PHÂN LOẠI & HƯỚNG DẪN GIẢI CÁC CHUYÊN ĐỀ HÓA HỌC 11 - HUỲNH VĂN ÚT

8/10/2019 PHN LOI & HNG DN GII CC CHUYN HA HC 11 - HUNH VN T

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HUNH VN T - HUNH NHIN QUYN

PHM TH TI - PHM TH HNG THMGii thng Sch h ay Vit Nam

GV. Bi dng hc sih gii

PHN LOI & HNG DN

GII cc CHUYN

H O H C

w ^ e . 7 0 3

NH XUT BN I HC QUC GIA H NI

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W.DAYKEMQUYNHON.UCOZ.COM

g gp PDF bi GV. Nguyn Thanh T


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^ \ i ni u

Cng vi vic i mi chng trnh v sch gio khoa, vic i

mi phng php hc i vi hc sinh l mt trong nhng vn c

bn, l mt trong nhng mc tiu phn u ca nhiu thy c gio

v hc sinh. Nhn thc c iu tc gi xin chn thnh gii

hiu n bn c quyn sch:

P H N L O I V H N G D A N G I I C C C H U Y N H H C 1 1 .

Ni dung quyn sch gm 9 chuyn ng vi 9 chng trong

ch gio khoa hin hnh. Mi chng gm:

A . L T H U Y T

B . C C D N G B T P T H E O C H

Trong mi chuyn u phn dng cc loi ton theo ch

phng php gii cho mi ch , trong mi ch u c bi

p minh ha. Cc bi tp trong mi ch c khai thc nhiuha cnh khc nhau v c sp xp t c bn n nng cao. Sau

mi bi c li gi chi tit, r rng nhm gip cc em i chiu

i k t qu sa u k h i t g i i b i ton .

Chng ti c gng khai thc trit cc dng ton khc nhau

l thuyt cng nh bi tp v c nhiu cch gii khc nhau nhm

m phong ph kin thc cho cc em.

Vi khong hn 2 0 0 trang sch, cha chng ti th hin

t c tng v mong c ca mnh v c th cn mt s sai st.

Rt mong quyn sch c qu thy c gio v cc em hc sinh n

hn, gp .

Tc g



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CHUYN 1.

S IN LI

A. THUYT

1. S BI N LI: L qu Mnh phn cc cht trong nc hoc nng chyton ion (ion m v ion dng).

S in i c biu din bng phng trnh in i.

2. PHN LO I CC CHT IN LI

a) nh ngha: in a (anpha) ca mt cht n i t s cas phn t phn i ra ion. (n) v tng s phi t ha tan (n0)

Cng thc: a = n0 I

b) Cht in i mnh: L cht kh tan trong nc, cc phn t hatan u phn ra oh.c) Cht in li yu: L cht khi tan trong nc ch c mt phn s

phn ha tan phn h ra ion, phn cn i vn tn ti di dngphn t trong dung dch.

3. AXT, BAZ V MUI) A xitr baz theo thuyt A -r- ni-u t

+) Axit cht k hi tan trong nc phn li ra cation T

+) Baz cht khi tan trong nc phn i ra anion 'OrT

b) Axit nhiu nc: L nhng axit m mt phn t phn i nhiu nc ra ion H*.c) Baz nhiu nc: L nhng baz m m t phn t phn i nhiu ncra ion OH~.d) Hroxit lng tnh: L nhng cht kh an trong nc va c th

phn h nhaxit , va c th phn i nh baz.

e) Khi nim v axit v baz theo thuyt Jron -stt

Axit cht nhng proton (H*). Baz l cht nhn proton.

f ) Hng s phn i axt:Xt cn bng: CH3COOH H* CH3COO~

[H*3[CH3COO~][CH3CGOH]

Kq l hng s phn li axit. i vi axit xc nh, Ka ch ph thuc vo nh it . Gi tr Ka ca axit cng nh, lc axit ca n cng yu.

PL & HD GIi CC CHUYN D HA HC 11 5



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g) Hng s phn i baz: Xt cn bng: NH3+ H- O < >NH4++ OH~

r _ [NH4+]OH']Ab "

[NH3] -

K, hng s phn i bay.. Kb ca mt baz xc nh ch ph thuc vonhit . Gi tr Kt, ca baz cng nh , c baz ca n cng yu.

h) Mui: Mui hp cht, kh tan trong nc phn i ra cation kimloi (hoc cation NH^j v anion gc axit. Mui c hai o: Muitrung ha v mui axit.

4. S IN LI CA NC, pH, CHT CH TH AX IT- BAZ

a) Tch s ion ca nc:

K= [HS F =* K * =K^ = m o H - ] = l -4 n

T (*) T ]= [OH~J = VlO 14 = 0~7mo/.

b) ngha tch s ion ca nc:- Mi-trng axit: /H+] > 10~7M

- Mi trng -ung tnh: [H* = 10~7M

-M i trng kim: r t ] < 10~7 M.

c) Khi nim v pH: Nng ion [ht] = 10^ mo/ th a tr s pHca dung dch.

Cng thc tnh pH v pH: pH= -glrt; pOH- -g[OH~

Vi dung dch axit: [ t f > icr7=>pH pH >7.p t +pOH = 14.

d) Cht ch th axit bazo:

- Axit Trung tnh Kim

Qu tm

Phenophtaein

Khng mu

Tm

Khng mu

Xanh

Hng

Ch : C th trn ln mt s cht ch th c khong pt i mu ktip nhau, Q c hn hp cht ch th- baz vn nng.

5. PHN NG TRAO ION TRONG DUNG DCH CC CHAT IN LIa) iu kin xy ra phn ng trao i ion trong dung dch cc chtin i.

~ Phn ng to thnh cht kt a:

Na2S 0 4+ BaC ---------> 2NaCl+ BaS04

Phng trnh ion: Ba2*r SO/~ ------ > BaS04

6 PL & HO GII CC CHUYN HA HQC 11



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- Phn ng to thnh cht in i yu:

+) Phn ng to thnh nc:

NaOH + HC --------- NaC + H20

=>Phng trnh ion: -T + 0H~ --------- > H20+) Phn ng to thnh ion phc:

AgCl(r)+ 2NH-J-------- [Ag(NH3)2]C+) Phn ng to thnh axit yu:

HC+ CHjCOONa --------- > CH,COOH + NaC

=>Phng trnh ion: CH3COO~ + H * --------- > CH3CO.OH-Phn ng to thnh cht kh:

CaC03(r) + 2HC --------- C02 f+ H20 + CaC2.

Phng trnh ion: CaC03(rj+ 2FT ---------> Ca^ + c o 2T+H20

b) Khi nim S fiiphn ca muiPhn ng trao i iou gia mui ha tan v nc lm cho pH bini phn ng thy phn ca mui.

c) iu ki n thy phn ca m t s mui:

+) Mui trung ha to bi gc baz mnh v gc axt yu, khi tantrong nc gc axit yu b thy phn, mi trng ca dung dch kim (pH > 7) nh: CHsCOONa, K2S, Na2C03 .

+) Mui trung ha to bi gc baz yu v gc axit mnh, khi tantrong nc gc baz yu b thy phn, m cho dung dch c tnh axit

(pH < 7} nh: Fe(NO^}3, NH4C, ZnBr2.+) Mui trng ha to bi gc baz m nh v gc a xit m nh, kh antrong nc khng b thy phn, mi trng ca dung dch vn trungtnh (pH =7) nh- NaCI, KN03, K.

C C D NG B TP THEO CH

CCH XC NH NNG MGL/L

CA I0N TRONG DUNG DCH N Li MNH

V DIN LI YU DA VO D N n G

PHNG PHP- Vit phng trnh in i- Da vo t mo ca phng trnh in , t nng ca chtin li msuy ra nng ca ion cn xc nh

Ch : Cht in li mnh phn li hon ton.HO GII CC CHUYN HA HC 11 7



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- T cng thc tnh:

a= s phn t A phn li _ nAphanu _ C phnlis phn t A ha tan CA

r A ha tan A ha tan

(vi A cht in li yu ang xt)- Suy ra CA phn h:

+) Vit phng trnh in li ca cht A

+) Da vo t mo ca phng trnh in i, t Hng phn cht A, suy ra nng moi ca on cn xc nh

Ch : Cht in i yu phn i khng hon ton, trong phng

in dng du ^ '

2. V D MINH HAV d 1. C mt dung dch axit sunuhiric H2S 0,1M. Bit rng H

mt axit hai chc (iaxit) c th phn li theo hai giai on:

- Giai on 1:H2S H+ + HS ; Kl = 10"7- Giai on 2:HS~ ;==s ET + ; K2 = 1,3.1013

Tnh nng mol/1 ca ion H+v suy ra pH ca duug dch.

V d 2. Tnh nng mol/1 ca cation v anion trong cjffc dung dch s

a)N3PO4 0,1M b) HNO3 0,02M c) KOH 0,01M

V d 3. Cho dung dch axit CH3COOH 0,1M

Bit Kch?cooh =1,75.10 ; 1E -CHJCOOH= ~4j757

- Tnh nng ca cc ion trong dung dch v tnh pH.

- Tnh in li a ca axit trn.V d 4. Ha tan 0,585 gam NaCl vo nc, thnh 0,5 lt dung dch,

nh nng raol cua cc ion trong dung dch thu c.VI d 5. Dung dch 0,043M c in li l 2%.Xc nh nng ca

ion trong dung dch axit .V d 6. Dung dch ca mt axit yu mt nc c nng 0,0IM v

li l 31,7%. Xc nh [H+] trong dung dch axit .HNG DN G

V d 1 . Cch l Ta thy Ki K2nn s phn li ch yu xy ra giai on 1, do th b qua s phn li giai on 2.

H2S H+ + HS" ; Ki = 107

Ban u: 0,1M

Phn li: xM xM xM

Cn bng: (0,1 - x)M xM xMg PL & HD GII CC CHUYN HA H



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rH+l[HS-] x2 ,M: K, = L- rJ-L . J = = 10'7. (*)

[H2S] ( 0 , 1 - x )

Gi s X 0,1 => 0,1 - X 0,1.

T (*) => X 2 = 10"s => X = KT4= [H+3=> pH = -IglO4= 4.

Cch 2:V H2S l axit yu nn c th dng cng thc:

pH = (pK a - lgC)i

Vi Ka Ka = l 3Na+ + P0^

T t l mol ca-phng trnh in li ta c:

[Na+] = 3 CM = 3 X0,1= 0,3M

[POj- ]=CMv = 0,1M

b) Phng trnh in li: HN03 ----- -> H++ NO3

T t l mol ca phng trnh in li ta c:

[H+] = [N O -]= C =0,02M- MHNC>3

c) Phng trnh in i: KOH ------ > K++ OH~

T ti l mol ca phng trnh in li ta c:

[1C] = [ o m = CM = 0,0IMKOH

V d 3. Phng trnh din li:

CH3COH ------ > CH3COO' + H+ (1)

Ban u: CM

in li: Ca Ca Ca

Cn bng: C - C t x . Ca C

Gi a l in li ca axit "

^ TX [h+]|cH 3COO 1 [h+Hng s Cn bng K,: Ka = [ch ,c o H [ c i f c o o s ]

+; CcA K . - | ^ =[CH3COOH] C -Cu C (l- a ) 1 - a

V CH3COOH l axit yu nn a nh, gi s 1 - a 1PL & HD GI CC CHUYN HA HQC 11 Q



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=>Ka = Ca2 => Ca = ,/C.Ka

[K+] = ^/C.K, = -jo,l X 1,75 X10"5 = vl,75.10 ' 3 = 1,323. l^(moI/l)

=> pH = -gV C = I (- gKa - IgC) = I (4,757 + 1) = 2,88

in li a: Ca = ^C-K* => a = ?

