bÀi tẬp hÓa hỌc Ở trƯỜng phỔ thÔng - nguyỄn xuÂn trƯỜng

8/10/2019 BI TP HA HC TRNG PH THNG - NGUYN XUN TRNG

1/244

WWW.FACEBOOK.COM/DAYKEM.QU

WWW.FACEBOOK.COM/BOIDUONGHOAHOCQU

B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N

W.DAYKEMQUYNHON.UCOZ.COM

g gp PDF bi GV. Nguyn Thanh T


2/244

PGS.TS. NGUYN XUN TRNG

Bi tp ha hc TRNG PH THNG

TI BN LN I

NH XUT BN S PHM



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




3/244

Chng U.CC PHNG PHP GII BI TON HO HC 48

1. Phng php bo ton khi lng. 48

2. Phng php tng gim khi lng. 52

3. Phng php bo ton electron. 56

4. Phng php dng cc gi tr trung bnh. 67

5. Phng php tch cng thc phn t. 89

6. Phng php ghp n s. 91

7. Phng php t chn lng cht. 92

8. Phng php bin lun tm cng thc phn t ca cc cht. 97

9. Phng php ng cho. 106

Chng IV.NHNG lM CN CH KHI GII BI TPV IN PHN V P SUT 111

I- Nhng im cn ch khi gii bi tp v in phn. 111

II. Nhng im cn ch khi gii bi tp v p sut. 123

Chng V. BI TP HO Hu c 138

I. ng phn cis-trans. 138

II. Xy dng v s dng cng thc tng qut trong

gii bi tp ho hu c. 148

Chng VI. BI TP TRC NGHIM 161

Chng VII. s DNG BI TP TRONG QU TRNH

DY HC HO HC 229

1. S dng bi tp iu khin kn trong qu trnh dy hc ha hc 229

2. Rn t duy k nng ha hc 230

3. S dng nhiu bi ton c ni dung bin lun 237

4. Xy dng nhiu bi tp thc nghim nh lng 238

5. S dng s trong vic gii, cha bi tp 242



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




4/244

LI M U

Bi tp ho hc l phng tin c bn nht dy hc sinhtp vn dng kin thc vo thc hnh. S vn dng cc kinthc thng qua cc bi tp c nhiu hnh thc rt phong ph va dng. Chnh nh s vn dng kin thc gii cc bi tp

m kin thc c cng c, khc su, chnh xc ho, m rng vnng cao.

Nh vy bi t p va l ni dung, phng p hp , va lphng ti n day tt, hoc t t mn Ho hoc.

gip sinh vin khoa Ho hc cc trng i hc S phmv Cao ng c nhng hiu bit c bn v bi tp ho hc trng ph thng, chun b cho ngh nghip, tng lai, chngti bin son cun sch ny.

Cun sch ny cng c th l ti liu tham kho i vi giovin ph thng v dng n, luyn thi vo cc trng i hcv Cao ng.

Chng ti xin c n nhn nhng kin ng gp ca bnc ln ti bn sch c hon thin.

Tc gi



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




5/244

QUY C CC CH VIT TT

PP: Phng php

p.: phn ng dd: dung dch

ctpt: cng thc phn t ctct: cng thc cu to

cttq: cng thc tng qut

ctg: cng thc n gin

ktc: iu kin tiu chun

askt: nh sng khuch tn

e: electron

p: in phn

pnc: in phn nng chy

pdd: in phn dung dch

HTTH: h thng tun hon

hh: hn hp

KLNT: khi lng nguyn tKLPT: khi lng phn t

ptp: phng trnh phn ng

nh lut BTKL: nh lut bo ton khi lng

S: p s'



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




6/244

Chng I

I CNG V BI TP HO HC

I. NGHA, TC DNG CA BI TP i VI VIC DY

HCHOHCBi tp Ho hc l phng tin c bn dy hc sinh tp

vn dng cc kin thc ho hc vo thc t i sng, sn xutv tp nghin cu khoa hc. Kin thc hc sinh tip thu cch c ch khi c s dng n. Phng php luyn tp thngqua vic s dng bi tp l mt trong cc phng php quan trng nht nng cao cht lng dy hc b rrn.i vi hc

sinh, gii bi tp l mt phng php hc tp tch cc. Bi tpHo hc c nhng tc dng gio dc tr dc v c dc to lnsau y:

1. Rn luyn cho hc sinh'kh nng vn dng c cc kinthc hc, bin nhng kin thc tip thu c qua cc biging ca thy thnh kin thc ca chnh mnh. Khi vn dngc mt kin thc no , kin thc s c nh lu.

2 . o su v m rng kin thc hc mt cch sinh ngphong ph, hp dn. Ch c vn dng kin thc vo vic gii bitp, hc sinh mi nm vng kin thc mt cch su sc. V d



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




7/244

sau khi hc v cc hp cht ca nhm, hc sinh u bit A1(0H)3c tnh cht lng tnh: n va tan c trong cc dung dch

axit mnh, va tan c trong cc dung dch kim. Cc em snh rt lu khi vn dng kin thc gii cc bi tp thuccc dng sau y:

- Nhn bit hay chng minh s tn ti ca cc cht tronghn hp (Nhn bit cc dung dch NaCl, CaClj., AICI3 ch dngdung dch NaOH; chng minh s c mt ca mi kim loi tronghn hp gm Al, Mg, Ag).

- Tch cc cht ra khi hn hp. Chng hn tch ring tngcht ra khi hn hp gm CuCl2, KC1, AICI3.

- iu ch ring tng cht i t mt hn hp. Chng hniu ch ring tng oxit i t hn hp cc kim loi Cu, Fe, Al.

- D on cc hin tng xy ra trong cc th nghim. Chnghn d on cc hin tng xy ra khi th dn tng mu Nakim loi vo dung dch mui nhm.

- Gii thch cc hin tng thc t nh ti sao khng nndng chu nhm ng vi? Ti sao khng nn dng ninhm nu qun o vi x phng? Ti sao phn chua li nhtrong c nc c v.v...

3. n tp, cng c v h thng ho kin thc mt cch thnli nht. Trong khi n tp nu ch n thun nhc li kin thc,hc sinh s chn v khng c g mi, hp dn. Thc t cho thy

hc sinh kh, gii ch thch gii bi tp trong gi 1tp.4. Rn luyn c nhng k nng cn thit v ho hc nh k

nng cn bng phng trnh phn ng, k nng tnh ton theo



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




8/244

cng thc ho hc v phng trnh ho hc; k nng thchnh nh un nng, nung, sy, ho tan, lc... k nng nhn

bit cc ho cht gp phn vo vic gio dc k thut tnghp cho hc sinh.

5. Pht trin nng lc nhn thc, rn tr thng minh cho hcsinh. Mt bi tp c nhiu cch gii c cch gii thng thngtheo cc bc quen thc nhng cng c cch gii c o,thng minh, rt ngn gn, m li chnh xc. a ra mt bi tpri yu cu hc sinh gii bng nhiu cch, tm nhng cch gii

ngn nht, hay nht l mt cch rn luyn tr thng minh chocc em.

6 . Gio dc t tng, o c tc phong nh rn luyn tnhkin nhn, trung thc, sng to, chnh xc, khoa hc. Nng caolng yu thch hc tp b mn. Rn luyn tc phong lao ng cvn ho, lao ng c t chc, c k hoch, gn gng, ngn np,sch s ni lm vic thng qua vic gii cc bi tp thc

nghim.

II. PHN LOI BI TP

1. C s phn loi

C nhiu cch phn loi bi tp tu thuc vo c s phn loi. C th da vo cc c s sau y:

- Da vo hnh thi hot ng ca hc sinh khi gii bi tp

c th chia thnh: bi tp l thuyt v bi tp thc nghim.- Da vo tnh cht ca bi tp c th chia thnh: bi tp



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




9/244

nh tnh v b tp nh lng.

- Da vo ki bi hoc dng bic th chia thnh: bi tpxc nh cng thc phn t ca hp cht; tnh thnh phn phntrm ca hn hp; nhn bit, tch cc cht ra khi hn hp;iu ch v.v...

- Da vo ni dungc th chia thnh: bi tp nng , inphn, p sut v.v...

- Da vo chc nng c th chia thnh: bi tp kim tra shiu v nh; bi tp nh gi cc kh nng v s , tm ti

liu, tng kt... Bi tp rn luyn t duy khoa hc (phn tch,tng hp, quy np, din dch...).

- Da vo khi lng kin thc hay mc n gin hocphc tpc th chia thnh: bi tp c bn v bi tp tng hp.

2. Phn loi chi tit bi tp ho hc trng ph thng

a. B i tp l thuyt, n h tn h ,gm cc dng chnh sau:

- Vit cng thc electron, cng thc cu to, cng thc ngphn...

- Vit phng trnh phn ng biu din dy bin ho ca cccht.

- Bi tp bng hnh v.

- Nhn bit hay phn bit cc cht.

- Tch cc cht ra khi hn hp.- iu ch mt cht



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




10/244

(Cc bi tp nhn bit, tch cc cht, iu ch mt cht cth ch gii l thuyt hoc phi lm thc nghim).

- Xc nh cu to ca mt cht da vo tnh cht ca n.

- Trnh by tnh cht ho hc ca mt cht.

- Trnh by cc nh lut v hc thuyt ho hc, cc khinim ho hc c bn.

b. B i tp l thuy t n h lng hay bi tp tnh ton:gm cc dng chnh sau:

- Tnh khi lng phn t ca mt cht. i t gam sang

mol nguyn t hoc mol phn t v ngc li.- Tnh theo cng thc ho hc:

+ Tnh t l khi lng gia cc nguyn t".

+ Tnh thnh phn phn trm v khi lng ca cc nguynt' trong hp cht.

+ Tnh khi lng nguyn t ca mt nguyn t khi bit cngthc phn t ca hp cht v thnh phn phri trm ca ccnguyn t cn li.

- Tnh theo phng trnh ho hc (cn bng phng trnh).

- Hon thnh cc phng trnh phn ng ht nhn. Tnhchu k bn hu v.v...

- Tnh lng cht tan v lng dung mi pha ch mtdung dch c nng cho trc hoc tnh nng ca dung

dch.- Xc nh nguyn t" ho hc.



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




11/244

- Xc nh cng thc phn t ca hp cht.

