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7/29/2019 Gii thiu 1 s bi vit ca thy Vn Long

1/26

TThhSS.. LLUUHUHUNNHHVVNN LLOONNGG

((00998866..661166..222255))

((GiGinngg vviinn TTrrnnggHHThTh DDuu MMtt))--------

GII THIEUMOT SO BAI VIET LTH CUA

THAY VAN LONG ANG TRENBAO HOA HOC & NG DUNG

(HOI HOA HOC VIET NAM)

Phng phap la Thay cua cac Thay

Tally Rand

LU HNH NI B12/12/2012


2/26

Cc chuyn trc nghim LTH ng trn Bo Ha hc & ng dng (Hi Ha hc Vit Nam)

ThS. LU HUNH VN LONG (Ging vin Trng H Th Du Mt) 23Lp bi dng kin thc 10, 11, 12 v luyn thi tt nghip THPT, Cao ng i hc mn HA HC

tm hiu v ng k hc, hy gi in ti s 0986.616.225 (gp Thy Vn Long)Ghi danh hng tun v hc gn Trng THPT Tn Phc Khnh vo lc 17h30-19h cc ngy trong tun (CN nguyn ngy)

GII THIU MT S BI VIT CC CHUYN

TRC NGHIM LTH CA THY

LU HUNH VN LONG NG TRN

BO HA HC & NG DNG(HI HA HC VIT NAM)


3/26




PHNG PHP TNH NHANH HIU SUT CA

PHN NG CRACKINH

Thc s LU HUNH VN LONG

Trng H Th Du Mt -Bnh Dng

I- CS L THUYT

Bi ton tnh hiu sut caphn ng crackinh ankan l mt dngbi tp kh i vi hc sinh.Hc sinh thng rt lng tng khi gpbi ton ny v thng gii rt di dng khng cn thit. Vy lm tt vn ny, chng ta cn phi tm hiu r bn cht ca phn ng crackinh ankan v raphngphp gii nhanhph hp vi hnh thc thi trc nghim hin nay:

Di tc dng ca nhit cao v xc tc thch hp th ankan c th xy ra nhiu loi phnng crackinh sau:

Ankanot C

Ankan + anken [1]

VD: CnH2n+2ot C CmH2m+2 + Cn-mH2(n-m)

Ankanot C Anken + H2 [2]

VD: CnH2n+2ot C

CnH2n + H2

Ankanot C

Ankin + 2H2 [3]

VD: CnH2n+2ot C

CnH2n-2 + 2H2

c bit: 2CH4o1500 C

lln C2H2 + 3H2 [4]

Gi s ta c s sau:

Hn hp ankan Xot C Hn hp khY

Ta thy trong ccphn ng crackinh trn th s mol kh sauphn ng lun tng nn: nX < nY

Mt khc theo LBTKL : mX = mY nn suy ra M MX Y>

T : dX/Y =

X

X X Y Y

Y X Y X

Y

m

n m n nM.

m n m nM

n

X

Y

= = = = (do : dX/Y > 1)

Vit gn li: dX/Y =Y

X

nM

nM

X

Y

= = [5]

Da vo biu thc ta tnh c nY, t tnh Hiu sutphn ng.

Nhnxt :

* Da vo h s phn ng ca ccphn ng [1], [2], [4] ta rt ra kt qu quan trng:

Th tch (hay smol) khtng sauphn ng bng th tch( hay smol) ankan tham giaphn ng

crackinh [6]


4/26




** Trong tnh ton ta thng dng phngphp t chn lng cht xem nh hn hp ban u l 1mol.

*** T cc kt qu trn ta c th p dng lm mt s dng Bi tp lin quan n phn ngcrackinh.

II- BI TP P DNG

Bi 1: Nhit phn CH4 thu c hn hp X gm C2H2, CH4 v H2. T khi hi so vi H2 bng 5.Hiu sut qu trnh nhit phn l:

A. 50% B. 60% C. 70% D. 80%

Hng dn gii

Gi s ban u c 1 mol CH4

2CH4 C2H2 + 3H2Trc p: 1mol

P: x(mol) 0,5x 1,5x

Sau p: 1 x 0,5x 1,5x mol = 1 + x

Da vo [5] ta c:X

X

n16n = 1,6 (mol) = 1 + x x = 0,6

2*5 1=

Hiu sut =0,6

*100% 60%1

= . Chn B

*Ta c th gii nhanhbi ton ny da vo nhn xt [6]:

mX = 16 g nX =16

1,6( )

5*2

mol= H =1,6 1

*100% 60%

1

=

Bi 2: Nhit phn 8,8 gam C3H8, gi s xy ra haiphn ng sau:

C3H8ot C

CH4 + C2H4

C3H8ot C

C3H6 + H2

Ta thu c hn hp X, bit MX = 23,16. Hiu sut caphn ng nhit phn trn l:

A. 60% B. 70% C. 80% D. 90%

Hng dn gii

Da vo [5] Y Yn44

n 0,38( )23,16 0,2 mol= =

Da vo [6] H =0,38 0,2

*100% 90%0,2

= Chn D

Bi 3: Crackinh C4H10c hn hp ch gm 5 hirocacbon c M =36,25vC.

Hiu sut phn ng crackinh l :

A.60% B.20% C.40% D.80%

Hng dn gii

Xt 1 mol C4H10


5/26




Da vo [5] Y

Y

n58n 1,6( )

36,25 1mol= =

Da vo [6] H =1,6 1

*100% 60%1

= Chn A

Bi 4: Crackinh V lit C4H10 thu c 35 lit hn hp A gm H2, CH4, C2H4, C3H6, C4H8 v mtphn C4H10 chab crackinh. Cho hn hp A i qua t t quabnh ng brom d thy th tch cn li20 lit. Tnh hiu sut caphn ng crackinh:

A. 25% B. 60% C. 75% D. 85%

Hng dn gii

t x, y, z, t ln lt l th tch C3H6, C2H4, C4H8, C4H10 d:

C4H10ot C

CH4 + C3H6

x x x

C4H10ot C

C2H6 + C2H4

y y y

C4H10ot C

C4H8 + H2

z z z

Khi dn hn hp i qua dung dch brom th ankenb gi li cn H2 , CH4, C2H6 v C4H10 dthotra:

Ta c: x + y + z = 35 20 = 15 (1)

Mtkhc: V(C4H10 ban u) = V(C4H10 p) + V C4H10 cn li = x + y + z + t = 20 (2)

T (1) v (2) t = 5

H =15

*100% 75%20

= Chn C

Bi 5: Crackinh 560 lit C4H10 thu c 1036 lit hn hp kh X khc nhau. Bit cc th tch kh uo ktc. Hiu sutphn ng crackinh l:

A. 75% B. 80% C. 85% D. 90%

Hng dn gii

Theo [6] H =1036 560

*100% 85%

560

= Chn C

Bi 6: Hn hp kh A gm etan v propan c t khi so vi H2 l 20,25 c nung trongbnh vicht xc tc thc hinphn ng hiro ha. Sau mt thi gian thu c hn hp kh B c t khiso vi H2 l 16,2 gm cc ankan, anken v hiro. Tnh hiu sutphn ng hiro ha bit rng tc phn ng ca etan v propan l nh nhau ?

