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12

 
20-06-2008   #1
 
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12

 

1-


(1)
:
(m)
NVMr=mf
(p) :
NV2Mr=V1fP
(γ )p :
100NV2Mr=V1fpγ

( : V1 ml V2L


(2)

: ( (N1V1= N2V2( )


(3)

:
( (N1V1= N2V2( )
m1f1\Mr1=m2f2\Mr2

(4)
:

= ()+ ()
(1) ( )
(2) ( )
:
=NV( (NV+ NV ( )
mf\Mr+mf\Mr=m2f2\Mr2

(5)
(m0)= (m)+
( ) = m\m0100%
= \m0
100%
.

( 6)
: =
:
( (N1V1= N2V2( )
m1f1\Mr1=m2f2\Mr2

(7)
=

:
( (N1V1= N2V2( )
m1f1\Mr1=m2f2\Mr2


**************************
2-

(1)
1) (0.25N)( (800) (10.6),
Ϳ
NVMr=mf
Mr=mf\NV
=10.6*2\0.25*0.8=106


2) (0.098\)
(50ml) (500ml) (0.2N (
ֿ

NV2Mr=V1fP
Mr=V1fP\NV2
=50*2*0.098\0.5*0.2=98


3) (1.2g\ml)
(17.5%) , (500ml)
(0.1N)
100NV2Mr=V1fpγ
100*0.1*0.5*63=V1*1*1.2*17.5
315=V1*12
V1=315\12
V1=15ml

 
 
 
 

 
20-06-2008   #2
 
   : 1884
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(2)
1) (100ml) , (20ml) (200ml) (0.3N)
(60)
(2)

( (N1V1= N2V2( )
N1*0.02=0.3*o.2
N1=3N
m=Mr\f*V*N=60\2*100\1000*3=9g

2) (5.6g\ml) , (10ml) , .


= \
=
m=5.6*10=56g
m=Mr\f*V*N
N=mf\MrV
=56*1\56*1=1N



N



N=mf\MrV
56*1\56*0.01
=100N
( (N1V1= N2V2( )
100*0.01=N2*1
N2=1N

****************************
(3)

1) (50ml) Na2CO3
(0.1M) HCl (0.1M) . :
Ý) HCl ( )
HCl
N1V1(HCl)=N2V2(Na2CO3)
f1M1V1=f2M2V2
1*0.1*V1=2*0.1*50
V1=100ml
) HCl .
m=CVN
C
=Mr\f*V*N
36.5\1*100\1000*0.1
=0.365g

2) ((5.3g Na2CO3
(500ml)
(50ml) (30ml) HCl .
Ý) Na2CO3
m=MVMr
M=m\VMr
=5.3\0.5*106
=0.1M
ȝ) HCl
N1V1(HCl)=N2V2(Na2CO3)
f1M1V1=f2M2V2
1*M1*0.03=2*0.1*0.05
M1=0.3M

3) (10ml) H3PO4 (250ml) (25ml) (34.8ml) NaOH(0.102N)
.
N1V1(NaOH)=N2V2(H3PO4)
0.102*34.8=N2*25
N2=0.141N
(eq)
=V*N
=250\1000*0.141
=0.03525eq
m=eq*Mr\f
=0.03525*98\3
=1.15g
= \
1.15\10
=0.115g\ml

4) ( (.8g (50ml) HCl (1N) ,
(25ml) (21.4ml) (0.1N)NaOH .
N1V1(NaOH)=N2V2(HCl)
0.1*0.0214=N2*0.025
N2=0.0856N
=
V*N
=0.25*0.0856
=0.0214eq
=
(N*V)-0.0214
(1*50\1000)-0.0214
=0.0286eq
eq=eq

0.0286=eq
= m\eq
=0.8\0.0286
=27.9


5) ( 1.26g ) (90g\mol) (25ml) (19.95ml) (0.1N)NaOH . .

eq(HCl)=eq(NaOH)
N1V1=N2V2
N1*25=0.1*19.95
N(HCl)=0.0798N
Mr=f*m\N*v
Mr=2*1.26\0.0798*0.25
Mr=126.315
= +
90+nH2O=126.315
90+nH2O=126.315
n=36.315\18
n=2
= 2
*************************************
(4)

1) (50ml) (0.2N) (200ml) (200ml) (0.05N) , ֿ
=
eq(KOH)+eq(NaHCO3)=eq(HCl)
N1V1+N2V2=NV
0.2*0.05+0.05*0.2=N*0.2
0.02=0.2N
N=0.1

2) (500ml) (0.1N) (200ml) (0. 5N) (0.125N)


=
eq(HCl)+eq(HCl)=eq(NaOH)
N1V1+N2V2=NV
0.1*0.5+0.5*0.2=0.125*V
0.05+0.1=0.125V
0.15=0.125V
V=1.2L

3) (200ml) (0.1N) (100ml) (0.05N)


=
eq(Na2CO3)+eq(NaOH)=eq(HCl)
eq(Na2CO3)+N2V2=NV
eq(Na2CO3)+0.05*0.1=0.1*0.2
eq(Na2CO3)+0.005=0.02
eq(Na2CO3)=0.02-0.005
eq(Na2CO3)=0.015
eq=mf\Mr
0.015=m(2)\106
m=0.759g

N M
mf\Mr+mf\Mr=mf\Mr





4) (10ml) (NH4)2SO4 NaOH (50ml) HCl(0.1N) (20ml) (0.1N)NaOH
(NH4)2SO4
eq(HCl)=N*V=0.1*0.05=0.005
eq(NaOH)=N*V=0.1*0.02=0.002
= NaOH
=0.002
=
0.005-0.002=0.003eq
N=eq\v
=0.003\0.01
=0.3N

 
 
 
 

 
20-06-2008   #3
 
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(5)
1) (1g) (250ml) , (50ml) (50ml)
HCl(0.1N) (10ml) NaOH(0.16N)
, .
N1V1(NaOH)+N2V2(Na2CO3)=NV(HCl)
(0.16*10)+(N*50)+0.1*50
1.6+50N=5
50N=5-1.6
50N=3.4
N=0.068
eq=N*V
0.068*250\100=0.017
m=eq*Mr\f()
=0.017*106\2
=0.901g
=
0.901\1*100=90.1%


2) (10g) Na2SO4.10H20 NaHSO4 (275ml) Na2CO3(0.1N) . .

