cong thuc giai nhanh hoa hoc

Hy s huTuyn tp100 thi th C-H gii chi tit v 6 tp cha kha vng gii nhanh ha hc.

Thc s: Nguyn Vn Ph: T 098 92 92 117. Email: [email protected]

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MT S CNG THC GII NHANH TRC NGHIM HA HC

C TNG HP BI Thc s : Nguyn Vn Ph: 098.92.92.117 Facebook: trieu phu nguyen

Vic nm vng cc cng thc ny s gip gii nhanh cc bi ton .Nu gii theo cch thng thng th

mt rt nhiu thi gian.Vy hy hc thuc nh.

1. Cng thc tnh s ng phn ancol n chc no, mch h : Cn H2n+2O

S ng phn Cn H2n+2O = 2n- 2

( 1 < n < 6 )

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

a. C3H8O = 23-2 = 2 ( 1 bc 1+ 1 bc 2+0 bc 3)

b. C4H10O = 24-2 = 4 ( 2 bc 1+ 1 bc 2+1 bc 3)

c. C5H12O = 25-2 = 8 ( 4 bc 1+ 3 bc 2+1 bc 3)

d. C6H14O = 26-2 + 1= 17 ( 8 bc 1+ 6 bc 2+3 bc 3)

S ng phn ca dn xut cng tng t nh ancol n chc c cng thc phn t l :

a. C3H7X = 23-2 = 2 ( 1 bc 1+ 1 bc 2+0 bc 3)

b. C4H9X = 24-2 = 4 ( 2 bc 1+ 1 bc 2+1 bc 3)

c. C5H11X = 25-2 = 8 ( 4 bc 1+ 3 bc 2+1 bc 3)

d. C6H13X = 26-2 + 1= 17 ( 8 bc 1+ 6 bc 2+3 bc 3) trong X l : Cl, Br, I, OH, ....

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 ( 1 ng phn este tham gia phn ng trng gng)

b. C3H6O2 = 23-2 = 2 ( 1 ng phn este tham gia phn ng trng gng)

c. C4H8O2 = 24-2 = 4 ( 2 ng phn este tham gia phn ng trng gng)

d. C5H10O2 = 25-2 +1= 9( 4 ng phn este tham gia phn ng trng gng)

CH : S ng phn ca este n chc khng no(mt ni i), mch h c cng thc phn t l :



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a. C3H4O2 = 1 ( 1 ng phn este tham gia phn ng trng gng)

b. C4H6O2 = 5 ( 3 ng phn este tham gia phn ng trng gng)

c. C5H8O2 = 16 ( 8 ng phn este tham gia phn ng trng gng)

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

S ng phn Cn H2n+2O = 2

)2).(1( nn ( 2 < n < 5 )

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

a. C3H8O = 2

)23).(13( = 1

b. C4H10O = 2

)24).(14( = 3

c. C5H12O = 2

)25).(15( = 6

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

S ng phn Cn H2nO = 2

)3).(2( nn ( 3 < n < 7 )

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

a. C4H8O = 2

)34).(24( = 1

b. C5H10O = 2

)35).(25( = 3

c. C6H12O = 2

)36).(26( = 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

= 2. ( 1 bc 1+ 1 bc 2+0 bc 3)

b. C3H9N = 23-1

= 4 ( 2 bc 1+ 1 bc 2+1 bc 3)

c. C4H11N = 24-1

= 8 ( 4 bc 1+ 3 bc 2+1 bc 3)

d. C5H13N =17 ( 8 bc 1+ 6 bc 2+3 bc 3)

e. C6H15N = 39 ( 17 bc 1+ 15 bc 2+7 bc 3)

8. Cng thc tnh s trieste ( triglixerit ) to bi glixerol v hn hp n axt bo : S tri este = 2

)1(2 nn

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 = 2

)12(22 = 6



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9. Cng thc tnh s ng phn ete to bi hn hp n ancol n chc : S ete = 2