A. 30% B. 250% C. 50% D. 40%

Hng dn gii

M 20,25.2 40,5 ; M 16,2.2 32,4A B= = = =

Xt s mol hn hp A l 1 mol:


6/26




Da vo [5] B

B

n40,5n 1,25( )

32,4 1mol= =

Da vo [6] H =1,25 1

*100% 25%1

= Chn B

Bi 7: Thc hinphn ng crackinh 11,2 lit hi isopentan (ktc) thu c hn hp A ch gm ccankan v anken. Trong hn hp A c cha 7,2 gam mt cht X m khi t chy th thu c 11,2 litCO2 (ktc) v 10,8 gam H2O. Hiu sut caphn ng crackinh l:

A. 80% B. 85% C. 90% D. 95%

Hng dn gii

5 12C Hn ban u =

11,20,5( )

22,4mol=

t chy X c2 2H O CO

10,8 11,2n = 0,6( ) > n = 0,5( )

18 22,4

mol mol= = X l ankan

Do : nX = 0,6 0,5 = 0,1 (mol) MX =7,2

720,1

= = 14n + 2 n = 5 (C5H12)

H =5 12

5 12

C H p

C H ban au

n 0,5 0,1*100% *100% 80%

n 0,5

= = Chn A

PHNG PHP TM NHANH CTPT FeXOY



I- CS L THUYT

xc nh nhanh CTPT ca oxit FexOy trong ccbi tp trc nghim Ha hc ta c th davo ni dung nh lut thnh phn khng i:

Vi mt hp cht cho trc, d c iu chtheo phngphp no th t l vsmol, t lvkhi lng hay t l vth tch gia cc thnh phn nguyn t to nn hp chtl nhng hng s

tigin

Xt hp cht FexOy th ta lun c:

Fe

O

m 56

m 16

x

y= v Fe

O

nx

y n=

Khi :

- Nu y

x

=1 FexOy l: FeO


7/26




- Nuy

x=

3

2 FexOy l: Fe2O3

- Nuy

x=

4

3 FexOy l: Fe3O4

Mt slu:

- Nu oxit st (FexOy) tc dng vi H2SO4c, HNO3 khng gii phng kh l Fe2O3.

- i vi FeO v Fe3O4 c c im l 1 mol phn t th nhng ng 1 mol electron:

+2 +3

Fe(FeO) Fe + 1e

+8+33

3 43Fe(Fe O ) 3Fe + 1e

- Khi gii bi tp dng ny, ta thng kt hp cc phng php: bo ton electron, bo tonnguyn t,bo ton khi lng v tng gim khi lng,

- i khi ta c th gii bng cch xt 3 kh nng ca FexOy l: FeO, Fe2O3, Fe3O4 ri da vo dkin cabi tm p n ph hp.

II- MT S BI TP

Cu 1: Kh hon ton 16g bt oxit st bng CO nhit cao, sau khi phn ng kt thc, khilng cht rn gim 4,8g. Cng thc oxit st dng l:

A. FeO B. Fe3O4 C. Fe2O3 D. Tt c u sai

Hng dn

Khi lng gim i chnh l khi lng ca oxi trong oxit FexOy.

Ta c: mO(FexOy) = 4,8 (gam) nO(FexOy) =4,8 0,3( )16

mol=

nFe(FexOy) =16 - 4,8

0,2( )56

mol=

2 30,2 2

0,3 3

xFe O

y= = . Chn C

Cu 2: Ha tan hon ton 6,4 gam mt hn hp Fe v FexOy vo dung dch HCl dth thu c2,24 lit H2(ktc). Nu un hn hp trn kh bng H2 dth thu c 0,2 gam H2O. Cng thc oxit stl:

A. Fe2O3 B. FeO C. Fe3O4 D. FexOy

Hng dn

Ch c Fe tc dng vi dung dch HCl mi giiphng kh nn:

nFe = n(H2) = 0,1 (mol) mFexOy = 6,4 0,1.56 = 0,8 (g)

Khi kh hn hp bng H2 th: nO(FexOy) = n(H2O) =0,2 1

( )18 90

mol=

nFe(FexOy) =

16*10,8 -

190

( )56 90 mol=

1

x

FeOy=

. Chn B


8/26




Cu 3: ha tan 4 gam FexOy cn 52,14 ml dd HCl 10%(D=1,05g/ml). Xc nh cng thc phnt FexOy.

A. Fe2O3 B. FeO C. Fe3O4 D. Fe2O3 v FeO

Hng dn

Ta c: +HClH

52,14*1,05*0,1n = n = 0,15( )

36,5mol=

Phn ng thc cht l: O2- + 2H+ H2O

0,075 0,15(mol)

nFe(FexOy) =4 - 0,075*16

0,05( )56

mol= 2 30,05 2

0,075 3

xFe O

y= = Chn A

Cu 4: Ha tan 10gam hn hp gm Fe v FexOy bng HCl c 1,12 lt H2(ktc). Cng lnghn hp ny nu ha tan ht bng HNO3c nng c 5,6 lt NO2(ktc). Tm FexOy?

