2011 _ Công thức dùng để áp dụng giải nhanh bài tập trắc nghiệm

8/2/2019 2011 _ Cng thc dng p dng gii nhanh bi tp trc nghim

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TT LUYN THI & BI DNG KIN THC NGY MI 18A/88 INH VN T

MT S CNG THC KINH NGHIM DNG GII NHANH BI TON HO HC

CHEMISTRY CLUB - HOAHOC.ORG NG XUN QUNHMi thc mc v thi V cc vn lin quan ti b mn Ho hc xin vui lng lin h vi mr. QUNH

theo 09798.17.8.85 09367.17.8.85 (ngoi gi hnh chnH) WEB: HOAHOC.ORG - E mail: [email protected]

n thii hc

HA I CNG

I. TNH pH

1. Dung dch axit yu HA: pH = 12

(log Ka + logCa) hocpH = log( Ca) (1)vi : l in li

Ka : hng s phn li ca axitCa : nng mol/l ca axit ( Ca 0,01 M )

V d 1:Tnh pH ca dung dch CH3COOH 0,1 M 250C . Bit KCH 3 COOH = 1,8. 10

-5

Gii

pH = -2

1(logKa + logCa ) = -

2

1(log1,8. 10-5 + log0,1 ) = 2,87

V d 2:Tnh pH ca dung dchHCOOH 0,46 % ( D = 1 g/ml ). Cho in li ca HCOOH trong dung dch l = 2 %Gii

Ta c : CM =M

CD %..10=

46

46,0.1.10= 0,1 M => pH = - log ( . Ca ) = - log (

100

2.0,1 ) = 2,7

2. Dung dch m (hn hp gm axit yu HA v mui NaA): pH = (log Ka + log a

m

C

C) (2)

V d : Tnh pH ca dung dch CH3COOH 0,1 M v CH3COONa 0,1 M 250C.

Bit KCH 3 COOH = 1,75. 10-5 , b qua s in li ca H2O.

pH = - (logKa + logm

a

C

C) = - (log1,75. 10-5 + log

1,0

1,0) = 4,74

3. Dung dch baz yu BOH: pH = 14 +12

(log Kb + logCb) (3)

vi Kb : hng s phn li ca bazCa : nng mol/l ca baz

V d : Tnh pH ca dung dch NH3 0,1 M . Cho KNH 3 = 1,75. 10-5

pH = 14 +2

1(logKb + logCb ) = 14 +

2

1(log1,75. 10-5 + log0,1 ) = 11,13

II. TNH HIU SUT PHNNG TNG HP NH3 :

H% = 2 2 X

Y

M

M(4)

3

XNH trong Y

Y

M%V = ( -1).100

M

(5)

- (X: hh ban u; Y: hh sau) K: t l mol N2 v H2 l 1:3V d : Tin hnh tng hp NH3 t hn hp X gm N2 v H2 c t khi hi so vi H2l 4,25 thu c hn hp Y c t khihi so vi H2 l 6,8. Tnh hiu sut tng hp NH3 .Ta c : nN 2 : nH 2 = 1:3

H% = 2 - 2Y

X

M

M= 2 - 2

6,13

5,8= 75 %


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n thii hc

HA V C

I. BI TON V CO21. Tnh lng kt ta khi hp th htlngCO2vo dung dchCa(OH)2 hocBa(OH)2

iu kin:

2CO

n n Cng thc: - 2COOH

n = n - n (6)

V d : Hp th ht 11,2 lt CO2(ktc ) vo 350 ml dung dch Ba(OH)2 1M. Tnh kt ta thu c.Ta c : n CO 2 = 0,5 mol

n Ba(OH) 2 = 0,35 mol => nOH

= 0,7 mol

nkt ta = nOH - nCO 2 = 0,7 0,5 = 0,2 mol

mkt ta = 0,2 . 197 = 39,4 ( g )

2. Tnh lng kt ta khi hp th htlngCO2 vo dung dch cha hn hp gmNaOH v Ca(OH)2 hocBa(OH)2

iu kin: 2-23

COCOn n Cng thc: 2- -

23COCO OH

n = n - n (7)

(Cn so snh 2-3CO

n vi nCav nBa tnh lng kt ta)

V d 1 : Hp th ht 6,72 lt CO2( ktc) vo 300 ml dung dch hn hp gm NaOH 0,1 M v Ba(OH)2 0,6 M. Tnh khilng kt ta thuc .

nCO 2 = 0,3 mol nNaOH = 0,03 mol n Ba(OH)2= 0,18 mol

=> nOH = 0,39 molnCO

23

= nOH - nCO 2 = 0,39- 0,3 = 0,09 mol

M nBa 2 = 0,18 mol nn nkt ta = nCO 23 = 0,09 mol

mktta = 0,09 . 197 = 17,73 gamV d 2 : Hp th ht 0,448 lt CO2( ktc) vo 100 ml dung dch hn hp gm NaOH 0,06 M v Ba(OH)20,12 M thu cm gam kt ta . Tnh m ? ( TSH 2009 khi A )

A. 3,94 B. 1,182 C. 2,364 D. 1,97nCO 2 = 0,02 mol nNaOH = 0,006 mol n Ba(OH)2= 0,012 mol

=> nOH = 0,03 molnCO

23

= nOH - nCO 2 = 0,03 - 0,02 = 0,01 mol

M nBa 2 = 0,012 mol nn nkt ta = nCO 23 = 0,01 mol

mktta = 0,01 . 197 = 1,97 gam

3. Tnh th tch CO2 cn hp th ht vo dung dchCa(OH)2hocBa(OH)2 thu clng kt ta theo yu cu (Dngny c 2 kt qu)

Cng thc: 2COn = n (8)

hoc2 -OH

COn = n - n (9)

V d : Hp th ht V lt CO2( ktc) vo 300 ml dung dch v Ba(OH)21 M thu c 19,7 gam kt ta . Tnh V ?Gii

- n CO 2 = nkt ta = 0,1 mol => VCO 2 = 2,24 lt

- n CO 2 = nOH - nkt ta = 0,6 0,1 = 0,5 => V CO 2 = 11,2 lt

II. BI TON V NHM KM1. Tnh lngNaOH cn cho vo dung dchAl3+ thu clng kt ta theo yu cu (Dng ny c 2 kt qu)

Cng thc: OHn = 3n (10)

hoc 3 -OH Aln = 4n - n (11)


