bai-tap-nhieu-cach-giai3

8/6/2019 bai-tap-nhieu-cach-giai3

1/15


2/15

3Fe3O4 + 28HNO3 9Fe(NO3)3 + NO

+ 14H2O

Fe2O3 + 6HNO3 2Fe(NO3)3 + 3H2O

S mol: 1,04,22

24,2y3

1x3

1tnNO ==++= 3t + x + y = 0,3 (II)

T (I), (II) 10x + 30y + 20z + 10t = 1,8 x + 3y + 2z + t = 0,18

Trong m gam Fe c s mol: 18,0tz2y3xnFe =+++=

m = 56.0,18 = 10,08 (gam).

Cch 2:

Theo nh lut BTKL ta c: BO mmm =+

m12mO = )m12.(0625,0nO =

Qu trnh nhng e Qu trnh nhn e

Fe 3e Fe3+

O + 2e O2-

56

m

56

m.3 0,0625.(12 m) 0,125.(12 m)

4H+

+ 3NO + 3e NO + 2H2O

3.0,1 0,1

56

m.3E = )m12.(125,03,0E +=+

Bo ton s mol e ta c: + = EE )m12.(125,03,056

m.3 +=

m = 10,08 (gam).

Cch 3:

Theo nh lut BTKL ta c: OHNO)NO(FeHNOB 233)p(3 mmmmm ++=+

12 + OH)NO(FeHNO 233)p(3 n.181,0.30n.242n.63 ++=


3/15

Trong :56

mnn Fe)NO(Fe 33 == ; 1,056

m.3nn.3n NO)NO(FeHNO 33)p(3 +=+=

)1,056

m

.3.(2

1

n.2

1

n )p(32 HNOOH+==

Suy ra: 12 + 63.(3.56

m+ 0,1) = 242.

56

m+ 30.0,1 + 18.

2

1(3.

56

m+ 0,1)

m = 10,08 (gam).

Cch 4:

Coi Fe3O4 FeO.Fe2O3 B gm {Fe ; FeO ; Fe2O3 }

t s mol cc cht trong 12 gam B { Fed

: x ; FeO: y ; Fe2O

3: z }

56x + 72y + 160z = 12 7x + 9y + 20z =1,5 (III)

B + HNO3: Xy ra cc phn ng

Fe + 4HNO3 Fe(NO3)3 + NO

+ 2H2O

3FeO + 10HNO3 3Fe(NO3)3 + NO + 5H2O


+14H2O

Fe2O

3+ 6HNO

3 2Fe(NO

3)

3+ 3H

2O

S mol: 1,0y3

1xnNO =+= 3x + y = 0,3 (IV)

T (III), (IV) 10x + 10y + 20z = 1,8 x + y + 2z = 0,18

Trong m gam Fe c s mol: Fen = x + y + 2z = 0,18

m = 56.0,18 = 10,08 (gam).

Cch 5:

V B ch gm cc nguyn t Fe v Oxi.

t s mol cc cht trong B { Fe: x ; O: y }

56x + 16y = 12 (VII)



4/15

Fe 3e Fe3+ O + 2e O2-

x 3x y 2y

4H

+

+

3NO + 3e

NO + 2H2O3.0,1 0,1

x3E = y23,0E +=+

Bo ton s mol e ta c: + = EE 3x = 2y + 0,3 (VIII)

T (VII), (VIII) x = 0,18 ; y = 0,12

Trong m gam Fe c: 18,0xnFe == m = 56.0,18 = 10,08 (gam).

Cch 6:

t cng thc chung ca 3 oxit l: mnOFe

Trong 12 gam B gm { Fed: x ; mnOFe : y }

56x + (56n + 16m)y = 12 7x + (7n + 2m)y = 1,5 (V)


Fe + 4HNO3 Fe(NO3)3 + NO+ 2H2O

3 mnOFe +(12n-2m)HNO3 3nFe(NO3)3 + (3n-2m)NO+ (6n m)H2O

S mol: 1,0y).m2n3.(3

1xnNO =+= 3x + (3n 2m)y = 0,3 (VI)

T (V), (VI) 10x + 10ny = 1,8 x+ ny = 0,18

Trong m gam Fe c: Fen = x + ny = 0,18

m = 56.0,18 = 10,08 (gam).

Cch 7:

Gi s lng Fe tham gia phn ng vi O2 ch to ra Fe2O3.

Theo nh lut BTKL ta c: BO mmm 2 =+


5/15

m12m2O

= 32

m12n

2O

=

T phn ng: Fe + 3O2 2Fe2O3

3.32

)m12.(4

32

m12

Cht rn B gm: { Fe ; Fe2O3 }



+ 2H2O

0,1 0,1


Bo ton nguyn t: )p(Fe)B(Fe)A(Fe nnn += 32.3

)m12.(41,0

56

m +=

m = 10,08 ( gam ).