= V0 1 X = J l ,75.10'" = 1,32-KT2hay 1,32%0,1

CdcA 2; Gi [H+] = [CH3COO] = x; [CH3COOH] = C -x

__*K [ h * ][c H,COQ- } _ x

[CH3COOH] C -x

^ X2 = K a( C- x) ^ X2 + K a - K a. c = 0o X2+1,75.1Q~'SX- 1,75.10'6 =0 (1)

Gii phng trnh (1) v loi nghim m, ta c:

X= 1,3.103=[H+] ^ pH - - l g l ,3.10"'*2,88

V d 4. Trc ht tnh s mol ri suy ra nng moi ca NaCl:

nN.c> = ^ = 0,01 (mol) => CM(.,cl, = = 0.02M

- Phng trnh in i: NaC ------ Na++ c rSuy ra: [Na+] = [Cl'] = Cw = 0,02M.

*NaCl

V d 5.

- T CCHC00H phn li => CCH phn li = 0,043 X2%= 0,86 X 10_3M.

- Phng trinh in li ta c:

[CH3COO'J = IH+] = CcfxX)H phn li = 0,86 X 10"3M.

V d 6.

- t cng thc ca axit yu mt nc l HA.CH,COOH phn l

a =Ct,COOH ka an

^ Cch:cooh Phn u = 0,01 X31,7% = 3,17 XO^M.

- Phng trnh in li: H ^ = H++ A" r -

T phng trnh^in li ta c: [H+] = Cha pHn li ='3,17 X 10"3M.10 PL &no 6II CC CHUYN HA HOC 11



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N XC NH NNG B MOL/L CA ION TRONG DUNGCH 2. \ DCH CHT IN Li YU DA VO HNG s PHN

-------------- / LI AXIT V BAZ. IN LI a

. PHNG PHP- Vit phng trnh n li ca cht in i yu

- t n s nng phn li ca cht in i ang xt, t suy ranng ca cc ion r cht in li ti cn bng.

- p dng biu thc hng s cn big, p phig trnh rgii, tm n.

- p dng cng thc:

s phn t A phn li s" mol A phn lia = ----------- ----------: ~ --------

s phn t A ha tan s mol A ha tan

(Vi A cht in li yu c xt)

- a thng c biu din di dng %.

- Gi tr ca a thuc khong: 0< a < hav 0< a < 100%

V D MINH HAd 1 . Tnh nng mol cion H+ca dung dch CH3COOH 0,1M, bithng s phn li axit Ka = l,7-10-5.

d 2. Tnh nng mo/1 ca ion OH' c trong dung dch NH3 0,1M,bit hng s phn li baz = 1}8.10SM.

d 3. Trong 1 lt dung dch CH3COOH 0,01M c 6,26.1021 phn t cha

phn li ra ion. Bit s' Avgaro l 6,023.1023. Hy tnh in li cadung dch axit trn.

d 4. Tnh in ca axt axetic trong dung dch 0,01M; nu trong500ml dung dch c 3,13.1021 ht phn t (hoc ion). Ly s" Avgaro= 6,02.1023

d 5. Trong dung dch CH3COOH 0,043M, c 100 phn t ha tan chc hai phn t phn li r ion. Vy in i a l bao nhiu?

d 6. Dung dch CH3COOK 0,1M c [H+] = 1,32 .10_3M. Tnh in li

a ca axit nng . d 7. Nng ca ion H+ trong dung dch CH3COOH 0,1M l 030013M.

Xc nh in li ca dung dch CH3COOH trn.

d 8 . Bit dung dch axit CH3COOH c KCIi t;oH= 1,8-105. Tnh din li

a ca CH3COOH 0,1M cho trn.

d 9. Cho dung dch axit fomic HCOOH 0,46% (d = 1 g/ml) c pH = 3.Hi in li ca axit fomic l bao nhiu?

& HO SI CC CHUYN HA HC 11 --}



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V d 10 . Tnh in li a trong cc trng hp sau:

a) Dung ch HC00H1M c Ka= 1,77.10^.

b) Dung ch CH3COOH IM, bit pH ca dung dch l 4.

HNG D N GII

V d 1 . CH3COOH CH3COO" + H+

Ban u: 0,1

Phn li: X X XCn bng: 0,1 - X X X

^ H W ! o i , 75.10-[CHgCOOHl 0,1-X

V CH3COOH axit yu nn X X2 = 1,75.1CT6 =>X= 1,32.10~30,1

Vy [H+j = 1,32.10"3M.

V d 2. NH3+ H20 ^ NH; + OH

Ban u: 0,1

Phn li: X XXCn bng: 0,1 - X X X

1,8 X10-' = -JEL,[NHs] 0,1-X

Tng t gii c X= 1,34 X 10'3. Vy [OHT] = 1,34 X 10SM.

V d 3. S" phn t CHsCOOH c trong 1 lt dung dch CH3COOH 0,0l: 0,06023.1023.S" ion c trong 1lt dung dch CH3COOH:Ca = 0,0626.1023 - 0,06023.1023 = 0,00237.1023

= 0,00237^.0^ = 3,93%.0,06023.10 J *

V d 4. Cch I .Trong 1 lt ung eh CH3COOH c 6,26.1021 ht.

Trong 1 lt dung dch CH3COOH 0,01m c 6,02.1021 phn t.

Phng trnh in li: CH3COOH CHgCOO' + H+ (

Ban u: 6,02.1c?1 X Xin Li: X X XCn bng: (6,02.1021 - x) X XTa c: 6,02.1021 - X+ 2x = 6,26.1021 X= 0,24.1021

0 24 1o21Vy in li: a = - - - V X 100 = 3,99%

6 ,02 .10s1J2 PL & HD GII CC CHUYN HA HO



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Cch 2. Phng trnh in i ca axit axetic:

c h 3c o o h - CH3COor + H+

Ban u: 0,01

Phn ng: Xmol Xmol X moi

Cn bng: (0,01 - x) mol Xmol X mol

=> 0,01 + X = 6,2 - 1(C = 0,010399 => X = 0,0003996 ,02.10

Vy in li: a = Q>0Q399 x \Q(y% - 399%.J 0,01

V du 5. a = = 0,02 hay 2%100

V d 6. CH3CO OH = C H 3 C 00 + H +

. T phng trnh in li, ta c: B==[H+] = 1;32.10"SM.

, _ 1,32.10"3 ^ 1Vy a = ----- - 1,32%.

0,1

V d 7. Phng trnh in li;

CH3COOH CH3COO' + H+

Ban du: 0,1MCn bng: (0,1 - 0,0013)M 0.0013M

^ a = _ 0.00i i - = 0,13 hay 1,3% (0,1 - 0,0013 0,1)0,1-0,0013

V d 8 . CH3COOH CH3C00 +

(mol) (0,1 - x) X X

j~H+~| JCH3COO~l R X2Ta c: K = L J = 1,8.10=

[CH3COOH] 0,1-X

Gi s X 0,1 => 0,1 X%0,1 => X2= 1,8-106=s>X= 1,34.10-3.

1,34.10~3

10'V d 9 .Phng trnh in li: HCOOH -~= s H++ HCOO.

Ta c: pH = -]g[H+J= 3 => +l = 10~\n 0,46X1000CM__= -------- = 0,1 (mol/1)Mhcooh 100 X46

M Cion = a .c ^ [H+] = c .a => a = = = 102hay 1%.u u, 1

PL & HD GII CC CHYN HA HOC 11

Vy a = = 1,34%.



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12/253

HCOOH > HCOCT + H

IM 0 0a a a(X - a) a a

V d 10. a) Phng trnh (lin li:

Ban u:in li:

Trng thi cn bng: (1 - a)

Suy ra: Ka = = 1,77.l)4=> a-= 0,0131 - a

b) Ta c: pH =4 =>[H+] = 1CT4M.CH3COOH ^=r-> CH3COO' + H+

Ban u: . IM 0 0

in li: 10"4M /f7 = 10~pH- Xc nh OT] ca dvng dch

p dng/T]J0T ]= 10"u suy ra [H*], t tnh pH.Hoc t [OH~] pOH bang cng thc:

pOH= -gOH- =>[OH~] = 10~pOH.Sau t tng pH + pO =14. Suy ra pH.Vi axit v baz yu cn kt hp vi hng s cn bng Ka hay Kfr.

2. V D MINH HAV d 1 . Tnh pH ca cc dung dch sau:

a) Dung dch H2S040,05M;b) Dung dch Ba(0H)20,005M;

c) Dung dch CH3COOH 0,1M c in li a = 1%;

d) Tnh pH ca ung dch hn hp CH3COOH 0,2M v CHgCOONa0,1M. Cho Ka ca CH3COOH l 1,75.10-5.

V d 2 .Tnh pH ca ung dch cha 1,46 gam HC1 trong 400ml.

V d 3. Dung dch CH3COOH 0,043M c in li a l 2%. Xc nh pHca dung dch .

1 4 PL & HD Gtl CC CHUYN D HA HQC 11



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13/253

V d 4. Ly 2,5ml dung dch CH3COOH 4M ri pha long vi nc thnh1 lt dung dch A. Hy tnh d in li a ca axit axetic v pH cadung dch A, bit rng trong lml A c 6,28.1018 ion v phn t axitkhng phn li.

V d 5. Mt dung dch baza mnh Ba(OH)2 c nng ion [Ba2+] = 5.10"5mol/1. Xc nhgi tr pH ca dung dch cho.

V d 6. Tnh pH ca dung dch thu c sau khi trn 40ml dung dchH2SO4 0,25M vi 60ml dung dch NaOH 0,5M.

V d 7. Cho 40ml dung ch HC1 0,75M vo 160ml dung dch cha ng

thi Ba(OH)2 0,08M v KOH 0,04M. Xc nh gi tr pH ca dung dchthu c.

VI d 8. Cho lOOml dung dch Ba(H)2 0,009M vi 400ml dung dchH2SO40,002M, Tnh pH ca dung dch thu c sau khi trn.

V d 1 .a) Dung dch H2SO40,05M c: [H+] = 0.1M = 10_1M => pH = 1.

b)Dng dch Ba(OH>20,005M c:

[OH] = 0,0 IM = 10_2M => [H+] = = 10' 12M => pH = 1210

c) Phng trnh in li; HCOOH >HCOCT.+ H+

Cho v uag dch = 1 lt => [H+] = 0,00 IM = 103M ^ pH = 3.

Lu : Kt qu bi ton ny khng thay i nu ta chn th tch l V lt.

d)Cc phng trinh in li:

HNG DN GII

CHsCOONa ?

0,1

CH3COO' + Na

0,1

Ban u:

in li:

Cn bng:

CH3COOH 4

0,2X

0,2 - X

* CHsCOO + H+

0,1

X

X + 0,1

X

X

Ta c: Ka - xWOjoO = 1 75 .10"50,2 - X

V Ka qu nh nn: X 0,1 => 0,1 + X ~ 0,1 v 0,2 - X w0,2.

T (*) => = 1,75.10"5 => X = 3,5.1050,2y

=> pH = -lg3,5.10_5) = 4,46.&HD GII C C CHUYN HA Hc11 15



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14/253

V d 2. Ta c: nHci = ~ r ~ = 04(moi) =* Chci = = 0,1M36,5 0,4

Phng trnh in li: HC1 = H++ c rT phng trnh in li ta c: [H+] = Chci = 0,1 = 10_iM => pH = 1

c1 _ J _____ CHXOOH phn iV d 3. Ap dng: a = ----- -------

CHjCOOHhatan

CH3C00H Phn li = a x C'CHXOOH ha tan ~ 2% X0,043 = 8,6 X 10

[H+] = CCI3COOH phn li = 8,6.10^M = lO3065 (M)

=> pH = -lgl(T3065= 3,065.

