2020届福建高三化学大题练-化工流程题

SiF

I____________

______3+________________________mm______

______

______A B

XCl

XX

______

Cl

X

______

KCl

Clb

2+K+

______

L

L

L

______

L

______

______

K

LK

L SiF

eq

\o\ac(△,?)____________

____________

c(SO4

n(KOH)

c(SO4

n(KOH)2−3c(CO23KKL

KF

+______cF

KF______4. I

II__________________________________________X_____________________AlCl__________I

________________

II___________________________ gggeq

\o\ac(△,）；)

eq

\o\ac(△,H)

______

______

______

______

______

=H+

+______

______

______

(SO43−xm2

BBB

n(OH3n(Fe×n______

______eq

\o\ac(□,2) eq

\o\ac(□,2)Feeq

\o\ac(□,3)

Fe4eq

\o\ac(□,2) eq

\o\ac(□,2)______m

gKF

3+

3+

V

V

n

B=______

g/100g

______

______

I

______

______

______

______

______

______

3+

______

______

AlAlAlAlF

______

I______）将

AlCl

F

AlCl

F

______F______

______K[AlAl++H

L

Al3+=________________________bAld MgB

gBClIB

______Mg

X

______

______IIE

______

B

______

E

______

__________________A____________

______

ClClCl

Cl______

________________________

+H

______4+______

______

A____________

__________________//g

______

3+

3+ 3+ +3O A

45m269m1

I

I

3+

3+3+3+3+

3+3+

+3O

A+3OA45m245m2

mmm2138m1

69m1

45m269m1

×I

3+3+II

3+

3+++3O+Cl+H++ClKaqaq【解析】解：（++AAB

B

X

X

Cl

Cl+H+Cl+H+nCl=n=a

n=n=b

Cl

nKCl=

K+=(a−2b)mol=10

L0.1LL

L+Cl=10KL

KK

aqaq+ClKaqaqI

X

X

Cl

nnCl=n=amoln=n=bmol

Cl

nKCl=

c=VnKKKSiF K SiF

SiF KF = +3O【解析】解：（SiFKSiFKc(CO

Ksp(BaCO

c(CO

Ksp(BaCO

3

SiF

SiFSiF

SiFn[Ba(OH)

KFn(KOH)KFKK

ksp(BaSO4

c(SO4 ksp(BaSO4

−9

L

KF

++cK+=c+cFK+=c+cF

+=cF=

KF+3O+3OSiFSiF

KKF

SiF

KFKF Al Al IXIIYI

II

AlAlXIAlII

Al

Cl4+

KCl

4+

eq

\o\ac(△,V)

V V=142×14 3mw

3mw

K

3+2++

2+

3+2++3+

0.056c(V0−V)3+2++3+2++2+

3+2++3+

n=

mw

n=

mw n(OH0.056c(V−V) 3n(Fe3+×2+

3+2++3+4xm+xx

nVV −3c(V−V)56 3×mw 3mw560.056c(V−V)3mwIIII+2+3+4xm+x+2+2+

+2++2+

2+AlAlClFAlFClAl3++3eAlabdAlAB

AlAlFAlAl

IAlClAlFCl

L

Al L

Al+H++H

+=

1.0×10−13=10pH=7=10

K[Al

Al

= 3+ 1.3×10 Mg

B

B

̲̲̲̲̲

̲̲̲̲̲

(10−73

LAlClAlFClAl+3+5

+3N1.0×10−6−33

Al3++3eAlAl3++3eAlbbddabdAlA

BAlAlF

Al+H++HLK[Albd̲̲̲̲̲̲B̲̲̲̲̲̲BClB

dIMgBB

BI

A

B

MgB

B

MgB3+BB̲̲

BCl

A

B

gBBBClBClBClEBClE

BClEBClBClClCl

LClL

+

BBB

g

B

gB

BClE

BClE

BClddIMgBB

B

A

BMgB

B

MgB3+BCl

A

B

gBBBClBClBClEBClE

BCl+32+++H3+

3+

2+4+4++H+H++3+32+++H3+

2+++H3+

3+

2+

2+

3+

2+53.3

Cl

4+4++H+H+4+4++H+H+2+Cl2+3+A

3+

2+2+

ClL

+2+ e