)1( nn

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

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 = 22

2

COOH

CO

nn

n

( Vi nH 2 O > n CO 2 )

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 t ca A ?( Vi nH 2 O = 0,7 mol > n CO 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

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 = mH 2 O - 112CO

m hoc

2CO

ancol H2O

Vm = m -

5,6

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 = mH 2 O - 112CO

m = 7,2 -

11

4,4 = 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 m

ab

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 = 751

3,05,0 = 15 gam



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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 n

ab

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 1

375,0575,0 = 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 ctNi

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

S n ca anken (CnH2n ) = )(14

)2(

12

12

MM

MM

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 = )105,12(14

10)25,12(

= 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 ctNi

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

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

)2(2

12

12

MM

MM

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

Mx

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

Mx

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

A

M

M - 1

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

hhX MV

V



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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 2

V d : Cho 8 gam hn hp kim loi gm Mg, Fe tc dng vi dung dch HCl thu c 4,48 lt kh H2 ( ktc).

Tnh khi lng mui thu c . mMui clorua = mKL + 71 nH 2 = 8 + 71. 0,2 = 22,2 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 2

V d : Cho 8,0 gam hn hp kim loi gm Mg, Fe tc dng vi dung dch H2SO4 long thu c 4,48 lt kh

H2 ( ktc). Tnh khi lng mui thu c . mMui Sunfat = mKL + 96. nH 2 = 8 + 96. 0,2 = 27.2 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 + 2

96.( 2nSO 2 + 6 nS + 8nH 2 S ) = mKL +96.( nSO 2 + 3 nS + 4nH 2 S )

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

* n H 2 SO 4 = 2nSO 2 + 4 nS + 5nH 2 S

24.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 2 + 3nNO + 8nN 2 O +10n N 2 +8n NH 4 NO 3 )

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

* n HNO 3 = 2nNO 2 + 4 nNO + 10nN 2 O +12nN 2 + 10nNH 4 NO 3

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 2

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 2

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 2

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 SO2

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

nO (Oxit) = nO ( H 2 O) = 2

1nH ( 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 H 2 SO 4



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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 H 2 O = mOxit + 27,5 n HCl

32.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 2 = n CO 2 = n H 2 O

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= a

2nH 2 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 2 = 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 2 ( vi nkt ta nCO 2 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 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 )

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 23 = nOH - nCO 2 ri so snh nCa

2 hoc nBa 2 xem

cht no phn ng ht suy ra n kt ta ( iu kin nCO 23 nCO 2 )

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 2 = 0,3 mol nNaOH = 0,03 mol n Ba(OH)2= 0,18 mol => nOH = 0,39 mol

nCO 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 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 2 = 0,02 mol nNaOH = 0,006 mol n Ba(OH)2= 0,012 mol => nOH = 0,03 mol

nCO 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

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 2 = nkt ta

- n CO 2 = nOH - nkt ta



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

- n CO 2 = nOH - nkt ta = 0,6 0,1 = 0,5 => V CO 2 = 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 3 - nkt ta

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 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 3 - 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 3 - 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 4)(OHAl xut

hin mt lng kt ta theo yu cu .Ta c hai kt qu :

- nH = nkt ta

- nH = 4. nAlO 2 - 3. nkt ta

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 39 gam kt ta .

Gii Ta c hai kt qu : nH = nkt ta = 0,5 mol => V = 0,5 lt

nH = 4. nAlO 2 - 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 4)(OHAl xut hin mt lng kt ta theo yu cu .

Ta c hai kt qu : nH = nkt ta + n OH

nH = 4. nAlO 2 - 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 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



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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 - 2.nkt ta

V 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 2 = 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 - 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 = 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 mui v 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

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 = 80

242( mhn hp + 8 nNO 2 )

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 = 80

242( mhn hp + 8 nNO 2 ) = 80

242( 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 = 80

242( mhn hp + 24. nNO + 8. nNO 2 )

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 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

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 = 160

400( mhn hp + 16.nSO 2 )

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.



