai r^c liTu huynh, luc 66

t^o thanh H2S khong bay hdi, ta c6 the thu gom de dang:

1 C6 hai dung dich loang FeCl2 v^ FeCl3 (gan nhiT khong mau) Ta c6 the diJng

dung djch NaOH ho|c nxXdc brom, hoac dong kim loai de phSn biet 2 dung

dich dd Hay giai thich bling cic phiTdng trinh phan tfng

2 Co 5 ong nghiem diTdc d^nh so thi? tiT 1, 2, 3, 4, 5 M o i ong diTng 1 trong 5

dung dich sau day: Na2C03, BaCh, HCl, H2SO4, NaCl Neu lay ong 2 66 vko

ong 1 thay c6 ket tua; lay ong 2 do vao ong 3 tha'y c6 khi tho^t ra, lay ong 1

do vao ong 5 thay cd ket tua Hoi ong n^o difng dung dich gi?

Bai giai

1- C6 the dilng NaOH de phan biet FeCh FeCU (rat loang) v i Itic d6 tao

thknh Fe(0H)2 mau tr^ng dnh luc v^ Fe(0H)3 mau nau do

C6 the dilng nurdc brom (mau nau do), vi FeCh lam ma't mau niTdc brom, c6n

PeCl3 khong phan iJng:

6 F e C l 2 + 3Br2 > 4 F e C l 3 + 2FeBr3

Cd the dfing dong kim loai vi luc dd FeCh khSng phan tfng vdi Cu, nhiftig f'eClj hoa tan di/dc Cu thanh CuCb mau xanh

2FeCl3 + Cu > CUCI2 + 2FeCl2

• "^rong 5 dung dich: Na2C03, BaCb, HCl, H2SO4, NaCl ta nhan thay chi cd

^aClz tao thanh ket tua vdi Na2C03 va H2SO4:

BaCl2 + Na2C03 • BaC03 >l + 2NaCI

iO) 111

BaCl2 + H2SO4 > BaS04i + 2HC1

Nhif v3y ong 1 phSi 1^ BaCh; ong 2 phii Ik NazCOs \i cho v^o ong 3 cjj

khi bay ra, va ong 3 p M i Ih H C l

NazCOj + 2HCI > 2NaCl + H2O + CO21 '

^- Ong 4 m H2SO4 va ong 5 m NaCl

Bai 10 Bpt dong oxit bi Ian bpt than (hon hdp A )

1 Trinh b^y phiTdng phap vat l i de lay rieng CuO

2 Lay mot it hon hdp A nung n6ng trong chan khong (khong c6 mat cua oxit)

tdi khi cac phan uTng xay ra hoan toan Giai thich sif bie'n doi m ^ u ciia hon

* hdp b^ng cac phiTdng trinh phan tfng Neu nung n6ng hon hdp A trong khong

khi thi hien tifdng xay ra nhiT the nao?

Bai giai

1 V i bpt CuO ra't nang, con bpt than ra't nhe nen ta c6 the dilng phiTdng phap

l^ng gan de tach lay CuO: cho hon hdp A vao coc, them ni^dc vao, khuay

deu roi l i n g gan, bpt than nhe se troi theo niTdc ra ngoai, lap di lap lai vai ba

Ian ta c6 CuO sach

2 K h i nung trong chan khong thl xay ra phan tfng

Neu it than: 2CuO + C > 2Cu + CO2

Na'u nhieu than: CuO + C > Cu + CO

(hoac: CO2 + C > 2C0)

Neu t i I9 so mol C: CuO tuT 1 : 1 den 1 : 2 thi mku den ciia hon hdp bien

thanh m ^ u do v^ng cua Cu; con neu nhif diT C hoac dif CuO thi h6n hdp c6

mau do Ian den (tuy ti le Cu va CuO hoac C)

Neu nung hon hdp U-ong khong khi thi c6 the coi than chay he't thanh CO2

con lai CuO mau den (vi Cu bi oxi thanh CuO)

, ^ Chu de 4 B d tuc phan tifng - Dieu ch6'

Bai 1 Viet cdc phiTdng trinh phan uTng theo cac sd do bien hoa sau:

Fe2 (S04)3 < > Fe(0H)3 Cu < > CuCh

FeCb CUSO4

to 1.1a Bai giai

1 Fe2(S04)3 + 6NaOH > 2Fe(OH)3 i + 3Na2S04

2Fe(OH)3 + 3H2SO4 > Fe2(S04)3 + 6H2O

112

2

Fe2(S04)3 + 3BaCl2 2FeCl3 + 3Ag2S04 Fe(OH)3 + 3HC1

FeCl3 + 3NaOH

Cu + CI2

Fe + CUCI2 (C6 the thay Fe h\ng Zn, Mg, v.v

C6 the dien phan

Y i CuCb

(Co the thay Fe bang Zn, Mg, v.v ) - i J j B + eOiS^

CUCI2 + Ag2S04 > 2AgCl i + CUSO4

CUSO4 + BaCl2 > BaS04i + CUCI2

u 2 A

(latO >JU X>? W^' } t

I C6 the dieu chc Fe b^ng each khur sit oxit theo cic phan ufng sau:

a) b) c) d) e)

Fe.Oa + CO FczOj + H2 FejOj + A l Fc.Oy + A l

B

+ Y FeCl2

• + T Trong do A, B, X, Y, Z, T la cac cha't khac nhau

113

Bai giai

1 Hoan thanh cac phan uTng:

a) , FejOa + 3C0 — ^ 2Fe + SCOj

• ' - • • If,

b) FeaOj + 3H2 — ^ 2Fe + 3H2O

d) 3FexOy + 2yAl — ^ 3xFe + yAlzOj

e) FezOa + 3C — ^ 2Fe + 3C0

phan iJng a) difcJc dung de san xua't gang

2 Cic phiTdng trinh phan tfng:

a) FezOj + 3C0 — ^ 2Fe + 3CO2 (X c6 the' la H2, Al, C )

b) Fe + CUCI2 > FeCh + Cui

c) FezOj + 6HCI > 2FeCl3 + 3H2O

d) 2FeCl3 + Fe > BFeClj (T c6 the \h Cu, KI)

Bai 3 Cho sd do bien hoa

+x t"

> Fe D — C

Biet ring A + HCl > D + C + H2O

Tim cac chat tiTdng tfng vdi cac chuT cai A, B, va viet cac PTPl/

Bai giai

Nhin sd do ta thay A phai Ih oxit sat, va vi A + HCl tao ra hai loai muoi ncn

A phai Ih Fe304:

Fe304 + 4C0 — ^ 3Fe + 4CO2

5 J ^ Fe304 + 4H2 — ^ 3Fe + 4H2O

' 3Fe304 + 8A1 9Fe + 4AI2O3

Fe + 2HC1 > FeCl2 + Hat

(B) (D) 2FeCl2 + Cl2 > 2FeCl3

(E) (C)

Fe304 + 8HC1 > FeCl2 + 2FeCl3 + 4H2O

pjay vie't 3 loai phan uTng tri/c tiep tao thanh muoi tit kim loai va 3 loai phan

^jig tT\ic tiep tao thanh kim loai ttf muoi < ,

j^ay vie't 4 loai phan ufng tao thanh NaOH ^

2 C^c phan tfng tao thanh NaOH

2Na + 2H2O > 2NaOH + H2 T ' ^ ^ Na20 + H20 > 2NaOH

Ca(0H)2 + Na2C03 > CaC03i + 2NaOH NaHC03 + Ba(0H)2 v^adu hoacdu > BaCOs i + NaOH + H2O

v.v

Bai 5 Cho 4 mau Na vao 4 dung dich sau: ZnCb, FeCl2, KCl, MgS04. Viet cac PTPU xay ra

Bai giai

2Na + 2H2O > 2NaOH + Hjt ' ' ••^ ' ;v /

-Sau do: 2NaOH + ZnClj > Zn(OH)2 i + 2NaCl ,

2NaOH du- + Zn(0H)2 > Na2Zn02 + 2H2O

2NaOH + FeCl2 > Fe(0H)2 i + 2NaCl , ,, Neu de trong khong khi:

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

KCl + NaOH > khong phan urng

2NaOH + MgCl2 > Mg(0H)2 i + 2NaCl

115

B a i 6 Tic cac chat A, B, C thich hdp viS't cdc PTPLf theo sd do bien hộ

B > C > D

B -> G

M phai la FeS Cac PTPLf each 1 la:

2FeS + 7/2O2 > FciOi + 2SO2

(A) (E) (B) SO2 + 2H2O + Br2 > 2HBr + H 2 S O 4

(hoac CI2, O 3 , H2O2 ) (C)

Fe203 + 3H2 — ^ 2Fe + 3H2O

FeCb + H.T (G)

2 A g C l i + F C S O 4

(H)