A. FeO B. Fe3O4 C. Fe2O3 D. Khng xc nh c

Hng dnCh c Fe tc dng vi dung dch HCl mi giiphng kh nn:

nFe = n(H2) =1,12

0,0522,4

= (mol) mFexOy = 10 0,05.56 = 7,2 (g)

Khi cho hn hp tc dng vi HNO3:

+8+33

3Fe 3Fe + 1e

0 3+

Fe Fe + 3e

+5 +4

N + 1e N

0,05

0,25

a

Bo ton electron: 0,05.3 + a = 0,25 a = 0,1

M(FexOy) =7,2

720,1

FeO= . Chn A

Cu 5:(TSC Khi A 2007)

Cho 4,48 lt kh CO (ktc) t ti qua ng s nung nng ng 8 gam mt oxit st n khiphn ng xy ra hon ton. Kh thu c sau phn ng c t khi so vi hiro bng 20. Cng thc caoxit st v phn trm th tch ca kh CO2 trong hn hp kh sau phn ng l:

A. FeO;75% B. Fe2O3;75% C. Fe2O3;65% D. Fe3O4;75%

Hng dn

Kh sauphn ng l hn hp:2

CO : x (mol)

CO : 0,2 - x (mol)

Quy tc ng cho:


9/26




44

28

40

12

4

CO2

COx

0,2-x

2CO

CO

n 0,2 123 0,05

n 4

xx

x

= = = =

%V(CO2) =0,15*100

75%0,2 =

nO(FexOy) = n(CO2) = 0,2 0,05 = 0,15 (mol) nFe(FexOy) =8 - 0,15*16

0,1( )56

mol=

2 30,1 2

0,15 3

xFe O

y= = Chn B

Cu 6: Cho 0,01 mol mt hp cht ca st tc dng ht vi H2SO4c nng (d) thot ra 0,112lit kh SO2(ktc) (l sn phm kh duy nht). Cng thc ca hp cht st l:

A. FeS B.FeS2 C. Fe2O3 D. Fe3O4

Hng dn

Ta c:

+6 +4

S + 2e S

0,1120,01 0,005( )

22,4mol

=

S mol hp cht = s mol electron trao i 1 mol hp cht ch nhng 1mol electron.

Do ta chn Fe3O4 Chn D

Cu 7: Ha tan han ton mt oxit FexOy bng dung dch H2SO4c nng(va ) thu c 2,24lt kh SO2 (ktc) v 120 gam mui. Xc nh cng thc oxit kim loi?

A. FeO B. Fe3O4 C. Fe2O3 D.Khng xc nh c

Hng dn+6 +4

S + 2e S

2,240,2 0,1( )

22,4mol

=

Nhnxt: s mol oxit FexOy l 0,2 (mol) nFe(FexOy) = 0,2.x

Ta c:2 4 3Fe (SO )

120n = 0,3( ) ( ) 0,3*2 0,6( )

400mol n Fe mol= = =

Bo ton nguyn t Fe: 0,2.x = 0,6 x = 3 Fe3O4 Chn B

Cu 8: Dng CO d kh hon tan m gam bt st oxit (FexOy) dn tan b lng kh sinh ra itht chm qua 1 lt dung dch Ba(OH)2 0,1M th va v thu c 9,85gam kt ta. Mt khc ha tantan b m gam bt st oxit trn bng dd HCl d ri c cn th thu c 16,25gam mui khan. m c gitr l bao nhiu gam? V cng thc oxit (FexOy).

A. 8 gam; Fe2O3 B. 15,1gam, FeO

C. 8 gam; FeO D. 11,6gam; Fe3O4

Hng dn

nBa(OH)2 = 0,1 (mol)


10/26




n(BaCO3) =9,85

0,05( )197

mol=

Ta thy: n(BaCO3) < nBa(OH)2 C 2 pxy ra:

CO2 + Ba(OH)2 BaCO3

0,1 0,1(mol) 0,1

BaCO3 + CO2 + H2O Ba(HCO3)2

0,05 0,05

S mol CO2 p: 0,1 + 0,05 = 0,15 (mol)

Hoc ta nhm: nCO2 = 2nBa(OH)2 nBaCO3 = 2.0,1 0,05 = 0,15 (mol)

nO(FexOy) = nCO2 = 0,15 (mol)

Khi oxit FexOy tc dng vi HCl to mui clorua:

O2- 2Cl-

0,15 0,3(mol)

mFe = 16,25 0,3.35,5 = 5,6 (g) m = mFe + mO = 5,6 + 0,15.16 = 8(g)

2 30,1 2

0,15 3

xFe O

y= = Chn A

Bi 9: kh hon ton 6,4g mt oxit kim loi cn 0,12mol kh H2. Mt khc ly lng kim loito thnh cho tan hon ton trong dung dch H2SO4 long th thu c 0,08 mol H2. Cng thc oxitkim loi l:

A. CuO B. Al2O3 C. Fe3O4 D. Fe2O3

Hng dn

2 2 4H / CO HCl/H SO 2(1) (2)Oxit KL KL H

Nhnxt: y l dng BT quen thuc trong cc k thi nn chng ta cn ch :

+ Oxit KL b kh bi H2/COphi l oxit ca KL ng sau Al

+ KL tc dng vi dung dch HCl/H2SO4 H2 phi ng trc H trong dy hot ng ha hc

+ S mol H2 (hoc CO) (1) S mol H2(2) Oxit ca KL a ha tr.

Do : Ta loi A v B oxit l FexOy

2 3

6,4 16 * 0,12256

0,12 3

xFe O

y

= = Chn D

Bi 10: Kh hon ton 4,06g mt oxit kim loi bng CO nhit cao thnh kim loi. Dn tonb kh sinh ra vobnh ng dung dch Ca(OH)2 d thy to 7g kt ta. Nu ly lng kim loi sinh raho tan ht trong dung dch HCl dth thu c 1,176lit H2(ktc). Cng thc oxit kim loi l:

A. FeO B. Fe3O4 C. Fe2O3 D. ZnO

Hng dn

Ta c: nCO2 = nCaCO3 =7

0,07( )100

mol= nH2 =1,176

0,0525( )22,4

mol= Oxit FexOy

nO(FexOy) = nCO2 = 0,07


11/26




nFe(FexOy) =4,06 16*0,07

0,0525( )56

mol

=

3 40,0525 3

0,07 4

xFe O

y= = Chn B

Bi 11 : Ha tan hon ton a gam FexOy bng dd H2SO4m c nng va , c cha 0,075

mol H2SO4 thu c b gam mt mui c 168 ml kh SO2 (ktc) duy nht thot ra .a/ Tr s ca b l :

A. 9g B. 8g C. 6g D. 12g

b)Tr s a gam FexOy l :

A. 1,08g B.2,4g C.4,64g D. 3,48g

c) Cng thc ca FexOy l :

