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oân thi ñaïi hoïc. Moân: Hoaù Hoïc. GV. NGUYEÃN TAÁN TRUNG (Trung Taâm Luyeän Thi Chaát Löôïng Cao VÓNH VIEÃN). Boå trôï kieán thöùc hoaù Voâ cô - Ñaïi cöông. Baøi 7. CO 2 phaûn öùng Vôùi dung dòch bazô. CaCO 3 . ?. TH1. TH2. Ca(HCO 3 ) 2. ?. Ca(HCO 3 ) 2.

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O n thi a i ho c

oân thi ñaïi hoïc

Moân: Hoaù Hoïc

GV. NGUYEÃN TAÁN TRUNG

(Trung Taâm Luyeän Thi Chaát Löôïng Cao VÓNH VIEÃN)


O n thi a i ho c

Boå trôï kieán thöùc

hoaù Voâ cô - Ñaïi cöông


O n thi a i ho c

Baøi 7

CO2 phaûn öùng

Vôùi dung dòch

bazô


O n thi a i ho c

CaCO3

?

TH1

TH2

Ca(HCO3)2

?

Ca(HCO3)2

CaCO3

CO2 phaûn öùng vôùi dd Ca(OH)2

  • Giaùo khoa

CO2+ Ca(OH)2 CaCO3+ H2O (1)

Sau (1) coøn CO2 thì:

CaCO3 + H2O + CO2 Ca(HCO)3 (2)

Toùm laïi:

CO2 + Ca(OH)2


O n thi a i ho c

nCO2

nCa(OH)2

1

2

  • Trong ñònh löôïng:

  • Phaûn öùng: (1), (2) neân vieát laïi

CO2 + Ca(OH)2 CaCO3  + H2O (1’)

2CO2 + Ca(OH)2 Ca(HCO3)2 (2’)

  • Baûng toùm taét saûn phaåm:

(1’), (2’)  baûng TTSP:

Ca(HCO3)2

Saûn phaåm

CaCO3

Ca(HCO3)2

Ca(OH)2 dö

CaCO3

CaCO3

Ca(HCO3)2

CO2 dö


O n thi a i ho c

  • CO2 phaûn öùng vôùi dd Ca(OH)2

  • Giaùo khoa

CO2+ Ca(OH)2 CaCO3+ H2O (1)

Sau (1) coøn CO2 thì:

CaCO3 + H2O + CO2 Ca(HCO)3 (2)

Toùm laïi:

CaCO3

CO2 + Ca(OH)2

?

TH1

TH2

Ca(HCO3)2

?

Ca(HCO3)2

CaCO3


O n thi a i ho c

  • Trong ñònh löôïng:

  • Phaûn öùng: (1), (2) neân vieát laïi

CO2 + Ca(OH)2 CaCO3  + H2O (1’)

2CO2 + Ca(OH)2 Ca(HCO3)2 (2’)

  • Baûng toùm taét saûn phaåm:

(1’), (2’)  baûng TTSP:

nCO2

nCa(OH)2

TH2

TH1

1

2

Ca(HCO3)2

Saûn phaåm

CaCO3

Ca(HCO3)2

Ca(OH)2 dö

CaCO3

CaCO3

Ca(HCO3)2

CO2 dö


O n thi a i ho c

nmax = nCO2 hñ

nCO2

nCa(OH)2

TH1

TH2

  • Ñöôøng bieåu dieãn löôïng keát tuûa

löôïng

m=ñeà

2

1


O n thi a i ho c

Khi cho CO2 vaøo dd Ca(OH)2 thaáy coù , suy ra baøi toaùn coù hai tröôøng hôïp:

TH1: Baøi toaùn chæ coù phaûn öùng

CO2 + Ca(OH)2 CaCO3 + H2O

TH2:Baøi toaùn goàm caùc phaûn öùng

CO2 + Ca(OH)2 CaCO3 + H2O (1)

CaCO3+ CO2+ H2O Ca(HCO3)2 (2)

Hoaëc:

CO2 + Ca(OH)2 CaCO3 + H2O (1’)

2CO2 + Ca(OH)2 Ca(HCO3)2 (2’)


O n thi a i ho c

300ml

ddCa(OH)2 0,1M

2 caùch giaûi

CO2

1,12lit

(ñkc)

  • Aùp duïng 1:

Cho 1,12 lít CO2 (ÑKC) vaøo bình chöùa 300 ml dd Ca(OH)2 0,1M. Tính khoái löôïng keát tuûa thu ñöôïc

1,12 lít CO2 (ÑKC)

300 ml dd Ca(OH)2 0,1M

Khoái löôïng Keát tuûa:?