Cch 8:

Gi s tt c Fe u phn ng vi O2 to ra Fe2O3.

4Fe + 3O2 2Fe2O3

Khi lng Fe2O3 ti a l:7

m10

2.56

m.160m

32OFe==

Khi lng O2 thiu l: )127

m10(m

2O= )12

7

m10.(

32

1n )Thiu(O2 =

Ta phi c s mol e do O2 thiu nhn phi bng s mol e do N+5

nhn to ra NO.

1,0.3)12

7

m10.(

32

1.4 = m = 10,08 ( gam ).

Bi 2: Cho hn hp A gm 3 oxit st Fe2O3, Fe3O4, FeO vi s mol bng nhau. Ly m ga

A cho vo ng s chu nhit, nung nng ri cho lung kh CO i qua. Cht rn B cn li trong n


6/15

s sau phn ng c khi lng l 19,20 gam gm Fe, FeO, Fe3O4. Cho B tc dng ht vi dun

dch HNO3, un nng c 2,24 lt kh NO (ktc) duy nht. Tnh m.

Li gii

S mol:4,22

24,2nNO = = 0,1 mol.

Cch 1:

t s mol cc cht trong m gam A { Fe2O3: x ; Fe3O4: x ; FeO: x }

m = 160x + 232x + 72x = 464x

A + CO: Xy ra cc phn ng

3Fe2O3 + CO 2Fe3O4 + CO2

Fe3O4 + CO 3FeO + CO2

FeO + CO Fe + CO2

t s mol cc cht trong 19,2 gam cht rn B { Fe: y ; FeO: z ; Fe3O4: t }

56y + 72z + 232t = 19,2 7y + 9z + 29t = 2,4 (I)


Fe + 4HNO3 Fe(NO3)3 + NO + 2H2O

3FeO + 10HNO3 3Fe(NO3)3 + NO

+ 5H2O

3Fe3O4 + 28HNO3 9Fe(NO3)3 + NO +14H2O

S mol: NOn = y +3

1z +

3

1t = 0,1 3y + z + t = 0,3 (II)

T (I), (II) 10y + 10z + 30t = 1,8 y + z +3t = 0,27

Ta c: )B(Fe)A(Fe nn = 6x = y + z + 3t x = 0,045.

Vy m = 464x = 464.0,045 = 20,88 (gam).

Cch 2:



7/15

m = 160x + 232x + 72x = 464x

Theo nh lut BTKL ta c:2COB)p(COA

mmmm +=+

464x + )p(COn28 = 19,2 + )p(COn44 )p(COn = 29x 1,2


Fe2+

1e Fe3+

4H+

+ 3NO

+ 3e NO + 2H2O

x x 3.0,1 0,1

3Fe8/3+

- 1e = Fe3+

3x x

C2+ - 2e = C4+

(29x-1,2) 2(29x-1,2)

E- = x + x + 2(29x-1,2) E+ = 0,3

Bo ton s mol e ta c E - = E+ x + x + 2(29x 1,2) = 0,3 x = 0,045

m = 464x = 464.0,045 = 20,88 (gam).

Cch 3:

t s mol cc cht trong m gam A {Fe2O3: x ; Fe3O4: x ; FeO: x }Theo nh lut BTKL ta c: OHNO)NO(FeHNOB 233)p(3 mmmmm ++=+

19,2 + OH)NO(FeHNO 233)p(3 n.181,0.30n.242n.63 ++=

Trong : x6nn )A(Fe)NO(Fe 33 ==

1,0x6.3nn.3n NO)NO(FeHNO 33)p(3 +=+=

)1,0x6.3.(2

1n.2

1n )p(32 HNOOH +==

Suy ra: 19,2 + 63.(18x + 0,1) = 242.6x + 30.0,1 + 18.2

1(18x + 0,1)

x = 0,045 m = 464.x = 44.0,045 = 20,88 (gam).


8/15

Cch 4:


Coi FeO.Fe2O3 Fe3O4 A ch c Fe3O4: 2x m = 2.232x = 464x


56y + 72z + 232t = 19,2 7y + 9z + 29t = 2,4 (III)



3FeO + 10HNO3 3Fe(NO3)3 + NO + 5H2O


+14H2O

S mol: NOn = y +3

1z +

3

1t = 0,1 3y + z + t = 0,3 (IV)

T (III), (IV) 10y + 10z + 30t = 1,8 y + z +3t = 0,27


Vy m = 464x = 464.0,045 = 20,88 (gam).