A. FeO B. Fe2O3 C.Fe3O4 D. Khng xc nh c

Hng dn

a/ Gc SO42- trong axit sau phn ng nm dng gc SO42- trong mui v dng SO2 nn:Bo ton nguyn t lu hunh: nS(H2SO4) = nS(mui) + nS(SO2)

nS(mui) = 0,075 -168

1000.22,4= 0,0675 (mol)

2 4 3Fe (SO )

0,0675n = 0,0225( )

3mol=

Khi lng mui = 0,0225*400 = 9 (g) Chn A

b/ Ta c: S mol H2SO4 = s mol H2O = 0,075 (mol)

Theo LBTKL: m(FexOy) + m(H2SO4) = m(mui) + m(SO2) + m(H2O)

m(FexOy) = 9 + 0,0075.64 + 0,075.18 98.0,075 = 3,48 (g) Chn D

c/ nFe(FexOy) = 0,0225.2 = 0,045 (mol)

nO(FexOy) =3,48 0,045*56

0,06( )16

mol

=

3 40,045 3

0,06 4

xFe O

y= = Chn C


12/26




PHNG PHP XC NH NHANH SN PHM TRONG CC

PHN NG CA HP CHT PHOTPHO



Bi ton v axit H3PO4 hay P2O5 tc dng vi dung dch bazcng nhccphn ng ca muiphotphat i vi hc sinh THPT cn kh phc tp. a s hc sinh khng nm r bn chtphn ngnn vn cha t tin khi lm ccbi tp dng ny. lm tt dngbi tp ny th trc tin taphixcnh ng c sn phm sinh ra trong cc phn ng, ri da vo s liu tnh ton. C th phnthnh mt s dng sau:

I- Axit H3PO4 tc dng vi dung dch baz

1/ Trng hp dung dch bazlNaOH hay KOH

Ta c ccphn ng sau:H3PO4 + NaOH NaH2PO4 + H2O

H3PO4 + 2NaOH Na2HPO4 + 2H2O

H3PO4 + 3NaOH Na3PO4 + 3H2O

xc nh mui g c to ra th a phn hc sinh da vophngphp i sl lp t l

molSo mol baz

T =So mol axit

hay ngc li ri lpbng ghi nh 7 trng hp, sau so snh vi t l mol

m bi cho kt lun. C th l:

3 4

NaOH

H POnn

San pham

1 2 3

2 4

3 4

NaH PO

D H PO 2 4NaH PO

2 4Na HPO 3 4Na PO3 4Na PO

D NaOH2 4

2 4

NaH PO

Na HPO

2 4

3 4

Na HPO

Na PO

2/ Trng hp dung dch bazl Ca(OH)2 hay Ba(OH)2

Ta c ccphn ng sau:

2H3PO4 + Ca(OH)2 Ca(H2PO4)2 + 2H2O

H3PO4 + Ca(OH)2 CaHPO4 + 2H2O

2H3PO4 + 3Ca(OH)2 Ca3(PO4)2 + 3H2OTng t:

San pham

0,5 1 1,5

2

3 4

Ca(OH)

H PO

n

n

2 4 2Ca(H PO ) 4CaHPO 3 4 2Ca (PO )2 4 2

3 4

Ca(H PO )

D H PO2 4 2

4

Ca(H PO )

CaHPO4

3 4 2

CaHPO

Ca (PO )

3 4 2

2

Ca (PO )

D Ca(OH)

Hc sinh thng lm bng cch lp bng trn, do cc em c lm quen cch lm ny t

THCS. Tuy nhin hc sinh cn nhmt cch my mc cc h s t l trn (14 trng hp) m khnghiu r bn cht caphn ng. Do i vibi ton hn hp bazth cc em s rt lng tng.


13/26




* nm r bn cht ca phn ng cng nh hn ch li mn ta vn dng kin thc vChng S in ly lp 11 ta giibi ton trn nh sau:

Axit H3PO4 l mt triaxit thuc loi trungbnh nn trong dung dch s phn ly ra 3 nc. Gi sta c a mol H3PO4 th s phn ly ca H3PO4 nh sau:

H3PO4 H+ + H2PO4

- 2H+ + HPO42- 3H+ + PO4

3-

a(mol) a 2a 3a (mol)Da theo s ta thy c a mol axit s giiphng 3 ln H+ l a, 2a, 3a mol H+. Khi cho dung

dch bazvo th bn chtphn ng l s trung ha:

H+ + OH- H2O

Do nu bit s mol NaOH hay KOH ta bit: - NaOHOHn = n

Bit s mol Ca(OH)2 hay Ba(OH)2 ta bit: -2Ca(OH)OH

n = 2n

Ta so snh s mol ca OH- vi s mol H+ s bitphn ng to ra mui g:

- Nu nOH- < a to mui H2PO4- v H3PO4 d

- Nu nOH- = a to mui H2PO4-

- Nu a < nOH-< 2a to mui H2PO4

- v HPO42-

- Nu nOH- = 2a to mui HPO4

2-

- Nu 2a < nOH- < 3a to mui HPO4

2- v PO43-

- Nu nOH- = 3a to mui PO4

3-

- Nu nOH- > 3a to mui PO4

3- v OH- d

Tm tt cc trng hp trn bngbng sau:

San pham

1 2 3

-

3 4

OH

H PO

n

n

2 4H PO 2

4HPO 3

4PO

2 4

3 4

H PO

D H PO

2 4

24

H PO

HPO

24

34

HPO

PO

34

-

PO

D OH

Ch : Ta c th dng s n gin sau ghi nh:

- - -OH OH OH- 2- 3-3 4 2 4 4 4H PO H PO HPO PO [1]

3/Dng P2O5 tc dng vi dung dch baz

Ta xem P2O5 tc dng vi H2O to axit H3PO4 ri gii tng t nh trn. Ch h s mol:

P2O5 + 3H2O 2H3PO4x(mol) 2x (mol)