- Tnh thnh phn phn trm v khi lng hoc th tch

ca hn hp.- Tnh tc phn ng ho hc.

- Tnh in li; hng s axit, baz.

- Tnh hiu ng nhit ca phn ng.

- Tnh tinh khit ca mt cht hoc hiu sut ca phnng.

- in phn.- p dng cc nh lut v cht kh (tnh p, V, t).

- Bin un theo ho tr, theo khi lng hoc the tnh chtca cc nguyn t' v cc cht.

Dng hn hp hay tng hp.

c. B i tp thc nghim nh tnh:gm cc dng chnh sau:

- Lp dng c th nghim.

- Quan st, m t, gii thch hin tng th nghim.

- Lm th nghim nghin cu tnh cht ca mt cht hocca mt phn ng ho hc.,

- Nhn bit cc cht.

- Tch cc cht ra khi hn hp.

- iu ch cc cht.

Cc bi tp nhn bit, tch cht ra khi hn hp, iu ch cth ch gii l thuyt, hoc phi lm thc nghim. Mun cho hcsinh phi lm thc nghim, v d bi tp nh sau: "Cho mt



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




12/244

hn hp bt 3 kim loi c tn trong dy Bektop. Hy dngnhng dng c, ho cht cho xc nh tn cc kim loi ".Khi ra bi tp nh vy, hc sinh khng th ch gii l thuyt

c m phi thc hnh.

d. Bi tp thc nghim nh lng:

Khi nghin cu ho hc, ta khng ch nghin cu mt nhtnh m cn cn nghin cu c mt nh lng ca n. Tronglch s pht trin ca ho hc, nh nhng nghin cu nhlng cc phn ng ho hc m cc nh bc hc pht minhra nh lut bo ton khi lng cc cht, nh lut thnh phnkhng i v.v...

nh gi tm quan trng ca vic nh lng trong hohc, nh bc hc Nga D.I. Mendlp tng ni: "Khi m mtc tnh no c th cho php o lng c th n mt tnh cht tu tin ch quan, m mang tnh cht khch quan so

snh".

Trong chng trnh Ho hc ph thng, ngay t gio trnhm u v Ho hc, hc sinh c lm quen vi nhng khinim v nh lut nh lng quan trng ca Ho hc nh khilng nguyn t, khi lng phn t, mol nguyn t, mol phnt, mol nguyn t, mol phn t, nh lut bo ton khi lngcc cht, nh lut thnh phn khng i, khi lng ring,nng dung dch v.v... V vy, ngi dy hc Ho hc phi cnhim v lm cho hc sinh nm vng mt nh tnh v nh

lng ca cc bin i ho hc.Trong nhng nm gn y cc nc pht trin nh Nga,



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




13/244

Anh, Php, c... ngi ta thng xy dng nhng bi tp thcnghim nh lng (tnl) lm bi thi hc sinh gii Ho hc vng thi thc nghim cho cc lp hc sinh ph thng. Bi tptnl cng bt u c s dng trong cc gi hc thc hnh nikho, trong gi hc cc gio trnh t chn v trong cc gi hcngoi kho.

Tu theo ni dung hay phng php tin hnh thc nghimta c th phn bi tp tnl thnh cc dng chnh sau y:

Dng 1. Xc nh khi lng, th tch, khi lng ring, nhit

si, nhit nng chy ca cc cht.V du:Cho cc kim loi ch, thic, km, nhm dng nhng

vin nh. Hy tm cch xc nh khi lng ring ca mi kimloi bng thc nghim.

Li gii

Rt nc vo ng ong n ng mt vch no tu , sau thn trng th cc vin kim loi vo ng cho nc dngln ng 1cm3. nc i, ly cc vin kim loi sy kh riem cn. Kt qu thu c chnh l khi lng ring ca n(khi lng ring ca mt cht l khi lng ca mt n v thtch cht . n v thng dng l g/cm3).

Mt s'v d khc:

- Dng cn ly nhng lng bt cc cht sau y: 0,2 mol

magie; 0,1 mol st; 0,3 moi lu hunh; 0,1 mol ng (II) oxit; 0,2mol natri clorua.

- Dng ng ong ly nhng lng nc sau y: 0,5 mol; X



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




14/244

mol; 2,5 mol; 3 mol; 5 mol.

- Trong 3 bnh, mi bnh cha mt trong 3 cht lng sau y:nc, du ho, nc mui.. Hy dng ph k xc nh ng

no ng cht g?

Dng 2. Xc nh t khi ca mt cht kh ny so vi mtcht kh khc hay khi lng phn t ca mt cht kh-

V d:

- Xc nh t khi ca oxi so vi hio bng thc nghim, bitrng 1 lt H2 iu kin tiu chun c khi lng l 0,09g.

- Xc nh khi lng phn t ca kh C02 bng thcnghim, cc ho cht v dng c cn thit u c .

Dng 3. Xc nh lng nc cha trong cc cht v cngthc phn t ca mui ngm nc.

V d: Hy xc nh cng thc ca mui ng sunfat ngmnc bng thc nghim?

Li gii

C th xc nh cng thc ca mui ng sunfat ngm ncbng cch nung mui ng sunfat ngm nc cho mt nc ntrng lng c nh, da vo phng trnh phn ng sau y;

CuS04. nH20 1 > CuS04 + nH20

(160 + 18n)g------------------- * 18g

a g -------------------- b g

a - Khi lng ca mui ngm nc trc khi nung (mu xanh).

b - Khi lng ca mui khan thu c sau khi nung (mui



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




15/244

trng). Cn c vo cc s liu thc nghim, ta tnh c gi trca n.

Dng 4. Xc nh tan ca cc cht v nng ca dung dch.

V d: Hy xc nh bng thc nghim tan ca muiNaN03 cc nhit 30, 40, 50 v 60c. Dng cc kt qu thuc v ng cong biu din s ph thuc tan ca muivo nhit .

Li gii

- Cn xc nh khi lng ca chn s: mr

- Pha dung dch NaN03bo ho nhit ln hn nhit cn xc nh tan t 5-10C. Sau lm lnh dung dch nnhit cn xc nh tan.

- Rt vo chn s ( xc nh khi lng trn) t 5-10 cm3dung dch mui bo ha.

- Cn xc nh khi lng ca chn s v dung dch mui: m2.

- un nng chn s lm bay ht nc ca dung dch.

- Cn xc nh khi lng ca chn s v mu: m3.Sau th nghim thc hin cc php tnh sau:

+ Tnh khi lng dung dch: m2- nil = m4(gam)

+ Tnh khi lng mui: m3m, = m5

+ Tnh khi lng nc: m4- m5= m6

+ Tnh tan ca mui nhit th nghim ( tan ca

mui l lng mui c th ho tan ti mc ti a vo 100 gamnc nhit cho).



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




16/244

+ V ng cong biu din s ph thuc tan ca mui vonhit .

Dng 5. Xc nh thnh phn phn trm v khi lng cahn hp cc cht.

V d: Cho hn hp mui BaCl2.2H20 v Na2C03.10H20.Hy tm cch xc nh thnh phn % v khi lng ca hn hp

bng thc nghim.

Li gii

C th xc nh thnh phn % v khi lng ca hn hpmui trn theo lng nc kt tinh.

- Cn ly mt lng hn hp: m, (gam).

- Sy lm m t nc kt tinh, cn li: m2(gam).

- X l cc kt qu th nghim: .

+ Tnh khi lng nc kt tinh: a = IU] - m2.

+ Gi khi lng mui BaCl2.2H20 c trong lng hn hp

ly l Xgam, khi khi lng mui Na2C03.10H20 l mi - x-+ Gi khi lng nc kt tinh ca BaCl2.2H20 l y gam, khi

lng nc kt tinh ca mui Na2C03.10H20 l a - y.

+ Ta c cc t l sau:

BaCl2. 2HzO 2H20

244g ->36 g

X >

x= (!)36 9



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




17/244

Na2C03.10H20 -> 10H20

286g -> 180g

H i j - X - a - y _ xJ 86(a -y ) = 143(a-y)

180 90

rrvi_ w o, , , 90m, -143aThay (1) v (2)ta c y =---------467

Tnh c X ta d dng tnh c % khilng ca cc mui.

Dng 6. iu ch cc cht v tnhhiu sut ca phn ng

hoc tinh ch mt cht ri tnh tinh khit.V d:Hy iu ch CuO t lg CuC03.Cu(0H)2 (malakhit)

v tnh hiu sut ca phn ng iu ch.

L gii

CuC03.Cu(0H)2 = 2CuO + C0 2t + H2Ot

222 g - 2. 80 g

1 g -> Xg

x = -= 0,72 g222

Hiu sut = - .100 %0,72

trong a l khi lng CuO thc t thu c qua thnghim.

Phn thc nghim kh n gin: cn ly lg malakhit (muxanh) nung cho phn hu hon ton thnh CuO (mu en) ricn xc nh CuO thc t thu c.



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




18/244

Cng c th cho hc sinh iu ch CuS0 4 t CuO hoc phoing v H2S0 4 ri tnh hiu sut ca phn ng iu ch.

Qua cc v d trn, ta thy vic gii bi tp tnl gip cho hcsinh nm vng khng ch mt nh tnh m c mt nh lngca cc qu trnh ho hc, hiu su sc hn cc nh lut hpthc ca ho hc, lp c cng thc phn t ca cc cht trnc s cc d kin thc nghim nh lng. Trong qu trnh giicc bi tp loi ny, hc sinh c lm quen vi vic nh gichnh xc mt nh lng ca cc phn ng ho hc, hiu r ngha ca cc h s trong cc phng trnh phn ng phn nhmi quan h nh lng gia cc cht tham gia v cc cht tothnh. Tt c nhng iu ni trn v thc cht s lm thay itnh cht hot ng nhn thc ca hc sinh.

Mt khc, bi tp tnl c tc dng rt ln trong vic rn knng thc hnh cho hc sinh, lm cho hc sinh bit s dngnhng dng c o lng nh: cn, ng ong, bnh chia , nhitk, ph k, v.v...

Bi tp tnl c tc dng gio dc hc sinh tnh cn thn,chnh xc, tinh thn trch nhim, lng tin vi cng vic; rnluyn tnh kin nhn, tc phoig lao ng c vn ho: lm vicc k hoch, tit kim ho cht, s dng hp l thi giari v ccdng c, thit b; t nh gi c kt qu cng vic ca mnh.