Cch 5:

t s mol cc cht trong m gam A {Fe2O3: x ; Fe3O4: x ; FeO: x }

m = 160x + 232x + 72x = 464x

Coi Fe3O4 FeO.Fe2O3 B gm { Fe, FeO, Fe2O3 }


56y + 72z + 160t = 19,2 7y + 9z + 20t = 2,4 (V)




+ 5H2O



9/15

S mol: NOn = y +3

1z = 0,1 3y + z = 0,3 (VI)

T (V), (VI) 10y + 10z + 20t = 1,8 y + z +2t = 0,27


Vy m = 464x = 464.0,045 = 20,88 (gam).

Cch 6:

t s mol cc cht trong m gam A {Fe2O3: x ; Fe3O4: x ; FeO: x }

m = 160x + 232x + 72x = 464x

t cng thc chung ca 2 oxit trong B l FenOm

Trong 19,2 gam B gm { Fed: y ; FenOm: z }

56y + (56n + 16m)z = 12 7y + (7n + 2m)z = 1,5 (VII)



3FenOm +(12n-2m)HNO3 3nFe(NO3)3 + (3n-2m)NO+ (6n m)H2O

S mol: nNO

= y +3

1(3n-2m)z = 0,1 3y + (3n-2m)z = 0,3 (VIII)

T (VII), (VIII) 10y + 10nz = 1,8 y+ nz = 0,18

Ta c: )B(Fe)A(Fe nn = 6x = y + nz x = 0,045.

Vy m = 464.0,045 = 20,88 (gam).

Bi 3: Nung m gam Fe trong khng kh, sau mt thi gian thu c 104,800 gam hn h

rn A gm Fe, FeO, Fe2O3, Fe3O4. Ho tan hon ton A trong dung dch HNO3 d, thu c dundch B v 12,096 lt hn hp kh X gm NO v NO2 (ktc) c t khi so vi heli l 10,167. Tnh m

Li gii


10/15

S mol: nX =4,22

096,12= 0,54 mol.

C He/Xd = 10,167 M = 10,167.4 = 40,668.

2NO

NO

n

n=

30668,40

668,4046

=

2

1 18,0nNO = ; 36,0n 2NO =

Cch 1:

T cc phn ng: 2Fe + O2 2FeO

3Fe + 2O2 Fe3O4

4Fe + 3O2 2Fe2O3

Trong 104,8 gam A gm { Fe: x ; FeO: y ; Fe2O3: z ; Fe3O4: t}

56x + 72y + 160z + 232t = 12 7x + 9y + 20z + 29t =13,1 (I)

A + HNO3: Xy ra cc phn ng


+ 2H2O

Fe + 6HNO3 Fe(NO3)3 + 3NO2

+ 3H2O


+ 5H2O

FeO + 4HNO3 Fe(NO3)3 + NO2

+ 2H2O


+14H2O

Fe3O4 + 10HNO3 3Fe(NO3)3 + NO2

+ 5H2O


Bo ton s mol e ta c: 3x + y + t = 3.0,18 + 1.0,36 = 0,9 (II)

T (I), (II) 10x + 10y + 20z + 30t = 14 x + y + 2z +3t = 1,4

Trong m gam Fe c s mol: Fen = x + y + 2z + 3t = 1,4

m = 56.1,4 = 78,4 (gam).

Cch 2:


11/15

Theo nh lut BTKL ta c: OHX)NO(FeHNOA 233)p(3 mmmmm ++=+

104,8 + OH)NO(FeHNO 233)p(3 n.1854,0.668,40n.242n.63 ++=

Trong :56

mnn Fe)NO(Fe 33 ==

54,056

m.3nnn.3n

233)p(3 NONO)NO(FeHNO+=++=

)54,056

m.3.(

2

1n.

2

1n

)p(32 HNOOH+==

104,8 + 63.(3.56

m+ 0,54) = 242.

56

m+ 40,668.0,54 + 18.

2

1(3.

56

m+ 0,54)

m = 78,4 (gam).

Cch 3:

Theo nh lut BTKL ta c:

AO mmm =+ Om = 104,8 - m On = 6,55 0,0625m


Fe 3e Fe3+

O + 2e O2-

56

m3.

56

m(6,55 - 0,0625m) 2(6,55 - 0,0625m)

4H+

+ 3NO

+ 3e NO + 2H2O

3.0,18 0,18

2H+

+ 3NO

+ 1e NO2 + H2O

1.0,36 0,36

E- = 3.56

mE+ = 0,9 + 13,1 0,125m

Bo ton s mol e: E - = E+ 3.56

m= 0,9 + 1,31 0,125m m = 78,4g.