Bi tp 1: Cho bit sn phm to thnh trong cc trng hp sau:

a/ 0,06 mol H3PO4 tc dng vi 0,08 mol NaOH

b/ 1 mol H3PO4 tc dng vi 1 mol Ca(OH)2

c/ 0,08 mol H3PO4 tc dng vi 0,06 mol Ca(OH)2

d/ 2mol H3PO4 tc dng vi 1 mol Ca(OH)2

e/ 0,24 mol H3PO4 tc dng vi hn hp gm 0,009 mol NaOH v 0,006 mol Ca(OH)2

Gii


14/26




Ta c th da vo-

3 4

OH

H PO

n

n d on sn phm

a/-

3 4

OH

H PO

n 0,081,33

n 0,06= = to NaH2PO4 v Na2HPO4

b/-

3 4

OH

H PO

n 22

n 1= = to mui CaHPO4

c/-

3 4

OH

H PO

n 0,06*21,5

n 0,08= = to Ca(H2PO4)2 v CaHPO4

d/-

3 4

OH

H PO

n 21

n 2= = to mui Ca(H2PO4)2

e/-

3 4

OH

H PO

n 0,009 0,006*2 0,0875n 0,24

+= = to mui NaH2PO4 v Ca(H2PO4)2

Bi tp 2(TSH B 2008):Cho 0,1 mol P2O5 vo dung dch cha 0,35 mol KOH. Dung dch thuc c cc cht:

A. K3PO4; K2HPO4 B. K2HPO4; KH2PO4

C. K3PO4; KOH D. H3PO4; KH2PO4

Gii

P2O5 + 3H2O 2H3PO4

0,1(mol) 0,2 (mol)

-

3 4

OH

H PO

n 0,351,75

n 0,2= = K2HPO4; KH2PO4

Chn B

Bi tp 3: Cho dung dch a mol H3PO4 tc dng vi dung dch cha b mol NaOH, thu c dungdch A

a/ Bin lun xc nh thnh phn cc cht trong dung dch A theo mi quan h gia a v b

b/p dng vi trng hp a = 0,12 mol v b = 0,2 molBi tp 4: Cho a mol P2O5 tc dng vi dung dch cha b mol KOH. Bin lun mi quan h gia

a v b xc nh cc cht trong dung dch sauphn ng (xem P2O5 tc dng vi H2O chuyn honton thnh H3PO4)

Hng dn: Da vo s trn bin lun

II- Mui photphat tc dng vi cc hp cht khc

Da vo s [1] ta thy mui photphat c th tn ti 3 dng: ihirophotphat H2PO4-;

hirophotphat HPO42-; photphat trung ha PO4

3-. Trong mi trng OH- H2PO4- c th chuyn thnh

HPO42- v PO43-. Do photphat l mui ca axit trungbnh c nhiu nc nn chng khng bn trong mi


15/26




trng H+ v chng c th chuyn v dng axit ban u H3PO4. Cc qu trnh chuyn i trn phthuc vo mnh yu ca axit v t l mol cc chtphn ng:

- - -

+ + +

OH OH OH- 2- 3-3 4 2 4 4 4

H H HH PO H PO HPO PO [2]

T s [2] rt ra mt s nhn xt:

+ Mui PO43- c th tc dng vi H3PO4 to ra mui HPO42- v H2PO4- ty theo t l PO43-/H3PO4( khng th to thnh H3PO4):

Nu t l l 1:1 CaHPO4 do H3PO4 nhng i 1 H+

Nu t l l 1:2 Ca(H2PO4)2 do H3PO4 nhng 2 H+

+ Mui PO43- c th tc dng vi axit mnh (HCl, H2SO4) to ra HPO4

2-; H2PO4- hay H3PO4 ty

theo t l PO43-/HCl:

PO43- + H+ HPO4

2-

PO43- + 2H+ H2PO4

-

PO4

3- + 3H+ H3PO

4

+ Hai cht khng ng cnh nhau trong 1 dy trn th khng bao gicng tn ti trong 1 dungdch, chng s tc dng ht vi nhau to ra sn phm l cht ng gia hai cht :

PO43- + H2PO4

- 2HPO42-

H3PO4 + HPO42- 2H2PO4

-

H3PO4 + PO43- HPO4

2- v H2PO4- ( ty t l xt trn)

Bi tp 1(SGK 11CB): Hy d on sn phm to thnh (mui)trong cc trng hp sau:

a/ 1 mol H3PO4 + 1 mol K2HPO4 ?

b/ 1 mol Ca(H2PO4)2 + 1 mol Ca(OH)2 ?

Gii

Da s [2]:

a/ HPO42- + H3PO4 theo t l 1:1 NaH2PO4

b/ H2PO4- + OH- theo t l 2:2 (hay 1:1) CaHPO4

Bi tp 2(BT 3 trang 66 SGK 11NC):

in cht thch hp vo ch c du ? trong cc s sau:

a/ H2PO4- + ? HPO4

2- + ?

b/ HPO42- + ? H2PO4

- + ?

Gii

Da s [2]:

a/ H2PO4- + OH- HPO4

2- + H2O

b/ HPO42- + H3O

+ H2PO4- + H2O

Bi tp 3: Cho P2O5 tc dng vi NaOH ngi ra thu c mt dung dch gm hai cht. Hai cht l :

A NaOH v NaH2PO4 B. NaH2PO4 v Na3PO4

C. Na2HPO4 v Na3PO4 D. Na3PO4 v H3PO4

Gii


16/26




Da vo s [2]:-A. NaOH v NaH2PO4 cn NaOH tc l d NaOH th khng th tn ti NaH2PO4c m phi lNa3PO4- B. NaH2PO4 v Na3PO4 hai mui ny khng tn ti ng thi v chngphn ng vi nhau Na2HPO4- C. Na2HPO4 v Na3PO4

- Tn ti

- D. Na3PO4 v H3PO4 cn d H3PO4 th khng th sinh ra Na3PO4c.p n l C

BI TP TNG T

Bi 1: Cho 100 ml dung dch H3PO4 3M tc dng vi 200 ml dung dch NaOH 2,5M. Khi lngmui to thnh v nng mol/l ca dung dch to thnh ?

A. 12g; 28,4g v 0,33M; 0,67M B. 12g; 28,4g v 0,36M; 0,76M

C. 21g; 24,8g v 0,33M; 0,67M D. 18g; 38,4g v 0,43M; 0,7M

Hng dn gii

Ta c: nNaOH = 0,2.2,5 = 0,5 (mol)

3 4H POn = 3.0,1 = 0,3(mol)

Lp t l:3 4

NaOH

H PO

n 0,2.2,51 1,67 2

n 0,1.3< = = < To ra hai mui NaH2PO4 v Na2HPO4

t s mol NaH2PO4 l a

t s mol Na2HPO4 l b

Cch 1: Em c th vit 2 phng trnh ri lp h gii

Cch 2: Dng phngphpbo ton nguyn t cho nhanh:Bo ton nguyn t Na c: a + 2b = 0,5

Bo ton nguyn t P c: a + b = 0,3

Gii h phng trnh trn c: a = 0,1 v b = 0,2

2 4

2 4

NaH PO

Na HPO

m = 0,1.120 = 12(g)

m = 142.0,2 = 28,4(g)

Nng mol/ca cc mui:[ ]

[ ]

2 4

2 4

0,1NaH PO 0,33M

0,3

0,2Na HPO = 0,67M0,3

= =

=

p n A

Bi 2: Cho 3,55 g P2O5 vo 500 ml dung dch c cha 7,28g KOH. Gi s th tch ca dung dchthay i khng ng k. Tnh nng mol/l ca cc mui trong dung dch thu c?