Nhng iu ny gp phn to ln vo vic gio dc k thut tnghp v hng nghip cho cc em.

Ngoi nhng tc dng k trn, bi tp tnl cn lm phttrin hc sinh nng lc nghin cu v hng th hc tp b



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




19/244

mn v thc cht bi tp tnl l nhng bi tp nghin cu dngph thng.

Bi tp tnl l loi bi tp c nhiu tc dng tt nhng cn tc nghin cu. Chng ta cn xy dng nhiu bi tp loi nyv tch cc s dng chng nng cao cht lng dy hc hohc trng ph thng.

III. BN CHT CA VIC LI GII MT BI TON HOHC

1. Cu trc ca bi ton ho hc nc ta trong nhng nm gn y, do yu cu ca cng tc

tuyn sinh i hc, bi tp ho hc kh pht trin. Ngi ta xy dng nhiu bi tp tng hp cha ng nhiu ni dung kinthc ho hc v phi s dng nhiu thut ton phc tp giinh h phng trnh bc nht nhiu n s, phng trnh bc 2 ,

phng trnh v nh, cp s' nhn, v.v... Mt s" bi tp khc i

hi kt hp vic s dng cc thut ton vi vic bin lun hohc gii (bin lun theo tnh cht, theo ho tr, theo khilng, v.v...). V vy, mt bi ton ho hc thng c cu trcnh sau:

- Ni dung ho hc (cc dng phng trnh phn ng hohc).

- Tnh ton theo cc dng phng trnh phn ng ho hc

(ton ho).- Cc thut ton(ton ton).



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




20/244

Trong phn ln cc bi ton ha hc c ni dung ho hc ttcng cn mt s' bi c nhc im l mt ton hc qu rc ri,qu cng knh lm ln t mt bn cht ho hc, bin vic rn

luyn t duy v k nng ho hc thnh rn t duy v k nngton hc.

2. Bn ch't ca vic gii mt bi ton ho hc

Khi gii mt bi ton ho hc, ta phi cn c vo cc d kin cho vit tt c cc phng trnh phn ng xy ra. Nhngcht vit trong cc phng trnh phn ng ny l nguyn cht.Hiu sut phn ng c coi l 100%. Th tch cc kh tham giahay thu c u quy v iu kin tiu chun (0c v 1 at).

Cc d kin cho trong u bi thng l nhng gi thitkhng c bn (cht) khng nguyn cht hoc dng dung dch,hiu sut phn ng nh hn 100%, th tch cc kh cho iukin thng, V.V.... Trc khi tnh ton theo cc phng trnh

phn ng xy ra trong bi ta phi a cc gi thi t khng cbn sang gi thit c bn (tnh lng nguyn cht jnu c tpcht hoc dng dung dch, thng i khi lng cc cht tgam sang mol, i th tch cc kh iu kin thng v ktc,tnh lng cht c trong dung .dch ra mol, v.v...). Sau khi dngcc gi thit c bn tnh ton theo phng trnh phn ng,kt qu thu c cn phi chuyn ngc li t dng c bn sangdng khng c bn theo yu cu ca u bi (th tch kh iukin thng, th tch hoc khi lng ca dung dch, V.V.).

Ta c th biu din bn cht ca vic gii mt bi ton hohc theo s sau:



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




21/244

Ni dung ho hc ca bi ton th hin cc phng trnhphn ng. Lng cht tham gia hoc thu c trong cc phn

ng c tnh theo phng trnh, cn p s ca bi ton nhiukhi phi phi hp cc php tnh theo phng trnh v s dngn cc thut ton nh gii h phng trnh bc nht nhiu n,

phng trnh bc 2 , phng trnh v nh kt hp vi bin lun,v.v... mi tm ra c.

3. Cc bc gii mt bi ton ho hc tng hp

+ Bc 1: Vit tt c cc phng trnh phn ng c thxy ra.

+Bc 2\i cc gi thit khng c bn sang gi thit c bn.

+Bc 3:t n s cho lng cc cht tham gia v thu ctrong cc phn ng cn phi tm. Da vo mi tng quan giacc n trong cc phng trnh phn ng lp ra cc phngtrnh i s.

+ Bc 4: Gii phng trnh hay h phng trnh v binlun kt qu (nu cn), ri chuyn kt qu thuc dng c bn



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




22/244

sang dng khng c bn.

Mun chuyn i cc g i th it khngc bn sang ccg i th i t c bn ta s dng 4 cng th c ch nh . l cc

cng thc biu th:1. Quan h gia khi lng (m), khi lng mol phn t hy

nguyn t (M) v s mol (n) ca cht.

2 . Quan h gia th tch kh iu kin tiu chun (V0) vis mol kh (nk).

3. Quan h gia nng mol (CM) vi s mol cht tan (nct) vth tch dung dch (V).

4. Quan h gia nng phn trm (C%) vi khi lng hays mol cht tan (mct, nct) v khi lng hay th tch dung dch(mdd) V).

Cng thc S mol eht

1. m = M.n _ mn M

2 . v 0= 22,4.nkh _ v0n = 22,4

3- CM= ~ - nc, = V.Cm

4. c% = m,:t .100%mdd

1 c%nct Mct ' m


23/244

Ch : Trong cng thc (3) V tnh bng lt, cn trong cngthc (4) V tnh bng ml v d (kh lng ring ca dung dch)tnh bng g/ml (hay g/cm3).

V d 1: Cho 2,16g hn hp 3 kim loi Na, Al, Fe ho tanhon ton trong dung dch H2S04 long thu c 1,568/ mtcht kh ktc v dung dch B. Thm dung dch NaOH vo

dung dch B cho ti d. Lc ly kt ta em nung ngoi khngkh n khi lng khng i thu c l,6g mt cht rn. Xc

nh khi lng ca cc kim loi c trng lng hn hp cho.

Li gii

+Bc 1:Vit cc phng trnh phn ng xy ra:

2Na + H2S04= Na2S04+ H2T (1)

2A1 + 3H2S04 = FeS04+ 3H2T (2)

y3

2Fe +H2S04= FeS04+ H21 (3)

z z z

A12(S04)3+ 6NaOH = 2Al(OH)3 + 3Na2SO,

Al(OH)3+ NaOH = NaA102+ 2H20

FeS04+ 2NaOH = Fe(OH)2 + Na2S0 4

(4)

(5)

(6)z z



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




24/244

4Fe(OH)2+ 0 2+ 2H20 = 4Fe(OH)3 (7)

z zt _

2Fe(OH)3 = Fe20 3+ 3H20 (8)

zz 2

Cht kh sinh ra l H2, cht rn thu c l Fe20 3.

+Bc 2:i gi thit khng c bn sang gi thi t c bn:

1,568 n Ar7nH. --- - - - =007 ; nFe0 =^=0,01H, 22,4 FeA 160

+Bc 3:t n s, lp h phng trnh:t s"mol ca Na, Al, Fe trong hn hp u ln lt l X, y, z.

Theo khi lng ca hn hp ta c:

23x + 27y + 56z = 2,16 (a)

Theo s" mol H2 (1) (2) (3) ta c:

-+ -^ + z =0,07 (b)

2

2Theo s chuyn ho ca Fe qua cc phn ng (3) (6) (7) (8) ta c:

Fe -> FeS04- Fe(OH)2-> Fe(OH)3-> Fe20 3 ;

zz z z z

2

Vy: - = 0,01 (c)Zi



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




25/244

+ Bc 4: Gii h phng trnh ri chuyn kt qu thucdng c bn sang dng khng c bn:

Phi hp (a) (b) (c) gii ra ta c:

X = 0,01; y = 0,03; z = 0,02mNa = 23.0,01 = 0,23g; rriAi = 27.0,03 = 0,81g;

mFe = 56.0,02 = l,12g

Qua v d trn ta thy bi ton khng cho bit nHm cho

bit th tch H2 ktc, khng cho bit nF0 m cho bit kh

lng Fe20 3... Mc ch bi ton cng khng phi l xc nh smoi ca cc kim loi (x, y, z) m l xc nh khi lng ca cckim loi .

V d 2: Ly 60ml dung dch NaOH th tc dng va (trung ho hon ton) vi 20ml dung dch H2S04. Mt khc nucho 20ml dung dch H2S04 ho tan hon ton 2,5g CaC03 thmun trung ho ht lng H2SO,| d cn 10ml dung dch NaOHni trn.

Tnh nng M ca cc dung dch.Li gii

+Bc 1:Vit cc phng trnh phn ng xy ra:

2NaOH + H2S04= Na2S0 4+ 2HsO

CaCOa + H2S04= CaS04+ C02T + H20

2NaOH + H2S0 4= Na2S0 4+ 2H20

(1)

(2)

(3)



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




26/244

+Bc 2:i gi thit khng c bn sang gi thit c bn:

S mol CaCO j cho l: = 0,025.100

+Bc 3:t n s, lp h phng trnh:Gi nng ca dung dch NaOH l Xmol/1

Gi nng ca dung dch H2S04l y mol/1. Ta c:

- S mol NaOH c trong 60ml dung dch l: 0,06x

- S" mol H2S04c trong 20ml dung dch l: 0 ,02y

- Theo (1) ta c: = 0 ,02y2

- Theo (2) v (3): 0 ,02y = 0,025 + 0,005x

+Bc 4:Gii h phng trnh trn c:

X= 1 v y = 1,5

V du 3: Cho 76,8g hn hp X gm CH3COOC2H5 vCH3COOH tc dng va vi 200g dung dch KOH c% thuc hn hp Y. Cho kali d vo Y thy thot ra 100,81 H2

ktc. Tnh khi ing cc cht trong X v c% ca dung dchKOH.