12/15

Cch 4:

Coi Fe3O4 FeO.Fe2O3 A gm {Fe ; FeO ; Fe2O3 }.

t s mol cc cht trong 104,8 gam A {Fed: x ; FeO: y ; Fe2O3: z }

56x + 72y + 160z = 104,8 7x + 9y + 20z =13,1 (V)

A + HNO3: Xy ra cc phn ng


+ 2H2O


+ 3H2O


+ 5H2O

FeO + 4HNO3 Fe(NO3)3 + NO2

+ 2H2O


Bo ton s mol e ta c:

3x + y = 3.0,18 + 1.0,36 3x + y = 0,9 (VI)

T (V), (VI) 10x + 10y + 20z = 14 x+ y + 2z = 1,4

Trong m gam Fe c: Fen = x + y + 2z = 1,4 m = 56.1,4 = 78,4 (gam).

Cch 5:

t cng thc chung ca 2 kh l nNO

C M X = 14 + 16n = 40,668 n = 1,66675

Theo nh lut BTKL ta c:

AO mmm =+ Om = 104,8 - m On = 6,55 0,0625m

A+ HNO3: Xy ra cc phn ng

(5-2n) Fe + (18-6n)HNO3 (5-2n Fe(NO3)3 +

nNO3 + (9-3n)H2O

(5-2n) FeO + (16-6n)HNO3 (5-2n)Fe(NO3)3 +nNO + (8-3n)H2O

(5-2n) Fe3O4 + (46-18n)HNO3 3(5-2n)Fe(NO3)3 +nNO + (23-9n)H2O


13/15



Fe 3e Fe3+ O + 2e O2-

56

m3.

56

m(6,55 - 0,0625m) 2(6,55 - 0,0625m)

(6-2n)H+

+ 3NO + (5-2n) e NOn + (3-n)H2O

(5 - 2n).0,54 0,54

E- = 3.56

mE+ = 2(6,55 0,0625m) + (5 2n).0,54

Bo ton s mol e ta c: E - = E+ 3.56

m= 2(6,55 0,0625m) + (5-2n).0,54

Thay n = 1,6667 m = 78,4g.

Cch 6:

V B ch gm cc nguyn t Fe v Oxi.

t s mol cc cht trong B { Fe: x ; O: y }

56x + 16y = 104,8 (IX)Qu trnh nhng e Qu trnh nhn e

Fe 3e Fe3+

O + 2e O2-

x 3x y 2y

4H+

+ 3NO

+ 3e NO + 2H2O

3.0,18 0,18

2H+

+3NO

+ 1e NO2 + H2O

1.0,36 0,36

E- = 3x E+ = 2y + 0,36 + 3.0,18

Bo ton s mol e ta c E - = E+ 3x = 2y + 0,36 + 0,18.3 (X)


14/15

T (IX), (X) x = 1,4 ; y = 1,65

Trong m gam Fe c: 4,1xnFe == m = 56.1,4 = 78,4 (gam).

Cch 7:

t cng thc chung ca 3 oxit l: mnOFe

A + O2: Xy ra phn ng

2nFe + mO2 2 mnOFe

Trong 12 gam B gm { Fed: x ; mnOFe : y }

56x + (56n + 16m)y = 104,8 7x + (7n + 2m)y = 13,1 (III)



+ 2H2O


+ 3H2O

3 mnOFe + (12n-2m)HNO3 3nFe(NO3)3 + (3n-2m)NO

+ (6n m)H2O

mnOFe + (6n-2m)HNO3 nFe(NO3)3 + (3n-2m)NO2+ (3n m)H2O

Bo ton s mol e ta c:

3x + (3n-2m)y = 3.0,18 + 1.0,36 3x + (3n-2m)y = 0,9 (IV)

T (III), (IV) 10x + 10ny = 14 x+ ny = 1,4

Trong m gam Fe c: Fen = x + ny = 1,4 m = 56.1,4 = 78,4 (gam).

Cch 8:

Gii s lng Fe tham gia phn ng vi O2 ch to ra Fe2O3.

Theo nh lut BTKL ta c: BO mmm 2 =+

m8,104m2O

= 32

m8,104n

2O

=

T phn ng:

4Fe + 3O2 2Fe2O3


15/15

3.32

)m8,104.(4

32

m8,104

Cht rn B gm: { Fe ; Fe2O3 }



+ 2H2O

0,18 0,18


+ 3H2O

0,12 0,36


Bo ton nguyn t: )p(Fe)B(Fe)A(Fe nnn +=

32.3

)m8,104.(412,018,0

56

m ++= m = 78,4 ( gam ).

Cch 9:

Gi s ton b Fe u phn ng vi O2 to ra Fe2O3.

4Fe + 3O2 2Fe2O3

Khi lng Fe2O3 ti a l:7m10

2.56m.160m

32OFe==

S mol O2 thiu l: )8,1047

m10.(

32

1n

2O=

Ta phi c s mol e do O2 thiu nhn phi bng s mol e do N+5

nhn to ra NO, NO2.

36,0.118,0.3)8,1047

m10.(

32

1.4 += m = 78,4 ( gam ).

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