A. 0,04M; 0,06M B. 0,05M; 0,06M

C. 0,04M; 0,08M D. 0,06M; 0,09M

Hng dn gii

2 5P O 3,55n = = 0,025(mol)142


17/26




KOH

7,28n = = 0,13(mol)

56

P2O5 + 3H2O 2H3PO4

0,025 0,05(mol)

Lp t l:3 4

KOH

H PO

n 0,132 2,6 3n 0,05

< = = < To ra hai mui Na2HPO4 v Na3PO4

t s mol Na2HPO4 l a

t s mol Na3PO4 l b

Cch 1: Em c th vit 2 phng trnh ri lp h gii

Cch 2: Dng phngphpbo ton nguyn t cho nhanh:

Bo ton nguyn t K c: 2a + 3b = 0,13

Bo ton nguyn t P c: a + b = 0,05

Gii h phng trnh trn c: a = 0,02 v b = 0,03

Nng mol/ca cc mui:[ ]

[ ]

2 4

3 4

0,02K HPO 0,04 M

0,5

0,03K PO = 0,06 M

0,5

= =

=

p n A

Ch : Em nn chuyn P2O5 thnh H3PO4 ri tnh nh trn. Khng nn vit phng trnh P2O5 tcdng trc tip vi dung dch KOH

Bi 3: Cho 1,42g P2O

5vo dung dch cha 1,12g KOH. Khi lng mui thu c l:

A. 2,72g B. 2,27g C. 2,3g D. 2,9g

Hng dn gii

2 5P O

1,42n = = 0,01(mol)

142

3 4H POn = 0,02 (mol)

KOH

1,12n = = 0,02(mol)

56

to mui KH2PO4 2 4KH POn = 0,02 (mol)

2 4KH PO

m = 0,02.136 = 2,72 (g)

p n A

Bi 4: S ml dung dch NaOH 1M trn ln vi 50ml dung dch H3PO4 1M thu c muitrung ha l:

A. 100ml B. 150ml C. 200ml D. 112ml

Hng dn gii

3 4H POn = 0,05.1 = 0,05(mol)

Phng trnh to mui trung ha:


18/26




3NaOH + H3PO4 Na3PO4 + 3H2O

0,15 0,05 (mol)

VNaOH =0,15

= 0,15 (lit) = 150 ml1

p n B

Bi 5: Oxi ha hon ton 6,2g P ri cho sn phm vo 25ml dung dch NaOH 25% (d = 1,28g/ml). Mui to thnh c cng thc nh th no ?

A. NaH2PO4 B. Na2HPO4

C. Na3PO4 D. c A v B u ng

Hng dn gii

4P + 5O2 2P2O5

2 5 3 4P P O H PO

6,2n = =0,2(mol) n = 0,1 (mol) n = 0,2 (mol)

31

NaOH25.1,28.25n = = 0,2(mol)

100.40

To mui NaH2PO4

p n A

Bi 6: Cho 20g dung dch H3PO4 37,11% tc dng va vi NH3 th thu c 10g muiphotphat amoni A. Cng thc ca mui A l:

A. (NH4)2HPO4 B. NH4H2PO4

C. (NH4)3PO4 D. Khng xc nh

Hng dn gii

3 4H PO

20*37,11m = = 7,422 (g)

100

H3PO4 + xNH3 (NH4)xH3-xPO498g 98 + 17x7,422 10

98 98 17

27,422 10

xx

+= = Cng thc mui: (NH4)2HPO4

p n A


19/26




VN DNG NH LUT BO TON IN TCH GII NHANH MT

S BI TON HA HC DNG TRC NGHIM

( ng Bo Ha Hc & ng Dng S 12/2008)



I- CS L THUYT

nh lutbo ton in tch c pht biu dng tng qut: in tch ca mt h thng c lpth lun lun khng i tc l c bo ton.

T nh lut trn ta c th suy ra mt s h qu p dng gii nhanh mt s bi ton ha hc:

H qu 1: Trong dung dch cc chtin ly hoc chtin ly nng chy th tng s in tch

dng ca cc cation bng tng s n v in tch m ca cc anion.(H qu 1 cn c gi l nh lut trung ha in)

V d 1: Dung dch A c cha cc ion sau: Mg2+, Ba2+, Ca2+, 0,1mol Cl- v 0,2 mol NO3-. Thm

dn V lit dung dch K2CO3 1M vo A n khi c lng kt ta ln nht. V c gi tr l:

A. 300 ml B. 200 ml C. 250 ml D. 150 ml

Gii

thu c kt ta ln nht khi cc ion Mg2+, Ba2+, Ca2+ tc dng ht vi ion CO32-:

2+ 2-3 3

2+ 2-

3 32+ 2-

3 3

Mg + CO MgCO

Ba + CO BaCOCa + CO CaCO

Sau khiphn ng kt thc, trong dung dch cha cc ion K+, Cl- v NO3- ( kt ta tch khi dung

dch ). Theo h qu 1 th:

+ - -2 33Cl NO

n = n + n 0,1 0,2 0,3( ) 0,15( )K COK mol n mol= + = =

2 3ddK CO

0,15V = 0,15( ) 150

1lit ml= = Chn D

V d 2: (TSH A 2007):Ho tan hon ton hn hp gm 0,12 mol FeS2 v a mol Cu2S voaxit HNO3(va ), thu c dung dch X (ch cha hai mui sunfat) v kh duy nht NO. Gi tr caa l

A. 0,04. B. 0,075. C. 0,12. D. 0,06.

Gii:

FeS2 Fe3+ + 2SO4

2

0,12 0,12 0,24

Cu2S 2Cu2+ + SO4

2

a 2a a

p dng nh lut trung ho in (h qu 1):