V d 4. +) Cch : Smol CH3COOH l: 0,0025 X 4 = 0701 (mol)Trong 1 lt dung dch sau kh pha long c 6,28.1018.103 - 6,28.1021 hm (ion v phn t axit khng phn li).Phng trnh phn li:

CHaCOOH------ CH3COO" + H+Ban u: 0,01 (mol)Phn ng: Xmol Xmo XmoiCn bng: (0,01 - x) mol X mo X moiTrong 1 lt dung dch c: nCHC00H = 0,01 -X (mol)

^ ncH3cocr = ntr = x (mo^

M tng s mol sau khi cn bng: 0,01 - x + x + x = 0, 01+x (mol)

=> 0,01 + X = = 1,0432.(T2 (mol) X = 0,0432.icr2 (m6 ,02.10

Vy: a = ----- X 100% = 4,32%0,0432.10~2

10'

V pH = ~g(4,32.10"4) = 3,36.+) Cch 2: 1 moi c 6.02.1023phn t; 0,01 mol c 6,02.1021phn t

CH3COOH ------ > CHsCOCT + H+Ban u: 6,02.1021 X XPhn ng: X X X

Cn bng: (6,G2.021- x) X mol X moi=> 6,02.1021- X + X+ X = 6,28.1021 => X= 0,26.102i

0,26.1021

6,02.10sVy: a = - X 100% = 4,32%.J-* 21 . 9

V pH = ~lg[H+] = -g^o^e.io21^

= 3,36.6,02 .1023

16 P- &HOGIICCCHUYNHAHO



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15/253

V d 5. Ta c: Ba(0 H)2 ------ > Ba2+ + 20H' (1)5.105 l(r 4

T (1) => [OH"] = 10~4 => pOH = 4 => pH = 14 - 4 = 10

V d 6. Ta c: 1Hso = 0,25 X0,04 = 0,01 (moi)

v nNaOH= 0,5 X0,06 = 0,03 (mol)

Phn ng: H2S0 4 + 2NaOH > Na2S04.+ 2H20

T t l mol ca H2SO4 v NOH theo phn ng v theo gi thuyt tathy: NaOH d

M: nNa0Hphn ng = 2 nHso = 0,02 (mo)

=> IlNaOH d = 0,03 0,02 = 0,01 (moi)

^ C (N,0> = 0,06 = 0 ,1 (M)

NaOH ------ Na++ OH

Ta c: [OH'] = CM(Na0H) = 01M => pOH = 1 => pH = 14 - 1 = 13.V d 7. Ta c: nr = 0,04 X0,75 = 0,03 (mol)

v n0H, = 0,1 X(0,08 X 2 + 0,04) = 0,032 (moi). ,

Phn ng:

H+ + OH" ------ H20 (1)

(mol) 0,03 0,03

T (1) K_ . = 0,032 - 0,03 = 0,002 (mol).Un d

=> [OH] = 02- = 0,01 (mcl/1)0,04 + 0,16

=> [H+] = = 10~12=> pH = 12.10

V d 8. Ta t: I W , = = BaSC^-l + 2H2O(1)(moi) 0,0008 HBa(OH) d = .0,0009 0,0008 = 0,0001 (mo)

=> [OH~] = 2 X 0,0001 = 2.10"4

=> [H*] = 10 . = 0,5.10"10=> pH = 10,3.2.10

PL & HO GII CC CHUYN HA HC 11 ^ , J -Z^ 17



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16/253

CH 0 4. TON V HN HP CC ON

1. PHNG PHP- Cho tng cation kt hp vi tng anion r xem xt tnh tan trong

nc ca cht to thnh tng ng (nn lp bng).-Nn xem xt t ion no m to c nhiu kt ta khi kt hp vicc ion khc trc... San o tr dn.

Ch : Nhng ion tn ti trong mt dug dch phi nhng ion khng phn ng'vi nhau, ngha l cht to thnh t s kt hp nhng ion phi khng cht kt ta, hay bay hi hay in li yu.

- Tng s mo in tch (+) = tng s mo in tch (-);

- Tng khi ng cht tan trong dimg dch = tng khi ng cc ion

(b qua s in i ca nc).2. V D MINH HAV d 1 . C bn cation K+, Ag+, Ba2\ C2+ v bn anion c r , NO.J, so* ,

CO* . C th hnh thnh bn dung dch no t cc ion trn, nu mi

dung dch ch cha mt cation v mt anion (khng trng lp).

V d 2. Trong 3 dung dch c cha cc ion sau: Al3+, Pb2+, Ba2+, NO;J, c r

v so* . l cc dung dch mui no, bit rng trong mi dung dch

ch c mt mui.V d 3. C 3 'ng nghim, mi ng ch cha hai cation v hai anion

(khng trng lp gia cc ng nghim) trong s' cc ion sau: NH*,

Na", Ag+, Ba2\ Mg2+, A]3+, c r , Br~, N Q - so;- , PO* , co* . Hy xc

nh cc cation v anion trong tng ng.nghim.

V d 4. C th pha .ch dung dch ng thi cha cc ion sau khng?

V d 5- C th tn ti cc dung dch c cha ng thi tng nhm cc ionsau y khng (b qua s in li ca cht in li yu v cht t tan)?

a) Na+, A g\ Cl' . '

c) Mg2+, H+, SO*-, NO*

b) Ba2\ K+, S042

c) Mg2+, Na+, s - , co -

a) N03 , so* , NH;, Pb2+

c) Br~> NH ;, Ag+, Ca2"

e) OH", HCO;i, Na+, Ba' +

b) c r , HS", Na+, Fe3+

d) cr, N0 3 , s 2, Fe2+

f) HCO, H+, K+, Ca2+

18 PL & HD GII CC CHUYN B HA KQC 11



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17/253

d 6. Khi c cn dung dch cha hn hp gm: 0,2 moi Na+; 0,1 mol

Mg2+; Xmol Cl v y mol SO4" thu c 23,7 gam mui. Tnh Xv y.

d 7- Cho lOOml dung dch X cha cc cht: Fe 2(SC>4)3 0,12M; A12(S04)3

0,15M; H2SO4 0,4M. Thm 200ml dung dch NaOH 0,13M vo dung

dch X.

a) Tnh khi lng kt ta thu c.b) Tnh nng mol cc cht trong dung dch sau phn ng.

d 8. Dung dch X cha cc mui NaHCC>3 0,1M; (NH4)2C03 0,2M;

K2SO4 0,3M. Tnh th tch dung dch Ba(0 H)2 0,1M ti thiu cn cho

vo 100ml dung dch X e lng kt ta ln nht.

HNG DN GII

d 1 . - Ta c bng tng hp v kh nng kt hp ca cc on sau:

- Da vo bng tng hp c th xc nh bn dung dch cha cc mui

d tan sau: Dung dch AgN03, dung dch BaCl2, dung dch CuS04,

dung dch K2CO3.

d 2. - Ta c bng tng hp v kh nng kt hp ca cc ion sau:

- Da vo bng tng hp c th xc nh ba dung dch cha cc mui

tan sau: Dung dch Pb(N03)2, dung dch BaCl2, dung dch AI2(S0 4)3-

d 3.

- Nhng xon cng tn ti trong cng mt ung dch phi l nhng ion

khng phn ng vi nhau. Xt kh nng phn ng kt hp ca cc ion

ta c bng tng hp.& HO GI! CC CHUYN B HA HOC 11 19



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18/253

- Vy, cc cation v anion cha trong tng ng nghim c th l

+) ng (1): POf , CO32- , Na+, NH'

+) ng (2): SO2; , NO', Ag\ Mg2*

+) ng (3): c r , Br", Ba2+, AI3+

V d 4. Nhng ion cng tn ti trong dung dch phi l nhng ion kphn ng vi nhau. Xt tng trng hp:

a) Khng, v to kt ta AgCl : Ag+ + Cl" ----- - AgCl

b) Khng, v to kt ta BaS04 : Ba2+ + so^' ------ > BaSG

c) c, v cc ion khng phn ng vi nhau

d) Khng, v to kt ta MgC03 : Mg2+ + COj' ------ > Mg

V d 5.

a) Khng, v to kt ta: Pb2+ + SO* ------ * PbS04>l

b) Khng, v c phn ng: 2Fe3+ -- HS' ------ 2Fe2+ + S-i' + H+.

r c) Khng, v to kt ta: Ag++ Br~ --------------> AgBri.

d) Khng, v to kt ta: Fe2+ + s 2" ------ > FeS

e) Khng, v c phn ng: OH" + HCO' ------ > H20 + CO3'

f) H+ nu nhiu s khng th cng tn ti cng vi HC03 v x

phn ng: H++ HCO ------ COT + H20.

V d 6. Theo nh lut bo ton in tch:

X + 2y = 0,2 + 0,1 X2 = 0,4

Tng khi lng cc mui:

0,2.23 + 0,1-24 + x.35,5 + y.96 = 23,7 35,5x + 96y = 16,7

Gii (1) v (2), ta c: X= 0,2; y = 0,1.20 M. *>'h o g i i c c c h u y n 0 h a h



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19/253

V d 7.

a) Ta c: nFvS0 0,012 (mol) => nF3. = 0,024 (moi);

n Ai (SO ) = 0,015 (mol) => 1 1 ^ 3 , = 0,03 (moi);

nHso = 0,04 (mol) => n = 0,08 (moi);

nNaOH = 0,26 (mol) nOH. = 0,26 (mol).

H+ + OH" -> h 20(mol) 0,08 0,08

Fes+ + 30H" ------ >. Fe(0H)3(mol) 0,024 0,072 0,024

Al3+ + 30H" -> A1(0H)3(mol) 0,03 0,09 0,03

Suy ra: n . , . = 0,26 - (0,08 + 0,072 + 0,090 =0,018 (moi).v OH cn

Al(OH)s' + OH" ----- - [Al(OH)4r(mol) 0,018 0,018 0,018Do : nMlMh^ = 0,03 - 0,018 = 0,012 (moi).

Khi lng kt ta: 0,024 X108 + 0,012 X78 = 3,528 (gam).

b) Tnh nng mol cc cht trong dung dch sau phn ng.Dung dch sau nhn ng c:

n 02. = 0,036 + 0,048 + 0,04 = 0,3 24 (moi)

=> C 0,124 moi NaS.&.

n = 0,018 (mol) => C 0,018 moi Na[Al(OH)4].[Al{OH)j

Th tch dung dch sau phn ng: 0,1 + 0,2 = 0,3 (lt).0 124

Vy: CM(NasS0


20/253


21/253

V D MINH HA

d 1 . Cho 8 gam CuO tc dng vi dung dch H2S04 19,6% c ly d

20% so vi lng tham gia phn ng. Tnh c% ca mui trong dung

dch sau phn ng.

d 2. Cho 34,8 gam Fe34 phn ng vi dung dch HC1 1,5M. Tnh thtch dung dch HC1 cn dng va nng mol ca mui thu c. Bit

rng s thay i th tch dng dch sau phn ng khng ng k.

d 3. Cho 10 gam oleum cha 20% H2SO4 v . khi lng vo 50 gam

dung dch H2SO4 12,25% thu c dung dch X. Tnh th tch dung dch

Ba(OH)2 0,5M cn dng trung ha dung dch X v khi lng kt

ta sinh ra.

d 4. Cn dng bao nhiu gam tinh th CUSO4.5H2O v bao nhiu gamdung dch CuS04 % thu c 280 gam dng dch C11SO416%? Tnh th

tch dung dch Ba(OH)2 IM cn kt ta ht ion Cu2+ trong 280 gam

dung dch. Lc ly kt ta v sy kh, tnh khi lng kt ta thu c.

d 5. Cho 6,4 gam hn hp gm Mg v MgO phn ng va vi

200ml dung dch H2SO4 IM thu c dung dch X v 2,24 lt kh (ktc).

C cn dng dch X thu c 49,2 gam mui Y. Xc nh cng thc ca

mui Y.d 6. Ho tan III! gam Na vo m2 gam H20 thu c dung dch B c

khi lng ring d. Phn ng xy ra:

2N a .+ 2H20 --------- > 2NaOH + H2T

a) Tnh nng phn trm ca dung dch B theo mi v m2.

b) Tnh nng mol ca dung dch B theo mi v

c) Cho c% = 16%. Hy tnh t s" cho Cm = 3,5M, tnh d.m2

d 7. Khi lm ngui 1026,4 gam dung dch bo ha R2S04.nH20 (trong

R l kim loi kim v n nguyn, tha mn iu kin 7 < n < 12) t

80c xung 10c th c 395,4 gam tinh th R2S04.n20 tch ra khi

dung dch. Xc nh cng thc phn t ca hirat ni trn. Bit tan

ca R2SO4 80c v 10c ln lt l 28,3 gam v 9 gam.