Na2CO3+NaHSO4====Na2SO4+H2O+CO2
eq2(NaHSO4)=eq1(Na2CO3)
eq2(NaHSO4)=N1V1
eq(NaHSO4)=0.1*0.275
=0.0275eq
= eq*Mr\f
0.0275*120\1=3.3g
= 3.3\10*100=33%
= 100-33=67%

3) (4g) Na2CO3 NaCl (250ml).
(25ml) (50ml) HCl(0.1N) , .
. .
N1V1(HCl)=N2V2(Na2CO3)
0.1*50=N2*25
N2(Na2CO3)=0.2N
250 =
M=N*V*Mr\f
0.2*250\1000*106\2=2.65g
= 2.65\4*100=66.25%
= 100-66.25=33.75%



4) CaCO3 MgCO3 (1.64g) , (50ml) HCl(0.8N) (16ml) (0.25N)NaOH , .
NaOH=NaOH
eq(NaOH)=N*V=0.25*0.016=0.004eq
=N*V=0.8*0.05=0.04
( )=0.004-0.04=0.036
eq(CaCO3)+eq(MgCO3)=eq(HCl)
=m
= 1.64-m
eq(CaCO3)+eq(MgCO3)=eq(HCl)
mf\Mr+mf\Mr=m2f2\Mr2
m(2)\100+1.64-m(2)\84=0.036
m\50+1.64-m\42=0.036
m=1.25
= 1.25g
= 1.64-1.25=0.39g
=1.25\1.64*100=76.21%
0.39\1.64*100=23.7%
= 100-76.21=23.7%

**********************
(6)
1) (3.04g) , (8ml) , .
= \
= mf\Mr
3.04(1)\152=0.02eq
=
N1V1=N2V2
0.02=N2*8\1000
N2=2.5N
m=Mr\f*V*N
=158\5*8\1000*2.5=0.632g
=m\V=0.632\8=0.079g\ml

2) (4g) (0.5N), (200ml) , . .

=
N1V1=N2V2
N1V1=0.5*200\1000=0.1eq
NV=0.1
(m)=
N*V*mr\f=0.1*63.5\2=3.17g
= 3.17\4*100=79.3%
= 4-3.17=0.83g
=0.83\4*100=20.7%
=100-79.3=20.7%

3) (6.7g) , 20 (25ml) :
C2O4(2-)+MnO4(-)=====CO2+Mn(2+)
:
Ý)

M=n\V=m\Mr\V
6.7\134(1)=0.05M
=
N1V1=N2V2
f1M1V1=f2V2M2
2*0.05*20\1000=5*M2*25\1000
M2=0.016M
N=fM=5*0.016*5=0.08

:
eq=m(f)\Mr=6.7(2)\134=0.1eq
eq=NV
N=eq\V=0.1\1=0.1N
=
N1V1=N2V2
0.1*20=N2*25
N2=0.08N

ȝ) : .
( )

 
 
 
 

 
20-06-2008   #4
 
   : 1884
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0
0
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(7)
1) 16ml) ) NaOH(1\8N) (10ml) . (20ml) (0.3501g) .
:
BaCl2(aq)+H2SO4(aq)=====BaSO4(s)+2HCl(aq)
98g------------233.4g
x--------0.3501g
(X)
=98*0.3501\233.4
=0.147g

eq=m(f)\Mr=0.147(2)\98=0.003eq
eq=NV
N=eq\V=0.003\0.020=0.15N
N1V1(H2SO4)+N2V2(HCl)=NV(NaOH)
=10

(N1+N2)V=NV
(0.15+N2)10\1000=1\8*16\1000
N2=0.05N
= m\V
m=Mr\f*V*N
=
=Mr\f*V*N\V

= Mr\f*N=36.5\1*0.05=1.825g\L


2) ) KCl NaCl (5.4892g) (12.7052g)AgCl . .
AgCl=
eq(NaCl)+eq(KCl)=eq(AgCl)
mf\Mr+mf\Mr=mf\Mr
=m
=5.4892-m
(5.4892-m)(1)\58.9+m(1)\74.5=12.7052(1)\143.5
6.99015-1.2735m+m\74.5=0.0885
-0.02735m=-0.397268
m=1.4525g
KCl= 1.4525g
= 1.4525\5.4892*100=26.46%
NaCl= 100-26.46=73.5%
.

3) (5g) KClO3 (1\2L) (25ml) (20.41ml) (0.1N)AgNO3 . KClO3.

AgNO3
m=Mr\f*V*N
m=170\1*20.41\1000*0.1=0.34

2KClO3(s)======2KCl(aq)+3O2(g)

KCl(aq)+AgNO3(aq)====AgCl(s)+KNO3(aq)
( Mr) 74.5g------------170g( Mr)
m-------------0.34g

(m)=0.34*74.5\170=0.152g
2KClO3(s)======2KCl(aq)+3O2(g)
2*122.5g--------2*74.5g
m------------0.152g
m(KClO3)=0.152*245\149=0.2499g

122.5(KClO3) 35.5(Cl)
0.2499g X
X=0.072
0.2499g = 0.072g
= 0.072\0.2499*100=28.99%

 
 
 
 

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