20/26




3.0,12 + 2.2a = 0,24.2 + 2a a 0,06= Chn D

V d 3: (TSC A 2007): Mt dung dch cha 0,02 mol Cu2+, 0,03 mol K+, x mol Cl- v y molSO4

2-. Tng khi lng cc mui tan c trong dung dch l 5,435 gam. Gi tr ca x v y ln lt l:

A. 0,03 v 0,02 B. 0,05 v 0,01

C. 0,01 v 0,03 D. 0,02 v 0,05

Gii:

p dng nh lut trung ho in:2.0,02 + 0,03 = x + 2y hay x + 2y = 0,07 (1)

Khi lng mui: 0,02.64 + 0,03.39 + 35,5x + 96y = 5,435 (2)

Gii h phng trnh (1) v (2) c: x = 0,03 v y = 0,02 Chn A

H qu 2: Trong ccphn ng oxi ha kh th tng smol electron do cc cht khnhng bngtng smol electron do cc cht oxi ha nhn.

( Da vo h qu 2 ny ta c phngphpbo ton electron)

V d 1: (TSH B 2007): Nung m gam bt st trong oxi, thu c 3 gam hn hp cht rn X.

Ha tan ht hn hp X trong dung dch HNO3( d), thot ra 0,56 lit ( ktc) NO( l sn phm kh duynht). Gi tr ca m l:

A. 2,52 B. 2,22 C. 2,62 D. 2,32

Gii

NO

Fe

0,56n = 0,025( )

22,4

mn = ( )

56

mol

mol

=

Da vo nh lutbo ton khi lng, ta c:

mO = 3 m(g) O3-m

n = ( )16

mol

3+Fe Fe + 3e

m 3m

56 56

2-

5 2

O + 2e O

3-m 2(3-m)

16 16

N + 3e N0,075 0,025

+ +

Da vo h qu 2 ta c:3m

56= 0,075 +

2(3-m)

16 m = 2,52 Chn A

V d 2: (TSH A 2008): Cho 11,36 gam hn hp gm Fe, FeO, Fe2O3 v Fe3O4 phn ng htvi dung dch HNO3 long(d), thu c 1,344 lit ( ktc) NO( l sn phm kh duy nht) v dungdch X. C cn dung dch X thu c m gam mui khan. Gi tr m l:

A. 49,09 B. 34,36 C. 35,50 D. 38,72

Gii


21/26




NO

Fe

1,344n = 0,06( )

22,4

mn = ( )

56

mol

mol

=

Da vo nh lutbo ton khi lng, ta c:

mO = 11,36 m(g) O11,36-mn = ( )

16mol

3+Fe Fe + 3e

m 3m

56 56

2-

5 2

O + 2e O

11,36-m 2(11,36-m)

16 16

N + 3e N

0,18 0,06

+ +

Da vo h qu 2 ta c:3m

56= 0,18 +

2(11,36-m)

16 m = 8,96

mmui = mFe + mNO3-= 8,96 + 62.3nFe = 8,96 + 62.3.8,96

56= 38,72gam Chn D

H qu 3:Mt hn hp gm nhiu kim loi c ha tr khng i v c khi lng cho trcsphi nhng mt smol electron khng i cho btk tc nhn oxi ha no.

V d 1: Chia 1,24 gam hn hp hai kim loi c ha tr khng i thnh hai phn bng nhau:- Phn 1:b oxi ha hon ton thu c 0,78 gam hn hp oxit.- Phn 2: tan hon ton trong dung dch H2SO4 long thu c V lit H2

( ktc). Gi tr V l:

A. 2,24 lit B. 0,112 lit C. 5,6 lit D. 0,224 lit

Gii

Khi lng mi phn: 1,24 : 2 = 0,62 gam

S mol O kt hp vi 0,62 gam hn hp kim loi:0, 78 0, 62

0,01( )16

mol

=

Qu trnh to oxit: O + 2e O2-

0,01 0,02(mol)

Theo h qu 3 th phn 2 hn hp kim loi kh H+ ca dung dch axit cng nhng 0,02mol electron:

2H+ + 2e H2

0,02 0,01(mol)

Vy th tch H2 thu c l: 0,01 . 22,4 = 0,224 lit Chn D

V d 2: Chia hn hp 2 kim loi A,B c ha tr khng i thnh 2 phn bng nhau:

- Phn 1: tan ht trong dung dch HCl to ra 1,792 lit H2( ktc)


22/26




- Phn 2: nung nng trong khng kh n khi lng khng i thu c 2,84 gam cht rn.Khi lng hn hp 2 kim loi trong hn hp u l:A. 2,4g B. 3,12g C. 2,2g D. 1,8g

Gii

Xt phn 1:

2H+

+ 2e H2

0,16 1,792

22,4= 0,08 (mol)

Theo h qu 3 th phn 2: O + 2e O2-

0,08 0,16(mol)

mKL = moxit mO = 2,84 0,08.16 = 1,56 gam

Khi lng hn hp ban u: 2.1,56 = 3,12 gam Chn B

V d 3: Ly 7,88 gam hn hp A gm hai kim loi hot ng X,Y c ha tr khng i, chia thnh haiphn bng nhau:

- Phn 1 nung trong oxi d oxi ha hon ton thu c 4,74 gam hn hp 2 oxit- Phn 2 ha tan hon ton trong dung dch cha hn hp hai axit HCl v H2SO4 long thu

c V lt kh (ktc). Gi tr V l:A. 2,24 lit B. 0,112 lit C. 1,12 lit D. 0,224 lit

Gii

Khi lng mi phn: 7,88 : 2 = 3,94 gam

S mol O kt hp vi 3,94 gam hn hp kim loi:4, 74 3, 94

0,05( )16

mol

=

Qu trnh to oxit: O + 2e O2-

0,05 0,1(mol)

Theo h qu 3 th phn 2:

2H+ + 2e H2

0,1 0,05 (mol)

Vy th tch H2 thu c l: 0,05 . 22,4 = 1,12 lit Chn C


23/26




PHNG PHP GII NHANH BI TON HN HP KIM LOI

Al/Zn V Na/Ba TC DNG VI NC

(ng Bo Ha Hc & ng dng s 12/2009)

Thc s LU HUNH VN LONGTrng H Th Du Mt -Bnh Dng

I- CS L THUYT

Bi ton hn hp Al/Zn v Na/Ba tc dng vi nc d sinh ra kh th chng ta cn lu :