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n thii hc

V d :Cn cho bao nhiu lt dung dch NaOH 1M vo dung dch cha 0,5 mol AlCl3 c 31,2 gam kt ta .Gii Ta c hai kt qu :

n OH = 3.nkt ta = 3. 0,4 = 1,2 mol => V = 1,2 lt

n OH = 4. nAl 3 - nkt ta = 4. 0,5 0,4 = 1,6 mol => V = 1,6 lt2. Tnh lng NaOH cn cho vo hn hp dung dch Al3+v H

+ thu c lng kt ta theo yu cu (Dng ny c 2kt qu)

- +minOH H

n = 3n + n (12)

3 - +maxOH HAl

n = 4n + n- n (13)

V d :Cn cho bao nhiu lt dung dch NaOH 1M ln nht vo dung dch cha ng thi 0,6 mol AlCl3v 0,2 mol HCl c 39 gam kt ta .Gii

n OH ( max ) = 4. nAl 3 - nkt ta+ nH = 4. 0,6 - 0,5 + 0,2 =2,1 mol => V = 2,1 lt3. Tnh lngHCl cn cho vo dung dch Na[Al(OH)4] (hoc NaAlO2) thu c lng kt ta theo yu cu

(Dng ny c 2 ktqu)

Cng thc: Hn = n (14)hoc

2 +H AlO

n = 4n - 3n (15)

V d : Cn cho bao nhiu lt dung dch HCl 1M vo dung dch cha 0,7 mol NaAlO2 hoc Na 4)(OHAl thu c 39gam kt ta .Gii Ta c hai kt qu :

nH = nkt ta = 0,5 mol => V = 0,5 ltnH = 4. nAlO 2 - 3. nkt ta = 4.0,7 3.0,5 = 1,3 mol => V = 1,3 lt

4. Tnh lng HCl cn cho vo hn hp dung dch NaOH v Na[Al(OH)4] (hoc NaAlO2)thu c lng kt ta theoyu cu (Dng ny c 2 kt qu)

Cng thc: -H OHn = n n (16)

hoc2

+H AlO OHn = 4n - 3n n (17)

V d : Cn cho bao nhiu lt dung dch HCl 1M cc i vo dung dch cha ng thi 0,1 mol NaOH v 0,3 mol NaAlO2hoc Na 4)(OHAl thu c 15,6 gam kt ta .Gii Ta c hai kt qu :

nH (max) = 4. nAlO 2 - 3. nkt ta + n OH = 4.0,3 3.0,2 + 01 = 0,7 mol => V = 0,7 lt

5. Tnh lngNaOH cn cho vo dung dch Zn2+ thu c lng kt ta theo yu cu (Dng ny c 2 kt qu):

-OHn = 2n (18)

hoc - 2+OH Znn = 4n - 2n (19)V d : Tnh th tch dung dch NaOH 1M cn cho vo 200 ml dung dch ZnCl22M c 29,7 gam kt ta .Gii

Ta c nZn 2 = 0,4 mol nkt ta= 0,3 molp dng CT 41 .

n OH ( min ) = 2.nkt ta = 2.0,3= 0,6 =>V ddNaOH = 0,6 lt

n OH ( max ) = 4. nZn 2 - 2.nkt ta = 4.0,4 2.0,3 = 1 mol =>V ddNaOH = 1lt

II. BI TON V HNO31. Kim loi tc dng vi HNO3d

a. Tnh lng kim loi tc dng vi HNO3d: . . KL KL spk spk n i n i (20)

- iKL=ha tr kim loi trong mui nitrat - isp kh: s e m N+5 nhn vo (Vd: iNO=5-2=3)


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n thii hc

- Nu c Fe dtc dng vi HNO3th s to mui Fe2+, khng to mui Fe3+

b. Tnh mmui nitrat thu c khi cho hn hp kim loi tc dng viHNO3d(Sn phm khng c NH4NO3)

Cng thc: mMui = mKim loi + 62nsp kh . isp kh= mKim loi + 62 2 2 2NO NO N O N

3n + n + 8n + 10n (21)

c. Tnh lng mmui nitratthu c khi cho hn hpst v oxit st t/d viHNO3d (Sn phm khng c NH4NO3)

mMui =

hh spk spk 242

m + 8 n .i80

=2 2 2

)

hh NO NO N O N

242m + 8(3n + n 8n 10n

80(22)

+) Cng thc tnh khi lng mui thu c khi cho hn hp st v cc oxt st t/d vi HNO3 long d gii phng kh NO.

mMui =80

242( mhn hp + 24 nNO )

V d :Ha tan ht 11,36 gam cht rn X gm Fe, FeO, Fe2O3, Fe3O4 trong dung dch HNO3 long d thu c m gam muiv 1,344 lt kh NO ( ktc ) l sn phm kh duy nht . Tm m ?.

Gii

mMui =80

242( mhn hp + 24 nNO ) =

80

242( 11,36 + 24 .0,06 ) = 38,72 gam

+) Cng thc tnh mmuithu c khi ha tan ht hn hp st v cc oxt st bng HNO3c nng, d gii phng kh NO2 .

mMui =80242

( mhn hp + 8 nNO 2 )

V d : Ha tan ht 6 gam cht rn X gm Fe, FeO, Fe2O3, Fe3O4 trong HNO3c nng, d thu c 3,36 lt kh NO2(ktc).C cn dung dch sau phn ng thu c bao nhiu gam mui khan.

mMui =80

242( mhn hp + 8 nNO 2 ) = 80

242( 6 + 8 .0,15 ) = 21,78 gam

d. Tnh s mol HNO3 tham gia:

3 2 2 2 4 3HNO NO NO N N O NH NO

= n .(i +s N ) =spk sp kh trong sp kh n 4n + 2n +12n +10n +10n (23)

2. Tnh khilngkim loi ban u trong bi ton oxh 2 ln

R + O2 hn hp A (R d v oxit ca R)3HNO R(NO3)n + SP Kh + H2O

mR= .hh spk spk M m + 8. n i80R = ) 2 2 4 3 2hh NO NO N O NH NO N

M m + 8(n 3n 8n + 8n +10n80

R (24)