Li gii

+Bc 1:Vit cc phng trnh phn ng:

C H 3COO C2H 5 + KO H * CH3CO O K + C2H 5OH (1)

CH3COOH + KOH CH3COOK + H20 (2)

2C 2H 5OH + 2 K 2C2H 5O K + H 2 t (3)

2H20 + 2K = 2KOH + H2 (4)



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




27/244

+Bc 2:i cc gi thit khng c bn sang gi thit c bn:

- S mol H2 l: 100,8: 22,4 = 4,5

- Smol H,0 c trong 200g dung dch KOH c% l ~2c18

+Bc 3:t n s v lp h phng trnh:

- Gi s mol CH3COOC2H5l X

gi sm ol CH3COOH l y

- Theo khi lng hn hp ta c:

88x + 60y = 76,8 (I)

Theo s mol H2gii phng (3) v (4) ta c:

X 200-2 c. 1 ,TT.+ (y+ ----- ).= 4,5 (II)2 18 2

Theo nng ca dung dch KOH ta c:

(x+y).56.100=c (III)200

+Bc 4:Li gii h 3 phng trnh (I) (II) (III) c:X= 0,6; y = 0,4; c = 28

Chuyn kt qu t dng c bn sang dng khng c bn:

- Khi lng CH3COOC2H5l: 88 . 0,6= 52,8 g

- Khi lng CH3COOH l: 60 . 0,4 = 24 g

Xt s tng tc ca cc cht trong cc bi ton ho hc, ta

thy c 3 trng hp c th xy ra:1- Cc phn ng xy ra hon ton v cc cht phn ng ly

va : Trong trng hp ny cc cht phn ng u phn ng



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




28/244

ht, ngha l s mol cht phn ng bng s mol cht c banu v vic tnh ton c th da vo s mol c ban u ca bt

k cht no.2- Cc phn ng xy ra hon ton v cc cht phn ng c

cht d, cht thiu.

Trong trng hp ny, chic cht thiu phn ng ht, nghal s mol cht thiu phn ng chnh bng s mol ca n banu v vic tnh ton c th da vo smol c ban u ch cacht thiu.

3Cc phn ng xy ra khng hon ton do hiu sut phnng khng bng 100% hoc do thi gian phn ng cha .

Trong trng hp ny, th khng nhng cht d m c chtthiu cng cha phn ng ht v vic tnh ton khng th davo s mol c ban u ca bt k cht no. Lc ny cr t s"mol ca mt cht no phn ng l Xv mi tnh ton phi

da vo gi tr X.

cc trng hp 2 v 3, vic bin lun bit r cht d,cht thiu, bit r phn ng c xy ra hon ton hay khng lrt quan trngkhi gii cc bi ton ho hc.

V d:

Cho hn hp kh A gm H2v mt olefin c t l y s" mol l1:1i qua ng ng Ni nung nng thu c hn hp kh B c t

khi (hi) so vi H2l 23,2; hiu sut phn ng l b%.Tm cng thc phn t ca olefin v tnh hiu sut ca phn

ng.



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




29/244

Li gii

Gi a l s mol olefin trong hn hp kh A.

Gi Xl s mol olefin phn ngNi

CH2n + H2 ~ CnH2n+2t"

Ban u: a a

Phn ng: X-> . X X

Sau p: (a - x) (a - x) X

, 14n(a-x) + 2(a -x ) + (14n + 2)xB/H, 2(2a-x)= 23,2

Ta c: 23,2x = a(45,4 - 7n); - = -4-~,-4~ - -a 23,2

, n , X -,_ 45,4 -7nV b = nn b = -----a 23,2

T (2): 0< b < 1 vn > 2nn

45 4 7n _ 0 < ^ ~


30/244

Chng IICC PHNG PHP CN BNG

PHN NG HO HC

gii ng v nhanh cc bi ton ho hc, ta cn bit cnbng ng v nhanh cc phn ng c trong bi ton .

C nhiu phng php cn bng phn ng ho hc. Tu theotrnh kin thc ho hc nhng giai on hc tp ho hckhc nhau m ta chn nhng phng php cn bng thch hp.Sau y l mt s phng php.

1. Phng php nguyn t nguyn t"

y l mt phng php kh n gin. Khi cn bng ta c vit cng thc ca cc n cht kh (H2, 0 2, c12,N2...) di dngnguyn t ring bit ri lp lun qua mt, s bc.

V d:Cn bng phn ng p + 0 2- P20 5.

Ta vit p + o P20 5. to thnh mt phn t P9O5 cn 2nguyn t p v 5 nguyn t O: 2P + 50 P20 5. Nhng phn toxi bao gi cng gm 2 nguyn t, nh vy nu ly 5 phn t 0 2tc l s nguyn t oxi tng gp 2th s nguyn t p v s" phnt P20 5cng tng gp 2(tc l 4 nguyn t p v 2P90 5) (tc 10nguyn t O). Do : 4P + 502= 2P20 5.



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




31/244

2. Phng php ho tr tc dng

Ho tr tc dng l ho tr ca nguyn t hoc nhm nguynt ca cc nguyn t trong cc cht tham gia v to thnh trong

phn ng ho hc.

p dng phng php ny, cn tin hnh cc bc sau:

-tXc nh ho tr tc dng

Ghi ho tr tc dng ln pha trn cng thc cc cht:

V d:

I I I III II II II III I

BaCl2+ Fe2(S04)3-> BaS04+ FeCl3

Tm bi s chung nh nht ca cc ho tr cht tc dngBSCNN (3,2,1) = 6

+ Ly BSCNN chia ch cc ho tr ta c cc h s:

=3; - - =2; 5=6II III I

3BaCL> + Fe,(S04)3= 3BaS04+ 2FeCl3Dng phng php ny s cng c" c khi nim ho tr,

cch tnh ho tr, nh ho tr ca mt s nguyn t thng gp.

3. Phng php dng h s phn s

t cc h s vo cc cng thc ca cc cht trong phn ngkhng phn bit s nguyn hay phn s sao cho s' nguyn tca mi nguyn t v u bng nhau. Sau kh mu schung ca tt c cc h s.

V d: p + 0 2 P2O5



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




32/244

- t h s cn bng: 2P + 5/202= P20 5

- Nhn cc h s vi mu s kh phn s:

2 .2P + 2. 0 2= 2P20 5

24P + 502 = 2P20 5

4. Phng php "chn - l"

Mt phn ng sau khi cn bng th s nguyn t ca mtnguyn t v tri phi bng s nguyn t ca nguyn t v phi. V vy nu s' nguyn t ca mt nguyn t mt v lschn th v kia n cng phi l s" chn. Nu mt cngthc no , s nguyn t ca nguyn t cn l s l th phinhn i.

V d: FeS2+ 0 2- Fe20 3+ S02

v tri snguyn t o l s chn vi bt k h s no. vphi trong S02, o l s chn nhng trong Fe20 3li l s" l nnphi nhn i. T cn bng tip cc h s' cn li. Ch s

trong du ngoc ch th t cc ng tc cn bng:(2) (4) (1) (3)

4FeS2+ 1102= 2Fe20 3+ 8S02

Hoc c th biu din th t cn bng theo s sau:

2F69O3>4F6S2 8SO2 I IO 9

5. Phng php xut pht t nguyn t chung nht

Chn nguyn t c mt nhiu hp cht nht trong phnng bt u cn bng h s cc phn t.



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




33/244

V d: Cu + H N03-> Cu(N03)2 + NO + H20

Nguyn t" c mt nhiu nht l oxi; v phi c 8nguyn toxi, v tri c 3. Bi s" chung nh nht ca 8v 3 l 24. Vyh s ca HNO3l 24/3 = 8 .

Ta c 8HNO3 4H20 - 2NO (v s nguyn t N v trichn) -> 3Cu(N03)2 [v (8- 2):2= 3] - 3Cu.

Vy: 3Cu + 8HNO3 = 3Cu(N03)2+ 2NO + 4H20.

6. Phng php cn bng theo "nguyn t tiu biu"

"Nguyn t tiu biu" l nguyn t .c nhng c im sau:

C mt t nht trong cc cht phn ng .Lin quan gin tip n nhiu cht nht trong phn ng.

Cha thng bng v s nguyn t 2v.

Phng php cn bng ny tin hnh qua 3 bc:

(a) Chn "nguyn t' tiu biu".

(b) Cn bng nguyn t tiu biu.

(c) Cn bng cc nguyn t khc theo nguyn t ny.

V d: KMn04+ HC1 - KC1 + MnCl2+ Cl2+ H20(a) Chn nguyn t tiu biu: 0

(b) Cn bng nguyn t tiu biu: KMn04- 4H20

(c) Cn bng cc nguyn t khc:

+ Cn bng H: 4H20 8HC1

+ Cn bng Cl: 8HC1 -> KC1 + MnCl2+ 5/2C1, ta c:

KMn04+ 8HC1 = KC1 + MnCl2+ 5/2Cl2 + 4H20



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




34/244

Sau cng nhn t t c cc h s vi mu s chuyn h s vdng nguyn, ta c phng trnh phn ng ho hc l:

2KMn04+ 16HC1 = 2KC1 + 2MnCl2+ 5C12+ 8H20

7. Phng php cn bng theo trnh t kim lo i - phi kim

Theo phng php ny, u tin cn bng s nguyn t kimloi v phi kim ri n nguyn t hiro, sau cng a vo cch s' bit cn bng nguyn t oxi.

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

V d 1: 2NH3+ - 0 2 2NO + 3H20 2

phn ng trn khng c kim loi. Nguyn t phi kim N cn bng nn u tin cn bng nguyn t H. Sau cn bngnguyn t N. Cui cng cn bng nguyn t 0 ri nhn tt cvi 2 a v s nguyn. (Ch s' trong du ngoc ch th tcn bng).

4NH3+ 502= 4NO + 6H20

V d 2:(5) (1) (5) (4) (5) (2) (5) (5) (3)

2 . 2CuFeS2+ 2 . 13/202 -> 2 ,2 CuO + 2Fe20 3+ 2 . 4S02

Nguyn t Cu cn bng. Vy u tin cn bng nguyn tFe. Tip theo cn bng theo th t cc nguyn t Cu, s v o rinhn i cc h s':

4CuFeS2+ 1302= 4CuO + 2Fe20 3+ 8S02



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




35/244

8. Phng php cn bng phn ng chy ca cht hu c

a. Ph n ng chy ca hrocacbonNn cn bng theo trnh t sau:

- Cn bng s nguyn t hiro. Ly s nguyn t H cahirocacbon chia cho 2 , nu kt qu l s l th nhn i phnt hirocacbon, nu l s' chn th nguyn.

- Cn bng s' nguyn t cacbon.

- Cn bng s nguyn t oxi.

V d: C2H6+ 0 2 C02+ H20

Ly s nguyn t H chia cho 2c: 6/2= 3l s l, nn cn

nhn i:(1) (4) (2) (3)2C2H6+ 702-> 4C02+ 6H20

b. P h n ng chy ca hp cht cha oxi

Gn bng theo trnh t:

- Cn bng s nguyn t c.- Cn bng s nguyn t H.