& HD GI CC CHUYN HA HC 11 23



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22/253

V d 8. Thm dn dung kch KOH 33,6% vo 40,3m dung dch HN37,8% (D = 1,24 g/ml) n khi trung ha hon ton thu c dung dA. a A v 0c thu c ung dch B c nng 11,6% v lng mtch ra i m gam.

a) Dung dch R bo ha cha?

b) Tnh tr s ca m.

HNG DN GII

V d 1. Ta c: nc 0 = = 0,1 (mol)80

CuO + H2S0 4 --------- CuS04 + H20T phn ng: nHjS0( = nCuS0< = n Cu0 = 0,1 (mol).

_ , _ 120Do ly d 20% nn: nHSOitmsimsath = 0 , l x ^ = 0,12 (mol).

=> inf.fS0 = 0,12 X98 = 11,76 (gam)

=> mdungdchH?so, = x 10= 6 (gam).

T : mt,ongichCuso. = 60.+ 8 = 68 (gam);

mCiiS0 = 0,1 X 160 = 16 (gam).

Vy: 0 % ^ = i f X 100% 23,53%-4 DO

V du 2. Ta c: np n = 0,15 (mol).3 < 232

Fe30 4+ 8HC1 --------- FgC12+. 2FeCl3+ 4H20

(mol) 0,15 1,2 0,15 0,3

Suy ra: VdungdichHci = | = 0,8 (lt); c FeC1 = = 0,1875M;1 ,0 2

C M ,r .c v = - O .S 7 5 M '

V 3. rong 10gam oleum c 8gam SO v 2gam H2SO4-

Smol SO3 0,1 mol. Do , khi lng H2SO4dc to ra t SOs l0,1 X98 = 9,8 (gam)

Trong 50 gam dung dch H2SO435,2% c 17,5 gam H2SO4.

T : E 0(1 30 = 2 + 9,8 + 17,6 = 29,4 (gam)

99 4= v . = =0.3

24 PL & HD GIi CC CHUYN B HA HOC



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23/253

(2)

H2SO4+ Ba(OH)2 ---------------- BaS04- + 2H2O

(mol) > 0,3

Vy: \lvngdchB(OH)j = ~ (1ft);

miiso = 0,3 x 233 = 69,9 (gam).

V d 4. Gi X v y ln lt khi lng CUSO4.5H2O v dung dchCuS04 8% cn pha ch, ta c: X+ y = 280 (i)

K h i l n g c a C u S t >4 t r o n g X g a m C U S O 4 . 5 H 2 O : f f i - k ji = 0 , 6 4 x ( g a m ) .

250

8vKhi cmg ca CuS4trong y gam dung dch CuS048%: = 0,08y (gam).

16x 8

Suy ra: 16 = - Q-------- Q X 100 => 0,64x +0,08y = 44,8280Gii h phng trnh (1) v (2) ta c: X = 40 (gam) v y = 240 (gam).

Khi lng cua OUSO4trong 280 gam dng dch CuS04 16%:

28-- X = 44,8 (gam) => nc s0 = ^ = 0,28 (mol).100 CuS0 160

C11SO4 + Ba(OH)2 ---------- > Cu(O)2>l + BaS04i

(moi) 0,28 0,28 0,28 0,28

Ta c: nBa(0H) = 0,28 (mol) =5 XiungdchBa(OH)2 = (lt).Khi lng kt ta: 0,28 X98 + 0,28 X233 = 92,68 (gam)

V d 5. Ta c: nlt s0 = 0,2 (moi); nH = 0,1 (mol).

H2SO4 + M g ---- MgS4 + H2

(moi) 0,1 0,1 0,1 0,1

=> nr, A n = 0,2 - 0,1 = 0,1 (moi).H2S{)4 phn MO 7

=> mMg = 0,1 X24 =' 2,4 (gam) => mMgo = 4 (gam) => nMg0 = 0,1 (moi).

H2SO4 + MgO -------1 MgS04 + H2.0

(mo) 0,1 0,1 0,1

=> Sn Mso = 0,2 (mol) => HiMgS0 = 0,2 X 120 = 24 (gam) < 49,2 (gam).

Do , mui Y c dng MgS04.xH20.

Ta c: (120 + 18x)0,2 = 49,2 (gam) => X= 7.

Cng thc mui l MgS04.7Hi>0.PL & HD GII CC CHUYN HA HC 11 25



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24/253

V d 6. a) Phn ng:

2Na + 2H20 ------- 2NaOH + H2T (1)

(mol) i - a . . . 23 23 46

Ta c: nNa = (mol) l* nH = (mol)23 46 .

T T , m, 22m. + 23m,Khi lngdung dch B: mi +m2 = J^1----------------- (gam)

23 23

T (1) => nNa0H= 7 (rnol) => mNa0H= (gam)23 23

~ 40m,

v^ ' c%B i i T k r 100% (S)

b) Th tch dung dch B l: VB= ? ^ l+ (ml).23d

^ m, X d X 100=> CM= ------- - M . (**)

M* 22m 1 + 23m2

c) +) Khi c% =16%,th vo (*), ta c: - ^ m-------- x 100% = 16%22mj+23m,

o 22mi + 23m2 = 250nii 23i2 = 228mi m2 228

+) Khi Cm = 3,5M, th vo (**), ta c:

m, X d X 1000 = 3,5 22mx + 23m2 = 285,743mid.

22m1 + 23m2

o 250m: = 285,7143mjd => d = 0,875

V d 7. 80c, tan cua R2S04 l 28,3 gam => Trong 128,3 gam dng

dch bo ha c 28,3 gam R2SO4v 100 gam nc

Vy 1026,4 gm dung dch bo ha c 226,4 gam R2SO4v 800 gam nc.

Khi lng dung dch bio ha ti thi im 10c l:1026.4 - 395,4 = 631 (gam)

10c, tan ca R2SO4l 9 gam, nn:

Trong 109 gam dung dch bo ha cha 9 gam R2SO4

Vy 631 gam dung dch bo ha cha 52,2 gam R2SO4

Khi lng R2SO4b tch ra di dng hirat:

226.4 - 52,1 = 174,3 (gam)26 PL & H8 GII CC CHUYN HA HC 11



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25/253

. A 395>4 _ 174>3frmui h i drat ^mu khan n s n :

b) Ta C: nHN0 = - = 0,3 (mol)

2R + 96 + 18n ' 2R + 96o ' 442,2R - 3137,4n + 21225,6 = 0 => R = 7,In - 48

chp R l kim loi kim, 7 < n < 12 , n nguyn

=> Nghim hp l: n = 10, R = 23: Natri (Na)

Vy,cng thc hirat cn tm l Na2S04.10H20 -d 8. a) Kh lm lnh dung dch A thu c dung dch B c khi lngmu KNO3 tch ra nn dung dch B bc ha

37,8 X 40,3 X 1,24100 X 63

Phn ng: HNO3 + KOH --------- > KNO3+ H20 (1)(.raol) ' 0,3 - 0,3 0,3T (1) => riKOH= 0,3 (mol) => mKOH= 0,3 X56 = 16,8 (gam)

=* nWgdichKOH = 1 1 4 X 100 = 50 (gam)33,6V nijgjo = 0,3 X101 - 30,3 (gam)

+) Khi h nhit :Khi lngKNO3cn li trong dung dch l: 30,3 - m (gam)

n i d i m g d c h c n l i = m d d (H N 03 ) + m H+].

pH= a =>[T = 10 ^M .

10~u- Vi dung dch baz: T pH =>[H*~ =>(OFF]: [OT'J= [H+]

- p dng quy tc ng cho:Dung dch (1): Khi ng (gam), nng Ci%.Dung dch (2): Khi ng m2 (gam), nng c2%

Duig dch thu dc c% => = m2 c - c ,

&H Gii CC CHUYN HA HQC 11 27



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26/253

Dung dch (1): Th tch v l (lt), nng c 2(M).

Dung dch (2): Th tch v 2 (lt), nng Cz(M).

V c - cDung di ch thu c CM => - =

* v2 c - Cx

Ch : Nu s pha trn c xy ra phn ng ha hc th khng n

dng quy tc dng cho.

2. V D MJNH HAV d 1 . Cn thm bao nhiu ml dung dch HC1 c pH = 2 vo dung

H2S 040,Q5M thu c dung dch c pH-1,2?

V d 2. Cn thm bao nhiu t dung dch NaOH 1,5M vo 1 lt dung

H2SO4 IM thu c:

a) Dung dch c pH = 1?

b) Dung dch c pH = 12?

V d 3. Trn Vi lt dung dch c pH = 12 vi V2 lt dung dch c pHth thu c 2 lt dung dch c pH = 10. Tnh VTv V2.

V d 4. Cn trn dung dch HC1 c pH = 5 vi dung dch NaOH c pH

theo t l no th c dung ch c pH = 8? Bit rng th

dung dch mi bng tng th tch cc dung dch phn ng.

V d 5. C hai dung dch axt HNO3 40% (d = 1,25 g/ml) v HNO3

( = 1,06 g/m). Cn ly bao nhiu lt mi dung dch ph thnhdung dch HNO3 15% (d = 1,08 g/ml)?

V d 6- Cn thm dung dch NaCi 10% vi dung dch NaCl 20% theo

khi lng no c dung dch c nng 18%?

V d 7. Cn trn lOm dung dch NaOH 0,5M vi bao nhiu ml dung

NaOH IM c dung dch c nng 0,8M?

HNG D N GI I

V d 1. Ta c: nfr = 2nHgQ = 2 X0,1 X0,05 = 0,01 (mol).

V lt dung dch HC1 c pH = 2 => [H+] = 10"2M =>. n . = 0,01V (m

Do : n H. = 0,01 + 0,0IV (mol).

Dung dch mi c: V' = (V + 0,1) lt.

pH = 1,2 => [H+j = 10_12M = 0,063M.

T suy ra: 0,063 = -1- - ?,Q-y => V = 0,07 (1) = 70 (ml).V + 0,1

28 PL & HD GI r CC CHUYN KA H



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27/253

V d 2. Phn ng: H+ 4- OH"--------- > H20

Gi V l th tch dng dch NaOH, suy ra:

n = n N a0 H = 1 , 5 V ( m o i ) ; n = 2 n H ~n = 2 X 1 X 1 = 2 ( m o i ) .OH' H 2 ^ 4

Th tch dung dch mi: V' = (V + 1) lt.

Ta c: [H+3 = 21 -M X = 0,1 (M) => V s 1,19 (lt).V + 1

b) pH = 12 => [H+] = 10_12M => [0H~] = 10~2M, suy ra OH" d.

S' mol H~ trong dung dch mi l: {1,5V - 2) mol.

Suy ra: [OUT]= - 2 = 0,2 (M) => V 1,69 (lt).V + 1

V d 3 . Ta c: Vi + v 2= 2 (lt). (1)

Lp s ng cho:Vi lt dung dch pH = 12

pH = 10

v 2 lt dung dch pH = 2

V 9>Suy ra: ~ = = 4. (2)

Gii h phng trnh (1) v (2) suy ra: Vj =1,6lt; v 2 = 0,4 lt.

V d 4. Gi Vi v v 2 ln lt lth tch cadung dch HC1 vdung dch

NaOH cn trn. Th tch dung dch mi l: V = Vi + V-2-

+) pH - 5 => [H+] = 105M nH, = 10~5V1 (mol).

+) pH = 9 => [H+] = 109M =* [0H] = 10"5M n0H. = 10'5V2 (moi).

Phng trnh h hc: H++ 0H --------- > H2O

Dung dch thu c c pH = 8, sy ra OH' d.

pH = 8=> [H+] = lO^M => [OHj = lO^M = n0H. = lO^Vi + v2) moi.