+) Cng thc tnh khi lng st dng ban u, bit oxi ha lng st ny bng oxi c hn hp rn X . Ha tan ht Xvi HNO3c , nng ,d gii phng kh NO2.

mFe =80

56( mhn hp + 8 nNO 2 )

V d :t m gam st trong oxi thu c 10 gam hn hp cht rn X . Ha tan ht X vi HNO3c nng, d gii phng10,08 lt kh NO2( ktc) . Tm m ?Gii

mFe =

80

56( mhn hp + 24 nNO 2 ) =

80

56( 10 + 8. 0,45 ) = 9,52 gam

+) Cng thc tnh khi lng st dng ban u, bit oxi ha lng st ny bng oxi c hn hp rn X . Ha tan ht Xvi HNO3 long d gii phng kh NO.

mFe =80

56( mhn hp + 24 nNO )

V d :t m gam st trong oxi thu c 3 gam cht rn X . H a tan ht X vi HNO3 long d gii phng 0,56 lt kh NO (ktc). Tm m ?Gii

mFe =80

56( mhn hp + 24 nNO ) =

80

56( 3 + 0,025 ) = 2,52 gam

+) Cng thc tnh mmuithu c khi ha tan ht hn hp st v cc oxt st bng HNO3 d gii phng kh NO v NO2 .

mMui =80242 ( mhn hp + 24. nNO + 8. nNO 2 )


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n thii hc

V d : Ha tan ht 7 gam cht rn X gm Fe, FeO, Fe2O3, Fe3O4 trong HNO3d thu c 1,792 lt (ktc ) kh X gm NOv NO2 v m gam mui . Bit dX/H 2 = 19. Tnh m ?

Ta c : nNO = nNO 2 = 0,04 mol

mMui =80

242( mhn hp + 24 nNO + 8 nNO 2 ) = 80

242( 7+ 24.0,04 + 8.0,04 )= 25,047 gam

V. BI TON V H2SO41. Kim loitc dng viH2SO4c, nng d

a. Tnh khilng mui sunfat mMui = 96

m + n .iKL spk spk2= m + 96(3.n +n +4n )KL S SO H S2 2 (25)

a. Tnh lng kim loi tc dng vi H2SO4c, nng d: . . KL KL spk spk n i n i (26)

b. Tnh s mol axit tham gia phnng:2

2 4 2 2H SO S SO H S

isp kh= n .( +s S ) =spk trong sp kh n 4n + 2n +5n (27)

2. Hn hp st v oxit st tc dng vi H2SO4c, nng d

mMui =400

160

22H Sm + 8.6n + 8.2n +8.8n

hh S SO

(28)

+ Cng thc tnh mmuithu c khi ha tan ht hn hp Fe, FeO, Fe2O3, Fe3O4bng H2SO4c, nng, d gii phng kh SO2 .

mMui =160

400( mhn hp + 16.nSO 2 )

V d : Ha tan ht 30 gam cht rn X gm Fe, FeO, Fe2O3, Fe3O4 bng H2SO4c nng, d thu c 11,2 lt kh SO2(ktc). C cn dung dch sau phn ng thu c bao nhiu gam mui khan.Gii

mMui =160

400( mhn hp + 16.nSO 2 ) = 160

400( 30 + 16.0,5 ) = 95 gam

3. Tnh khi lng kim loiban u trong bi ton oxh 2 ln

R + O2

hn hp A (R d v oxit ca R)2 4dacH SO

R(SO4)n + SP Kh + H2OmR= .hh spk spk

Mm + 8. n i

80R = 6 10 ) 2 2hh SO S H S

Mm + 8(2n n n

80R (29)

- n gin: nu l Fe: mFe = 0,7mhh + 5,6ne trao i; nu l Cu: mCu = 0,8.mhh + 6,4.ne trao i (30)V. KIM LOI (R) TC DNG VI HCl, H2SO4 TO MUI V GII PHNG H2

tng (gim) khi lng dung dch phn ng ( m) s l:2KL H

m = m - m (31)

Kim loi R (Ha tr x) tc dng vi axit thng: nR.x=22H

n (32)

1. Kim loi + HCl Mui clorua + H22clorua KLp Hmuoi

m = m + 71.n (33)

2.

Kim loi + H2SO4 long

Mui sunfat + H22sunfat KLp Hmuoi

m = m + 96.n (34)

VI. MUI TC DNG VI AXIT: (C th chng minh cc CT bng phng php tng gim khi lng)

1. Mui cacbonat + ddHCl Mui clorua + CO2 + H2O2COmuo i c lorua muo i cacbonat

m = m + (71- 60).n (35)

2. Mui cacbonat + H2SO4 long Mui sunfat + CO2 + H2O2COmuoi sunfat muoi cacbonat

m = m + (96 - 60)n (36)

3. Mui sunfit + ddHCl Mui clorua + SO2 + H2O2SOmuoi clorua muoi sunfit

m = m + (80 - 71)n (37)

4. Mui sunfit + ddH2SO4 long Mui sunfat + SO2 + H2O2SOmuoi sunfat muoi sunfit

m = m + (96 - 80)n (38)

II. OXIT TC DNG VI AXIT TO MUI + H2O:

c th xem phn ng l: [O]+ 2[H] H2O 2O/oxit O/ H O H1

n = n = n2 (39)


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n thii hc

1. Oxit + ddH2SO4 long Mui sunfat + H2O2 4oxit H SOmuoi sunfat

m = m + 80n (40)

2. Oxit + ddHCl Mui clorua + H2O2oxit H O oxit HClmuoi clorua

m = m + 55n = m + 27,5n (41)

II. CC PHN NG NHIT LUYN1. Oxit tc dng vi cht kh

TH 1. Oxit + CO :R

xO

y+ yCO xR + yCO

2(1) R l nhng kim loi sau Al.

Phn ng (1) c th vit gn nh sau: [O]oxit + CO CO2TH 2. Oxit + H2 : RxOy + yH2 xR + yH2O (2) R l nhng kim loi sau Al.