F c S i +Na2S04 (M)

2 F e O + CO2

(F) -> Fe2(S04)3 + S02T + 4 H 2 O

(B)

- > Fc + H2O (G)

-> C U S O 4 + H2O

(hoac Cu, C U C O 3 ) (D)

116

J V i ^ t l?i '^^"S ^^^^ ^^^^ thanh phan cho dvldi day v£k goi ten c'liing,

nd'u c6ng thỉc nao sai dUdc phep chinh lai chi só cua mpt nguyen tó:

1 Viet lai cong thuTc dung va goi ten:

N2HSCO3 > (NH4)2C03 amoni cacbonat

H2P20xCa > CăH2P04)2 canxi dihidrophotphat

BaC4H,„06 > BăCH3C00)2.2H20 tinh the bari axetat ngam

hai phan tur H2O

2 Fe + 2HC1 > F e C l 2 + H 2 t

Hoac Fe + C u C b > FeCh + Cui

2FeCl2 + CI2 > 2FeCl3

(1)

FeCl3 + 3 N a 0 H > Fe(0H)3 i + 3NaCl

(CO the thay N a O H bang K O H , Bă0H)2 v.v )

2Fe(OH)3 Fe203 + 3 H 2 0 t

Fe203 + 3H2SO4 > Fe2(S04)3 + 3H2O Fe2(S04)3 + 3BăN03)2 > 3 B a S 0 4 i + 2Fe(N03)3

(2) (3)

,ị if!

(4) (5) (6)

Chu de 5 T i n h theo phi/cftig trinh phan ufng

H i ^ u suS't phan ufng - Nong dp dung dich ^

1 De mieng AI nSng 5,4g trong khong k h i mot thdi gian thu di^dc chát r^n

^- Hoh tan hoan toan A b^ng dung dich H C l tháy bay ra 6,5856 lit H2 (dktc)

Tinh khói lu^dng ran A va %A1 bi oxi hoa thanh oxit

-117

Ne'u khong bi oxi hoa thi 5,4 gam Al tao ra: —-—x — x22,4 = 6,72 l i t H 2

Theo (1, 2) so mol nguyen tuT oxi phan iJng bkng so mol H2

6,72-6,5856 22,4 • = 0,006mol Vay khoi liTcJng oxi da tham gia phan iJng b^ng: 0,006 x 16 = 0,096 gam

Do 66 khoi liTdng A = 5,4 + 0,096 = 5,496g

B a i 2 Cho a mol bot Fe vao dung dich chtfa b mol CUSO4. Sau khi ket iho"^

phan iJng ta thu diTdc dung dich X va chat r^n Y Hoi trong X, Y c6 nhiWr

chat gi? Bao nhieu p i o l ?

B a i giai

Phan tfng: Fe + CUSO4 > FeSO^ + Cui (1)

Ta can xet cac triTdng hdp:

1/ a = b ttfc cac cha't tdc dung vdi nhau vijfa du

Dung dich X chi c6 a mol FeS04; chat r^n Y chi c6 b mol Cu

3/ a < b tiJc du- CUSO4

Dung dich X c6 a mol FeS04 va (b - a) mol CUSO4; chat r^n Y c6 a m o l Cu

pai 3- Nhung mot mieng nhom kim loai nSng 10 gam vao 500 ml dung dich

CUSO4 0,4M Sau mot I h d i gian lay mieng nhom ra, ruTa sach, say kho, can

1 Tinh khoi li/dng dong tho^t ra bam vao micng nhom (gia suT ta 't ca dong thoat

ra bam vao mieng nhom)

2 Tinh nong dp mol cua cac chat sau phan vlng (gia suT the tich dung dich van

500 ml)

B a i giai , gnrdv:

1 Phan tfng: 2A1 + 3CuS04 > Al2 (S04)3 + 3Cu>l' "f'^^ "f ' Goi X la so mol Al2 (S04)3 tao t h a n h , nhu* vay l i f d n g Cu tao t h a n h la 3x, luTdng

Al p h a n iJng la 2x, lifdng CUSO4 p h a n iJng la 3x

Dya treh nguyen tic l a Al phan tfng thi kho'i liTdng mieng Al bi giam, con Cu tao thanh bam vao mie'ng Al nen khoi l i f d n g tang len, ta c6 phu'dng trinh:

10- 2x X 27 + 3x X 64=ll,38gam Giai ra ta di/dc: x = 0,01 mol

Khoi li/dng Cu thoat ra bkng 3x x 64 = 3 x 0,01 x 64 = l,92g

2 Tinh nong do mol

n 0,01 ^ 0,5x0,4-3x0,01 ^ ^ '

CA12(SO4)3 = "5y = " ' ^ ^ ' ^ '^CuS04 ^ = 0,34M Bai 4 Cho V lit khi CO (dktc) di qua ong stf diTng a gam CuO nung nong Sau khi ket thiic thi nghiem cho khi di ra khoi oag si? hap thu vao dung dich Ba(0H)2 tha'y tao thanh m gam ket tua

1- Viet cac PTPl/xayra

2 Tinh hieu suat ciia phan i ?ng khur CuO theo V, a, m

B a i giai ' '

^- Cac phu'dng trinh phan uTng:

CuO + CO — ^ C u + C02 (1) C02 + Ba(OH)2 - ^ B a C O j i + H j O (2) ' f 2- Tinh so mol cac chat:

^ n, xlOO„

Trirdng hcJp 2: ni > 0 2 hi?u suat h% = — %

Trirdng hdp 3: n, < nz hieu suat h% =

" i

Bai 5 Cho 13,44 gam bot d6ng kim loai vao 1 coc difng 500 m l dung dich

AgNOs 0,3M, khuay deu dung dich mot thcJi gian sau d6 dem Ipc ta thu diroe

22,56 gam chat r^n A va dung dich B

1 Tinh nong do mol cua cha't tan trong dung dich B Gia suf the tich ciia dung

dich khong thay ddi

2 Nhiing mot thanh kim loai R nSng 15 gam vao dung dich B, khuay deu de

c^c phan iJng xay ra hoan toan, sau do lay thanh kim loai R ra khoi dung

djch, can nang 17,205 gam (gia suT ta't ca kim loai thoat ra deu bam vko thanh

R) Hoi R la kim loai gi trong so cac kim loai cho sau: Na = 23, M g = 24,

Gpi X la so mol Cu tham gia phan uTng (1) (nhd r^ng Cu chiTa phan tfng het)

ta CO phufdng trinh ve khoi IiTdng cha't r^n A:

Kho'i lufdng thanh R tang = 17,205 - 15,00 = 2,205 g

Theo cac phan utag (2, 3) ta c6 phifdng trinh ve suf thay doi khoi liTdng thanh R:

pai 6 Cho 27,4 gam bari vao 400 gam dung dich C U S O 4 3,2% thu diTdc khi A,

Icd't tija B va dung dich C

J Tinh the tich khi A (dktc)

2 Nung ke't tua B d nhiet do cao den khoi liTdng khong doi thi thu diTdc bao

nhieu gam cha't ran?

3 Tinh nong do % cua cha't tan trong dung dich C

Bai giai Cac phan iJng xay ra:

Ba + 2H2O

Ba(0H)2 + C U S O 4

BaS04 Cu(OH)2 27,4

(3)

nga = = 0,2mol; np,,<-o = = 0,08mol

137 100x160

Theo phan ufng(l): n^^ =nB,(0H)2 " " B U =0,2mol

The tich H2: VA = 0,2 x 22,4 = 4,48 lit Theo cac phan uTng (2, 3) chat ran gom BaS04 va CuO, vi Ba(0H)2 diT nen

nBaS04 =ncu(OH)2 = n c u O =0,08mol ' i,v

Kho'i lircfng cha't r^n = 0,08 x 233 + 0,08 x 80 = 25,04 gam Trong dung dich C chi con Ba(0H)2, khoi liTdng dung djch C b^ng tdng kho'i

lifdng cac cha't ban dau truf lufdng H2 bay ra va li/dng ke't tiia

Bai 8. Cho biet do tan cua dong sunfat 6 5"C la 15 gam va d 80"C la 50 gam,

Hoi khi lam lanh 600 gam dung dich bao hoa dong sunfat d 80"C xuong 5"c

•Y thi CO bao nhieu gam tinh the CUSO4.5H2O thoat ra

B a i g i a i

Tinh tong khoi lifdng C U S O 4 (khan)

Ctf 100 + 50 = 150 gam dung dich M o hoa d 80"C c6 50 gam C U S O 4

Nhir vay trong 600 gam c6 ^ ^ ^ ^ ^ "^^^^ ^"^^^ ~ ^ "^^^^

5x18 Gpi X la so gam tinh the CUSO4.5H2O thoat ra trong 66 c6 -^j^ x x = 0,36x

gam H 2 O va 0,64x gam CUSO4. Vay liTdng H 2 O con lai trong dung dich la

400 - 0,36x va li/dng CUSO4 con lai la 200 - 0,64x, do do ta c6 ti 16 (d 5"C):

1^9 = 1^ rut ra X = 238,9 gam

400-0,36x 200-0,64x

B a i 9 Mot loai da X chiJa 10,2% A I 2 O 3 , 16% Fe203 c6n lai la hon hdp CaCO,,

MgCOj Lay 100 gam X, nung mot thdi gian thay con lai 82 gam chat r^n Y

De hoa tan hoan toan 10 gam Y can 173 ml dung dich HCl 2M Neu lay lOHg

X nung tdi phan huy hoan toan muoi cacbonat thi con lai bao nhieu gam cha!