Ta c: n , = (10-sv 2 - 10'5V,) = 10-6CVt


28/253

V 5. Cch2. S dng qui tc ng cho:

Khi lng dung dch 40%

Khi ng dung dch 10%

Khi lng dung dch 40% 5%

15%

Khi lng dung dch 10% 25% 5

Khi lng dung dch sau khi trn:

dung dch = V X d = 2000 X 2,08 = 2160 (gi)

Vy:Khi lng dung dch HNO340% l: x * _ 360 (gm)

HNO, 40% = 288 (ml) v25 HN310%

(5 + 1)

18001,06

= 1698 (ml)

Cch 2:Phng php i s:

Gi Xv y l khi lng dung dch HNO340% v HNO3 10%Khi lng dung dch HNO3 thu c sau khi pha l:

md = x + y = V x d = 2000 X 1,08 = 2160 (gam)

. 15 X 2160n i t M n

10040x

( 1)

= 324 (gam)

(2)

Trong Xgam dung dch c: = 0,4x (gam) HNOg 40%

lOxTrong y gam dung dch c: - = 0,lx (gam) HNO3

=> 0,4x + 0,ly = 324

Gii (1) v (2) => X = 36, y = 1800

Vy ^HN0340%= 1 = 288 (ml) v VHN0 10%= = 1698 (ml)3 1,25 3 1,06

V 6. Gi mi l khi lng dung dch NaCl 10%

m2 l khi lng dung dch NaC 20%

m, 20-18 14Ta c: m0 1 8 - 1 0

Vy cn trn phn trm dung dch vi t .l khi lng tng

ng l 1 : 4

V d 7. Gi V l s mol dung dch NaOH IM cn ly

1 - 0,8T . 10 _Ta c: =V 0,8 -0 ,5

2 t . 3 x 1 0 iBr/_ 1N= - => V = ---- = 15 (ml)

3 2

30 PL&HDGII CCCHUYNHAHC11



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29/253

HUYEN E 2.

NHM NIT

L THUYTNIT

a) Tc dng vi hiro:

N2 + 2 N Hg, H= -92 k/.

b) Tc dng vi kim loi:

nhit thng nit ch tc dig vi kim oi iti, to inh lti trua:

6Li+ N2 --------- > 2Li3N .

nhit cao, nit tc dng vi mt s kim loi: Ca, Mg, A, ...

3Mg+N2 ---- : > Mg3N2 (Magie nitrua)

2A+ N2 ---------> 2N (Nhm nitrua)Trong cc phn ng vi hiro v kim oi, s oxi ha ca nt gim: nit th hin tnh oxi ha.

c) Tnh kh

Tc dng vi oxi:

N2 + ' 2 NO (Kh khng mu); AH = +180 kj.

+2 +12 NO + 0,, ^ 2N 0 2 (Kh mu nu ).

N + - 0 ,, ; ^ = == '2N0 .2 3 :i

Ch : Cc oxit khc ca nit nh N2OtW2Oj, N2Os khng iu ch ct phn ng trc tip gia nit v ox.

AM O N IA C (NH 3)

a) Tnh baz yu

+) Tc dng vi nc: NH.3+ HyO - - NHjOH NH/ + Oi~.+) Tc dng vi axi:

2NHj + H2S O ,--------- > (NH4),S04

NH, + t --------- N H / .

NH3(k)+ HC(k) --------- > NH.C(r)

Ch : Phn ng ny dng nhn ra kh amoniac.& HD GII CC CHUYN 0 HA HOC 11 31



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30/253

+) Dung dch amoiac c th m kt ta nhiu hdroxit kim loi:

A 3*+ 3NH3 + 3H20 --------- A(OH),yl+ 3NH/ .

Fe2++ 2NH3+ 2H20 --------- >Fe(O H)J+ 2 N H / .b Kh n ng to phc:

Dung dch amoniac c kh ng ha tan hiroxit hay mui t tan cmt s.kim oi, to thnh cc dung dch phc cht.

Cu(OHh + 4 N H j--------> [Cu(NHj)J(OH)>.

Cu(OH}2+ 4NH3 -------- (Cu(NH3)J2++ 20HT(Mu xanh Ihm

Ch : Phn ng vi NH3 gipta phn bit A vi Zn (c hai trong NaOH) nhng ch c Zn(OH)z tan trong NHj.c) Tnh kh:+) Tc dng vi oxi:

4NH, + 302 -------- 2 N, +6H 20

Nu c xc tc hp kim patin v irii 850- 900c th:

4NH3 + 502 --------> 4N 0 t + 6H.0.

+) Tc dng vi co: ,

2NH3 + 3CI2 ------- N2 T + 6HC1.

NH3 sinh ra phn ng vi HC to thnh khtrng.

8NH3+ 3 C2 ------ > N2T+ 6NH4Cl.+j Tc dng vi mt s oxit kim oi:

2NH3 + 3CuO 3Cu + N, t + 3Ha0 .

Phn g xy ra tng t ch oxitr st oxit.

3. MU I AM ON Ia) Phrt ng trao i ion

(NH4)2S 0 4 + 2NaOH 4 > 2NH. +Na2S4+ 2HzO

N H /+ OfT --------- > NH3 T+ HOb) Phn ng nh it ph n

+} Mui amoni to bi axit khng c tnh oxl ha kh un nng phn hy thnh amoniac v axit.

NH4C(r) ---- NH3(k) + HC(k)

(NHJ2C0 3 ---- > NH3T+ n h 4h c o3

NH4HCO3 --- > n h 3+ c o 2T+ h 2o.32 PL 4 HD GIt Ce CHUYN HA HC



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31/253

+) Mui amoni to bi axit c tnh, oxi ha nh axit nit, axit nitric

kh i b nh it phn cho ra Nz hoc N20 (nt oxit) v nc.

NH4N 0 2 ---- - >N2f+ 2H20 . - y.

NH4N 03 ---- > N20 + 2H20 .

NH4NOj > N2f+ ~ 0 2 + 2H20.

4. A X IT NITRIC (H \0 3)

a) Tnh axit:

+) Dung dch HN 03 m qu tm ha .

+) Tc dng vi oxit baz:

CuO +'2HNO3 --------- Cu(N03)2+H20

+) Tc dng vi baz:

Ca(OH)2 + 2HN03 ------------- Ca(N03)2+ 2HsO

+) Tc dng vi iui ca axt yu: -

CaCOs + 2HN3 ---------> Ca(N03). + C02t+ n 20

Ch :^'HN03 rt long v nh phn ng vi kim loi to mui nitrat

(kim o s oxi ha thp) v H2.

Fe + I0HNO3long. ]nh --------4F((NOs}2+ NH4NO3+ 5H20

Fs + 2NO3jf' ong. nh ------ >Fe.(N03) 2+H2fb) Tih oxi ha:

+) Vi kim loi:

Kh tc dng vi nhng kim loi c tnh kh yu nh Cu, A gr HNO3

c b kh n NO 2, c n HNO 3 lo ng b kh n NO :

Cu + 4HN03c---------> Cu(N03)2+ 2N 02T+ 2H20

3Cu + 8HNO30ng- 3Cu(N03)2+ 2NO+ 4H20

Khi tc dng vi nhng kim loi c tnh kh mnh hn nh Mg, Zn,A HNO3 long b k h n N20 hoc N2 v HNO3 rt ong b kh

n NH3 (dng NH4NO3).

8A+ 3OHNO3long-----------> 8A(N03)3+ 3N20 1 + 15H20.

5Mg + I2HNO3eig --------- > 5Mg(N03)2+ NT+ 6H2O.

4Zn + 10HNO3 ioIg------- > 4Zn(N03)2+NH4N03+ 3t20.Pt 4HD GII CC CHUYN HA HOC 11 3 3



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32/253

Ch : Fe, A d tantrong dung dch HNOj long nhng b th ng ha trongduig dch HN03 c, ngui v to nn mt mng oxit trn b mt cc kim lo ny.

Hn hp gm mt i ch HNO3 c v 3 th tch HC c c gi nc cng thy. N c th ha tqgi vng hay patin theo phn ng:

Au+HN03+ 3HC ------ > AuC3+ NOT+ 2H20

C ch: H N 03+ 3HC ----------- C2f+ NOC + 2HzO

2 N 0 C --------- 2N 0 +C2

3C2+ 2A u --------- 2AuC3+) Vi phi kim:

Axit nitric c c th tc dng c vi cc phi kim nh c, s, p, ...Khi , cc phi kim b oxi ha n mc oxi ha cao nht, cn HNO3b kh n NO2hoc NO ty theo nng ca axit

c+ 4HN03dc--------- CO2T+ 4N02f+ 2H2O

3P + 5HN03long+ 2H20 ---------> 3H3PO4+ 5NO T s + 6HN03 dc-----------> H2S 0 4 + 6NO u 2t 20

32+ IOHNO3 ----------> 6HO3+ 10NOT+ 2H2O+) V hp cht: Cc hp cht nh.H2S, H, S 0 2>FeO, mui st (l), ...tc dng dc vi dung dch HNO3.

3 FeO + 0HN3 ong ----- > 3 Fe(N03)3+ NO T+ 5H2O

2HNO3+ H ----- > H 03+ 2N0 P+ H20

2 03H2S + 2HNOs Jong ---------> '3S + 2NOf+ 4H2O

3FeS+ 12HMO3 -------->Fe(N3)3+ Fe2(S04) 3+ 9N0f+ 6H20

5. MU NIT R AT

a) Tnh cht ha hc- S thy phn: Mui nitrat ca kim loi kim, k im th th khng bthy phn, cn mui ca cc kim loi khc b thy phn to mi trng axit.

Fe(N03h + H20 ^= = [Fe(OH)]2+ 3H30+ + 3 N 0 fFe(OH)f++ 2H20 ^ = [Fe(OH)2r + H30 +

[Fe(OH)2r + 2H20 ^ = Fe(Ot)3 +H30 *

- Phn ngtrao i:Vi dung dch mui khc:

Pb(N 032 + K2S 0 4 ---------> P b S O j + 2KN033 4 PL & HD G11CC CHUYM HA HOC 11



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33/253

+J Vi dung dch axit:

Ba(N 03)2 + H2S0 4 ----------- > B a S O j + 2HNOj

+) Vi duig dch kim:

Fe(N 03)3+ 3KOH------ Fe(OH)3J- + 3KN03

- B nhit p hn hy:

+) Nhit phn mui ntrat ca kim loitrc Mg:To thnh mui nitr v k h Oz.

2KNOs ---- ^ > 2KN2 + 0 3f

Ring: Ba(N 03)2 ----------- > 2BaO+ 4N 02T+ 0 2 T

+} Nhit phn mui nitrat ca kim oi t Mg~7>Cu:

To thnh kim oi+N 0 2+ Oz.

2Cu (N03)2 ---- - 2CuO+ 4N 02T+ Oz T

+) Nhit phn mui nitrat ca kim oi sau Cu:To thih kim oi+N 0 2+ 0 2.

2AgNOs ---- > 2Ag+ 2N02 + 0 2t

PHOTPHO

a) Tnh oxi ha:

Photpho ch th hin r r t tnh oxi ha khi tc dng vi mt s kim

oi mnh (K, Na, Ca, to ra photphua kim loi.