Phn ng (2) c th vit gn nh sau: [O]oxit + H2 H2OTH 3. Oxit + Al (phn ng nhit nhm) : 3RxOy + 2yAl 3xR + yAl2O3 (3)

Phn ng (3) c th vit gn nh sau: 3[O]oxit + 2Al Al2O3

C 3 trng hp c CT chung:n = n = n = n =n[O]/oxit CO H CO H O2 2 2

m = m - mR oxit [O]/ oxit(42)

2. Th tch khthu c khi cho hn hp sn phm sau phn ng nhit nhm (Al + FexOy) tc dng vi HNO3:

x yspk

kh Al Fe O

in = [3n + 3x - 2y n ]3 (43)

3. Tnh lngAg sinh ra khi cho a(mol) Fe vo b(mol) AgNO3; ta so snh:

3a>b nAg =b 3a


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n thii hc

HA HU C

1. Tnh s lin kt ca CxHyOzNtClm: i i2 + n .(x - 2) 2 + 2 x + t - y - mk = =

2 2(n: s nguyn t; x: ha tr) (45)

k=0: ch c lk n k=1: 1 lk i = 1 vng k=2: 1 lk ba=2 lk i = 2 vng2. Da vo phn ng chy:

S C =2

CO

A

n

nS H=

2H O

A

2n

n

2 2Ankan(Ancol) H O COn = n - n

2 2Ankin CO H On = n - n (46)

* Lu : A l CxHy hoc CxHyOz mch h, khi chy cho:2 2CO H O A

n - n = k.n th A c s = (k+1)3. Tnh s ng phn ca:

- Ancol no, n chc (CnH2n+1OH): 2n-2 (1


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n thii hc

Loi trieste Cng thc (s loi tri este)Trieste cha 1 gc axit ging nhau = nTrieste cha 2 gc axit khc nhau = 4.C2nTrieste cha 3 gc axit khc nhau = 3.C2nCng thc chung (tng s trieste) = n + 4.C2n + 3. C

3n (n 3)

V d : un nng hn hp gm glixerol vi 2 axit bo l axit panmitic v axit stearic ( xc tc H 2SO4 c) th thu c baonhiu trieste ?

S trieste =2

)12(22 = 6

5. Tnh s n peptit ti a to bi x amino axit khc nhau xn (55)V d : C ti a bao nhiu ipeptit, tripeptit thu c t hn hp gm 2 amino axit l glyxin v alanin ?

S ipeptit = 22 = 4 S tripeptit = 23 = 8

6. Tnh s ete to bi n ancol n chc:2

)1( nn(56)

V d : un nng hn hp gm 2 ancol n chc no vi H2SO4 c 1400c c hn hp bao nhiu ete ?

S ete =2

)12(2 = 3

7. S nhm este = NaOH

este

n

n(57)

8. Amino axit A c CTPT (NH2)x-R-(COOH)y HCl

A

nx =

n NaOH

A

ny =

n(58)

9. Cng thc tnh s C ca ancol no, ete no hoc ca ankan da vo phn ng chy :

S C ca ancol no hoc ankan =22

2

COOH

CO

nn

n

( Vi nH 2 O > n CO 2 ) (59)

V d 1 : t chy mt lng ancol no n chc A c 15,4 gam CO2 v 9,45 gam H2O . Tm cng thc phn t ca A ?

S C ca ancol no =22

2

COOH

CO

nn

n

=

35,0525,0

35,0

= 2 => Vy A c cng thc phn t l C2H6O

V d 2: t chy hon ton mt lng hirocacbon A thu c 26,4 gam CO2 v 16,2 gam H2O . Tm cng thc phn tca A ?

( Vi nH 2 O = 0,7 mol > nCO 2 = 0,6 mol ) => A l ankan

S C ca ankan =22

2

COOH

CO

nn

n

= 6,07,0

6,0

= 6 => Vy A c cng thc phn t l C6H14

10. Cng thc tnh khi lng ancol n chc no hoc hn hp ankan n chc notheo khi lng CO2 v khi lng H2O :

mancol =2H O

m -11

2COm

(60)

V d : Khi t chy hon ton m gam hn hp hai ancol n chc no, mch h thu c 2,24 lt CO2( ktc ) v 7,2 gamH2O. Tnh khi lng ca ancol ?

mancol =2H O

m -11

2COm

= 7,2-11

4,4 = 6,8

11. Cng thc tnh khi lng amino axit A( cha n nhm -NH2 v m nhm COOH ) khi cho amino axit ny vo dung dchcha a mol HCl, sau cho dung dch sau phn ng tc dng va vi b mol NaOH.


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n thii hc

mA = MAm

ab (61)

V d : Cho m gam glyxin vo dung dch cha 0,3 mol HCl . Dung dch sau phn ng tc dng va vi 0,5 mol NaOH.Tm m ? ( Mglyxin = 75 )

m = 75

1

3,05,0 = 15 gam

12. Cng thc tnh khi lng amino axit A( cha n nhm -NH2 v m nhm COOH ) khi cho amino axit ny vo dung dchcha a mol NaOH, sau cho dung dch sau phn ng tc dng va vi b mol HCl.

mA = MAn

ab (62)

V d : Cho m gam alanin vo dung dch cha 0,375 mol NaOH . Dung dch sau phn ng tc dng va vi 0,575 molHCl . Tm m ? ( Malanin = 89 )

mA = 891

375,0575,0 = 17,8 gam

13. Cng thc xc nh cng thc phn t ca mt anken da vo phn t khi ca hn hp anken v H2trc v sau khi

dn qua bt Ni nung nng.

Anken ( M1) + H2 ctNi

o, A (M2) ( phn ng hiro ha anken hon ton )

S n ca anken (CnH2n ) =)(14

)2(

12

12

MM

MM

(63)

V d : Cho X l hn hp gm olefin M v H2 , c t khi hi so vi H2 l 5 . Dn X qua bt Ni nung nng phn ng xyra hon ton c hn hp hi Y c t khi so vi H2 l 6,25 .Xc nh cng thc phn t ca M.

M1= 10 v M2 = 12,5

Ta c : n =)105,12(14

10)25,12(

= 3

M c cng thc phn t l C3H614. Cng thc xc nh cng thc phn t ca mt ankin da vo phn t khi ca hn hp ankin v H2trc v sau khi dnqua bt Ni nung nng.