Cac phan tjTng hoa tan Y bkng dung dich HCl:

CaO + 2HC1 > CaCl2 + H2O (3)

S6 mol HCl can hoa tan 82g Y bing: mHci - 0,173 x 2 x — = 2,8372mol,

$6 mol do cua HCl cung b^ng so mol HCl hoa tan 100 g X, vi mot mol

CaC03 dil da bi phan huy thanh CaO hay khong deu can 2 mol HCl, tufdng tif

d6'i vdi MgCOj (xem cac phiTdng trinh phan iJng 3, 4, 5, 6) Goi x, y \k so mol

CaC03, MgC03 trong lOOg X

1 Cho 500 gam dung dich Na2S04 x% vao 400 ml dung dich BaCh 0,2M thay

tao thanh 10,485 gam ket tua Tinh x?

2 Dung dich A chtfa 24,4 gam hon hdp hai muoi Na2C03 v^ K 2 C O 3 Cho dung djch A tac dung vdi 33,3 gam CaCh thay tao thanh 20 gam ket tua va dung dich B Tinh so gam moi muo'i trong dung dich A, B

B a i giai

1 Phan ifng Na2S04 + BaCl2 > BaS04i + 2NaCl (1) 'T

So'molke'ttua: nBaso4 = ^ ^ ^ ^ = 0.045mol n g n <> V

SS'molBaCb: Ug^cia =0,4x0,2 = 0,08mol

Vi so' mol ket tua BaS04 nho hdn so mol BaCh nen so mol BaS04 phai b^ng s6' mol cua Na2S04 = 0,045 mol

Do do n6ng do cua dung dich Na2S04 b^ng: x = Q'Q45x 142x100 ^ ^^78%

500

Na2C03 + CaCl2 > CaCOji + 2NaCl (1) ,i K2C03 + CaCl2 >CaC03i+2KCl (2) a'5 ul^ ' mol CaCl2 = — = 0,3 mol So mol CaC03 = — = 0,2 mol Ill 100

so mol CaC03 it hdn so' mol CaCl2 chiJug to CaCb di^ 0,3 - 0,2 Ik so mol

ciia Na2C03 va K 2 C O 3 ta c6 he phiTdng trinh

123

flOex + nSy = 24,4

[x + y = 0,2

Giai he phifdng trinh ta c6: x = 0,1 mol y = 0,1 mol

Kho'i lifcfng cdc muo'i trong dung dich A:

NazCOj: 0,1 x 106 = 10,6 gam; K2CO3: 0,1 x 138 = 13,8 gam

jf Kho'i lU'ctng cac muoi trong dung dich B:

CaCl2:0,l X 111 = 11,1 gam t,

NaCl: 0,1 X 2 X 58,5 = 11,7 gam

KC1:0,1 X 2 X 74,5 = 14,9 gam

Bai 11

1 A la mot loai quSng sat chtfa 60% Fe203; B la mot loai quSng s^t khdc chiJa

69,6% Fe304 Hoi trong 1 tan quang A hoac B c6 chtfa bao nhieu kg s^t?

2 Tron quang A vdi quSng B theo ti le khoi lufdng mA : ma = 2 : 5 ta ductc

quang C Hoi trong 1 tan quang C c6 bao nhieu kg s^t?

Bai giai

1 V i trong 1 tan quang A c6 60% Fe203 va trong 160g Fe203 c6 2 x 56 = 112g

Fe nen trong 1 tan (1000kg) quang A c 6 : ^ ^ ^ " ^ ^ 0 ^ ^ ^ ^ ^

# s 100x160 *

Tinh tiTdng tir, 1 tan quang B co: ^000^69,6x3x56 ^

100x232 ^

2 Tron 2 tan quang A vdi 5 tan quang B ta dtfdc 7 tan quang C, lifdng s^t c6

trong 7 tan quang C la: 2 x 0,420 + 5 x 0,504 = 3,36 tan Fe

Vay 1 tan quang C c6 ^ = 0,48 tan Fe

B^i 12

1 De xdc dinh ham lifdng cacbon trong thep (khong c6 lifu huynh), ngifdi ta

cho mot d6ng oxi dif di qua o'ng stf difng 15 gam thdp (dang bot) dot ndng va

cho khi di ra khoi ong sif hap thu hoan toan vao dung dich Ba(0H)2 dif thay

tao thanh 1,97 gam ke't tua Tinh ham lifcJng % cua cacbon trong thep

2 * De san xua't th6p tijf gang ngifdi ta cd the loai bdt cacbon cua gang bSng

Fe203 theo phan urng: Fe203 + 3C > 2Fe + 3 C O t

Hoi muon loai bdt 90% lifdng cacbon cd trong 5 tan gang chtfa 4% cacbon thl

can bao nhieu kg Fe203?

Bai giai

CO2 + Ba(0H)2 > BaC03 4 + H2O (2)

1 97

Theo cic p h i n tfng (1, 2): H C = ncoj = nBacoj = 7 ^ = O.Olmol 197

, , V 0,01x12x100 „ , ; ,

2 Phan tfng khuT Fe203 bkng C: Fe203 + 3C — ^ 2Fe + 3COT (1)

^ A ^ v 5x1000x4 90 ,

Kh6i lifdng C can phai loai bang — x — = 180 kg '

100 100

Xheo phan urng (1) de loai 3 x 12 = 36 g cacbon can 160g Fe203

Vay de loai 180kg cacbon can: ^ ^ ^ ^ ^ = 800 kg Fe203

36

Bai 13 Hoa tan hoan toan 49,6 g hon hdp A gom Fe, FeO, Fe304, Fe203 bkng

dung dich H2SO4 dac ndng thu difdc 8,96 lit SO2 (dktc) va dung dich B chi

chtfa mot loai muo'i s^t

1. Via't cdc P T P U

2 Tinh % khoi lifdng ciia oxi trong A

3 Tinh khoi lifdng cua muoi s^t trong B

Bai giai

| £ M nmi fat) v<?

-> Fe2(S04)3 + 3S02t + 6H20 (1) Fe2(S04)3 + SO21 + 4H2O (2)

-> 3Fe2(S04)3 + S O 2 T + IOH2O (3) Fe2(S04)3 + 3H2O (4)

1 Cac phan tfng:

2Fe + 6H2S04dac

2FeO + 4H2S04dic 2Fe304+ 10H2SO4dac

S6' mol oxi trong A = y + 4z + 3t = no

Ta ( I , II) rut ra no = 0,65 mol

vay % kho'i hfdng oxi = ^ ' ^ ^ ^ ^ ^ ^ ^ ^ ^ = 20,96%

• ^ 49,6

3- Khoi lifcJng cua Fe trong A = 49,6 - 0,65 x 16 (khoi lifdng oxi) = 39,2 gam

ttfc 0,7 mol va so mol Fe2(S04)3 = — = 0,35mol

2

Khoi lifdng Fe2(S04)3 trong B = 0,35 x 400 = 140 gam

(Co the giai cau 2 theo phu'dng phap bao toan electron

125

"B<5/ amng hoc sittngwt nuu nut o, y - uuu

Bai 14 Cho 220 ml dung dich HNO3 nong do C mol/1 tdc dung vdi 5 gam hS^

hop Zn, A l Sau khi ke't thuc cac phan iSng thu diTOc 0,896 lit hSn hdp khi X

gom NO va N2O c6 ti kho'i so vdi H2 bkng 16,75 dung dich A va c6n lai 2,0l3

gam kim loai

1 Hoi CO can dung dich A thu diTdc bao nhieu gam muoi khan?

Giai ra ta co x = 0,03 mol (NO); y = 0,01 mol (N2O)

Theo cac phan tfng ( 1 , 2 ) so mol goc - NO3 tao muoi nitrat luon b^ng 3 so

mol NO, khong phu thuoc kim loai nao

Theo cac phan uTng (4, 5) so mol goc - NO3 tao muoi luon b^ng 8 l^n s6' mol

N2O Do do tong so mol goc - NO3 tao muoi b^ng:

0,03 X 3 + 0,01 X 8 = 0,17 mol

Khoi liTdng kim loai tan trong axit b^ng: 5 - 2,013 = 2,987 gam

Tong kho'i liTdng muoi = tong khoi liTdng kim loai + khoi liTdng goc nitrat:

m™,tikhan = 2,987 + 0,17 X 62 = 13,527 gam

2 Tong so mol HNO3 trong 220 ml dung dich HNO3 bing

0,22 0,17 + 0,03 + 0,01x2 =0,22

Bai 15 Nhiing 1 thanh sat va 1 thanh kem vao ciing mot coc chuTa 500 m l dung

dich CUSO4 Sau mot thcJi gian lay 2 thanh kim loai ra khoi coc, liic d6 tat

Cu thoat ra deu bdm het vao 2 thanh kim loai va khoi liTdng dung djch trong

coc bi giam 0,22 gam Trong dung dich sau phan iJng nong dp mol ciia ZnSO^

126

dn

I n C '

dn gap 2,5 Ian nong dp mol cua FeS04 Them dung dich NaOH du v^o coc,

Ipc lay ket tua roi nung ngoai khong khi tdi kho'i liTpng khong doi thu diTpc 14,5 gam cha't r^n

Tinh khoi lu'dng Cu bam vao moi thanh va nong dp mol cua dung dich CUSO4 ban dau

Bai giai

Cic phan tfng xay ra:

Zn + CUSO4

Fe + CUSO4 FeS04 + 2NaOH CuS04 + 2NaOH ZnS04 + 2NaOH Zn(0H)2 + 2NaOH

-+ ZnS04 + Cu

4 > FeS04 + Cui

-+ Fe(OH)2i +Na2S04 -+ C u ( 0 H)2i +Na2S04

• m it*.;

-> Z n(0H)2i +Na2S04 -> Na2Zn02 + 2H2O

I TiWng tir doi vdi Zn tang (65 + 96 - (64 - 96) = 1 gam

Do do ta CO phiTPng trinh: 8x - 2,5x = 0,22 => x = 0,04 mol

Vay khoi luTpng Cu bdm vao sat = 64 x 0,04 = 2,56 gam

bam vao kem = 64 x 0,04 x 2,5 = 6,4 gam

Theo cic phan iJug (2, 3, 7):

Kh6'i lirpng mpejoj = 160 x 0,02 = 3,2 gam

Vay khoi liTpng CuO = 14,5 - 3,2 = 11,3 gam

T^ng so mol CUSO4 ban dau bang:

0,5

= 0,04 + 0,1 +0,14125 = 0,28125

= 0,5625M :

127

B a i 16 Hoa tan hoan t o ^ n m o t m i e n g bac k i m l o a i v ^ o m o t ItfcJng diTdung dici^

H N O 3 15,75% thu diTcJc k h i duy nha't N O va a gam dung dich F trong do nong

do C % cua A g N O s bang C% cua H N O 3 dir T h e m a gam dung dich HC|

1,46% vao dung dich F H o i c6 bao nhieu % A g N O j tac d u n g v d i H C l

B a i g i a i

P T P U : 3 A g + 4 H N 0 3 > 3 A g N O j + N O t + 2H2O (1)

Cach 1: Gpi m la so gam dung dich HNO3 da dijng de hoa tan 3 m o l A g ,

so gam HNO3 = itix 15,75% = 0,1575m

2 0528

V a y % A g N 0 3 da phan lirng bang — x 100 = 68,40%

Cach 2: Gia suT l a y 100 gam dung dich HNO3 15,75%, luc do so mol Ag hi

hoa tan la x T a t i m difdc x = 0,062 m o l , a = 106,076 g a m

0 0424 X100 nHci = 0,0424 m o l va %AgN03 phan uTng bang — = 68,40%

0,062

B a i 17 Hoa tan h e t 5,94 gam A l bang dung dich N a O H diTdc k h i A Cho dung

dich H C l dac, diT tac dung v d i 1,896 gam K M n 0 4 di/dc k h i B N h i $ t phan

hoan toan 12,25 gam KCIO3 (xuc tac M n O j ) ihu diTdc k h i C Cho 3 khi A, B,

C vao binh k i n va tic'n hanh phan iJng no hoan l o a n , sau do l a m lanh binh

xuong nhiet dp thiTcfng, gia sijT luc do niTdc nguTng tu he't va chat tan het vao

nirdc diTdc dung dich D

1 V i e t cac PTPLf x a y ra

2 T i n h nong do C % cua cha't tan trong D

B a i giai

Cac phan ufng x a y ra:

; 2A1 + 2NaOH + 2 H 2 O - - - > 2NaA102 + 3H2 (1)

SO2 + Ca(0H)2 > C a S 0 3 i + H2O ^ '

S O 2 + H2O + CaSOj > Ca(HS03)2 ; <v} h'

^- Cho hon hdp k h i Ian lu-dt qua binh 1 diTng ni/dc b r o m du" va binh 2 difng niTdc Voi trong, dn tha'y mau nau cua ni/dc brom bi nhat do phan iJng:

SO2 + 2H2O + Br2 > 2HBr + H2SO4

trong binh 2 xua't hien ket tua: ?',,» rf ' • •

C02 + Ca(OH)2 > C a C 0 3 ^ + H 2 0

129

4 Khoi liTdng cua 1 mol hon hOp A, B = 2,1875 x 22,4 = 49 gam

Goi X va 1 - X la so' mol S O 2 va C O 2 trong I mol hon hcJp, ta c6:

64x + 4 4 ( l - x ) = 49

Giai ra ta c6 x = 0,25 mol va 1 - x = 0,75 mol; tiJc ty le so mol : nco^ = 1 : 3

Tir do theo cac phan tfng (1,2) suy ra ti le so mol: Up^s^ : n F e C 0 3 ^ ^ ' ^

(Cach tim: nhan phiTcJng trinh (2) vofi 6 de cho so' mol CO2 gap 3 so mol SO^)

Vay % khoi liTdng cua: %FeS2 = ^^^^"^^Q^ 25,64%

• ^ 1x120 + 3x116

Va %FeC03 = 100 - 25,64 = 74,36%

Bai 19 Tien hanh dien phan (dien cifc trd mang ngan xop) 500 ml dung dich

NaCl 4 M (d = 1,2 g/ml) Sau khi 75% NaCl bj dien phan thi dCfng lai

1 Tinh nong do % cua cac chat trong dung djch sau dien phan

2 La'y cac khi thoat ra cho tac dung het vdi nhau dufdc san pham A Hoa tan A

vao 500 g nirdc dU'cJc dung dich A Tinh nong do % cua dung dich A

Bai giai

1 Phan iJng dien phan:

2NaCl + 2H2O H21 + CI21 + 2NaOH (1)

nNacic6n = 0,5 X 4 X — - = 0,5mol; nH2 =nci2 = - n N a O H = - x 1,5 = 0,75mol

Khoi lifdng dung dich sau dien phan bang:

500 X 1,2 - 0,75 X 2 - 0.75 X 71 = 545,25 g

w XT r^u 1,5x40x100 0,5x58,5x100 , , „

Vay % NaOH = =^11% ; %NaCl = = 5,36%

545,25 545,25

2 Phan tfug: H2 + CI2 > 2HC1 (2)

Theo phan drug (2): HHCI = 2 X 0,75 = 1,5 mol

Vay nong do % cua dung dich A b^ng: %HC1 = ^'^^^6,5x100 ^ ^ ^^^^

500 + 1,5x36,5

Bai 20 Co 200 ml dung dich hon hdp AgNOj 0,1M va Cu(N03)2 0,5M Them

2,24 gam bot s^t kim loai vao dung dich do Khuay deu tdi phan uTng hoai^

toan thu diTdc chat r^n A v^ dung dich B

1 Tinh so gam chat ran A

2 Tinh nong do mol cua cac muoi trong dung dich B, biet r^ng the tich dung

dich khong doi

2AgN03 + Fe > Fe(N03)2 + 2Ag i (1)

Cu(N03)2 + Fe > Fe(N03)2 + Cu i (2) ^ ^4 HU:

Tinh: npe = ^ = 0,04mol; n^gNO:, = 0,20x0,1 = 0,02mol; ,

ncu(N03)2 ==0'20>< 0.5 = 0, Imol

S6'mol Fe tham gia phan tfug (1) bSng: ^^n^gNo,^ =rnAg = - ^ ^ = 0,01mol

So mol Fe tham gia phan tfng (2) bang: npc = 0,04 - 0,01 = 0,03 mol "

Do do: ncu(N03)2 ' ^ ^ ^ = 0,1 - 0,03 = 0,07 mol

Chat ran A gom Ag, Cu c6 kho'i li/dng b^ng: ^ ^ ^ j , ^ ^ , „ ^^.^ j ^ , ^

Cu + 4HNO3 > Cu(N03)2 + 2NO21 + 2H2O (4)

Theo phan tfng (3, 4): nN02 ^ " A g +2xncu =0,02 + 2 x 0 , 0 3 = 0,08mol Vay the tich N O 2 (theo dktc) = 0,08 x 22,4 = 1,792 lit

Bai 21 X la quang hematit chiJa 60% Fe203 Y la quSng manhetit chtfa 69,6%

Fe304

^- Hoi tir 1 tan quang X hoac Y c6 the dieu che diTcJc toi da bao nhieu kg sat kim loai?