O n thi a i ho c

300. 0,1

= 0,03 (mol)

1000

1,12

nCO2 bñ =

= 0,05 mol

22,4

  • Giaûi:

nCa(OH)2 bñ =

Caùch 1:Giaûi baèng phöông phaùp 3 doøng


O n thi a i ho c

(mol)

Bñ:

Bñ:

0,03

0,05

0

Pöù:

Pöù:

O,02

O,03

Sau:

Sau:

0,02

0,03

m

= 1 g

CaCO3

- Theo ñeà ta coù pöù:

Ca(OH)2 + CO2 

CaCO3+ H2O (1)

(mol)

0,03

0,03

(mol)

0

0,02

0,03

CO2 +CaCO3 +H2O  Ca(HCO3)2 (2)

(mol)

0,02

0,01

(mol)

0


O n thi a i ho c

Ca(HCO3)2

Saûn phaåm

CaCO3

Ca(HCO3)2

nCO2

nCO2

Ca(OH)2 dö

CaCO3

CaCO3

Ca(HCO3)2

CO2 dö

nCa(OH)2

nCa(OH)2

1

2

nCa(OH)2=0,03

0,05

= 1,67

=

nCO2 = 0,05

0,03

Caùch 2:Döïa vaøo baûng toùm taét saûn phaåm

  • Ta coù baûng toùm taét saûn phaåm:

  • Theo ñeà ta coù:

Neân baøi toaùn coù 2 phaûn öùng sau:


O n thi a i ho c

CO2 +Ca(OH)2 

CaCO3+ H2O (1’)

2CO2 + Ca(OH)2 Ca(HCO3)2 (2’)

m

= 1 g

CaCO3

Deã daøng tính ñöôïc:


O n thi a i ho c

NaHCO3

?

Na2CO3

?

Na2CO3

NaHCO3

CO2 phaûn öùng vôùi dd NaOH (hay KOH)

  • Giaùo khoa

CO2+ NaOH NaHCO3(1)

Sau (1) coøn NaOHthì:

NaHCO3 + NaOH  Na2CO3 + H2O(2)

Toùm laïi:

CO2 + NaOH


O n thi a i ho c

nNaOH

nCO2

1

2

  • Trong ñònh löôïng:

  • Phaûn öùng: (1), (2) neân vieát laïi

CO2 + NaOH NaHCO3(1’)

CO2 + 2NaOH  Na2CO3 + H2O (2’)

  • Baûng toùm taét saûn phaåm:

(1’), (2’)  baûng TTSP:

NaHCO3

Saûn phaåm

NaHCO3

Na2CO3

CO2 dö

NaHCO3

Na2CO3

Na2CO3

NaOH (dö)


O n thi a i ho c

300ml

ddNaOH 0,2M

2 caùch giaûi

CO2

1,12lit

(ñkc)

  • Aùp duïng 2:

Cho 1,12 lít CO2 (ÑKC) vaøo bình chöùa 300 ml dd NaOH 0,2M. Tính khoái löôïng muoái thu ñöôïc

1,12 lít CO2 (ÑKC)

300 ml dd NaOH 0,2M

Khoái löôïng muoái:?


O n thi a i ho c

300. 0,2

= 0,06 (mol)

1000

1,12

nCO2 bñ =

= 0,05 mol

22,4

  • Giaûi:

nNaOH bñ =

Caùch 1:Giaûi baèng phöông phaùp 3 doøng


O n thi a i ho c

(mol)

Bñ:

Bñ:

0,05

0,06

0

Pöù:

Pöù:

O,01

O,01

O,05

Sau:

Sau:

0,01

0,05

- Theo ñeà ta coù pöù:

CO2 + NaOH NaHCO3 (1)

(mol)

0,05

0,05

(mol)

0

0,01

0,05

NaOH + NaHCO3 Na2CO3 + H2O (2)

0,01

(mol)

0,04

0,01

0

(mol)

(1),(2)  m Muoái =0,04.84 +0,01.106 =4,42 gam


O n thi a i ho c

NaHCO3

Saûn phaåm

NaHCO3

Na2CO3

nNaOH

nNaOH

CO2 dö

NaHCO3

Na2CO3

Na2CO3

NaOHdö

nCO2

nCO2

1

2

nNaOH=0,06

0,06

= 1,2

=

nCO2 = 0,05

0,05

Caùch 2:Döïa vaøo baûng toùm taét saûn phaåm

  • Ta coù baûng toùm taét saûn phaåm:

  • Theo ñeà ta coù:

Neân baøi toaùn coù 2 phaûn öùng sau:


O n thi a i ho c

nNaOH=x+2y =0,06

x = 0,04; y = 0,01

m Muoái= 0,04.84 +0,01.106 =4,42 gam

nCO2 =x +y = 0,05

Caùc phaûn öùng:

CO2 + NaOH NaHCO3(1’)

x

x

x

CO2 + 2NaOH  Na2CO3 + H2O (2’)

y

2y

y

Theo (1’), (2’) ,ñeà coù:


O n thi a i ho c

  • Aùp duïngï 3:

    (Trích ñeà ÑH Sö phaïm TP HCM-2001)

Cho V lít khí CO2 ño ôû 54,6oc vaø 2,4 atm haáp thuï hoaøn toaøn vaøo 200 ml dd hh KOH 1M vaø Ba(OH)2 0,75M thu ñöïôc 23,64 g keát tuûa. Tìm V lít?

CO2 ño ôû 54,6oc vaø 2,4 atm

200 ml dd hh KOH

1M vaø Ba(OH)2 0,75M

V = 1,344 (l) ; 4,256 (l)

GV. NGUYEÃN TAÁN TRUNG

(Trung Taâm Luyeän Thi Chaát Löôïng Cao VÓNH VIEÃN)


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