2P+ 3a ---------- Ca3P2 (Canxi phophua)

3Zn + 2ptng --------- > Zn3P2 (Km photphua-thuc chut)

b) Ti kh:

+} Tc ng vi oxi:

Thiu oxi: 4P+ 302 -------- > 2P2O3 (iphotpho trioxit)

D i o x i : 4P + 5 O 2 > 2P20 5 ( p h o t p h o p e n t a o x i i)

+) Tc dng vi clo:

Thiu co: 2P+ 3C2 -------- > 2PC3 (Phtpho triclorua)

D co: 2P +5C2 -------- > 2PCS (Photpho pentacorua)

+) Tc dng vi mui

6Ptl-dg + 5K2Cr20 7 -------- > 5K20 + 5Cr23 + 3P20 5

6P+ 5KCO3 --------- 3P20 5 + 5KC&HO GI! C C CHUYN O HA HOC 11 35



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34/253

+) Tc dng vi axit nitric

3P + 5HN03 + 2H2O ;------ > 3H3P04+ 5MO T+) Tc dng vi hro

2P+ 3H2 ---- > 2PH3(Photphin)

PH3 cht kh, rt c. Trn 150c b bc chy trong khng kh

2PH3+ 402----

> p205+ 3HzOPH3snh m do s thi ra xc ng ic vt, nu c ln iphotphin Pth t bc chy ph t ra nh sng xanh ( hin tng ma tri),c) iu ch v ng dng

~ p kh hot ng, trong t nhin n tn ti dng hp cht nh qung photphorit-Ca3(P04)2, apatit~3Ca3(P04)z.CaF2.- p c dng ch to dim: Thuc gn u que dim gm cht oxi ha nhKCO.-, KN03, mt cht d chy nh s, ... v -dnh. Thuc qut bn cnh hp dim p v keo dnh. tng

c st cn trn thm bt thy tinh mn vo c 2 oi thuc trn.- p d dng s xut axit photphoric:

p --------p20 5 ------------- h 3p o 4

- Trong cng nghip, ngi ta iu ch p bng cch nung hn canxi photphat, ct (S1O2) v than:

2Ca3(P 04)2+ 6SO2"t"O ---------^ P4+ 6CqSC?3+ lOCO

7. H P C H T CA PHOTPHO

a) IPHOTPHO PENTAOXIT (P2Os)p?05 cht ri, mu trng, rt ho nc, tc ng mnh ht nc to thnh Qxit phophoric

P2O5+3H2O ---------> 2H3PO4Chnh v vy ngi ta dng P2Os m kh nhiu ch t.b) AXIT PHOTPHORC (H3POJ- Trong p205 v H3P04, photpho c s oxi ha +5. Khc vi n

photpho c m in nh nn bn hn mc+5. Do vy H3P4p20 5 kh b kh v khng c tnh 0X1ha nh HN 03.

- H3PO4 axit trung bnh, trong dung dch in i theo 3 nc: Trbnh nc th nht, yu v rt yu ccnc th hai, th ba.

h 3p o 4 f+ H2PO'

h 2p o ; / r + HPOJ-

HPO;- ; = f+ POJ-

30 PL & HD GII CC CHUYN Q HA HOC



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35/253

Dung dch H3P04 c nhng tinh cht chung ca axit: Lm qu m,

tc dng vi baz, oxit baz to thnh mui axit hoc mui trung ha

nh: NaH2P 04, Na2HP04, Na3P 04.

H3PO4 c th tc ng vi nhng kim o i ng trc H trong dy

hot ng kim oi, gii phng H2-

2H3PO4 + 3Zn ---------> Zn3(POJ2+ 3H2t- Tc dng bi nhit:

2H3P 04 ----H4p20 7+ H20

(Axit iphotphoric)

Tip tc un nng n khong 400-500 c, c phn ng:

* H4p20 7 --------- 2HP03 + H20

(Axit metaphotphoric)

Cc axit HPO3,H4p20 7c th kt hp vi nc to ra axit H3P 04.

- iu ch v ng dng ca H3P 04

+) : H3O4 C3( 0 43

a x i t H 2 S O 4 .

CQ3(P4)2+ 3H2SO4 ---------^ 2H.3PO4 + 3 C0SO4l

+) Trong phng th nghim: H3PO4 dc iu ch t p205ha tan voB 20 hay t p ha tan bng HN0.3. H3P04 ch yu c dng s

xut phn bn.

c) TNH CHT CA MUI PHOTPHAT.

+} Mui photphat: Na3P 04>Ca3(P04)2y (NH4)3P 04

+) Mui ihirophotphat: NaH2P04>Cq(H2POJ:; NHH2P04.

+) Mui hirophotphat: Na2HP04, CaHP04, (NH4)2HP04.

- Tnh tan:

+) Tt c cc mui ihrophotphat u tan trong nc,

+) Cc mu i photphat u khng tan trong nc tr Na3P04, K3P04 v.(NH4)3P 04 tan tt.

- Ph n ng th y ph n: Cc mui photpht tan b thy phn trong

dung dch v lm qu im ha xanh.

Na3P04 + H20 ^ Na2tP 04 +NaOH

P 0\- + n 2o ^ HPOJ- + O K

PL & H GI CC CHYN 0 HA KC 11 .. 3 7



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36/253

B. C C DNG BI P THEO CH

CH 1. > HON THNH sd PHN NG

1. PHNG PHPCn nm chc kin thc v tnh cht ha hc ,phng php iu ch

cc chu c bit v cc cht thuc/ihm nit nh N2, NO, NO2,

HN03, NH3, mui nitrai, mui amoni, H3P 04, m ui pho tphat...

Ch : Mi m tn trong s mt phn ng.

2. V! D MINH HAV d 1 . Cho s sau:

--> NO - N02 - ' x; > Y ^ -> C a (N 0 3)2

N. x2

M *x> NO NO> _________> Y15) * 1V (6) * . ^ ^ 7 ^ T in . x

H NH4NO3

Hy hon thnh cc phn ng trong s trn.

V d 2 . Vit cc phng trnh phn ng thc hin dy chuyn ha sau:

B CuO, t ^ A V t + 0 S V r * + 0 2 N T k + O j + H s O ^ - p------- U ) ------- A --------- > > b -------- (3 ) * u -------- (4) D ------------ i s ) -------- > *>NaOH G ---

(6) (?)

V d 3. Hon thnh s chuyn ha sau:

Fe(OH)2 -> Fe(N03)3 > Fe20 3

- > Fe(N03)3

(NH^COs ^ NH3 > Cu NO N 02

- ->HN03 A1(N03)s

L_HC1 >NH4C1 NH3

n h 4h s o 4

V d 4. Thc hin chui phn ngsau:

Nit > Kh amonac > Amonisunfat > Kh amon

* Amoni hirocacbonat Amoni cacbonat >C2

- >Cacbon monoxt(8)- > Nit monoxit >Nit38 PL & KD GII CC CHUYN D HA HC 11



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38/253

V d 3.

(1) (NH4)2C03 ---- - 2NH3 + C02+ H20

(2) 2NHs + 2H20 + FeCJ2 --------- Fe(OH)2 + 2NH4C1

(3) Fe(OH)2+ 4HN0, ---- - Fe(N03)s + N0 2T + 3H20

{hay 3Fe(OH)2+ 10HN03(1) ---------> 3Fe(N0 3>3+ NOT + 8H2O)

(4) 4Fe(N03)3 ----- 2Fe20 3+ 12N0 2t + 302

2Fe(N03)s + 3H20

(6) 2NH3+ 3CuO -------- 3Cu4 + N2+ 3H20

(7) 3Cu + SHNO3long ------- > 3Cu(N03)2+ 2NOt + 4H20

(8) 2NO + 0 2 ------------------------------> 2N02

(9) 3N02+ H20 ----------> 2HNO3+ NO

(10) AI20 3+ 6HNO.3 -------- > 2A1(N03)3+ 3E20

(11 ) 2NH3+ 3C2 --------- > N2+ 6HC1

(12) NH3+ HC1 ------------------- > NH4CI

(13) NH4CI + NaOH --------> NaCl + NH3t + H20

(14) NH3+ H2S0 4 ------- NH4HSO4

V d 4.

400(1C. Fe . .(1)N 2+ 3H2 (NH4)2S04 .

(3) (NH4)2S0 4+ 2NaOH --------- Na2S04+ 2NH3T + 2H20

(4) NHa + C02+ H20 ------- > NH4HCO3

(5) 2NH4HCO3 -------------------------------------- ----- > (NH4)2C03+

(6) (NH4)2C03 > 2NH3t + C02t + H20

hay (NH4)2C03+ 2HC1 ----- > 2NH4CI + C0 2t + H20

(7) C02+ c 400--00~ > 2CO

(8) NO2+ CO ---------- NO + C02t

(9) 2NO + 2H2S --------- 2S + N2t + 2H204JJ PL & H GII CC CHUYN HA HC 11



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V d 5.

(1) Ca3N2+ 6H20 ------> 3Ca(OH)2+ .2NH3

(2) 4NHS+ 50z -----*> 4N0 + 6H20soo c

(3) 2NO + 0 2 -------> 2N02

(4) 2N02+ H20 --------- HNO3+ HNOs

(5) 3HN2 -----> HNO3+ 2NO + II20 (phn ng phn hy HNO2)

(6) HNO3+ NH3 ------- - NH4NO3

(7) NH4NO3 c N20 + 2H20

V d 6.

(1) Cu + 4HNO3d c > Cu(N03)2 + 2N0 2 + 2H20

(2) 2NOs + 2NaOH ------> NaNOs + NaNO + H20

(3) 2NaN02 + 0 2 --------- 2NaN03(4) 2NaN3 ---- 2NaN02 + 0 2t

(5) 3NaN02+ HC10 --------- 3NaN03+ HC1( 0 ) 2 H C 1 , d i n p h n d u n g d ch ^ H j t + C l s T

40 0HC Fc(7) N2 + 3H2 - 2NHst + Q

(8) 2NH3+ 302 ---- - > 2N2t + 6H2O

(9) 2NH3+ ~ 0 2 - 8 c- > 2N 02t + 3H20

(10) N 02 + CO ------ NO + C02t

V d 7. Dy chyn ha:

C a3(P 0 4)2 > p * C2lz?2 P H 3> H 3PO 4 N a3p 0 4 > AgsPO.

Phn ng:

Ca3(P04)2+ 5C + 3Si020- > 2P + 3CaSi03+ 5C

2P + 3Ca ---- > Ca3P2Ca3P2+ 6H20 ------> 2PH3+ 3Ca(OH)2

P H 3 + 2 O 2 --------- > H 3 P O 4

H3PO4-+ 3NaOH -------- - Na3p0 4+ 3H20

Na3P04 + 3AgN03 --------- Ag3P 04 + 3NNOsPL & HD I CC CHUYN HA HOC 11 41



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40/253

CH D 2 > NH N BIT TCH TINH CH " ^U CH______ _____V GII THCH HIN .NG

1. PHNG PHPLa chn nhng ph r ng c du hiu c trng (s bin i mu

mi, kt ta, si bt kh ...} nhn bit

STTCht nhn

bitThc th Hn tng xv ra v phn ng

1 . NH3 [kh) Qa tm m Qu tm m ha xanh

2.n h ;

Dimg dch kim() hi nh)

Gii phng kh c mi khai:

NH: + OH" - NH3+ h 20'

3. h n o 3 Cu

Dung dch ha xanh, gii phng 1

kh khng mu v ha nu trong 1khng kh: 1

3Cu + 8HNO3 -> Cu(N03}2 1

+ 2NO + 4H20

v 2NO + 0 2 - 2NO2

4. NO- H2S04, Cu

Dung dch ha xanh, gii phngkh khng mu v ha nu irongkhng kh:

3Cu + 8H++ 2 NO3 3Cu2++ 2NO + 4H20

v 2NO + 0 2 2NO2 ;

5. PO fDng dch

AgNOsTo kt ta mu vng

3Ag+ + POf Ag3P0 4 1

2. V D MINH HA IV d 1. Cho 4 dung dch khng mu ng trong 4 i mt nhn: AgN0|

KOH; HC v NaNOa. Ch c dng mt kim loi, hy nhn bit cdung dch trn. :

V d 2. Ch c dng mt kim loi, lm th no phn bit nhng ua|

dch sau : NaOH, NaNOs, HgCl2, HNOs, HC1? IV d 3. Ch dng mt cht khc nhn bit tng dung dch sau.

N H 4N O 3 , N a H C O a , ( N U ) 2 S 0 4 , F e C l a v P e C l s - V i t p h n g t r n h c

phn ng xy ra. I42 PL & HD GII CC CHUYN D HA HOC 111

I



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41/253

d 4. Mi cc cha mt trong cc cht sau: Pb(N03)2, Na2S203 , MnCl2,NH4C17 (NH^aCOs, ZnS04> Ca3(P0 4)2 v MgS04. Dng nc, dung dchNaOH, dung dch HC1 nin bit mi cht trn.

5. Hy phn bit cc cht trong cc trng hp sau:

a) Cc l cha cc kh: N2, H2, c\2,S02, 0 2, NH3, NO.

b) Cc l cha cc dung dch: HNO3, HC], H3PO4, H2S, ch dngthmhai ha cht khc.

c) Cc l cha cc dung dch: NH4NO3, NH4CI, (NH4)2S04, NaCl,Na2S04, NaNOs.

d) Cc l cha cc dung dch: A1C13jMgCl, ZnCl2.

d 6. Trong phng th nghim lm th no d tch rng N2 vC02 rakhi hn hp kh gm: Ns, C02j CO, H2?