Can tron X va Y theo ti le khoi lu-dng nhu the nao de dirdc quang Z ma ttf 1

tSn Z CO the dieu che dtfcJc 0,5 tan gang chiJa 4% cacbon

2 LrfcJng Fe c6 trong 1 ta'n quSng Z hKng: mp^^2.) = ^ ' ^ ^ ^ ^ = 0,48T = 480kg

1$ mx : my = 2 : 5

B a i 22 De m gam bpt s^t nguyen chat trong khong khi mot thdi gian thu duoj,

chat r^n A nSng 12 gam gom Fe, FeO, Fe304 va FejOj Hoa tan hoan loan

cha't dn A bkng dung dich HNO3 loang thay c6 2,24 lit khi N O duy nhat

(dktc) va dung dich B chi chtfa mot muoi nitrat sat duy nhat

1 Viet cac PTPLf xay ra

2 Tinh khoi liTdng m

3 Tinh khoi liTdng muoi trong dung dich B

Fe(N03)3 + N O t +2H2O 3Fe(N03)3 + N O T + SH.O -> 9Fe(N03)3 + N O t + I4H2O 2Fc(N03)3 + 3H2O

(1) (2)

(3)

(4) (5) (6) (7)

2 Goi X, y, z, t la so mol cua Fe, FcO, Fe304, Fe203 trong chat ran A Thco dicn

kien bai loan va theo cac phiTcfng trinh phan tfng, ta c6 he phiTdng trinh:

Qlii chii: CO the giai theo phuTdng trinh bSo loan electron nghia la t^ng so'

electron cac chat khur cho bang long so electron cac cha't oxi hoa nhan, d day

pg la cha't khur, c6n O2 va N*^ la chat oxi ho^ nen ta c6 phu'dng trinh:

j n ^ 3 ^ i l z i ! ! X 4 + 0,1 X 3. Rut ra m = 10,08g

56 32

3 T6'ng khoi lirdng muo'i: npe = npe(N03)3 = = 0,18 m o l ^

p o d o : mpg(fjo^)^ = 0,18 X 242 = 43,56 gam pai 23- Hai mie'ng k e m c6 cung khoi li/dng 100 gam Mie'ng thi? nha't nhiing vao

100 ml dung dich CUSO4 diT, mie'ng thuT hai nhiing v^o 500 m l dung dich

AgN03 dir Sau mot thcfi gian la'y 2 mie'ng k e m khoi dung dich nhan thay mieng thiJ nha't giam 0 , 1 % kho'i liTdng, nong do mol cua cdc m u o i k e m trong hai dung dich bkng nhau H o i khoi liTdng mie'ng k e m thiir hai thay doi nhuTthe

nao? Gia si( cac kirn loai thoat ra deu bam vao mie'ng k e m

B a i giai ' '

C i c phan tfug: ' ' < ^

Zn + CuS04 > Z n S 0 4 + C u i (1) t t

Zn + 2AgN03 )• Zn(N03)2 + 2Ag>l (2) Gpi X la so' mol Z n tham gia phan tfng Theo phan lirng (1) Ihi:

So mol CuClj tham gia phan u"ng b^ng: 0,4 x 0,5 x = 0,05 mol

100

^ o i m lii khoi lUdng mie'ng nhom sau phan iJng, theo phan tfug (1) ta c6 bieu

thu-c: 20 - - X 0,05 x 27 + 0,05 x 64 = 22,3 gam

133

K h i k h i ngijfng thoat ra c d the x a y ra hai kha nang: hoac A l het hoac axit H C l het

N e u A l h e t t h i cha't r ^ n Y c h i c o n C u , nhU' v a y theo p h a n iJng ( 2 ) t i 1^ tang kh*'

liTdng - 5 ^ ^ = — = 1,25, dieu nay k h o n g phu h d p v d i d i e u k i e n cho (1,35) D i '

m Cu 6 4

d d a x i t p h a i h e t va A l c o n dir T h e o p h a n ufng (3) t i l e tang k h o i liTdng cua A l

mAi203 _ 102

m 2AI 2 x 2 7 = 1,89 D i e u d o p h i i h d p v d i d i e u k i e n b a i toan

G i a sur k h o i liTdng Y l a lOOg

G p i X, y la so m o l c u a C u v a A l trong chat r ^ n Y , ta c d h e phuTdng t r i n h

^ V a y t o n g t h e t i c h k h i : V = 0,15 x 2 2 , 4 = 3,36 l i t

4 M o t l o a i d a v o i chiJa 8 0 % C a C O j , p h a n c o n l a i la c a c cha't t r d N u n g m

^ g a m da m o t t h d i g i a n t h u diTdc cha't r ^ n c d k h o i lufdng b a n g 0 , 7 8 m g a m

^ T i n h h i e u sua't p h a n i^ng phaft h u y

Phan k h o i liTcJng h u l di chinh bang khoi liTcfng C O j = 100 - 78 = 22g

Theo l y thuyet k h o i liTdng COj toi da b a n g 0,8 x 44 = 35,2g

H i e u suat phan uTng: h % = ^ ^ ^ ^ ^ = ^ 2 , 5 %

2 So moi C O 2 tao thanh b i n g so moi CaO = 22 : 44 = 0,5 moi

78

B a i 5 Ngirdi ta c6 the dieu che CI2 bang each cho dung dich H C l dac tac dung

v d i K M n 0 4 hoac M n O j , KCIO,

1 Vie't cac phiTdng trinh phan i^ng

2 Neu lifcJng clo thu duTcJc trong 3 tri/dng hdp nhiT nhau thi ti le khoi li/cfng cua

2 Gia suT luTdng CI2 bay ra trong moi tn/dng hdp la 1 moi, liic do theo cac phan iJrng

( 1 , 2 3) can - moi KMn04, 1 moi M n O j va ^ moi KCIO3, tufc ty 1$ khoi lifdng:

5 3

m K M n 0 4 : m M „ 0 2 ^Kao, = | X 1 5 8 1 8 7 : i X 1 2 2 , 5 = 6 3 , 2 : 87 : 40,83

B a i 6 M o t loai da chiJa M g C O j , CaCOj, A I 2 O 3 bang 1/8 tong liTdng hai muoi

cacbonat Nung da d nhiet do cao tdi phan huy hoan toan hai muoi cacbonat

thu du-dc chat ran A c6 khoi lUdng bang 60% khoi lu^dng da triTdc k h i nung

1 T i n h % khoi lifdng moi chat trong da tru-ctc khi nung

2 M u o n hoa tan hoan toan 2 gam chaft ran A can toi thieu bao nhieu m l dung

(3)

Gia sur da ban dau c6 9 gam trong do co 1 gam A I 2 O 3 ; k h o i lufdng chat ra"^

A = 9 x — = 5,4 gam trong do c6 1 gam A I 2 O 3 va 5,4 - 1 = 4,4 gam 1"'"'

2 Hon hdp gom cac k i m loai M g , A l va Cu »' a) Oxi hoa hoan toan m gam hon hdp A thu dUdc 1,72 gam hon hdp 3 oxit

b) Hoa tan m gam hon hdp A bang dung dich HCl diT thu du-dc 0,952m lit H2 (dktc)

Tinh % khoi lifdng moi k i m loai trong hon hdp A

^' Ta CO the tif chon gia trj cua m la 100 gam

Cac phan ij-ng:

Cu + 0,502 -> CuO (3)

b) Mg + 2HC1 >MgCl2 + H 2 t (1)

Gpi X, y, z la so mol cua Mg, Al, Cu trong 100 gam hon hdp A, ta c6 cac

phiTdng trinh: 24x + 27y + 64z = 100

Khoi liTrtng oxit = 40x + 102 I + 80z = 1,72 X m = 172

3 0,952xm 95,2

So mol hidro = x + - y = = = 4,25 2 22,4 22,4

Giai he phu^dng trinh ta c6: x = 1,25 mol; y = 2 mol; z = 0,25 mol

Vay % khoi lu'dng:

Bai 8 Nung 500 gam da voi chuTa 80% CaCO., (phan con lai la cac oxit nhom,

s^t (III) va silic), sau mot thdi gian thu diTdc chat r^n X va V lit khi Y

1 Tinh khoi Ming chat ran X, biet hieu suat phan huy CaCOa la 75%

2 Tinh % khoi lu-rfng CaO trong chat ran X

3. Cho khi Y sue ra't tif lit vao 800 gam dung dich NaOH 2% thi thu dmc miioi

gi, nong do bao nhieu %?

