MT S CNG THC GII NHANH BI TI TRC NGHIM HA HC

1. Cng thc tnh s ng phn ancol n chc no, mch h : Cn H2n+2O2S ng phn Cn H2n+2O2 = 2n- 2 ( 1 < n < 6 )

V d : S ng phn ca ancol c cng thc phn t l :

a. C3H8O = 23-2 = 2

b. C4H10O = 24-2 = 4

c. C5H12O = 25-2 = 8

2. Cng thc tnh s ng phn anehit n chc no, mch h : Cn H2nO

S ng phn Cn H2nO = 2n- 3 ( 2 < n < 7 )

V d : S ng phn ca anehit n chc no, mch h c cng thc phn t l :

a. C4H8O = 24-3 = 2

b. C5H10O = 25-3 = 4

c. C6H12O = 26-3 = 8

3. Cng thc tnh s ng phn axit cacboxylic n chc no, mch h : Cn H2nO2

S ng phn Cn H2nO2 = 2n- 3 ( 2 < n < 7 )

V d : S ng phn ca axit cacboxylic n chc no, mch h c cng thc phn t l :

a. C4H8O2 = 24-3 = 2

b. C5H10O2 = 25-3 = 4

c. C6H12O2 = 26-3 = 8

4. Cng thc tnh s ng phn este n chc no, mch h : Cn H2nO2

S ng phn Cn H2nO2 = 2n- 2 ( 1 < n < 5 )

V d : S ng phn ca este n chc no, mch h c cng thc phn t l :

a. C2H4O2 = 22-2 = 1

b. C3H6O2 = 23-2 = 2

c. C4H8O2 = 24-2 = 4

5. Cng thc tnh s ng phn ete n chc no, mch h : Cn H2n+2O S ng phn Cn H2n+2O = ( 2 < n < 5 )

V d : S ng phn ca ete n chc no, mch h c cng thc phn t l :

a. C3H8O = = 1

b. C4H10O = = 3

c. C5H12O = = 6

6. Cng thc tnh s ng phn xeton n chc no, mch h : Cn H2nO S ng phn Cn H2nO = ( 3 < n < 7 )

V d : S ng phn ca xeton n chc no, mch h c cng thc phn t l :

a. C4H8O = = 1

b. C5H10O = = 3

c. C6H12O = = 6

7. Cng thc tnh s ng phn amin n chc no, mch h : Cn H2n+3N

S ng phn Cn H2n+3N = 2n-1 ( n < 5 )

V d : S ng phn ca anin n chc no, mch h c cng thc phn t l :

a. C2H7N = 22-1 = 1

b. C3H9N = 23-1 = 3

c. C4H12N = 24-1 = 6

8. Cng thc tnh s trieste ( triglixerit ) to bi glixerol v hn hp n axt bo : S tri este =

V d : un nng hn hp gm glixerol vi 2 axit bo l axit panmitic v axit stearic ( xc tc H2SO4 c) th thu c bao nhiu trieste ?

S trieste = = 6

9. Cng thc tnh s ng phn ete to bi hn hp n ancol n chc : S ete =

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

S ete = = 3

10. 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 = ( Vi nHO > n CO)

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 = = = 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 t ca A ?

( Vi nHO = 0,7 mol > n CO= 0,6 mol ) => A l ankan

S C ca ankan = = = 6

Vy A c cng thc phn t l C6H14

11. Cng thc tnh khi lng ancol n chc no hoc hn hp ankan n chc notheo khi lng CO2 v khi lng H2O : mancol = mHO -

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 gam H2O. Tnh khi lng ca ancol ?

mancol = mHO - = 7,2 - = 6,8

12. Cng thc tnh s i, tri, tetra..n peptit ti a to bi hn hp gm x amino axit khc nhau :

S n peptitmax = xn

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

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

mA = MA

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= 15 gam

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

mA = MA

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 mol HCl . Tm m ? ( Malanin = 89 )

mA = 89 = 17,8 gam

15. Cng thc xc nh cng thc phn t ca mt anken da vo phn t khi ca hn hp anken v H2 trc v sau khi dn qua bt Ni nung nng.

Anken ( M1) + H2 A (M2) ( phn ng hiro ha anken hon ton )

S n ca anken (CnH2n ) =

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 xy ra 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 = = 3

M c cng thc phn t l C3H6

16. Cng thc xc nh cng thc phn t ca mt ankin da vo phn t khi ca hn hp ankin v H2 trc v sau khi dn qua bt Ni nung nng.

Ankin ( M1) + H2 A (M2) ( phn ng hiro ha ankin hon ton )

S n ca ankin (CnH2n-2 ) =

17.Cng thc tnh hiu sut phn ng hiro ha anken.