+ Kim loi Na/Ba + H2O dung dch bazNaOH/Ba(OH)2 + H2

+ Kim loi Al/Zn + dung dch bazsinh ra mui + H2

Ta thy kh H2 thot ra do 2 qu trnh to nn. C th:

a/ Hn hp Na vAl tc dng vi H2O

Na + H2O NaOH + 12 H2

Al + NaOH + H2O NaAlO2 +3

2H2

Ta c: nNa = nNaOH = nAl(p)

b/ Hn hp Ba vAl tc dng vi H2O

Ba + 2H2O Ba(OH)2 + H2

2Al + Ba(OH)2 + 2H2O Ba(AlO2)2 + 3H2

Ta c: nBa = nBa(OH)2 =1

2 nAl(p)

Nhnxt: Da vo c im v d kinbi ton ta c th vit hai phng trnhphn ng ri tnhton. Tuy nhin vi dngbi tp ny ta thy l c 2 ptp trn u l phn ng oxi ha kh nn c thlm nhanh bngphngphp bo ton electron.Trng hp Zn lm tng t.

II- MT S V D

V d 1: Hn hp X gm Na v Al. Cho m gam hn hp X vo mt lng d nc thot ra V litkh (ktc). Nu cng cho m gam X vo dung dch NaOH d th thu c 1,75V lit kh (ktc). Phntrm khi lng ca Na trong X l:

A. 39,87% B. 77,31% C. 59,87% D. 29,87%

Hng dn gii

Cch 1: Giibnh thng (a s HS thng lm)

TN1: Na + H2O NaOH +1

2H2

x(mol) x 0,5x


2H2

x x 1,5x


24/26




TN2: Na + H2O NaOH +1

2H2

x(mol) x 0,5x


2H2

y(mol) 1,5y

So snh th tch 2 TN TN1 Al cn dv TN2 Al tan ht.

Ta c: 0,5x + 1,5y = 1,75(0,5x + 1,5x)

y = 2x

Xt 3 mol hn hp X th mNa = 23g v mAl = 54g

%Na =23

*100% 29,87%23 54

=+

Chn D

Cch 2: Theo phngphpbo ton electron

TN1: Cht kh l Na v Al nn:Na Na+ + 1e

x(mol) x

Do Al dv t l Na : Al l 1:1 nn:

Al Al3+ + 3e

x 3x

H2O l cht oxi ha nn: 2H+ + 2e H2

V*2

22,4

V( )

22,4mol

Bo ton electron: x + 3x =V*2

22,4(1)

TN2:

Na Na+ + 1e

x(mol) x

Do Al p ht nn:

Al Al3+ + 3e

y(mol) 3yV: 2H+ + 2e H2

1,75V*2

22,4

1,75V( )

22,4mol

Bo ton electron: x + 3y =1,75V*2

22,4(2)

T (1) v (2) y = 2x

Gii tng t Cch 1

V d 2: Cho hn hp X gm 0,1 mol Na v 0,05 mol Al tc dng vi lng nc dth thu cth tch kh ktc l bao nhiu ?


25/26




A. 2,8 lit B. 1,12lit C. 1,67 lit D. 2,24lit

Hng dn gii

Nhnxt: T l Na : Al l0,1

20,05

= nn Al p ht nn:

0,1 Na Na+ + 1e

0,05 Al Al3+ + 3e

x 2H+ + 2e H2

Bo ton electron: 0,1 + 0,05.3 = 2x x = 0,125 V = 0,125.22,4 = 2,8 (lit)

Chn A

V d 3(TSH A 2008): Cho hn hp gm Na v Al c t l s mol tng ng l 1:2 vo ncd. Sau khi ccphn ng xy ra hon ton thu c 8,96 lit H2(ktc) v m gam cht rn khng tan.Gi tr ca m l:

A. 43,2g B. 5,4g C. 7,8g D. 10,8gHng dn gii

T l Na : Al l1

2nn Al d

Ta c:

x Na Na+ + 1e

x Al(p) Al3+ + 3e

0,4 2H+ + 2e H2

Bo ton electron: x + 3x = 0,4.2 x = 0,2. Do : nAl = 2nNa = 0,2.2 = 0,4 (mol) mAl = 0,4 . 27 = 10,8 (g) Chn D

V d 4: Cho m gam hn hp Na v Al vo nc d, thu c 4,48 lit kh. Mt khc m gam hnhp trn tc dng vi dung dch Ba(OH)2 d thu c 7,84 lit kh (cc kh u o ktc). Tr s ca ml:

A. 5g B. 7,7g C. 6,55g D. 12,5g

Hng dn gii

Gi s mol Na l x

Gi s mol Al l y

TN1: Nhm d nn nNa = nAl(p) = x

x Na Na+ + 1e

x Al Al3+ + 3e

0,2 2H+ + 2e H2

Bo ton electron: x + 3x = 2.0,2 x = 0,1 mNa = 0,1.23 = 2,3 (g)

TN 2: Nhm p ht nn:

0,1 Na Na+ + 1e

y Al Al3+ + 3e


26/26


0,35 2H+ + 2e H2

Bo ton electron: 0,1 + 3y = 0,35.2 y = 0,2 mAl = 0,2.27 = 5,4 (g)

Vy m = 2,3 + 5,4 = 7,7 (g) Chn B

V d 5: Chia hn hp X gm kim loi Al v Ba thnh 2 phn bng nhau:

Phn 1: Cho vo nc dth thu c 448 ml kh (ktc)

Phn 2: Cho vo dung dch NaOH dth thu c 784 ml kh (ktc)

Khi lng hn hp trong mi phn l:

A. 0,685g B. 2,45g C. 1,225g D. 2,45g

Hng dn gii

Gi s mol Ba l x

Gi s mol Al l y

TN1: Nhm d nn nBa =1

2nAl(p) = x

x Ba Ba2+ + 2e

2x Al Al3+ + 3e

0,02 2H+ + 2e H2

Bo ton electron: 2x + 3.2x = 2.0,02 x = 0,005 mBa = 0,005.137= 0,685 (g)

TN 2: Nhm p ht nn:

0,005 Ba Ba2+ + 2e

y Al Al3+ + 3e

0,035 2H+ + 2e H2

Bo ton electron: 0,005.2 + 3y = 0,035.2 y = 0,02 mAl = 0,02.27 = 0,54 (g)

Vy m = 0,685 + 0,54 = 1,225 (g) Chn C

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