Ankin ( M1) + H2 ctNi

o, A (M2) ( phn ng hiro ha ankin hon ton )

S n ca ankin (CnH2n-2 ) =)(14

)2(2

12

12

MM

MM

(64)

15.Cng thc tnh hiu sut phn ng hiro ha anken: H% = 2- 2My

Mx (65)

16.Cng thc tnh hiu sut phn ng hiro ha anehit no n chc: H% = 2- 2My

Mx (66)

17.Cng thc tnh % ankan A tham gia phn ng tch: %A =X

A

M

M- 1 (67)

18.Cng thc xc nh phn t ankan A da vo phn ng tch: MA = XA

hhXMV

V (68)


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n thii hc

------------------------

Trong nhng nm qua, cc k thi quc gia: tt nghip THPT, tuyn sinh H & C, mn ha hc thi theo hnhthc trc nghim khch quan i hi th sinh phi gii quyt s lng cu hi v bi tp tng i ln. Trong bi tp ha hc nh lng chim mt t l khng nh vi khong thi gian tng i ngn: trung bnh 1,5 n 2pht/cu. Do vic tm nhanh ra cc phng php gip gii nhanh bi ton ha hc c ngha ht sc quantrng. gip cc bn hc sinh cng nh cc thy c gio trong trng c thm ngun t liu n tp ha gii

cc bi tp nh lng trc nghim hiu qu trong thi gian nhanh nht. Bng kinh nghim luyn thi nhiu nm,ca bn thn, t tp san s 2 ny, xin gii thiu vi cc thy c gio, cng cc em hc sinh 27 iu cn nhtrc khi i thi. Hi vng y s l ti liu hu ch phc v cho cc em hc sinh trong cc k thi ti . Mc d rtc gng nhng kh c th trnh khi sai st ngoi mun. Rt mong c s ng gp xy dng qu bu tbn c. Mi kin xin gi v ban bin tp tp san YP2. Chc cc bn c mt ma thi thnh cng!

iu 1: 32 cng thc gii nhanhVic nm cc cng thc ny s gip gii nhanh cc bi ton nh lng thng gp trong cc k thi, m nu giitheo cch thong thng s mt rt nhiu thi gian.I. C s xy dng h thng cng thc:Cc phn ng ha hc xy ra u tun th nghim ngt theo h thng cc nh lut bo ton: Khi lng,nguyn t, electron, in tch,

1 Bo ton khi lng:Tng khi lng cc cht trc phn ng = tng khi lng cc cht sau phn ng.Tng qut: A + B C + DmA + mB = mC + mD md nu choc: mA p + mB p = mC ms + mD msV d: Cho m gam Fe tc dng vi oxi, sau mt thi gian thu c 12g hn hp gm FeO, Fe3O4, Fe2O3 v Fed. Th:

2. nh lut bo ton nguyn tTng s mol nguyn t ca 1 nguyn t trc v sau phn ng lun c bo ton:

V d: Ha tan hn hp gm a mol Fe2O3, b mol Fe3O4, c mol FeO v d mol Fe trong dung dch HCl c dungdch B. Cho NaOH d vo dung dch B, lc ly kt ta nung ngoi khng kh n khi lng khng i c mgam Fe2O3. Tnh m theo a, b, c, d.


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n thii hc

n y bn c t tnh c ri!

3. nh lut bo ton electronTrong phn ng oxi ha kh lun c: Tng s mol e cht kh cho = tng s mol e cht oxi ha nhn.

V d: x mol Fe, y mol Cu tc dng vi HNO3 d thu c a mol NO v b mol NO2.Ta c qu trnh cho v nhn e:

Qu trnh cho e Qu trnh nhn eFe Fe3+ + 3e N+5 + 3e N+2 (NO)

x 3x 3a aCu Cu2+ + 2e N+5 + 2e N+4 (NO2)

x 2x 2b bSuy ra: 3x + 2y = 3a + bCh : Khi tnh s mol e ch cn quan tm n s oxi ha u v cui ca nguyn t tham gia qu trnh oxi ha kh. c bit cn lu n bo ton nguyn t.

V d: Fe3O4 3 Fe3+; 2N+5 + 8e N2O.4. nh lut bo ton in tchTng s mol ion dng = tng s mol ion mTng qut: Dung dch A cha x mol Al3+, y mol Fe2+, a mol SO42-, b mol Cl- v d mol NO3- th ta c:3x + 2y = 2a + b + dMi nh lut trn s c phng php gii i cng, iu ny s c ng trong cc s tip theo.II. H thng cng thcTrong s ny, xin gii thiu mt s cng thc lin quan n bi ton CO2 tc dng vi dung dch kim. Bi tonny chng ta gp c trong cc bi tp v c ln bi tp hu c.1. Cng thc tnh lng kt ta xut hin khi hp th ht lng CO2 vo dung dch Ca(OH)2 hocBa(OH)2

(1)iu kin:V d: Hp th ht 7,84lit CO2 (ktc) vo 300ml dung dch Ba(OH)2 1M. Tnh lng kt ta thu c.

Hng dn gii:

Lu : nn kt qu trn ph hp. Ta cn kim tra li kt qu v nu Ba(OH)2 d, khng ph thuc vo lngOH-.2. Cng thc tnh lng kt ta xut hin khi hp th ht mt lng CO2vo dung dch hn hp chaNaOH v Ca(OH)2hoc NaOH v Ba(OH)2 (C th thay NaOH bng KOH).

(2)Sau phi so snh vi s mol Ca2+ (hoc Ba2+) xem cht no phn ng ht. S mol OH- c tnh bngcng thc:V d: Hp th ht 6,72 lit CO2 (ktc) vo 300ml dung dch hn hp gm NaOH 0,1M v Ba(OH)2 0,06M. Tnhkhi lng kt ta thu c.