Bai giai

1 Phan u-ng nung da voi: CaCOi — ^ CaO + COjt

500x80 , ,

Tmh: nc.rn^ = = 4mol ^"^"^ 100x100

"cac03 phan huy = n c a o = nco2 = 4 X — = 3mol

Khoi li/ctng chat dn bang khoi lu'dng ban dau triif khoi luTdng CO2 bay di

Vi so mol NaOH < so mol CO2 nen ta chi thu difdc muoi axit:

CO2 + NaOH > NaHCOj

va nNaHCO.-, = n N a O H =0,4mol Nong do %NaHC03 = 0,4x84x100 = 4,1% 800 + 0,4x44

0ai 9* ^^"^ nhom kim loai trong khong khi mot thcJi gian thu diTcJc chat ran

/V CO kho'i lu'dng 2,802 gam Hoa tan chat ran A bang dung djch HCl di/ thay thoat ra 3,36 lit H2 ,

J Tinh % kho'i lu"dng ciaa Al va AI2O3 trong A

2 Tinh % Al bi oxi hoa thanh AI2O3

3 Neu hoa tan hoan toan cha't ran A bkng axit nitric dSc nong thi c6 bao nhieu

lit khi mau nau duy nhat thoat ra

Cho cac the tich khi do d dktc f

Bai giai

1 Cac phan ifng: MuA)

2A1+ 1,502 2A1 + 6HC1

2 Theo phan ijrng(l): nAihi.,xihoa = 2 x n^i^^^ =2x-^^zzO,002moI

Vay % Al bi oxi hoa bang = 0,002x100 = 1,96% 'ts; '

(4) (5)

0,1 + 0,002 3- Cac phan tfng hoa tan A bkng HNO3 dac:

AI2O3 + 6HNO3 > 2A1(N03)3 + 3H2O

Al + 6HNO3 > A1(N03)3 + 3NO21 + 3H2O

Vay the tich NO2 = 0,3 x 22,4 = 6,72 lit

^ai 10 Chia hon hcJp kim loai Cu - Al thanh hai phan bang nhau

Phan thu" nha't nung nong trong khong khi tdi phan tfng hoan toan thu difdc 18,2 gam hon hcJp hai oxit Hoa tan hoan toan phan thu" hai b^ng dung dich

H2SO4 dac nong thay bay ra 8,96 lit SO2 (d dktc) A; ',

^- Tinh so' mol moi kim loai trong hon hpp

^' Ne'u hoa tan hoan toan 14,93 gam kim loai X b^ng dung djch H2SO4 dac

nong va thu diTdc mot Itfdng SO2 nhiT tren thi X la kim loai gi? 139

a + - b = = 0,4

2 22,4

Giai he phiTcfng trinh ta co: a = 0,1 mol va b = 0,2 mol

2 Phan lirng hoa tan kim loai X hoa tri n:

1 Hoa tan m, gam A l va m2 gam Zn bang dung dich HCl diT thu diTdc nhffng the

tich nhu^ nhau H2 Tinh ti le mi : m2

2 Hoa tan hon hdp A l - Cu b^ng dung dich HCl cho tcfi khi khi ngiTng thoat ra ihav

con lai chat ran X Lay a gam chat r^n X nung trong khong khi tdi phan iJng

hoan loan thu dU'dc 1,36a gam oxit Hoi Al bi hoa tan hoan toan hay khong?

Bai giai

1 Cac phan iJng: 2A1 + 6HC1 > 2 A I C I 3 + 3H21 ( 0

Zn + 2HC1 > ZnClj + H21 (2)

Theo cac phan i?ng ( 1 , 2 ) , de c6 1 mol H2 bay ra can - mol A l hoac 1 m<''

Zn Vay ti le khoi liTdng: ^ 1 _ 3

Kg'u A l con: 2A1+ 1,502 > A I 2 O 3 , , (3)

jChi khi H2 ngijrng thoat ra, c6 the A l het hoac HCl he't

jsleu A l he't, chat ran con lai chi c6 Cu, va khi nung trong khong khi thu dU'dc CuO: nhu- vay ti le khoi lu'dng tang bKng '"cuO j 25 a gam thu

mcu 64

diTdc toi da 1,25a gam oxit d day khoi lu'dng thu dU'dc la 1,36a chiJng to

trong X phai c6 A l v i ty le tang khoi lu'dng cua A l (theo phan vlng 3) bang:

m j A , 2x27 '

Chu de 7 L a p c o n g thufc mpt c h i t ' ' l i I

Bai 1 Them tijT tiif dung dich HCl vao 10 gam muo'i cacbonat kim loai hoa tri I I , sau mot thdi gian thay lu'dng khi thoat ra da vu'dt qua 1,904 lit (dktc) va lifdng muo'i tao thanh da vu'dt qua 8,585 gam Hoi do la muo'i kim loai gi

trong so cac kim loai cho difdi day: Mg = 24, Ca = 40, Cu = 64, Ba = 137

Bai giai

Phan li-ng hoa tan: R C O 3 + 2HC1 > R C I 2 + H 2 O + C O 2 1

Khi khi thoat ra dung 1,904 lit thi:

nco2 =nRCi2 =nRC03 = ^ ^ = 0,085mol

ThiTc tc so mol R C O 3 Mn hdn 0,085 mol, do do ta c6: , ,

— ^ > 0 , 0 8 5 tu'cR<57,6

R + 60 Mat khac khoi Itfdng muoi (R + 71) x 0,085 > 8,585 gam ttfc R > 30

Ta CO 30 < R < 57,6

Vay do la muoi canxi cacbonat CaC03

^ai 2 Nung 25,28 gam hon hdp FeC03 va Fe.Oy trong khong khi (O2 dif) tdi

phan UTng hoan toan thu diTdc khi san pham A va 22,4 gam chat rj(n Cho khi

A ha'p thu he't vao 400ml dung dich Ba(0H)2 0,I5M thay c6 7,88 gam ket tua

2 Tinh: nQ ^^Oj, = ^ ^ = 0,04mol; nBa(OH)2 = 0,4x0,15 = 0,06mol

Trirdng hrtp CO2 diT: CO xay ra phan tfng (4)

Cich 1: Tong so mol CO2 = 0,06 + (0,06-0,04) = 0,08mol

P^O) pir(4)

Do do khoi lircfng FeC03 = 0,08 x 116 = 9,28 gam

' K h o l li/dng O2 tham gia cac phan tfttg (1, 2) b^ng

Do do O2 khong tham gia phan uTng (2) chuTng to FCjOy phai la FeaOs vi FeO

ho5c Fe304 thi bj oxi hoa thanh Fe203

Cach 2: Khoi liTdng Fe203 d phan uTng (1) bing ^ y ^ x 160 = 6,4gam

Do do khoi liTcJng Fe203 d phan Ung (2) b^ng 22,4 - 6,4 = 16 gam

Khoi liTdng FeC03 ban dau bang 9,28 g do do kho'i liTcJng Fe^Oy ban dau

bang 25,28 - 9,28 = 16 gam Dieu do chufng to Fe^Oy ban dau chinh la Fe203

Trirdng hdp CO2 thieu: khong c6 phan iJng (4)

Luc do n^oj = i^BaC03 ~ 0'04r"ol

TriTcfng hdp nay loai vi lufdng Fe^Oy Idn hdn Fe203 d phan tfng (2)

Bai 3 De hoa tan 3,9 gam kim loai X can diing V ml dung dich HCl va c6 LB^''

• lit H2 bay ra (d dktc) Mat khac de hoa tan 3,2 gam oxit cua kim loai Y cung

can diing V ml dung dich HCl d Iren. Hoi X, Y la cac kim loai gi?

Theo phan iirng (2), ta co ti le: — — = ^ " ^ g ^ => Y = 18,67m

fjghiem thich hdp: m = 3, Y = 56 (Y la Fe) Cdc kim loai X, Y la Zn va Fe

Bai 4

J C6 the coi sat tijf oxit la hon hdp c6 cilng so' mol ciia FeO va Fe203 dufdc

khong,tai sao?

2 De san xuat mot lifdng gang nhU' nhau ngifcfi ta da diing mi tan quSng

hematit chuTa 60% FczO^ va m2 ta'n quSng manhetit chiJa 69,6% Fe304 Tinh

ti le m, : m2

Baigiai ,

1 Ve hinh thtfc cong thtfc Fe304 c6 the viet thanh FeO.Fe203, nhiftig kh6ng the xem Fe304 nhiT hon hdp 2 oxit FeO va Fe203 cilng so' mol Vi hon hdp FeO

va Fe203 khong c6 cac tinh chat ciia Fe304 nhiT khong c6 til tinh

2 Vi san xua't mot liTcJng gang nhuT nhau nen li/dng Fe cung nhu* nhau Gia suf de

c6 1 tan Fe thi can bao nhieu tan quSng?