- Cn bng s nguyn t 0 bng cch tnh s nguynt o v phi ri tr i s nguyn t oxi c trong hp cht. Kt qu thuc em chia i s l h s' ca phn t0 2. Nu h s l

phnsth nhn i c 2v ca phng trnh kh mu s.

V d:

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

2C2H5COOH + 2 . 7/202 2 . 3C02+ 2. 3H202C2H5COOH + 7 0 2r* 6C 02 + 6H 20



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




36/244

9. Phng php xut pht t bn cht ho hc caphn ng

Phng php ny lp lun da vo bn cht ca phn ng cn bng.

V d: Fe20 3+ c o ->Fe + C02

Theo phn ng trn, khi c o b oxi ho thnh co, R kt hpthm oxi. Trong phn t Fe20 3c 3 nguyn t oxi. Nh vy bin 3 phn t c o thnh 3 phn t C02. Do ta cn t hs 3 trc cng thc c o v C02>sau t h s 2trc Fe:

Fe20 3+ 3CO = 2Fe + 3C0210. Phng php cn bng electron hoc ion -electron

y l phng php dng cn bng cc phn ng oxi hokh. Bn cht ca phng trnh ny da trn nguyn tc:Trong mt phn ng oxi hokh, s' electron do cht khnhng phi bng s electron do cht oxi ho thu.

Vic cn bng qua 3 bc:

a) Xc nh s thay i s oxi ho.b) Lp thng bng electron.

c) t cc h s tm c vo phn ng ri tnh cc h s"cn li.

V d:Cn bng phn ng:

FeS + HN03 Fe(N03)3+ N20 + H2S04+ H20

a) Xc nh s thay i s oxi ho:

+2-2 +5 +3 +1 +6

FeS + HN03 Fe(N03)3+ N20 + H2S0 4+ H2



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




37/244

b) Lp thng bng electron

Fet2~le = Fe+31 .. -9e 8

s - 8e = s 2N+5+ 8e = 2N+l 9

c) t cc h s tm c l 8v 9 vo phn ng ri tnh cch s cn li ta c:

8FeS + 42HN3 = 8Fe(N03)3+ 9N20 + 8H2S04+ 13H20

Phng php cn bng ion-electron cng ging nh phngphp cn bng electron, nhng khc l vit cc cht ox ho v

cht kh di dng ion, th hin ng s tn ti ca chngtrong dung dch.

C 3 trng hp c th xy ra:

a. Phn ng trong dung dch axit, nu qu trnh oxi ha hockh:

- Thiu oxi: mi o -2c b bng 1H20 v thm 2H+ v sau.

- Tha oxi: mi o -2 c ghi bng 1H20 v thm 2H+ vtrc.

b. Phn ng trong dung dch baz, nu qu trnh oxi hahoc kh:

- Thiu oxi: mi O-2c b bng 20H" v thm 1H20 v sau.

- Tha oxi: mi O"2 c ghi bng 20H" v thm 1H20 vtrc.

c. Phn ng trong dung dch c nc tham gia, nu qu trnhoxi ha hoc kh:

- Thiu oxi: mi o -2c b bng 1H20 v thm 2H+ v sau.



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




38/244

- Tha oxi: mi 0 2 c ghi bng 20H v thm 1H20 vtrc.

V d:Phn ng trong dung dch axit:+7 +3 +2 +5

KMn04+ KN02+ H2S04- MnS04+ K2S04+ KNOs + H20

MnO- + 5e + 8H+= Mn2++ 4H20

NO, - 2e + H20 = NO3+2H+

Cng tng v ca hai phng trnh ta c:

2Mn04- + 16H++ 5N02 + 5H20 = 2Mn2++ 8H20 + 5NO3+ 10H+Gin c nhng ion cng loi 2 v ta c:

2Mn04- + 6H++ 5N02" = 2Mn2++ 3H20 + 5NO3-

Thm 7K+ v 3S042 vo hai v s c phng trnh phnng dng phn t:

2KMn04+ 5KN02+ 3H2S04= 2MnS04+K2S04+ 5KNO3+ 3H20

V d:Phn ng trong dung dch baz:

+3 0 +6 -1NaCrO, + Br2+ NaOH -> NaCr04+ NaBr

+3 +62

3

CrO' - 3e + 40H ' = CrO; + 2H20

Br2+ 2e = 2Br

2Cr02- + 80H- + 3Br2= 2CrO/- + 6Br' + 4H20

Phng trnh phn t:2NaCr02+ 3Br2+ 8NaOH = 2Na2Cr04+ 6NaBr + 4H20



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




39/244

V d:Phn ng trong dung dch c nc tham gia:

+7 +4 +4 +6

KMn04+ K2S03+ H20 Mn02+ K2S042 Mnt, + 3e + 2H20 = M n02+ 40H"

3 S O j - 2e + H 20 = S O 4" + 2H +

2Mn 0 ' + 7H20 + 3S0- = 2MnOa + 20H" + 3 S 0f + 6H20

Phng trnh ion:

2Mn 0 ' + H20 + 3SO3" = 2Mn02 +20H~+ 3S0J-

Phng trnh phn t:2KMn04+ 3K2S03+ H20 = 2Mn02 + 3K2SO, + 2K0H.

11. Phng php cn bng bng i s

ng i s' xc nh h s phn t ca cc cht thamgia v thu c sau phn ng ho hc, ta coi cc h sl ccn s v k hiu bng cc ch a, b, c, d, v.v... ri da vo mi

tng quan gia cc nguyn t ca cc nguyn t theo nhlut bo ton khi lng lp ra mt h phng trnh bcnht nhiu n s. Gii h phng trnh ny v chn ccnghim l nhng s' nguyn, dng nh nht ta s xc nhc h s phn t ca cc cht trong phng trnh ho hc.

V d:cn bng phn ng ho hc:

Cu + HNOa -> Cu(N0 3)2+ NO + H20

K hiu cc h s' phi tm bng cc ch a, b, c, d, e v ghi vophng trnh ta c:

a Cu + b HNO3 = c Cu(N03)2+ d NO + e H20



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




40/244

Xt v s' nguyn t Cu ta c: a = c (1)

Xt vsnguyn t H ta c: b = 2e (2 )

Xt v s nguyn t N ta c: b = 2c + d (3)Xt v s nguyn t 0 ta c: 3b = 6c +d + e (4)

> Ta c 4 h phng trnh nm n v gii nh sau:

Rt e = t phng trnh (2) v d = b-2c t phng trnh (3)2

v thay vo phng trnh (4):

3b = 6c + b - 2c + 26b = 12c + 2b - 4c + b

6b - 3b = 12c 4c

3b = 8c hay b = 3

Ta thy cho gi tr ca b lnhng s' nguyn th gi trca c phi l nhng s' chia ht cho 3. Trong trng hp ny cho cc h s ca phng trnh ho hc l nhng s' nguyn nhnht ta cn ly c = 3. Lc a = 3; b = 8; d = 2; e = 4.

Vy phng trnh phn ng ho hc trn c dng:

3Cu + 8HNO3= 3Cu(N03)2+ 2NO + 4H20

Khi cho c = 6 v tnh cc h s khc ca phng trnh hohc ta thy nhng h s th hai ny ch l bi s ca cc h s

th nht (xem bng i).



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




41/244

a b c d e

3 8 3 2 4

6 16 6 4 8

v d trn trong phng trnh ho hc c 5 cht (Cu;HN03; Cu(N03)2; NO; H20) v 4 nguyn t (Cu; H; N; O) khi lp

phng trnh i s cn bng ta c mt h 4 phng trnhvi 5 n s. Hay ni mt cch tng qut ta c n n s v n-1phng trnh.

Nh vy khi lp h phng trnh i s cn bng mtphng trnh ho hc, nu c bao nhiu cht trong phng trnhphn ng ho hc th c by nhiu n s v nu c bao nhiunguyn t to nn cc cht thi c by nhiu phng trnh.

T nhn xt trn, chng ti th dng phng php ny cn bng hu ht cc phng trnh phn ng ho hc c trongchng trnh ph thng, mt schng trnh i hc v gp

cc trng hp sau y:1. Cc phng trnh phn ng ho hc trong s cc cht

tham gia v thu c sau phn ng bng s cc nguyn t tonn cc cht y. V d:

p20 5+ h 20 - h 3p o 4

Hoc s cc cht tham gia v thu c sau phn ng ln hns cc nguyn t' cu to nn chng 1 n v (i b phn cc

phng trnh phn ng ho hc u ri vo trng hp ny).Khi lp cc h phng trnh i s cn bng cc phng



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




42/244

trnh phn ng ho hc ny, ta c cc h phng trnh bc nhtc n n s v n phng trnh hoc n n s v n-1 phng trnh.

Gii nhng h phng trnh i s ny ta c v s nghimnhng bao gi ta cng chn c nhng nghim (nguyn,dng) nh nht ph hp vi phng trnh ho hc (nh v du tin).

2 . Cc phng trnh phn ng ho hc trong s cc chttham gia phn ng ho hc ln hn s ccnguyn t cu tonn cc cht y t 2n v tr ln, v d:

KI + H20 + 0 3 - KOH + I2+ 0 2(sch giokhoa Ha hc lp8) (6 cht, 4 nguyn t)

H2S + K2Cr20 7 + H2S 0 4 - s + K2S 0 4 + Cr2(S 0 4)3 + H20(7 cht, 5 ngyn t)

H20 2+ AgNOa + NH4OH -> 0 2+ Ag + NHjN03+ H20 (7 cht,4 nguyn t)

Khi lp h phng trnh i s cn bng nhng phng

trnh phn ng ho hc ny, ta c nhng h phng trnh cn n s' nhng ch c n2hoc n3 phng trnh.