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Gii mMui = 160

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

400( 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 = 80

56( 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 = 80

56( mhn hp + 24 nNO ) =

80

56( 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 = 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 HNO3 c nng, d

gii phng 10,08 lt kh NO2 ( ktc) . Tm m ?

GiimFe = 80

56( mhn hp + 24 nNO 2 ) = 80

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

48.Cng thc tnh pH ca dung dch axit yu HA. pH = -2

1(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 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 dch HCOOH 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

49.Cng thc tnh pH ca dung dch baz yu BOH. pH = 14 + 2

1(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 3 = 1,75. 10-5

pH = 14 + 2

1(logKb + logCb ) = 14 +

2

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

50. Cng thc tnh pH ca dung dch axit yu HA v mui NaA pH = - (logKa + logm

a

C

C)

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



10

51. Cng thc tnh hiu sut phn ng tng hp NH3 H% = 2 - 2Y

X

M

M

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 2 : nH 2 = 1:3 H% = 2 - 2Y

X

M

M = 2 - 2

6,13

5,8 = 75 %

52. Gp bi ton: Nhng mt thanh kim loi A ha tr a ( khng tan trong nc) nng m1 gam vo V lt dung dch B (NO3)b xM. Sau mt thi gan ly thanh A ra v cn nng m2 gam. Nu bi ton cn tnh khi lng m gam kim loi B thot ra th ta p dng nhanh cng thc:

2 1B B

B A

m mm a.M .

a.M b.M

53. Gp bi ton: Cho n mol( hoc V lt.) oxit axit CO2 ( SO2) tc dng vi dung dch Ca(OH)2, (Ba(OH)2 ) thu c a mol kt ta, sau un nng dung dch li thu c b mol kt ta na th ta ch cn p dng nhanh cng thc sau:

2COn a 2.b (*)

54. Gp bi ton: Nung m gam Fe trong khng kh, sau mt thi gian ta thu c a gam hn hp cht rn X gm Fe, Fe2O3, Fe3O4, FeO. Ho tan ht a gam hn hp cht rn X vo dung dch HNO3 d thu c V lt kh NO2 (ktc) l sn phm kh duy nht v dung dch mui sau khi lm khan c b gam . Nu bi ton cn tnh mt trong cc gi tr m, a, b, V th ta p dng nhanh cc cng thc i y. a. Trng hp 1: tnh khi lng st ban u trc khi b xi ha thnh m gam hn hp X gm: Fe, FeO, Fe2O3 v Fe3O4 .

eFe

7.a 56.nm (**)

10

, trong e

Vn mol

22,4 .

+ Nu sn phm kh l NO th eV

n 3. mol22,4

.

+ Nu sn phm kh l N2O th eV

n 8. mol22,4

.

+ Nu sn phm kh l N2 th eV

n 10. mol22,4

b. Trng hp 2: tnh khi lng a gam hn hp X gm: Fe, FeO, Fe2O3 v Fe3O4 .

Fe ehh10.m 56.n

a (2)7

trong ne cng tng t nh trn.

c. Trng hp 3: tnh khi lng b gam mui to thnh khi cho a gam hn hp X gm: Fe, FeO, Fe2O3 v Fe3O4 vo dung dch HNO3 nng d.

3 3 3 3

FeFe(NO ) Fe , Fe(NO )

mn n ymol b m 242.y gam(3)

56

d. Trng hp 4: tnh khi lng mui to thnh khi cho m gam hn hp X gm: Fe, FeO, Fe2O3 v Fe3O4 vo dung dch H2SO4 c, nng d.

2 4 3 2 4 3

FeFe (SO ) Fe Fe (SO )

m1n .n x mol ,m 400.x gam(4)

2 112

cong thuc giai nhanh hoa hoc

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