H% = 2- 2

18.Cng thc tnh hiu sut phn ng hiro ha anehit no n chc.

H% = 2- 2

19.Cng thc tnh % ankan A tham gia phn ng tch.

%A = - 1

20.Cng thc xc nh phn t ankan A da vo phn ng tch.

MA =

21.Cng thc tnh khi lng mui clorua khi cho kim loi tc dng vi dung dch HCl gii phng kh H2

mMui clorua = mKL + 71. nH

V d : Cho 10 gam hn hp kim loi gm Mg, Al, Zn tc dng vi dung dch HCl thu c 22,4 lt kh H2 ( ktc). Tnh khi lng mui thu c .

mMui clorua = mKL + 71 nH= 10 + 71. 1 = 81 gam

22.Cng thc tnh khi lng mui sunfat khi cho kim loi tc dng vi dung dch H2SO4 long gii phng kh H2

mMui sunfat = mKL + 96. nH

V d : Cho 10 gam hn hp kim loi gm Mg, Al, Zn tc dng vi dung dch H2SO4 long thu c 2,24 lt kh H2 ( ktc). Tnh khi lng mui thu c .

mMui Sunfat = mKL + 96. nH= 10 + 96. 0,1 = 29,6 gam

23.Cng thc tnh khi lng mui sunphat khi cho kim loi tc dng vi dung dch H2SO4 c to sn phm kh SO2 , S, H2S v H2O

mMui sunft = mKL + .( 2nSO+ 6 nS + 8nHS ) = mKL +96.( nSO+ 3 nS + 4nHS )

* Lu : Sn phm kh no khng c th b qua

* n HSO= 2nSO+ 4 nS + 5nHS24.Cng thc tnh khi lng mui nitrat khi cho kim loi tc dng vi dung dch HNO3 gii phng kh : NO2 ,NO,N2O, N2 ,NH4NO3

mMui Nitrat = mKL + 62( n NO + 3nNO + 8nNO +10n N +8n NHNO)

* Lu : Sn phm kh no khng c th b qua

* n HNO= 2nNO+ 4 nNO + 10nNO +12nN + 10nNHNO

25.Cng thc tnh khi lng mui clorua khi cho mui cacbonat tc dng vi dung dch HCl gii phng kh CO2 v H2O

mMui clorua = mMui cacbonat + 11. n CO

26.Cng thc tnh khi lng mui sunfat khi cho mui cacbonat tc dng vi dung dch H2SO4 long gii phng kh CO2 v H2O

mMui sunfat = mMui cacbonat + 36. n CO

27.Cng thc tnh khi lng mui clorua khi cho mui sunfit tc dng vi dung dch HCl gii phng kh SO2 v H2O

mMui clorua = mMui sunfit - 9. n SO

28.Cng thc tnh khi lng mui sunfat khi cho mui sunfit tc dng vi dung dch H2SO4 long gii phng kh CO2 v H2O

mMui sunfat = mMui cacbonat + 16. n SO

29.Cng thc tnh s mol oxi khi cho oxit tc dng vi dung dch axit to mui v H2O

nO (Oxit) = nO ( HO) = nH ( Axit)

30.Cng thc tnh khi lng mui sunfat khi cho oxit kim loi tc dng vi dung dch H2SO4 long to mui sunfat v H2O

Oxit + dd H2SO4 long ( Mui sunfat + H2O

mMui sunfat = mOxit + 80 n HSO

31.Cng thc tnh khi lng mui clorua khi cho oxit kim loi tc dng vi dung dch HCl to mui clorua v H2O

Oxit + dd HCl ( Mui clorua + H2O

mMui clorua = mOxit + 55 n HO = mOxit + 27,5 n HCl32.Cng thc tnh khi lng kim loi khi cho oxit kim loi tc dng vi cc cht kh nh : CO, H2 , Al, C

mKL = moxit mO ( Oxit)

nO (Oxit) = nCO = n H= n CO = n HO

33.Cng thc tnh s mol kim loi khi cho kim loi tc dng vi H2O, axit, dung dch baz kim, dung dch NH3 gii phng hiro.

nK L= nH vi a l ha tr ca kim loi

V d: Cho kim loi kim tc dng vi H2O:

2M + 2H2O 2MOH + H2

nK L= 2nH = nOH

34.Cng thc tnh lng kt ta xut hin khi hp th ht mt lng CO2 vo dung dch Ca(OH)2 hoc Ba(OH)2 .

nkt ta = nOH

- nCO

( vi nkt ta nCO hoc cho dd baz phn ng ht )

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= 0,5 mol

n Ba(OH)= 0,35 mol => nOH = 0,7 mol

nkt ta = nOH

EMBED Equation.3 - nCO = 0,7 0,5 = 0,2 mol

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

35.Cng thc tnh lng kt ta xut hin khi hp th ht mt lng CO2 vo dung dch cha hn hp gm NaOH, Ca(OH)2 hoc Ba(OH)2 .