Gii nhng h phng trnh phc tp, di, mt nhiu thigian v cho ta nhng kt qu khng duy nht. Khi phi cnc vo bn cht ho hc ca phn ng bin lun v chnnhng nghim s thch hp cho phng trnh ho hc. y lnhng vic lm kh i vi hc sinh, nht l cc lp 8v 9.

lm sng t nhng nhn xt trn, chng ta hy cn cbng phng trnh phn ng oxi ho - kh gia axit



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




43/244

sunfuhidric v bicromat kali trong mi trng axit sunfuriclong xy ra theo s sau:

aH2S + bK2Cr20 7+ cH2S04= dS + eK2S04+ fCr^SO^ + gH20

long

Xt v s" nguyn t H ta c: 2a + 2c = 2g (1)

xt v s nguyn t s ta c: a + c = d + e + 3f (2)Xt v s nguyn t K ta c: 2b = 2e (3)

Xt v s nguyn t Cr ta c: 2b = 2f (4)

Xt v s nguyn t 0 ta c: 7b + 4c = 4e + 12f + g (5)

T phng trnh () ta c: a + c = g

T phng trnh (2) ta c: d = a + c e 3f

T phng trnh (3) v (4): b = e = f

Thay cc gi tr ca e v f bng b v ca g bng a+c vo hnm phng trnh trn ta c:

7b + 4c = 4b + 12b + a + c

a = 7b + 4c - 4b - 12b - c

a = 3c - 9b

a > 0 hay 3c > 9b o c > 3b

Cho b v c mt s gi tr thch hp ri tm gi tr ca cc n

s cn li, ta c kt qu cc bng sau:



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




44/244

b = 1; 2; 3

c = 4; 7; 10

a b c d e f g

3 1 4 3 1 1 73 2 7 2 2 2 10

3 3 10 1 3 3 13

b = 1; 2; 3

c = 5; 8; 11

a b c d e f g

6 1 5 7 1 1 11

6 2 8 6 2 2 14

6 3 11 5 3 3 17

b = 1; 2; 3

c = 6; 9; 12

a b c d e f g9 1 6 11 1 1 15

9 2 6 10 2 2 18

9 3 12 9 3 3 21

Ghi vo phng trnh phn ng ho hc cc h s xcnh ta c nhiu phng trnh vi cp h s' cc cht khc

nhau:



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




45/244

3H2S + K2Cr20 7+ 4H2S04= 3S + K2S04+

+ Cr2(S04)3+ 7H20 (1)

6H2S + K2Gr20 7+ 5H2S04= 7S + K2SO, +

+ Cr2(S04)3+ 11H20 (2)9H2S + K2Cr20 7+ 6H2SO, = US + K2S04+

+ Cr2(S04)3+ 15H20 (3)

y ta thy rng cc h s (2) v (3) khng phi l bi sca cc h s ca (1). v mt ton hc, tt c cc phng trnhho hc trn u ng nhng vi quan im ho hc, ta ch

dng c nhng h s (1).Th nhng, nu phn ng oxi ho - kh trn xy ra trong

mi trng axit sunfuric c th (2) cng ng v H2S04c lcht oxi ho mnh, c th oxi ho H2S theo phng trnh:

H2S04+ 3H2S = 4S + 4H20

Cn (3) v bn cht ho hc khng khc (2), ch c s' lngcc phn t axit sunfuhidric v phn t axit sunfuric tham gia

khc nhau m thi.Qua nhng nhn xt v nh gi trn chng ti c my

ngh sau y:

1. Tuy c hn ch v mt ni dung ho hc nhng ta vn nndy phng php cn bng phng trnh ho hc ny cho hcsinh cc lp 8v 9 v:

- Trong khi hc sinh cha c hc phng php cn bng

electron (cn bng electron ch c gii thiu trong sch bi



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




46/244

tp lp 10) th phng php ny l mt trong nhng phngphp thun li v ni chung hc sinh c trnh gii thnhtho v nhanh chng cc h phng trnh bc nht.

- Khng phi ta ch dy phng php ny m song song vin cn dy cc phng php khc na. Khi gp mt phngtrnh ho hc c th, hc sinh c th t chn ly mt phngphp cn bng tin nht, nhanh nht.

- Hu ht cc phng trnh ho hc c trong chng trnhph thng u n gin. Khi gp h phng trnh i s" cnbng ta u c mt h phng trnh c n n s v n - phngtrnh. Gii nhng h ny d dng, cho kt qu dy nht, khngphi bin lun cho nghim s.

2. Khi dy phng php ny, ta nn lu cho hc sinh duhiu nhn bit mt phng trnh ho hc c nn dng i s

cn bng hay khng phn ng. Nu thy s cht ln hn snguyn t t 2n v tr ln th khng nn dng phng phpi s cn bng.

3 . cc phng trnh ho hc c cc nhm nguyn t", cc gcaxit chuyn nguyn vn t v tri sang v phi (chng khng bph v trong phn ng ho hc) th khi lp h phng trnh is cn bng ta nn tnh theo cc nhm, cc gc . Lc hphng trnh i s s n gin i nhiu. V d:

a FeCl3+ b KCNS = c KC1 + d Fe(CNS)3

khi so snh gi tr gia h s b v d ta nn tnh theo nhm

CNS, lc b = 3d.



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




47/244

Chng III

CC PHNG PHP GII BI TON HO HC

1. Phng php bo ton khi lng

Nguyn tc ca phng php ny l: "Tng khi lng cccht tham gia phn ng bng tng khi lng cc cht to

thnh sau phri ng".Ch : Khng tnh khi lng ca phn khng tham giaphn ng.

V d 1: Hn hp X gm Fe, FeO v Fe20 3. Cho mt lungCO i qua ng s dng m gam hn hp X nung nng. Sau khikt thc th nghim, thu c 64,0 g cht rn A trong ng sv 11,21 kh B ktc, c t khi so vi H, l 20,4. Tnh gi tr

ca m?L i g i i

Cc phn ng kh st oxit c th c:

3Fe20 3+ CO = 2Fe30 4+ C02 (1)

Fe30 4+ CO = 3FeO + C02 (2)

FeO + CO = Fe + C02 (3)

Nh vy cht rn A c th gm 4 cht: Fe, FeO, Fe20 3 v, Fe3Oj hoc t hn. Kh B c th l hn hp ca C02v co .



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




48/244


49/244

Gi s mol C02l a th s mol H,0 l a.v 4

Ta c: 44.a + 18. = 4,6 -> a = 0,08ml4

m c c o 2 , = mC(cste) = 0,08 . 12 = 0,96g

m H,H20 ) = m H(eSte) = I > 0 8 2 = > 1 2

() = 1,88 0,96 0,12 = 0,8

Gi cng thc n gin ca este l CxHyOz

Ta c:x ^ 9 6 : 2 : | = 8 : 1 2 : 5

12 1 16

Cng thc phn t: (C8H120 5)n M < 29.6,5 = 188,5

V C8Hi20 5= 188 nn n = 1. CTPT ca este l C8HI20 5

V d 3: Khi cho 0,1 mol este to bi axit 2 ln axit v

ru 1 ln ru tc dng vi NaOH thu c 6,4g ru vmt lng mui (gam) nhiu hn lng este l 13,56% (so vilng este). Tnh khi lng ca mui v xc nh cng thccu to ca este.

Li gii

Gi lng este l m gam:

R(COOR')2+ 2NaOH -> R(COONa)2+ 2R'OH

0,1mol 0,2 0,1 0,2

m(g) 40.0,2 m + 0,1356m 6,4



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




50/244

Ta c: m + 8= m + 0,1356m + 6,4

0,1356m = 1,6 - m = 11,8g

Vy khi lng mui l: 11,8+ 0,1356 . 11,8= 13,4g

0,1mi mui R(CONa)2bng 13,4g

Vy: R(COONa)2= 134 -> R = 0

M,.u= 32; ROH = 32; R = 15 - CH3. Vy cng thc0 ,2

V d 4:Ho tan 2,84g hn hp 2mui cacbonat ca 2kimloi thuc phn nhm chnh nhm II v thuc 2chu k lin tip

bng dung dch HC1 d, ngi ta thu c dung dch A v 0,6721kh ktc. Hi c cn dung dch A th thu c bao nhiu gammui khan?

Li gii

Phng trnh phn ng ho hc xy ra:

Gi m l khi lng mun khan thu c. Theo nh lut boton khi lng ta c:

cu to ca este l:


51/244

2,84 + 0,06.36,5 = m + 0,03.44 + 0,03.18l,32g ' " 0,54 '

m = 2,84 + 2,19 - 1,32 - 0,54 = 3,17g

V d 5:Ho tan lOg hn hp 2mui cacbonat kim loi hotr 2v 3 bng dung dch HC1 ta thu c dung dch A v 0,6721kh ktc. Hi c cn dung dch A th thu c bao nhiu gammu khan?

Li gii


XC03+ 2HC1 =XC12+ C02T + H20

Y2(C03)3+ 6HC1= 2YCI3+ 3C02 + 3H20

0,672npA Uj Uuc 2 22,4

Gi m l kh lng mun khan thu c ta c:

10 + 0,06.36,5 = m + 0,03.44 + 0,03.18

2 19 "l,32g 0,54

m = 10 + 2,19 - 1,32 - 0,54 = 10,33g

2. Phng php tng gim khi lng

Nguyn tc ca phng php ny l: "Khi chuyn t cht A thnh cht B (c th qua nhiu giai on trung gian), khilng tng hay gim bao nhiu gam (thng tnh theo 1 mo) v

da vo khi lng thay i ta tnh c s mol cht tham, gia phn ng hoc ngc li".



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




52/244

V d 1: C 11 dung dch hn hp Na2C03 0,1 mol/1 v(NH4)C0 30,25 mol/1. Cho 43g hn hp BaCl2v CaCl2vo dungdch . Sau khi kt thc phn ng, ta thu c 39,7g kt ta A.

Tnh %khi lng cc cht trong A.

Li gii

Phng trnh phn ng ho hc xy ra

Na2C03 = 2Na+ + C 032-

(NH4)2C03= 2 N H ; + C 0 r

BaCl2= Ba2++ 2C r

CaCl2= Ca2++ 2C1"

Ba2++ c O3' = BaCO (1)

Ca2++ c O3' = CaC03 i (2)

Theo (), (2) c 1mol BaCl2hoc CaCl2chuyn thnh BaC03hoc CaC03th kh lng mui gim 71 60 = llg

Tng s mol 2mui BaC03v CaC03l:

- 59 ,7 =0,311

nNa2co3= 0)1-1 - 0,1; n(Nl4)2C03 - 0,25.1 - 0,25

Tng s mol CO3" = 0,1 + 0,25 = 0,35 -> d CO3"

Gi X l smol BaC03, y l s' mol CaC03 trong A ta c hphng trnh:

X + y = 0,3

. 197x+ lOOy = 39,7



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




53/244

G ii ra X= 0,1

y = 0,2

% BaC03= 197-Q1 ,100% = 49,62%39,7

% CaC03= 100% - 49,62% = 50,38%

V d 2: Ho tan 2,84 hn hp 2 mui cacbonat ca 2 kimloi thuc phn nhm chnh nhm II v thuc" 2chu k lin tip

bng dung dch HC1 d ngi ta thu c dung dch A v kh B.C cn dung dch A th thu c 3,17g mui khan.

a. Tnh th tch kh B ktc?b. Xc nh tn ca 2kim loi?