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n thii hc

Hng dn gii:

Tng t nh cng thc (1), gia s mol CO32- v s mol CO2 c s rng buc:3. Tnh th tch CO2cn hp thu vo mt lng dung dch Ca(OH)2hoc Ba(OH)2 thu c lng ktta theo yu cu.Dng ton ny c 2 kt qu:

(3)V d: Hp th ht V lit CO2 (ktc) vo 300mol dung dch Ba(OH)2 1M c 19,7 gam kt ta. Tnh V.Hng dn gii:

4. Tnh th tch dung dch NaOH cn cho vo dung dch cha Al3+ xut hin mt lng kt ta theo

yu cu.Dng ny c hai kt qu:

(4)V d: Cn cho bao nhiu th tch dung dch NaOH 1M ln nht l bao nhiu vo dung dch cha 0,6mol AlCl3 v0,2mol HCl thu c 3g kt ta.

Hng dn gii:

Bn c th l gii ti sao bi ton ny ch c 1 kt qu!5. Tnh th tch dung dch HCl cn cho vo dung dch cha AlO2- xut hin mt lng kt ta theo yucu.Dng ny c hai kt qu:

(5)V d: Cn bao nhiu lit dung dch HCl 1M cn cho vo 700ml dung dch NaAlO2 1M thu c 39g kt ta.


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n thii hc

Hng dn gii:

BT p dng: Tnh th tch dung dch HCl 1M cc i cn cho vo dung dch cha ng thi 0,1mol NaOH v0,3mol NaAlO2 thu c 15,6g kt ta.6. Tnh khi mui nitrat thu c khi cho kim loi hay hn hp kim loi tc dng vi dung dch HNO3 d(khng to mui NH4NO3)

Ch : khng to kh no th s mol kh bng 0.V d: Ha tan hon ton 10g hn hp Mg, Al, Zn trong dung dch HNO3 d thu c dung dch cha m gammui nitrat v 5,6lit NO (ktc, sn phm kh duy nht). Tnh m

Hng dn gii:

BT p dng: Ha tan ht 10,71g hn hp Al, Zn, Fe trong 4lit dung dch HNO3 va thu c dung dch A v1,792lit hn hp kh (ktc) gm N2O v N2 c s mol bng nhau. C cn dung dch A thu c m gam muikhan. Tnh m7. Tnh khi lng mui sunfat thu c khi cho hn hp cc kim loi tc dng vi dung dch H2SO4 cnng d to thnh SO2

(6)

Ch : SO2 l sn phm kh duy nht.V d: Ha tan hon ton 10g hn hp gm Mg, Al, Cu bng dung dch H2SO4 c, nng d thu c dung dchcha m gam mui sunfat v 10,08lit SO2 (ktc, sn phm kh duy nht). Tnh m

Hng dn gii:

8. Tnh khi lng mui thu c khi ha tan hn hp gm Fe v cc oxit ca Fe bng dung dch H2SO4c nng hoc HNO3

(7)Ch : SO2, NO l sn phm kh duy nht.Bn c hy th chng minh hai cng thc trn s thy rt th v.V d 1: Ha tan 30g hn hp gm Fe, FeO, Fe2O3 v Fe3O4 bng dung dch H2SO4 c nng thu c dungdch cha m gam mui sunfat v 11,2lit SO2 (ktc, sn phm kh duy nht).

Hng dn gii:


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n thii hc

V d 2:t mt lng bt Fe trong khng kh mt thi gian thu c 12g hn hp rn gm Fe v 3 oxit ca Fe.Ha tan ht hn hp ny trong dung dch HNO3 thy thot ra 2,24lit kh NO (ktc, sn phm kh duy nht) vdung dch X. Tnh khi lng mui khan thu c khi c cn dung dch X.

Hng dn gii:

(xem tip cc s sau)

BBT: Trn y l mt s cng thc tnh nhanh m qua rt nhiu nm luyn thi thy hiu trng ca chng ta c kt li. H thng cng thc ny rt hu ch cho vic gii nhanh cc cu bi tp trong cc thi quc gia .BBT s tip tc ng h thng cc iu cn li trong cc s sau .

9. Tnh khi lng mui thu c khi ha tan hn hp gm (Fe, FeO, Fe2O3, Fe3O4) bng dung dch HNO3c nng gii phng NO2:

mmui = (8)

p dng: Ha tan ht 6g hn hp X gm Fe, FeO, Fe3O4 v Fe2O3 trong dung dch HNO3d thu c 3,36litNO2 (ktc, sn phm kh duy nht). C cn dung dch sau phn ng thu c m gam mui khan . Tnh m

S: 21,78g

10. Tnh khi lng mui thu c khi cho hn hp Fe, FeO, Fe3O4 v Fe2O3 tc dng vi dung dchHNO3thu c hn hp NO v NO2.

mmui = (9)

p dng: Dn lung kh CO qua ng ng Fe2O3nung nng thu c 9g hn hp rn X. Ha tan X trong dungdch HNO3d thu c 6,72lit hn hp kh Y gm NO v NO2 c mY=10,6g. Tnh khi lng mui khan thuc trong dung dch sau phn ng.

S: 18,4g

Ch :

Trong bi ton ny, hn hp X khng nht thit phi c c 4 cht: Fe, FeO, Fe3O4 v Fe2O3m c thch c 2 hoc 3 cht.

HNO3 phi cho l d v nu cho HNO3va th s c th pht sinh trng hp Fe kh Fe3+v

Fe2+.

11. Phn ng oxit kim loi tc dng vi axit c gc axit khng c tnh oxi ha: HX, H2SO4 long, H3PO4

TQ: MxOy + 2yH+ xM+2y/x + y H2O


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n thii hc

Ta c:

Hay (10)

p dng 1: Cho m gam hn hp 3 kim loi tc dng vi oxi d thu c (m+4,8) gam hn hp 3 oxit. Ha tanhon ton hn hp oxit ny cn V lit dung dch hn hp cha HCl 1M v H2SO4 0,5M. Tnh V

S: 300ml

Hng dn:

p dng 2: X l oxit ca mt kim loi c cha 27,586% khi lng l oxi. ha tan 37,12g X cn bao nhiu mldung dch HCl 1M?