Tnh nCO = nOH

EMBED Equation.3 - nCO ri so snh nCa hoc nBa xem cht no phn ng ht suy ra n kt ta ( iu kin nCO nCO )

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 khi lng kt ta thu c .

nCO= 0,3 mol

nNaOH = 0,03 mol

n Ba(OH)2= 0,18 mol

=> nOH = 0,39 mol

nCO = nOH

EMBED Equation.3 - nCO = 0,39- 0,3 = 0,09 mol

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

mkt ta = 0,09 . 197 = 17,73 gam

V 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)2 0,12 M thu c m gam kt ta . Tnh m ? ( TSH 2009 khi A )A. 3,94

B. 1,182

C. 2,364

D. 1,97

nCO= 0,02 mol

nNaOH = 0,006 mol

n Ba(OH)2= 0,012 mol

=> nOH = 0,03 mol

nCO = nOH

EMBED Equation.3 - nCO = 0,03 - 0,02 = 0,01 mol

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

mkt ta = 0,01 . 197 = 1,97 gam

36.Cng thc tnh th tch CO2 cn hp th ht vo mt dung dch Ca(OH)2 hoc Ba(OH)2 thu c mt lng kt ta theo yu cu .

Ta c hai kt qu :

- n CO= nkt ta

- n CO= nOH - nkt taV d : Hp th ht V lt CO2 ( ktc) vo 300 ml dung dch v Ba(OH)2 1 M thu c 19,7 gam kt ta . Tnh V ?

Gii

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

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

37.Cng thc tnh th tch dung dch NaOH cn cho vo dung dch Al3+ xut hin mt lng kt ta theo yu cu .

Ta c hai kt qu :

- n OH = 3.nkt ta

- n OH = 4. nAl - nkt taV 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 - nkt ta = 4. 0,5 0,4 = 1,6 mol => V = 1,6 lt

38.Cng thc tnh th tch dung dch NaOH cn cho vo hn hp dung dch Al3+ v H+ xut hin mt lng kt ta theo yu cu .

Ta c hai kt qu :

- n OH( min ) = 3.nkt ta + nH

- n OH( max ) = 4. nAl - nkt ta+ nH

V d : Cn cho bao nhiu lt dung dch NaOH 1M ln nht vo dung dch cha ng thi 0,6 mol AlCl3 v 0,2 mol HCl c 39 gam kt ta .

Gii

n OH( max ) = 4. nAl - nkt ta+ nH = 4. 0,6 - 0,5 + 0,2 =2,1 mol => V = 2,1 lt

39.Cng thc tnh th tch dung dch HCl cn cho vo dung dch NaAlO2 hoc Na xut hin mt lng kt ta theo yu cu .

Ta c hai kt qu :

- nH = nkt ta

- nH = 4. nAlO - 3. nkt taV d : Cn cho bao nhiu lt dung dch HCl 1M vo dung dch cha 0,7 mol NaAlO2 hoc Na thu c 39 gam kt ta .

Gii

Ta c hai kt qu :

nH = nkt ta = 0,5 mol => V = 0,5 lt

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

40.Cng thc tnh th tch dung dch HCl cn cho vo hn hp dung dch NaOH v NaAlO2 hoc Na xut hin mt lng kt ta theo yu cu .

Ta c hai kt qu :

nH = nkt ta + n OH

nH = 4. nAlO - 3. nkt ta + n OH

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 NaAlO2 hoc Na thu c 15,6 gam kt ta .

Gii

Ta c hai kt qu :

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

41.Cng thc tnh th tch dung dch NaOH cn cho vo hn hp dung dch Zn2+ xut hin mt lng kt ta theo yu cu .

Ta c hai kt qu :

n OH( min ) = 2.nkt ta

n OH( max ) = 4. nZn - 2.nkt taV d : Tnh th tch dung dch NaOH 1M cn cho vo 200 ml dung dch ZnCl2 2M c 29,7 gam kt ta .