Li gii

Gi X, Y l 2kim loi. Ta c phng trnh phn ng ho hcsau:

XC3 + 2H C 1= X C 12+ C 02t + H 20

YCOa + 2HC1= YC12+ C02t + H20

234 g 37g+ 3. 4 4 + 0 ^ 3 . 18T 1 mol mui cacbonat chuyn thnh mui clorua th khi

lng tng 71 - 60 = llg . Vy tng s" mol mui cacbonat bng

s" mol CO, v bng = 003 mol11

M = 1 = 94,660,03M = 94,66 60 = 34,66 > M, < 34,66 < M2



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




54/244

Hai kim loi thuc phn nhm chnh nhm 2, hai chu klin tip th l Mg (24) v Ca (40).

V d 3: Khi un 9,7g hn hp A gm 2 ng ng cabromua benzen vi dung dch NaOH, ri cho kh C 0 9i qua thuc 5.92g mt hn hp B gm 2cht hu c. Tnh tng s molca cc cht trong A.

Li gii

Phn ng ho hc xy ra:

CnH2n_7Br + NaOH - CnH2n_7ONa + HBr

CmH2m_7Br + NaOH -> CmH2m_7ONa + HBr

CnH2n_7ONa + C02+ H20 -> CnH2n_7OH + NaHCOa

CmH2m_7ONa + C02+ H20 -> CmH2m_7OH + NaHCOs

T hn hp A - hn hp B khi lng gim i: 9,7 5,92 = 3,78g

T 1mol A>1mol B khi lng gim i l: 80 17 = 63g

(CnH2n_7Br CnH2n_7OH)

.

80 17

Tng s" mol cc cht trong A : 3,78 : 63 = 0,06 mol

V d 4:Khi un 0,06 mol hn hp A c khi lng l 9,7ggm 2ng ng ca bromua benzen vi dung dch NaOH, richo kh C02i qua thu c hn hp B gm 2cht hu c! Tnhkhi lng ca B.



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




55/244


56/244

Nhc im:

- Ch p dng cho h phng trnh phn ng oxi ho - kh.

- Ch thng dng gii bi ton v c.

Nguyn tc:

"s mol cht nhng Xs' e nhng = s mol cht nhn Xs e nhn".

V d:

4M + n 0 2= 2M*nOn

Gi s c Xmol M tc dng vi y mol 0 2th:

- S' mol electron nhng i: x.n

- S mol electron thu vo: y.4

xn = y.4

H2- 2e = 2H+s mol electron l 2n.

0 2 + 4e = 202s mol electron l 4n.

V d 1: Cho 16,2g kim loi M, ho tr n tc dng vi 0,15mol 0 2. Cht rn thu c sau phn ng cho ho tan hon tonvo dung dch HC1 d thy bay ra 13,441 H2 ktc. Hi M lkim loi g?

Li gii

4M + n0 2= 2M2On (1)

Cc phn ng ho tan cht rn:

M20 + 2nHCl = 2MCln + 2H20 (2)



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




57/244

2M + 2nHCl = 2MC1 + nH21 (3)

13,44 A Jn - : = 0,6 molH 22,4

Theo (1), (3) th tng smol electron m kim loi M cho phibng tng s" mol electron m 0 2v H nhn.

Gi Xl s mol ca kim loi M: M - ne = M+n

- S mo electron ca kim loi M cho l: nx

- S mol electron 0 2nhn l: 0,15 . 4

- S mol electron H2nhn l: 0,6 . 2

nx = 0,15 . 4 + 0,6 . 2 = 1,8

X= ~ ~ =rr~ -> M = 9nn M

Bin lun M theo n ta thy ch c cp n = 3 v M = 27 l phhp. Vy M l Al.

V d 2:t mt lng AI trong 6,721 0 2, cht rn thu csau phn ng cho ho tan hon ton vo dung dch HC1 li thy

bay ra 6,721 H2. Cc th tch kh u o ktc.

Xc nh khi lng AI dng.

Li gii

6,72 nH = ~r~ = 0,3 molH* 22,4


4A1 + 302 = 2A120 3 (1)

2AI + 6HC1 = 2A1C13+ 3H2 (2)



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




58/244

Gi X l s' mol Al:

3x = 0,3 . 4 + 0,3 . 2 = 1,8

V d 3:

Cho hn hp Y gm Fe v kim loi M c ho tr n duy nht.

1. Ho tan hon ton 3,61g hn hp Y bng dung dch HC1thu c 2,1281 H2, cn khi ho tan 3,61g hn hp Y bng dung

dch HN03 long d th thu c 1,7921 kh NO duy nht. Hyxc nh kim loi M v tnh % khi lng ca mi kim loitrong hn hp Y.

2 . Ly 3,61g hn hp Y cho tc dng vi 100ml dung dchcha AgN03v Cu(N03)2, khuy k ti phn ng hon ton chthu c 8,12g cht rn gm 3 kim loi. Ho tan cht rn

bng dung dch HC1 d thy bay ra 0,6721 H2. Tnh nng mol

ca AgN03v Cu(N03)2trong dung dch ban ca chng.

Bit cc th tch kh o ktc v hiu sut cc phn ng l100%.

Li gii: Phng trnh phn ng ho hc xy ra:

F e + 2HC1=F eCl2+ H2 1 (1)

(2)

(3) /

(4)1 ;

02 M +2nHCl =2MCl + nH2T

F + 4H N 0 3= Fe(N03)s + N o t + 2H200 +5 +3 +2

0 + 5 + n3 M + 4nH N 0 3= 2 M(NOs ) + n N 0 + 2nH20



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




59/244

Gi X l smol Fe, y l s" mol M c trong 3,61g hn hp Y.

S mol electron Fe nhng i (1) l: 2x

S' mol electron M nhng i (2) l: ny

S' mol electron H thu vo (1) v (2) l: '^ .222,4

V s" mol electron nhng ra phi bng s mol electron thuvo nn:

9 192 + = ^ . 2 =0,19 ()

22,4

S mol electron nhng i (3) l: 3x (Fe - 3e = Fe+3)S mol electron nhng i (4) l: ny (M - ne = M+n)

1 792So mol electron thu vo to ra NO l: - - .3

22,4

(v N+5+ 3e = N+2)

Vy 3x + ny = 3 = 0,24 (b)22,4

Ly (b) tr i (a):3x + ny = 0,24

2x + ny = 0,19

X = 0,05; ny = 0,09 (c)

Mt khc theo khi lng hn hp:

0,05 . 56 + yM = 3,61; yM = 0,81 (d)

T (c): ny = 0,09 -> y = n



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




60/244

Thay vo (d): M =y

M = 9n

Nghim duy nht l n = 3; M = 27 - Al; y = 0,03

o/oFe = 00 5-56-1Q0%= 77,25/o3,5.1

%A1 = 100% - 77,25% =22,75%

2. V AI hot ng hn Fe v v cht rn gm 3 kim loi nnchng phi l Ag, Cu v Fe, nn c th xy ra cc phn ng:

AI + 3AgN03= A1(N03)3+ 3Ag (5)

2A1 + 3Cu(N03)2= 2A1(N03)3+ 3Cu (6)

Fe + 2AgN03= Fe(N03)2+ 2Ag (7)

Fe + Cu(N03)2= Fe(N03)2+ Cu (8)

Fe + 2HC1 = FeCl2+ H2T (9)

Cc phn ng (5) v (9) chc chn xy ra, cn (6), (7), (8) xy

ra hay khng tu thuc vo lng AI hay AgN03tha thiu.

Gi a l s mol AgN03

Gi b l s' mol Cu (N0 3)2. Ta c phng tr nh bo tonelectron:

la + 2b + 2.0,03 = 3. 0,03 + 2. 0,05Ag Cu H2 a i Fe

Theo (9): nH = ---0,03 = s mol Fe cn liH-' 22,4



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




61/244

Mt khc theo khi lng 3 kim loi ta c:

108a + 64b + 0,03. 56 = 8,12

T 2phng trnh trn rt ra: a = 0,03; b = 0,05= 0,5M

V d 4:Oxi ho hon ton 2,184g bt Fe ta thu c 3,048ghn hp cc oxit st (hn hp A). Chia hn hp^A thnh 3 phn

bng nhau.

1. Cn bao nhiu lt H2(ktc) kh hon ton cc oxit trongphn mt?

2 . Ho tan phn th hai bng dung dch HN0 3 long d.Tnh th tch kh NO duy nht bay ra ktc.

3. Phn th ba trn vi 5,4g bt AI ri tin hnh phnng nhit nhm (hiu sut 100%). Ho tan hn hp thu csau phn ng bng dung dch HC1 d. Tnh th tch kh bayra ktc.

Li gii

Cc phn ng c th c:

2Fe + 0 2= 2FeO

2Fe + 1,502= Fe20 3

3Fe + 202= Fe30 4

FeO + H2= Fe + H20Fe20 3+. 3H = 2Fe + 3H20

(1)

(2)

(3)

(4)(5)



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




62/244

Fe30 4+ 4H23Fe + 4H?0 ()

Theo cc phn ng (1 >6) th Fe khng thay i s" xi ho,hiro cho v oxi nhn electron nn t c phng trnh (k hiu nl s mol):

2nH2= 4n02

3Fe30 4+ 28HN03+ 9Fe(NC>3)3+ NO t + 14H20 (9)

Ta nhn thy t t c Fe t Fe b oxi ho thnh Fe3+, cn N+sb kh thnh N+2, cn 0 b kh thnh o -2nn phng trnh boton electron l:

3n + 0 ,0 0 9 .4= -^ i^ . - .3 =0,039

do n H =0,018 mol

v2= 0,018 . 22,4 = 0,4032 lt.