HNG DN GI I

d 1 . Trch mi cht mt t lm mu th. Cho bt ng ln t vo ccmu th.- Mu th to dung dch mu xanh l AgN03.

Cu + 2AgNO3 ---------- ^ Cu(N3)2+ 2AgCho dung dch AgN03vo cc mu cn li:- Mu to kt ta trng l HC1.

AgNOa + HC ------ -> AgCli + HNO3- Mu to kt ta trng, sau ha en l KOH.

KOH + AgNOs --------- > AgOHl + KNO32AgOH -------> Ag20 (en) + H20

- Mu th cn li l KNO3-

d 2. Dng kim loi Al, cho AI tc dng ln lt vi ccmu th

- Nu c kh mu nu bay ra l HNO3

AI + 4HNO3 Al(NOs)a+ NOt + 2H20

2NO + O2 > 2NO2 (mu nu)- Nu c kim loi trng sinh ra l HgC2

2AI + 3HgCl2 > 3Hg + 2A1C3

- C bt kh bay ra v c kt ta, kt ta tan ra l NaOH

AI + 2H20 + 2NaOH ---------> 2NaA102+ 3H2t

- C bt kh bay ra l HC1

2AI + 6HC1 --------- 2AICI3+ 3H2T

- Cn li l NaNOsHD Gli CC CHUYN D HA HC 11 43



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V d 3. Dng Ba(OH)2d nhn bit. Tm: tt theo bng sau:

n h 4n o 3 NaHCOg (NH4)2S 04 I FeCl2 FeCl

Ba(OH)2 NH3T mikhai

i trngBaC03

NH3T mikhai,

trngBaS04

' trng,hi xanhFe(OH)2

1 nFe(OH

V d 4. Cho nc vo cc mu th, tt c u tan, ch c mu th cCa3(P0 4)2khng tan.Cho t t dung dch NaOH vo cc mu th cha cc ha cht trn th+} Mu th c kh mi khai bay ra l NH4CI v (NH^COs

NH4C1 + NaOH --------- NHgt + H20 + NaCl

(NH4)2C03+ NaOH -------> 2NH3T + 2H20 + Na2COsCho dung dch HC1 ln lt vo hai mu th trn: Mu th no cbay ra l (NEU^COg, cn mu th khng c hin tng xy ra l NH

(NH4)2C03+ HCI -------- 2NH4CI + COsT + H20

+) Mu th to kt ta trng l: Zn(OH)2 Mg(OH)2, Pb(OH>2Mn(OH)2, nu tip tc cho NaOH vo Zn(OH)2 v Pb(OH)2 tan Mg(OH)2khng tan, nh vy ta bit c cc cha MgS04:

ZnS04+ 2NaOH ---------- Zn(OH)2 + Na2S0 4

Zn(OH>2 + 2NaOH --------> Na2Zn02 + 2H20

MgS04 + 2NaOH --------->Mg(OH)2 + Na2S 0 4

Pb(N03)2+ 2NaOH -------- Pb(OH)2i + 2N aN 03Pb(OH)2+ 2NaOH --------> Na2Pb02+ 2H20

MnCls + 2NaOH ---------- Mn(OH)2+ 2NaCl

phn bit Pb(N03)2 vi ZnS04 ta cho dung dch HC1 vo hai mth, mu th no cho kt ta mu trng l Pb(N03)2, cn mu khng tc dng l ZnS04:

Pb(N03)2+ 2HC1 --------- PbC U + 2HNO3Mn(OH)2 khng bn, d b ox ha thnh Mn(OH)4 mu nu

Mg(OH)2khng b oxi ha.2Mn(OH)s + 0 2(Idc) + 2H20 ---------> 2Mn(OH)4

+) Mu cui cng cn li l Na2S20 3C th cho dung dch HC1 vo mu th cn li ny, c kt ta mvng v c kh mi hc (S02):

Na2s 20 3 + 2HCI ---------> 2NaC + S 02t + s i +H2044 PL & HD GII CC CHUYN B HA HC



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43/253

V d 5. a) L kh c mu vng lc l clo.

Cho tip xc vi khng kh, l kh ha nu l NO:

2NO (khng mu) + O2 ---------> 2NO2(nu )

- a mu giy qu tm t vo cc mu th: Qu tm ha xanh l

NH3, ha hng l S02.

NH3+ H

2O N NH

4OH (Lm qu tm ha xanh)

SO2 + H2O -------- > H2SO3 (Lm u tm ha hng)

- a que m vo 3 mu th cn li: Lm que m chy bng l 2-

- t 2mu th cn li, mu th chy cho ngn la mu xanh nht l

kh H2j mu th khng chy l N'2.

2H2+ 0 2 ---- 2H20

b) Cho dung dch Ba(OH)2vo cc mu th, c kt ta l H3PO4:

2H3PO4+ 3Ba(OH)2 --------- > Ba3(P04)2^ + 6H20Cho dung dch AgNOs vo cc mu th:

C kt ta trng l HC1:

AgNOs + HC1 ----- AgCli + HN3-

C kt ta en l H2S:

2AgN0 3+ H2S " Ag2S-l- + 2HNO3.

Khng to kt ta l HNO3. ..

Ch : Khng dng dung dch AgN3 nhn ra H3P04 v AgNOs khngto kt ta vng (Ag3P04) vi H3P04.

c) Dng dung dch NaOH chia 6 mui thnh 2 nhm: Nhm I gm

cc mui NH* (to kh c mi khai vi NaOH) v nhm gm cc

mui Na+.

Ln lt dng Ba2+ nhn ra mui SO4' (do to kt ta trng BaS04)

v Ag+ nhn ra mui cn (o to kt ta trng AgCl) trong mi

nhm, cn i l mui NO3 khng to kt ta vi 2 thuc th ny.

Phn ng:

NH; + OH~ ---------- NHs + H20

Ba2+ + SO2; -------------- BaS04i

Ag++ Cl"---------* AgClPL & HD GII CC CHUYN o HA HC 11 45



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44/253

d) Cho dung dch NH 3 d vo 3 mu th:

Mt mu th to kt ta sau tan trong NH3 d l ZnC2.

Hai mu th to kt ta khng tan .trong NH 3 d l AICI3 , MgCl2.

Cho thm NaOH d vo 2 kt ta ny 7kt ta tan trong NaOH l Al(OH)3.

Cc phng trnh ha hc:

Z n C l * + 2 N E 3 + 2H20 > Zn(OH)2^ + 2 N H 4C

Zn(OH ) 2 + 4NH 3 > [Zn{NH 3 )4](OH )2 tan

MgCh + 2NH + 2HsO > Mg(OH)2>i + 2N H 4C1

AICls + 3NH 3 + 3 H 20 Al(O H )34. + 3 N H 4C I

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

V d 6 . Cho hn hp 4 kh qua ng s cha CuO d, un nng, sau

lm lnh, hn hp kh bay ra l N 2, co.

CO + C u O * Cu + C 0 2

H 2 + C u O Cu + H 20

Cho hn hp 2 kh li qua dung dch Ca(OH ) 2 d, thu k h bay ra l N 2.

Lc ly kt ta CaCC >3 ri cho tc dng vi H 2SO4 d t i to C 0 2.

CCOs + H2SO4 --------- C&SO4+ CO2+ H2O

CH 3 >CN B m PHki NG 0X1 HA - KH CA NHNG ------ / PHN NG C S THAM GIA CA HNO3 HOC NO3

]. PHNG PHPCn bng phn ng ox ha-kh theo phng php thng bng ion-

eectron cng phi m bo nguyn tc: Tng electron m cht kh

cho bng tng electron m cht oxi ha nhn (nh phng php:

thng bng electron). Ch khc cht oxi ha, cht kh vit d

dng on.

Ch : Cht k t t khng tan), cht kh (cht d bay hi), cht ;in i (H20) phi dng phn t.

Ty theo mi trng phn ng axit, baz hoc trung tnh m sau-

khi xc nh nhng, nhn electron ta phi cn bng thm in tclhai v. [p

4 6 PL & HO GIt CC CHUYN HA HOC 11 I



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45/253

Nu phn ng xy ra trong mi trng axt, ta thm JHT vo v nod oxi, v ci i thm H20. Nu phn ng xy ra trong mi trng baz, ta thm 0 H~ vo vno thiu 0X1, v cn i thm H20.~ Nu phn ng xy ra trong mi trng nc th nu to axit cn bng

nh mi trng axit, nu to baz ta cn bng nh mi trng baz.+) Nhn h s cho hai qu'trnh nhng v nhn electron sao cho: selectron nhng ra ca cht kh bng s electron nhn vo ca cht

oxi ha.

+) Kim tra s nguyn t ha v theo th :Kim oi Phi kimHro v oxi.

V D MINH HAd 1 . Cn bng phn ng sau y theo phng php thng bng ion electron:

Cu + HNOs * Cu(N0 3)2 + NO + h 2o

d 2 . Cn bng phn ng sau theo phng php thng bng ion electron

C u + N a N 3 + H 2 S O 4 > C u ( N 03 )2 + N O + N a 2 S 4 + H 2 O

d 3 . Cn bng ph n ng oxi ha kh sau:

Cu2S + HNO3 (long) ----- Cu(N03}2 +C11SO4 + NOT + H20

d 4 . Cho phng trnh ha hc sau:

AI + HNO 3 A1(N 0 3 )3 + N 2OT + NOt + H 2O

Bit hn hp k h to th nh c 25% N20 v th tch. T m h s ca NOv N20 kh i phn ng c cn. bng.

d 5 . Hy cn bng phn ng oxi hakh sau:

Fe30 4 + HNO 3 F e(N 0 3) 3+ N xOy + H 20

d 6 . Cn bng phn ng oxi hakh sau bng phng php thng

bng electron.

FeO + H + + NO 3 > F e3+ + N G 2 + NO + H 2 o

Bit t l s mol N 0 2: n o = a : bHNG DN G I I

0 +5 +2 +2d 1.D n g io n: Cu + H++ N 0 3 Cu + N o + H2O

' 0Qu trnh oxi ha: Cu > Cu + 2e

4-2Qu trnh kh: N O3 + 3e ----- > N o

&HO GII CCCHYN D HA HC 11 47



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46/253

(V mi trng axit nn thm H+vo v tri (d oxi) v thm nc v

v phi:

NO- + 3e + 4H+ ----- NO + 2H20

+2

Ta c: 3 X2 X

Cu-----

> Cu + 2eNO; + 3e + 4H+ ----- > NO + 2H20

3Cu + 2 NO; + 8H+ -> 3Cu2+ + 2NOT + 4H20

Dng phn t: 3Cu + 8HN03 ----- > 3Cu(NC>3)2 + 2NOt + 4H2O

V d 2,

Qu trnh ox ha: 3 X I Cu -- > Cu2+ + 2e

Qu trinh kh: 2 XNO' + 2e + 4H+ ------ > NO + 2H2O

Phng trrth dng ion rt gn:3Cu + 2 NO; + 8H+ ----- > 3Cu2+ + 2NOT + 4H20

Phng trnh dng phn t:

3Cu + 8NaN03 + 4H2S04 ----- > 3Cu(N03)2 + 2NO + 4Na2S04 + 4H

V d 3. Cu2S + H N 0 3l0ng -------- > C u(N 03)2 + C u S 0 4 + NO + H

+) -2 +2 +63 X Cu2 s - lOe ----- > 2 Cu + s : Qu trnh oxi ha

oX

+5 +;

N + 3e-----

> N Qu trnh kh

=> 3Cu2S + lOHNOg (long) --------- > 3Cu(N03)2 + 3CuS04

+ lONOt + 5H

V d 4. Ta c:%VN 0= 25% => %VN0 = 75%.