Theo cong thuTc Fe203 va theo thanh phan quang hematit, de c6 1 tan Fe can:

J Y ^ x ^ = 2,38 tafn quSng hematit j^,,i„^M riftii ?irf; n

TirOng tif doi vdi quang manhetit Fe304: j ,55

232 100 , ^„ ,

X = 1,98 tan quSng manhetit

168 69,6 ^ * , Vay ti le m, : m2 = 2,38 : 1,98 = 1,2 ,a, ^< , ' >

B&iS

^- Hoa tan hoan toan 6,66 gam tinh th^ Al2 (S04)3. nHjO v^o nvCdc thanh dung

dich A Lay 1/10 dung dich A cho tac dung vdi dung dich BaCb dU" thi thu diTcJc

0.699 gam ket tua Xac dinh cong thtfc cua tinh the muoi sunfat cua nhom

^- Hoa tan 24,4 gam BaCl2. xHjO vao 175,6 gam niTdtc thu dircfc dung dich 10,4%

• C6 can ra't txi tuf 200ml dung dich CUSO4 0,2M thu diTdc 10 gam tinh thi

CUSO4 PH2O. Tinh p

143

B a i giai

1 Hoa tan tinh the Al2 (S04)3. nH20 v^o nvCdc ta thu diTdc dung dich Al2(S0<,)

Khi cho BaCl2 vao xay ra phan v?ng:

Al2 (S04)3 + 3BaCl2 > 3BaS04>| + 2AICI3 (1)

Theo phan uTng (1) ta thay cvt 1 mol tinh the, tvtc (342 + 18n) gam tinh the th^

dtfdc: 3 X 233 = 699 gam ket tua BaS04 Theo dieu kien bai toan cho thv

= 0,666 g tinh the thu diTdc 0,699 gam ket tua, nen ta c6 ti le:

2 KhoiliTcfng BaCl2 nguyen chat

= (24,4 + 1 7 5 , 6 ) X ^ = 20,8gam, tfng vdi — = O,lmol '

100 208 >i

Dod6 nH2o=0.1xx = — ^ - r ^ : — ^24,4-20,8 18 = 0,2mol

Rut ra X = 2 Cong thtfc ciia tinh the 1^ BaCl2 2H2O

3 Tinh ncuso4 =0,2x0,2 = 0,04mol

Do d6 U H J O = 0.04p = = 0,2mol

18

Rut ra p = 5 Cong thuTc ciaa tinh the la CUSO4. SHjO

C6 the tinh M ii„h ,hd = = 250 Do do p = = 5

Bai 6 Hoa tan hoan toan 14,2 gam hon hdp A gom MgCOs va mot muoi cacbonat

kim loai R b^ng mot liTOng vfiTa du dung dich HCI 7,3% thu diTdc dung dich D

va 3,36 lit CO2 (6 dktc) Nong do MgClj trong dung djch D la 6,028%

1 Xac dinh kim loai R va tinh % khoi liTdng moi chat trong A

2 Cho dung dich NaOH duT vao D roi lay ket tua nung ngoai khong khi cte"

khoi liTdng khong doi thi thu dUdc bao nhieu gam cha't ran?

B a i giai

1 Cac phan i^ng xay ra:

MgCOs + 2HC1 > MgClz + H2O + CO21 (1)

R2 (C03)„ + 2nHCI > 2RCI„ + nC021 + nHzO (2)

144

Tinh: UHCI = 2 n^o^ = 2 x = 0,3mol

x x ^ , ^ 0,3x36,5x100 g^hoi Wdng dung dich HCI 7,3% = — = 150gam KhS'i liTdng dung dich D b^ng

jnp = fflA + mHci + mco2 = 14,2 + 150 - 0,15 x 44 = 157,6 gam

S6'molC02C[phanu'ng(l)= n^gco.^ =nMgCi2 " ^ ^ ^ j ^ ^ ^ J ^ =

Do do khoi Itfdng MgCOs = 0,1 x 84 = 8,4 gam

S6' mol CO2 d phan uTng (2) = 0,15 - 0,1 = 0,05 mol, do do so mol ,0?, • • R2 (C03)n = 0,05 0,05 (2R + 60n) = 14,2 - 8,4 - 5,8gam

Mg(0H)2 4Fe(OH)2 + O2

MgCl2 + H20 + C02t

•^FeCl2 + H20 + C02t Mg(0H)2>l, +2NaCl ->Fe(0H)2>l, +2NaCl -> MgO + HsOt

2Fe203 + 4H20t

(1) (2) (

(3)

(4)

(5) (6)

Chat ran gom MgO va Fe203, khoi lifdng chat r^n b^ng:

0,1x40 + — x i x 160 = 8gam

^ ^ T M ^ 1 1 6 _ ^

KL Fe203

Bai 7 Oxi hoa hoan toan p gam kim loai X thu dtfdc l,889p gam oxit, ho^ tan

mu6'i cacbonat kim loai Y (hoa tri II) bang mot liJdng vijfa du dung dich

H2SO4 9,8% thu diTdc dung dich muoi sunfat 14,18%

!• Hoi X, Y la kim loai gi? ^ «,,1 •

2- A \n hdp kim loai X, Y

^) Hoa tan hoan toan 3,61 gam A bang dung dich HCI thu diTdc 2,128 lit H2

(dktc) Tinh so' mol moi kim loai trong 3,61 gam A

Cho 3,61 gam A vao 200 ml dung dich hon hdp AgN03 va Cu(N03)2 Sau khi kS't tua cac phan tfng thu dtfdc chat ran B nSng 8,12 gam chiJa 3 kim loai

Hoa tan B bang dung dich HCI diTthay bay ra 0,672 lit H2 (dktc)

Tinh nong dp mol cua AgNOs va Cu(N03)2 trong dung dich ban dau

145

1. T i m X : 2 X + - O 2

2

Bki giai

X20„

Theo (1) ta CO ti le khoi luTdng cua kim loai oxit: 2 M ,

Rut ra Mx = 9n, chi c6 n = 3; Mx = 27 (Al) la diing

T i m Y: Y C O 3 + H2SO4 > Y S O 4 + H2O + CO21

(1)

2 M x + 16n l,889p

2 a) Cac phan ufng hoa tan kim loai trong HCl:

Giai ra ta difdc a = 0,03 mol va b = 0,05 mol

b) V i A l hoat dpng hdn Fe va vi AgNOs tinh oxi hod manh hdn Cu(N03)2 n c n

phan tfug so'mpt la: - A

Al + 3AgN03 > Al(N03 )3 + 3 A g i (5)

Sau phan uTng (5) neu A l het (hoac viTa du) thi nAg = 0,03 x 3 = 0,09

tifc khoi lirpng Ag = 0,09 x 108 = 9,72 gam trai vdi dieu kien bai toan cho,

^ do do sau phan iJng (5) se xay ra cac phan uTng:

4 , 2A1 + 3Cu(N03)2 > 2A1(N03)3 + 3Cu >!' (6)

Fe + Cu(N03)2 > Fe(N03)2 + Cu J' (7)

Phan u-ng hoa tan B: Fe + 2HC1 > FeClj + H21 (8)

Gpi X , y la so mol AgN03, Cu(N03)2 trong dung dich dau ta c6 cac phu'cfng trinh:

- Khoi lirpng B = 108x + 64y + 56 X 0 ^ =8,12

-J Cho a gam bpt kim loai M c6 hoa tri khong doi vao 500 ml dung dich hon

hpp gom Cu(N03 )2 va AgNOj deu c6 nong dp 0,4 mol/1 Sau khi cdc phan

ij-ng xay ra hoan toan, ta Ipc difdc (a + 27,2) gam chat r^n gom 3 kim loai va difdc mot dung dich chi chiJa mot muoi tan Hay xac dinh kim loai M va so mol muoi nitrat cua no trong dung dich

2 Hoa tan hoan toan 9,9 gam hon hpp kim loai A hoa trj n va kim loai B hoa tri

m bang dung dich HNO3 loang thu diTpc dung dich X va 6,72 lit khi duy nhat

NO (dktc) Tinh tong khoi lufPng muoi nitrat c6 trong dung dich X

Bai giai ?« o; lii* u

Gpi n la hoa tri cua kim loai, ta c6 cac phan i?ng: >i«i»*f

M + nAgN03 > M(N03)„ + n A g i (1)

2M + nCu(N03 )2 > 2M(N03)n + nCu 4 (2)