2 . Cc phn ng ho tan c th c:

3FeO + 10HN3= 3Fe(N03)3+ NO t + 5H20Fe20 3+ 6 HNO3= 2Fe(N03)3+ 3H20

(7)()

NO 02 Fe

Trong n lsmol NO thot ra. Rt ra n = 0,001

VN0= 0,001. 22,4 = 0,224 lt

3. Cc phn ng c th c:

2A1 + 3FeO = 3Fe + AI2O3

2A1 + Fe20 3= 2Fe + A120 3

(10)

(11)

63



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




63/244

8AI + 3Fe30 49Fe + 4A120 3

Fe + 2HC1 = FeCl2+ H21

2A1+ 6HC1 - 2A1C3 + 3H21

(12)

(13)

(14)

Xt cc phn ng (1, 2, 3, 10, 11, 12, 13, 14) ta thy Fe CUIcng thnh Fe+2, Al thnh Al+3, 0 thnh 20~2v 2H+thnh H2nn ta c phng trnh bo ton electron nh sau (gi s" mol H9l n):

Fe - Fe+2; Al -> Al+3; 0 20~2; 2H+->H2

Rt ra n = 0,295 mol. VHs = 0,295 . 22,4 = 6,608 lt.

V d 5:Cho 7,8g hn hp bt kim loi Mg v AI tc dng vidung dch HC1 d thu c 8,961 H2 ktc. Xc nh % khilng ca cc kim loi trong hn hp.

Li gii

Gi X l s" moi Mg, y l s mol AI trong 7,8g hh. Ta c:

24x + 27y = 7,8 (a) (theo khi lng hn hp)

=0,009.4 + n.2

Phng trnh phn ng xy ra:

Mg + 2HC1 = MgCl2+ H2t

2A1 + 6HC1 = 2AICI3+ 3H2t(1)

(2)

64



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




64/244

2x + 3y = 0,42 (b) (theo nh lut bo ton electron)

Gii h (a), (b) ta c

y = 0,2 ; X = 0,1

- M7,8

i7,8

mMg=24.0,1= 2,4g ; %Mg = . 100% = 30,77%

= 27 . 0,2 = 5,4; %A1 = . 100% = 69,23%

V d 6:Ho tan hon ton 19,2gCu bng dung dch HN03.Tt c lng kh NO thu c em oxi ho thnh N0 2 ri

chuyn ht thnh HNO3

. Tnh th tch kh 0 2 ktc thamgia vo qu trnh trn.

Li gii

Phng trnh phn ng xy ra:

3C u +8H N 0 3= 3 Cu(N03)2+2N o +4H20

0,3 0,2+ 2 0 + 4 - 2

2N0+2=2N0,

0,2 0,1 0,2

4 N o , + 2H ,0 + , = 4HN 0 9.

0,2 = 0,054 . .

nCu= ^ = 0,364

Theo cc phn ng trn th n0 =0,1 + 0,05 = 0,15 molv 0 = 22,4 . 0,15 = 3,361

65



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




65/244

Theo nh lut bo ton electron:

Cu nhng 2e thnh Cu+2

0 2 thu 4e thnh 20~20,3.2 = 4n0 -> n0 = = M =0,15> . * 4 4

Cu

v02 = 22,4 . 0,15 = 3,36 lt

V du 7: C22,41 N2 ktc iu ch HN03th cn phi c

bao nhiu lt H2v 0 2tham gia vo qu trnh iu ch trn?Li gii

Cch 1:Phng trnh phn ng ho hc xy ra:

N2+ 3H2= 2NH3

X 3x 2x

4NH3+ 302= 4NO + 6H20

2x 2 ,5x 2x2NO + 0 2= 2N02

2x X 2x

4N02+ 2H20 + 0 2= 4HN03

2x =0,5x4

VHa = 22,4 . 3 = 6,72 lt

n02 = 2,5x + X+ 0,5x = 4x; v 0 = 22,4 . 4 = 8,96 lt

66



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




66/244

Cch 2: N2 - lOe = 2N+5; H - 2e = 2H+

0 2 + 4e = 202

16nvr = 1 mol 1.10 + 2.3 = n.4 = 16 n = - = 42 4

4. Phng php dng cc gi tr trung bnh

. Phng ph p khi lng mol trung bnh (M)

Phng php khi lng mol trung bnh l mt phng phpgip gii nhanh cc bi ton v c v h c loi hn. hp 2 hay

nhiu cht.- i vi ton v c nh xc nh kh lng nguyn t ca 2kim loi thuc cng mt phn nhm chnh v nm 2 chu k ktip nhau; xc nh % s lng mi loi ng v mt nguyn t;tnh % th tch ca kh trong hn hp. '

- i vi ton hu c c bit thun tin khi gii cc bi tonxc nh cng thc phn t ca cc cht k tip nhau trong dy

ng ng... gii bi ton hn hp bng cch dng khi lng mol

trung bnh, ta cn nm vng nh ngha khi lng mol trungbnh v cc cng thc tnh khi lng mol trung bnh (k hiu lM). ^ :

- nh ngha M : Khi lng mol trung bnh l khi lngca mt mol hn hp.

- Cc cng thc tnh M :+ Hn hp nhiu cht:

67



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




67/244

M = =

INdjIlj + n1+n2 + n3+...

M = M,V,+M2V2 +M3V3+...Vt +V~+Vt+...

M], M2, M3... l khi lng phn t ca cht 1, 2, 3...

1], n2, n3... l smol ca cht th 1, 2, 3...

vlfV2, V3... l th tch ca kh th 1, 2 , 3...M = MX + M22+ M 3X3 +...

Xj, x2, x3... l % s lng (khng phi khi lng) hoc % thtch.

+ Hn hp 2cht:

T MV, +M2(V -V 1)

M = M 1x 1+M 2(1 Xj)

V d 1:Hai kim loi kim M v M' nm trong 2chu k ktip nhau trong bng h thng tun hon. Ho tan mt t hnhp ca M v M' trong nc c dung dch A v 0,3361 H2 ktc. Cho HC1 d vo dung dch A v c cn c 2,075g muikhan. Xc nh tn kim loi M v M'.

; M = MII+ M2(l - )n

68



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




68/244


2M + 2H20 = 2MOH + H2T (1)XX X

2

2M' + 2H20 = 2M'OH + H2T (2)

y y " 2

MOH + HC1 = MCI + H20 (3)

X XM'OH + HC1 = MC1 + H20 (4)

y y

Gi X, y l s" mol ca kim loi MvM'

Theo (1,2) = =0,015 ;x + y = 0,032 2 22,4

Theo (1, 2, 3, 4): n 2muS = X + y = 0,03

JV2 0,03

Ta c: M + 35,5 < 69 < M' + 35,5

Vy: M < 33,5 < M'

Do : M l Na = 23 v M l K = 39

V d 2:Ho tan vo nc 7,14g hn hp mui cacbonat vcacbonat axit ca mt kim loi ho tr I. Sau thm vo

dung dch thu c mt lng dung dch HC1 th thu c0,6721 kh ktc. Xc nh tn kim loi.

Li gii

69



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




69/244


K hiu kim loi l M

M2C03+ 2HC1 = 2MC1 + C021 + H20

X X

MHCO3+ HC1 = MCI + C0 21 + H20

y y

Gi X, y l s mol mui M2CO3v MHCO3trong 7,14g hh

Theo s mol co? ta c: X+ y = = 0,03

22,4 7 14M2 = = 238

0,03

M + 61 < 238 < 2M + 60

* M + 61 < 238; M < 238 - 61 = 177

* 2M + 60 > 238; 2M > 238 - 60 = 178

M > =89

2Vy 89 < M < 177

Do : M l Cs = 133

V d 3:Nguyn t lng ca 3 kim loi ho tr II t l vi nhautheo t s 3:5:7. T l nguyn t khi ca 3 kim loi tng ng ctrong hn hp l 4:2:1. Khi ho tan 2,32g hn hp 3 kim loi trn

trong axit HC1 d thu c 1,5681 H2 ktc. Xc nh 3 kim loitrn, bit rng chng u ng trc H trong dy Bktp.

Li gii

70



B

I

D

N

GT

O

N

-

L

-

H

A

C

P

2

3

1

0

0

0

B

T

R

N

H

N

G

O

T

P

.

Q

U

Y

N

H

N




70/244

Li gii


X + 2HC1 = XC12 + H2t

Y + 2HC1 = YC12 + H2t

z + 2HC1= ZC12+ Hat

(1)

(2)

(3)

Theo (1, 2, 3): nhnh


71/244

Gi Xl s mol Ba, y l tng s mol ca A v B.

Phng trnh phn ng ho hc xy ra:Ba + 2H20 = Ba(OH)2+ H2t (1)

X X X

2 A + 2H20 = 2 AOH + H2t (2 )

2B + 2H20 = 2BOH + H2t - (3)

y " I2

Ba(OH)2+ Na2S 04= BaS04 +2NaOH (4)Theo (1, 4): 0,18 < X < 0,21 (5)

137.0,18


72/244

V d 5:Trong thin nhin ng kim loi cha 2 loi ng v29Cu63 v 2 9CU65. Khi lng nguyn t trung bnh ca ng l64,4. Tnh thnh phn % s lng mi loi ng v.

Li gii

Gi Xl % s lng ca ng v 29C1165. Theo cng thc

M = MjX] + M2( lXi) ta c:

M = 65x + 63(l-x) = 64,4

Gii ra: X = 0,7 ngha l c 70% 29 CU65

(Ch : % y l % s nguyn t, ngha l trong 100ngu yn t Cu th c 70 nguy n t l 29CU65)

V d 6:Hn hp kh S 02 v 0 2 c t khi i vi CH4bng 3.Cn thm bao nhiu l t 0 2 vo 2 0 lt hn hp cho t khi

gim i 1/6 ngha l bng 2,5?

Li gii

Gi Xl % th tch ca S02 trong hn hp ban u. p dngcng thc M = M1x1+ M2(l - x j ta c:

M = 16.3 = 48 = 64x +'32 (1 - x).

G ii ra c X = 0,5 ng ha l mi kh chim 50% (mi k h c

10 lt).

Gi V l th tch 0 2 cn thm vo; hn hp mi gm c 101

S02v (10 + V) lt 0

bÀi tẬp hÓa hỌc Ở trƯỜng phỔ thÔng - nguyỄn xuÂn trƯỜng

Documents