S: 1280ml

Hng dn: moxit v %O mO nO nH+

12. Phn ng kim loi + axit HCl d

2M + 2nHCl 2MCln + nH2

Pt ion: 2M + 2nH+ 2Mn+ + nH2

Vi M l kim loi ngtrc H, ta c:

(11)

Ch :

Nu M l kim loi tc dng c vi nc nhit thng (IA, Ca, Ba, Sr) th khi cho vo dung dch axitn phn ng vi axit trc, ht H+ca axit nu cn d M n s phn ng vi H2O cng sinh ra H2

V d: K + dd HCl:

2K + 2HCl 2KCl + H2


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n thii hc

Ht HCl m cn K, s c phn ng:

2K + 2H2O 2KOH + H2

Do m rn = mKL + mgc axit + mOH-

Khi H2gii phng trong phn ng ca kim loi + axit dn qua ng s ng oxit kim loi ng sau Al th khilng ng s gim chnh bng khi lng oxi trong oxit kim loi phn ng:

p dng 1: Ha tan 14,8g hn hp bt gm Al, Zn, Fe bng dung dch HCl 0,5M va thu c dung dch Av V lit kh (ktc). Dn ton b kh thu c qua ng s ng CuO nung nng, sau phn ng khi lng ng s

gim 5,6g. C cn dung dch A thu c m gam mui. Tnh V, m v th tch dung dch HCl dng.

Hng dn:

p dng 2: Cho 7,8g hn hp bt gm mt s kim loi tan ht trong dung dch HCl d. Sau khi kt thc phnng thy khi lng dung dch tng 7 gam. Tnh khi lng mui trong dung dch thu c.

Hng dn:

p dng 3: Ha tan 9,14g hp kim Mg-Al-Cu bng dung dch HCl d thu c 7,84lit kh (ktc), dung dch A v2,54g cht rn khng tan. C cn dung dch A thu c m gam mui khan . Tnh m

S: 31,45g

p dng 4: Cho 3,9g K vo 100ml dung dch HCl x M thu c dung dch A. C cn dung dch A th thu c6,525g cht rn khan. Tnh x


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n thii hc

khng khN2O Kh khng muN2 Kh khng muNH4NO3 Dung dch

iu kin:

M Au, Pt

n: Ha tr ca kim loi.

Al, Fe, Cr, Ni, Co th ng ha khng tc dng vi dung dch HNO3c ngui.

M c tnh kh cng mnh, N+5b kh xung cng su.

(13)

= s e trao i

Cch biu din:


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n thii hc

Khi HNO3tham gia phn ng vi kim loi ni ring v vi cht kh ni chung th HNO3va ng vai tr l chtoxi ha va ng vai tr cht mi trng. V:

= s e nhn.s mol sp kh

p dng 1: Cho 1,35g hn hp bt Al, Mg, Cu tc dng vi dung dch HNO3thu c hn hp khgm 0,01molNO v 0,04mol NO2. Tnh khi lng mui thu c.


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n thii hc

Hng dn:

p dng 2: Ha tan ht m gam hn hp cc kim loi trong dung dch HNO 3thu c dung dch A v hn hpkh gm 0,12mol NO; 0,08mol N2O; 0,06mol N2. Cho NaOH d vo dung dch A thy c 3,36lit NH3 (ktc) thotra. Tnh s mol HNO3 phn ng

Hng dn:

15. Phn ng ca H2SO4c nng vi kim loi

M+H2SO4 c nngM2(SO4)n+H2O+SO2/S/H2S

M Au, Pt

n l s oxi ha cao nht ca M

(14)

=1/2(nsp kh .s e nhn)

= n nguyn t S trong sp kh

Cch biu din:


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n thii hc

p dng 1: Cho 15,82g hn hp Fe, Al, Cu tc dng ht vi dung dch H2SO4c nng thu c dung dch A v9,632lit SO2 (ktc). C cn dung dch A thu c bao nhiu gam mui khan .

S: 57,1g

Hng dn:

p dng 2: Cho 13,248g kim loi M tan trong dung dch H2SO4c nng d, thu c V lit H2S (ktc) v dungdch A. Khi lng mui khan trong A c khi lng l 66,24g. Tnh V.

S: 3,0912

Hng dn:

66,24=13,248+96x x = 0,55

V = 0,55/4.22,4=3,0912(lit)

17. Phn ng kh oxit kim loi bng CO

MxOy + yCO xM + yCO2

(M l kim loi ng sau nhm)

(15)

p dng 1: X l hn hp gm Fe2O3, FeO, Al2O3 v MgO. Dn kh CO i qua ng s nung nng ng 42,4g X,thu c 41,6g cht rn v hn hp kh. Cho hn hp kh ny qua dung dch Ba(OH)2d thy to ra m gam ktta. Tnh m.

S: 9,85g

Hng dn:


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n thii hc

p dng 2: Kh m gam hn hp gm MgO, FeO, CuO bng CO nhit cao, sau phn ng thu c 21,6ghn hp rn A v hn hp kh B. Cho B qua dung dch Ba(OH)2thu c 19,7g kt ta, dung dch C. Dung dchC, un nng dung dch C li thu c 19,7g kt ta na. Tnh m

S: 26,4g

Hng dn:

17. Khi t chy mt hidrocacbon A thu c CO2 v H2O

a) Nu A l ankan v:

(16)

Sn phm dn vo bnh ng Ca(OH)2hoc Ba(OH)2d:

(17)

p dng 1:t chy hon ton mt hidrocacbon A cn dng 38,4g O2, thu c 16,8lit CO2 (ktc). Khi cho A tcdng vi Cl2 (askt, 1:1) ch thu c 1 sn phm th monoclo. Gi tn A

Hng dn:


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n thii hc

A l ankan

A l C5H12

p dng 2:t chy hon ton m gam hn hp gm CH4, C2H6, C4H10thu c 3,3g CO2 v 4,5g H2O. Tnh m.

Hng dn:

p dng 3:t chy hon ton V lit hn hp X gm 1 s hidrocacbon thuc cng dy ng ng. Hp th tonb sn phm chy vo bnh ng Ca(OH)22ban u 14,4g. Tnh V d thy c 45g kt ta to ra v khi lng

dung dch gim so vi khi lng dung dch Ca(OH)

Hng dn:

X l hn hp cc ankan.

b)Nu th A l anken hoc xicloakan.

A l anken th mch C mch h: CnH2n (n2).

A l xicloankan th mch C l mch vng v CnH2n (n3).

c) Nu hoc v hoc th A l ankin hayankaien.