Gii

Ta c nZn = 0,4 mol

nkt ta= 0,3 mol

p 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.nkt ta = 4.0,4 2.0,3 = 1 mol =>V ddNaOH = 1lt

42.Cng thc tnh khi lng mui thu c khi cho hn hp st v cc oxt st tc dng vi HNO3 long d gii phng kh NO.

mMui = ( 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 mui v 1,344 lt kh NO ( ktc ) l sn phm kh duy nht . Tm m ?.

Gii

mMui = ( mhn hp + 24 nNO ) = ( 11,36 + 24 .0,06 ) = 38,72 gam

43.Cng thc tnh khi lng mui thu c khi ha tan ht hn hp st v cc oxt st bng HNO3 c nng, d gii phng kh NO2 .

mMui = ( mhn hp + 8 nNO )

V d : Ha tan ht 6 gam cht rn X gm Fe, FeO, Fe2O3, Fe3O4 trong HNO3 c 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 = ( mhn hp + 8 nNO ) = ( 6 + 8 .0,15 ) = 21,78 gam

44.Cng thc tnh khi lng mui thu c khi ha tan ht hn hp st v cc oxt st bng HNO3 d gii phng kh NO v NO2 .

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

V d : Ha tan ht 7 gam cht rn X gm Fe, FeO, Fe2O3, Fe3O4 trong HNO3 d thu c 1,792 lt (ktc ) kh X gm NO v NO2 v m gam mui . Bit dX/H= 19. Tnh m ?

Ta c : nNO = nNO= 0,04 mol

mMui = ( mhn hp + 24 nNO + 8 nNO ) = ( 7+ 24.0,04 + 8.0,04 )= 25,047 gam

45.Cng thc tnh khi lng mui thu c khi ha tan ht hn hp Fe, FeO, Fe2O3, Fe3O4 bng H2SO4 c, nng, d gii phng kh SO2 .

mMui = ( mhn hp + 16.nSO )

V d : Ha tan ht 30 gam cht rn X gm Fe, FeO, Fe2O3, Fe3O4 bng H2SO4 c 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 = ( mhn hp + 16.nSO ) = ( 30 + 16.0,5 ) = 95 gam

46.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 X vi HNO3 long d gii phng kh NO.

mFe = ( mhn hp + 24 nNO )

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

Gii

mFe = ( mhn hp + 24 nNO ) = ( 3 + 0,025 ) = 2,52 gam

47.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 X vi HNO3 c , nng ,d gii phng kh NO2.

mFe = ( mhn hp + 8 nNO )

V d : t m gam st trong oxi thu c 10 gam hn hp cht rn X . Ha tan ht X vi HNO3 c nng, d gii phng 10,08 lt kh NO2 ( ktc) . Tm m ?

Gii

mFe = ( mhn hp + 24 nNO ) = ( 10 + 8. 0,45 ) = 9,52 gam

48.Cng thc tnh pH ca dung dch axit yu HA.

pH = -(logKa + logCa ) hoc pH = - log (Ca )

vi

: l in li

Ka : hng s phn li ca axit

Ca : nng mol/l ca axit ( Ca 0,01 M )

V d 1: Tnh pH ca dung dch CH3COOH 0,1 M 250C . Bit KCHCOOH = 1,8. 10-5

Gii

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

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

Gii

Ta c : CM = =

EMBED Equation.3 = 0,1 M

pH = - log (Ca ) = - log (.0,1 ) = 2,7

49.Cng thc tnh pH ca dung dch baz yu BOH.

pH = 14 + (logKb + logCb )

vi Kb : hng s phn li ca baz

Ca : nng mol/l ca baz

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

pH = 14 + (logKb + logCb ) = 14 + (log1,75. 10-5 + log0,1 ) = 11,1350. Cng thc tnh pH ca dung dch axit yu HA v mui NaA

pH = - (logKa + log)

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

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

pH = - (logKa + log) = - (log1,75. 10-5 + log) = 4,74

51. Cng thc tnh hiu sut phn ng tng hp NH3H% = 2 - 2

vi MX : hn hp gm N2 v H2 ban u ( t l 1:3 )

MY : hn hp sau phn ng

V d : Tin hnh tng hp NH3 t hn hp X gm N2 v H2 c t khi hi so vi H2 l 4,25 thu c hn hp Y c t khi hi so vi H2 l 6,8. Tnh hiu sut tng hp NH3 .

Ta c : nN

: nH

= 1:3

H% = 2 - 2 = 2 - 2 = 75 %

ng

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