Trv ,v -.7 Vn 2 nN?0 25 1V l chtkh nn %v= %n => 2 =-- - = - ^NO nNO ^ 3

Phng trnh phn ng bi cho c tch ring thnh hai ph

trnh phn ng v cn bng nh sau:9 X AI + 4HNO3 ------- >A1(N03)3 + NO + 2H20

1 X 8A1 + 3OHNO3 ---- >8A1(N03)3 + 3N20 + 15H20

c t l trn ta nhn phng trnh (1) vi 9 ri cng vi phtrnh (2) suy ra:

17AI + 66HNO3 ---------> 17Ai(N03)3 + 9NOT + 3N2OT + 33H20.48 PL 6 HD GII CC CHUYN D HA HC



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47/253

V d 5. (5x - 2y)Fe30 4+ (46x - 18y)HN03 --------- > (15x - 6y)Fe(N0 .3)3

' + NxOy + (23x - 9y)H20.

V d 6. FeO + H++ NO' -------- Fe + N 0 2t + NO t + H20

F e > Fe : Qu trnh oxi ha

(a 4- b )N -(a- 3b)e > aNOj + bNO: Qu trnh kh(a + 3b)FeO + (4a + 10b)H++ (a + b) NO: --------> (a + Sb)Fe3+ + aN02

(a + 3b) X

1 X

+ bNO + (2a + 5b) H20

CH D 4. / L P CNG THC OXIT V HIU SUT PHN NG

1. PHNG PHP- Lp cng thc oxit ca nit thng qua cc bc sau:+Jt cng thc oxt ca nit NxO (


48/253

2. v D MINH HAV d 1 . a) Cho 2,24 lt N 2 v 3,36 lt H 2 (ktc) tc dng vi nhau, tnh

khi lng NH 3 to thnh bit hiu sut phn ng l 25%.

b) Cn ly bao nhiu lt N 2 v H 2 (ktc) iu ch 51 gam NH 3 vi

hiu sut 25%.

V d 2. Cn bao nhiu tn NH3 iu ch c Om3 dung dch HNO3?Bit hiu sut mi giai don l 80%.

V d 3. Mt bnh k n 10 lt cha N 2 v 10 lt H 2 0c, lOatm.

bnh phn ng x y ra r i a nh it bnh v 0 c th p suttrong bn h sau ph n ng l 9 atm. T nh hiu su t ph n ng.

V d 4. M t hn hp X cha 100 mol gm N 2 v H 2 ly theo tp sut ban u l OO atm. Sau phn ng to amoniac, p sut gimcn 285 atm. Nu nhit phn ng c gi khng i th hiu sutphn ng l bao nhiu?

V d 5. iu ch 63 gam NH 3 cn ly bao nhiu lt N 2 v H 2 ktc.Bit hiu sut phn ng l 20%.

V d 6 . Cn ly bao nhiu gam N 2 v H 2 (ktc) iu ch c 51 gam

NH3, bit hiu sut ca phn ng l 25%.

V d 7. Hn hp X gm 2 kh N 2 v H theo t'l moi l 1 : 4. Nung X vi

xc tc nhit cao thu c hn hp kh Y, trong 'NH 3 chim20% th tch. Tnh hiu sut ca phn ng trn.

V d 8 . Cn t (c xc tc) bao nhiu m3 kh amoniac (ktc) sn xut

700kg dung dch BNO 3 99%? Bit rng c 98,56% amoniac cchuyn thnh axit.

V d 9. Tron g bnh ph n ng c cha hn hp kh A gm 10 mol N 2 v40 mol H 2. p sut trung bnh c u l 400 atm, nhit bnh c

gi khng i. Khi ohn ng xy ra v t n trng thi cn. bng

th hiu sut ca phn ng tng hp l 25%.

a) Tnh s mol cc kh trong bnh sau phn ng.

b) Tnh p sut trong bnh sau phn ng.

V d 1 0 . Cn bao nhiu lt N 2 v H 2 (ktc) iu ch c 1 0 2 gamNH3? Bit hiu sut phn xig20%.

V d 1 1 . M t oxit A ca nit c cha 30 ,43%N v k hi lng. T kh i hica A so vi khng kh l 1,586 Xc nh cng thc phn t, cng

thc cu to v gi tn A.

V d 1 2 . Mt hn hp X gm CO2 v mt oxit nit c t khi i vi H 2

l 18,5. Hy xc nh cng thc oxit ca nit v % th tch cc khtrong hn hp X.

50 PL & HDGII CC CHUYN D HA HQC11



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d 13. M t hn hp k h X gm 3 oxit ca N l NO, N 0 2 v NxOy. Bit

phn trin th tch ca cc oxit trong X l %V N 0 = 45% , % V NOj = 15%;

% v = 40%, cn ph n trm theo kh i lng NO trong hn hp l

23,6%. Xc nh cng thc NxOy.HNG DN G I I

d 1 .a) Ta c: nN = 0,1 (moi); n H = 0,15 (mol).

Ph n ng: N 2 + 3H 2 < 2 N H 3 (1)

T ( 1 ) => H 2 thiu so vi Ns nn lng sn ph m tn h theo H 2.

23

25

V n MH = n H = 0 ,1 (m oi).

m = 0,1 X 17 X = 4,25 (gam).NH3thudc thc t 100

b) Ta c: nNR = 0,3 (mol).

3T i l l 1 n XIU = 0 , 4 5 ( m o l ) .AU phn ng 2 NH=

Do V , V., = 0,45 X 22,4 X = 40,32 (lt).L,O QU > v Ha cn dng thc t 2 5

d 2 . Ta c: 10m3 = 104 (lt) => n HN0 = 2 X 104 (mol)

+) Cch 1. Vit cc phng trnh ha hc xy ra, t phng trnh hahc suy r s" moi sn phm trung gian, suy ra so mol N H 3 ,suy ra khoilng NH 3 theo l thuyt, suy ra khi lng thc t.

Tuy nhin phng php ny tn nhiu thi gian.

+) Cch 2 . N H 3 NO N 0 2 > HNO 3

(mol) 2.104


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Th tch hn hp trc phn ng: v x = 20 (lt).

Th tch hn hp sau phn ng: v2= 20 2x = 18 (lt) => X = 1

Suy ra th tch H 2 tham gia phn ng l 3 lt.

So snh t l phn ng ta thy th tch H 2 cho l thiu so

Vy, h iu sut phn ng tnh theo H 2: H = X 100% = 30%

+) Cch 2. Gii nhanh.Ph n ng: N 2 + 3H 2 > 2N H 3

T ( 1 ) => v h5nhpgimi = VNH Einh = I VH

=> AV = 20 18 = 2 (lt ) = VHi I X 2 = 3 (lt)

Vy, H = X 100% = 30%.

VI d 4. Phn ng:

n 2 + 3H 2 F u 'p 2N H 3

Ban u: 25 mol 75 moi 2 moi

Phn ng: X moi 3x mol 2x moi

Cn bng: (25 x) moi (75 3x) m o l 2x mo

Tng s mol sau phn ng: 100 2x (moi)

Ta c t i l : ^ ^ ns = g 8 5 x 1 0 0 = 9 5 (mo)Ps ns 300

=> 100 - 2x = 95 => X= 2,5 (moi)

V y h i u s u t H = ~ X 1 0 0% = 1 0%25

V d 5. Phn ng:

N2 + 3H 2 2NH3

22,4 lt 3 X 22,4 lt 2 X 17 gam

V , lt VHi lt 6 8 gamV hiu su t phn ng (1) l 20% nn thct cn:

/ i u Mi 68 X 22,4 100 Th tch N 2(ktc) l: _ X =224 (lt)2 X 17 .20

_ 'VUi ' t u / i n 6 8 X 3 X 22,4 100 Th tch H 2 (ktc) l: . X = 672 (lt)2 x 17 20

52 PL & HD GII CC CHUYN HA



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V du 6 . Ta c: n NH = T3 =. 3 (moi)3 17

N 2 + 3H 2 . 2N E 3

(mol) I I 3

Khi lng N 2 v H 2 cn ly: mN; = X X 28 - 168 (gam)

m = t X ^ * 2 = 36 (gam).'* 2 25

V d 7 . Ph n ng: N 2 + 3H2 ~ 2 N H 3

Ban u: 1 mol 4 molPhn ng: X mol 3x moi 2x mol

Cn bng: ( x) mol (4 3x) moi 2x mo

Oy _ 1

Ta c'

:

x 100 = 20 => X = (1 x) + (4 3x) + 2x 2,4

51

2 x o aHiu sut: H = 2 ^ X 100% = 41,67%.

z

V d 8 . Phn ng xy ra:

4N H 3 + 5 0 2 > 4 N 0 t + 6 H 2O (1)

2 N 0 V 0 2 * 2 NO 2 (2)4N 02 + 2H20 + 0 2 > 4 HNO 3 (3)

Nhn xt t (1), (2) v (3) ta thy: C 1 ol NH 3 to thnh 1 mol

HNO3 hay 22,4 lt NH 3 to ra 63 gam HNO 3.Cch 1. Khi lng HN 0 3 trong 7G0kg dung dch HNOs 99% i:

99 X 700100

693

63

T nhn xt trn => I1HN0 = n NHa = 11 (km ol)

=> V NH = 22 ,4 X 11 = 2^6,4 m 3

Nhng ch c 98,56% NH 2 chuyn thnh KQSO3 nn ing NH 3 thc t

dng l: 246,4 X gg 250 (m 3>.

PLLHO GII CC CHUYN D HA HC 11 53

n HN 0 3 = = 1 1



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Cch 2. S : NHs : N O -----------N 02 - > HN03

Theo s : c 6,kg- HNO3cn 22,4m3kh NH3.Vy mun c 693kg HNO3cn xm3kh NH3

693 X 22,4 c , .=> X - --------- = 246,4 ( m 3 )

63V hiu sut phn iig 98,56% nn lng NH3 thc t em dng l:

= 250 (m3)98,56 .

V u 9. Phh ng tng hp NH3xy ra theo t l: nM : nw = 1 : 3* n z

Theo bi, ta c: nN : = 10 : 40 = 1 : 4. Vy H2d.

Phi da vosmol N2phn ng tnh s mo NH3:

a) Phn ng:

Ban u:Phn ng:

Sau phn ng:

m 25= 10 X 100

= 2,5 mol

n 2 + 3H2 2NHS u10 40 0 moi2,5 7,5 5 moi7,5 32,5 5 moi

n N = 1 0 , ^100= 2,5 (moi)

V y , s mol cc kh trong bnh sau phn ng l: 7,5 (moi) N 2; 32,0(mol) H2; 5,0 (mol) NH3

b) Tng s moi kh trong bnh ban u: O + 40 = 50 (mol)

v PV = nRT m y V b , Tb khng i, nn ta c:K n. , . n ^ 45= hay Ps = X p = ^ X 400 = 360 atm.

n102

pd n n 50

V d 10 . Cch 1. Ta c: - 6(mol)3 ' ^

Phn ng: N2 + 3H2 - > 2NH3 (1)

(mi) 3 9 nN = 3 (moi) => VN = 3 X 22,4 = 67,2 (t)

V nH = 9 (mo) => Vj^ - 9 X 22,4 = 201,6 (lt)

V hiu sut H = 20% nn lng N2v H2thc t em dng l:

cn dmgr5 67,2 X = 336 (lt) I

VH cn dng = 201,6 X = 1008 (lt). I2 20 I

54 PL & H GII CC CHUYN HA HC 11 I

\



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Cch 2, N2 + 3H2 --------- > 2NH3 (1)

22,4 (lt) 3 X 22,4 (lt) 2 x 1 7 (gam)

y (lt) X(lt) 102 (gam)

T (1) => Th tch Hs phn ng l: X = X2,i -X = 201,2 (lt)2 X 17

22 4 X 102V th tch N2phn ng l: y = = 67,2 (lt)

V H = 20% nn: Th tch H cn dng l: 201

PHÂN LOẠI & HƯỚNG DẪN GIẢI CÁC CHUYÊN ĐỀ HÓA HỌC 11 - HUỲNH VĂN ÚT

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Transcript of PHÂN LOẠI & HƯỚNG DẪN GIẢI CÁC CHUYÊN ĐỀ HÓA HỌC 11 - HUỲNH VĂN ÚT