18. bt bo ha ca hp cht hu c cho bit s lin kt hay s vng trong phn t

i: s nguyn t ca nguyn t trong phn t


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n thii hc

ni: ha tr ca nguyn t

Dng bt bo ho bin lun xc nh CTPT hp cht hu c .

p dng: Xc nh CTPT ca cc cht c cng thc n gin:

a) anehit no: (C2H3O)n

b) axit cacboxylic no: (C3H4O3)n

c) Ancol no: (C3H8O2)n

Hng dn:

a) (C2H3O)n = C2nH3nOn

V l andehit no nn s lin kt = = s nguyn t O trong phn t.

Tng t, b, c bn c t gii

19. S ng phn ancol n chc no ng vi cng thc CnH2n+2O l: 2n-6vi n6.

p dng: C3H8O c 2 ng phn; C4H10O c 4 ng phn, C5H12O c 8 ng phn.

20. S ng phn anehit no, n chc ng vi cng thc CnH2nO l: 2n-3vi n7.

p dng: Vit v tnh s ng phn ca cc anehit c cng thc: C4H8O, C5H10O, C6H12O

21. S ng phn axit no, n chc ng vi cng thc CnH2nO2 l: 2n-3vi n


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n thii hc

p dng: Vit cc ng phn ete ca C3H8O (1); C5H12O (3)

26. S ng phn xeton n chc no ng vi cng thc CnH2nO (3


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n thii hc

H2 + Cl2 2HCl

nn Cl2 thiu

Nu tnh theo H2, H = 60% (sai)

V d 2:un nng 28,5g etanol vi H2SO4c 1700C, ton b sn phm kh v hi thu c dn ln lt

quan bnh 1 ng CuSO4khan, d; bnh 2 ng NaOH c; bnh 3 ng nc brom d trong dung mi CCl4.Sau th nghim thy khi lng bnh 3 tng 10,5g. Tnh hiu sut phn ng oxi ha to sn phm oxi ha bnh3.

Hng dn

Khi lng bnh 3 tng chnh l khi lng C2H4

T d dng tnh c H.

Cch 2: Tnh theo sn phm

nth te tt: lng thc t to thnh, thng cho trong

ntt ly thuyet: lng to ra tnh theo l thuyt, thng c tnh theo phng trnh phn ng theo lng cht thiutrong phn ng.

V d 1: Nung 1kg vi cha 80% CaCO3thu c 112 dm3kh (ktc). Tnh hiu sut phn ng

Hng dn

Khi lng CaCO3trong 1kg vi:


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n thii hc

CaCO3 CaO + CO2

S mol CO2thu c theo l thuyt:

Hiu sut phn ng

Ch : C th tnh hiu sut ca qu trnh gm nhiu phn ng lin tip, th d

Hiu sut chung ca qu trnh A D:

H = H1%.H2%.H3%.100%

V d 1: Trong th chin th gii th 2, ngi ta phi iu ch cao su buna t tinh bt theo s sau

Tinh bt glucoz ancol etylic butan-1,3-ien cao su buna. T 1 tn khoai cha 80% tinh bt iu chc bao nhiu tn cao su buna, bit hiu sut ca c qu trnh l 60%?

S: 1,6 tn

V d 2: T t n cha 90% CaC2ngi ta iu ch anilin theo s

Tnh khi lng t n cn dng iu ch 50kg anilin c nguyn cht 98%

S: 217,49kg

28. Khi ru no, n chc mt nc to ete t0


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n thii hc

p dng: un nng 132,8g hn hp 3 ancol n chc vi H2SO4c 1400C cho n khi phn ng hon ton

thu c 111,2g hn hp cc ete c s mol bng nhau. Tnh s mol mi ete.

Hng dn

s: 0,2mol

Ch :ru A sn phm hu c B

+ Nu th B l ete

+ Nu th B l anken

29. Khi 1 cht hu c A tham gia phn ng trng gng vi Ag+/NH3d to Ag kt ta vi H = 100%:

- Nu nAg = 2nA th A l anehit n chc hay HCOOH; HCOOR

+ Nu nAg = 4nA th A l HCHO hay anhit hai chc R(CHO)2

Ch : Nu bi cho Ag+ dng va m nAg+>nAg khi H = 100% th A ngoi c chc CHO, phn gc R cn c

ni ba u mch:

CH C-R(CHO)x

p dng: Cho 4,2g anehit A mch h phn ng vi lng d AgNO3/NH3. Ton b lng Ag thu c ha tan

ht trong HNO3c, nng d thu c 3,792lit NO2 270C v 740mmHg. Xc nh CTCT ca A bit

Hng dn:

Theo bi MA < 4.28 hay 56x


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n thii hc

RCHO = 56 R = 27 (C2H3-)

30. Khi gii ton este A tc dng vi dung dch NaOH hay KOH:

Nu: nA=nNaOH pu=nmuoi=nancol

Th este l este n chc.

Ch : Cht rn khan thu c khi c cn dung dch sau phn ng x phng ha ngoi mui ca axit hu ccn c th c kim d.

V d: X l hn hp 2 este ng phn ca nhau A v B. X phng ha hon ton 66,6g X cn dng va ht900ml dung dch NaOH 1M. C cn dung dch sau phn ng thu c 65,4g hn hp hai mui khan. Xc nhCTCT ca A v B, bit rng khi ha hi 66,6g X thu c th tch bng th tch ca 28,8g O2 cng iu kinnhit , p sut.

31. Khi trng hp, ng trng hp hay trng ngng hoc ng trng ngng:

X thng l H2O

H s polime ha:

V nu H=100% th:

mpolime = mmonome

mpolime = mmonome-mX (phn ng trng ngng)

p dng: Tnh s mt xch ca mt mch t nilon-6,6 v on t capron. Bit khi lng ca mi on polimetrn tng ng l 27346u v 17176u

Hng dn:

PTK 1 mt xch t nilon-6,6: 226u

PTK 1 mt xch t capron: 113u

Nn:

2011 _ Công thức dùng để áp dụng giải nhanh bài tập trắc nghiệm

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Transcript of 2011 _ Công thức dùng để áp dụng giải nhanh bài tập trắc nghiệm