Tinh: n c u(N03)2 = n A g N 0 3 = 0 5 x 0 , 4 = 0,2mol^;^^_,.^ ,^

Vay chat ran chiJa 3 kim loai chiJng to dir M va cac phan iJng ( 1 , 2) xay ra ho^n toan Theo khoi lu'png tang d cac phan iJng ( 1 , 2) ta cd phu^dng trinh ve tong kho'i lu'dng tang sau day:

M 2 M (108 ) X 0,2 + (64 - — ) X 0,2 = 27,2 'R'V n •^'>

n n

Giai ra ta cd: — = 36 hay M = 12n -^'^

n M

Nghiem thich hdp n = 2 => M = 24 Mg Vay M la kim loai Mg ^'f'^^

So mol muoi Mg(N03 )2 bang - so mol AgNOj + so mol Cu(N03)2

Khoi iMdng goc: - N O 3 = 0,9 x 62 = 55,8 gam

Vay long khoi liTOng mu6'i = 9,9 + 55,8 = 65,7 gam

Bai 9 Cho cac k i m loai X hoa trj I , Y hoa tri I I va Z hoa tri 11, K L N T tiTdng iJng

la Mx, M Y va Mz Nhiing 2 thanh kim loai Z c6 cilng khoi liTcJng v^o 2 dung

dich muoi nitrat cua X va Y Ngifdi ta nhan thay khi so mol muoi nitrat cua 2

f^: trong 2 dung dich hlng nhau thi khoi liTdng thanh thtf nhat tang a% c6n thatih

jf^, thtf 2 tang b% Gia suf tat ca kim loai X, Y bam het vao thanh kim loai Z

•g^ Lap bieu thtfc tinh Mz theo Mx, My, a, b

Trong do p la khoi liTdng ban dau cua 2 thanh kim loai Z

Chia (I), (II) ve theo vS", ta c6: |M2LZMz_ = i

^ M y - M z ^

r ^ - - A , ^ J./, 6 b M x - 3 a M Y

Giai ra ta diTdc Mz =

2 ( b - a )

Bai 10 Cho dong khi C O di qua 6'ng stf nung nong, difng 8,12 gam m6t oxit cua

kim loai M khuT het o x i thslnh kim loai Khi di ra khoi ong stf cho Ipi tir t^^ qua

binh difng liTcfng dir dung dich Ba(0H)2 thS'y tao th^nh 27,58 gam ket tua

tr^ng Lay kim loai thu diTdc hoa tan hoan t o i n b^ng dung dich HCl thay bay

ra 2,352 lit H2 (d dktc) Hay xic djnh kim loai M v^ cong thtfc ciia oxit

Goi a la so mol oxit, theo cdc phan urng ( 1 , 2, 3) ta c6:

ay = nco2 = ne.coj = = 0,14mol (D

Khi n = 2 => - = - ttfc M3O4 luc dd a = 0,035 mol v£k KLPT cua M3O4 bling

^ = 232 ttfc 3 M + 32 = 232 hay M = 56, d6 la Fe va cong thtfc oxit la

Fe304 (s^t tuf oxit)

Bai 11 Hoa tan 3,2 gam oxit M20„, trong mot ItfcJng vifa du dung dich H2SO4

10% thu duTcfc dung dich muoi sunfat 12,9% C6 can dung dich muoi roi lam lanh dung dich thay thodt ra 7,868 gam tinh the muoi sunfat vdi hieu suat mu6'i ket tinh la 70% Hay xdc dinh cong thtfc cua tinh the

Bai giai Theo phan iJug M20n + nH2S04 > M2(S04)„ + nH20 (1)

Ci? 1 mol M20n c i n 98n gam H2SO4 nguyen cha't ttfc c i n = 980n (gam) dung dich H2SO4 10% va tao ra 1 mol muoi sunfat, n6n theo c6ng thtfc n6ng dp C% ta c6

2 M + 96n

C % = 12,9 =

2 M + 16n + 980n

R i J t r a M = 18,66n

Nghiem thich hdp: n = 3 => M = 56 (Fe) Vay cong thtfc cua oxit Ik

FCiOj-"Fe203 = np,2(so4)3 = ^ = 0,02mol 2

VI hipu suSft ket tinh chi 70% nen mol mu6i b i n g : ' ^

0,02 X 0,70 = 0,014 mol c6ng thtfc ciia muoi 1^: Fe2(S04)3 XH2O, ta c6:

(400 + 18x) X 0,014 = 7,868 Giii ra x = 9

^ c6ng thu-c cua muoi la Fe2(S04)3 9H2O ' " '

' ^2 Hoa tan hoan toan a gam kim loai R c6 hoa tri khong ddi n v£lo b gam

j^""g dich HCl diTdc dung dich D Them 240 gam dung dich NaHCOa 7% vao

^hi vtra du tac dung het vdi HCl dU thu difdc dung dich E trong do nong do

% cua NaCl la 2,5% va cua muoi RC1„ la 8,12% Them tiep liTdng dir du^

dich N a O H vao E, sau d6 loc lay ke't tua roi nung den khoi ItfOng khong

thi thu diTdc 16 gam cha't r^n

1 V i e t cac PTPLf xay ra >

i RCln + nNaOH )• R(OH)„ + nNaCl (3)

2R(OH)„ — ^ RjOn + nHaO (4) ) >•

2. Theo (2) n N a H c o 3 = " N a C i = „ = 0,2mol ^ ,_

K U i j lo-i j o u r j i 100x84

mm:-j^w'y,^ 1 ^ - U T T 0,2x58,5x100 ^ n i i •

Khoi liTdng dung dich E = — = 468,Ogam

Kho'i lifdng RC1„ = "^^^^^^'^^ = 38gam

Theo cic phiTdng trinh (3, 4) ta c6 ti le: ^^JlLlil = 2R + 16n

• 38 16

^ R u t r a M = 12n u^,

Chi CO n = 2, M = 24 (Mg) la diing

3 Xac dinh C% cua HCl

Theo (1, 2, 4): nMg = nwgo = = 0,4mol

Do d6 khoi liTdng Mg 1^ a = 0,4 x 24 = 9,6 gam

" H J = "Mg = 0,4mol; Uco^ - nN.ci = 0,2mol

Khoi Itfdng dung dich D = 9,6 + HCI - 0,4 X 2 = 8,8 + HCI

Bai 13 Hoa tan hoan to^n p gam kim loai R b i n g dung dich HCl thu diTcJc V li'

H2 (dktc) Mat khac hoa tan hoan toan p gam kim loai R bang dung di'^''

HNO3 loang thu diTdc V lit khi NO duy nhat

150

J 50 sanh hoa tri cua R trong 2 muo'i clorua va nitrat

2 R la kim loai gi? Biet khoi Itfdng muoi nitrat tao thanh gap 1,905 Ian khoi liTdng muo'i clorua

Bai giai -••ive:, '

j V i khong biet kim loai R chi c6 1 hod tri duy nhat nen ta goi n va m la hoa tri ' c^a R khi hoa tan trong HCl va trong HNO3 Cac PTPLf: j ^ j ^ , q ^ j,.,

2 V i so mol 2 muoi nhiT nhau (deu b i n g a) nen ta co: ; ih 1

M + 62m = 1,905 ( M + 35,5n), trong do M la K L N T cua R „ ^ , , Thay m = 1,5n ta rut ra M = 28n

Chi CO n = 2 ; M = 56 la dung Vay R la Fe f , i ^ « l ^ T J

^-Bai 14 Chia 8,64 gam hon hdp Fe, FeO va Fe203 thanh 2 phan b i n g nhau Phan

mot cho cac coc diTng lifdng dtf dung dich CUSO4, sau khi phan iJng hoan loan tha'y trong coc co 4,4 gam cha't ran Hoa tan het phan hai blng dung dich HNO3 loang, thu di/dc dung dich A va 0,448 lit khi NO duy nha't (dktc)

Co can tii tiS dung dich A thu diTdc 24,24 gam mot muo'i sat duy nhat B

1 Tinh % khoi luTdng moi chat trong hon hdp ban dau •

2 xac dinh CTPT cua muoi B iO-!

Bai giai

1- cac phiWng trinh phan liTng: "^^'l' *

Fe + CUSO4 > FeS04 + Cui ^^^^^^ (1) FeO + CUSO4 > khong xay ra

Fe203 + CUSO4 ' > khong xay ra

Fe + 4HNO3 > Fe(N03)3 + NO t + 2H2O (2) 3FeO + IOHNO3 > 3Fe(N03)3 + N O t + 5H2O (3) Fe203 + 6HNO3 > 2Fe(N03)3 + 3H2O (4)

I Goi X , y, z la so mol cua Fe, FeO, Fe203 ta c6 cac phiTdng trinh:

151