7 cach bai toan hidrocacbon
TRANSCRIPT
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8/3/2019 7 Cach Bai Toan Hidrocacbon
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BI HO HIDROCACBON 7 CCH GII
L Thanh PhongLp Ha 2010A - HSP Hu
Hin nay mn Ho trong trng ph thng c rt nhiu phng php gii hay nhanh. Nhiu bi Hokinh in c trnh lng nh b i ca thy Sao bng lnh gi ( V Khc Ngc ).... Bi vit ny hyvng c gp mt phn nh trong s pht trin ca mn Ho trong trng ph thng.. y cng l bivit u tay ca mnh , rt mong nhn c s ng h , ng gp kin ca qu bn c , thy c vnhng ai yu Ho..
Bi ton : Cho hn hp A gm anken X v H2 qua Ni ung nng , thu c hn hp B gm 2 kh. BitB khng lm mt mu dung dch Brom. T khi ca A v B so vi H2 ln lt l 6 v 8 . Xc nh CTPTca X v thnh phn phn trm th tch ca mi kh trong hn hp A..
S phn ng :
CnH2n CnH2n+2 B khng lm mt mu Brom . ( x mol ) t oN i, ( x mol ) B l hn hp ca ankan v H2 d
H2 H2 d Anken X phn ng ht( y mol ) ( y-x mol )
--------------------- ---------------------MA= 6.2= 12 MB= 8.2= 16
CnH2n + H2 CnH2n+2
x mol x x molCch 1 : Bo ton khi lng + Khi lng mol trung bnh :mA= mB 6.2 ( x + y ) = 8.2 ( x+y - x ) 6 ( x + y ) = 8y y = 3x
MA =yx
ynx
+
+214= 12 14nx = 12x + 10y = 42x 14n = 42 n = 3
CTPT ca anken X l C3H6 .Trong A ( % Vanken= 25 , % VH2 = 75 )
Hoc :
MB = xyxxyxn
+
++ )(2)214(
= xynx 214 +
= 16
14nx = 14y = 42x
14n= 42
n = 3 CTPT ca anken X l C3H6 .
Cch 2 :Khi lng mol trung bnh :
MA =yx
ynx
+
+214= 12 14nx + 2y = 12x + 12y 14nx = 12x + 10y ( 1 )
MB =xyx
xyxn
+
++ )(2)214(=
x
ynx 214 += 16 14nx = 14y ( 2 )
Gii ( 1 ) & ( 2 ) 12x + 10y = 14y y = 3x.Thay y = 3x vo ( 2 ) ta c n = 3. CTPT ca anken X l C3H6 .
Trong A ( % Vanken= 25 , % VH2 = 75 )
Cch 3 : Hiu 2 t khi + Khi lng mol trung bnh + Bo ton khi lng :
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dB/H2 - dA/H2 = 2 1/2 ( MA - MB ) = 2 Bn
mB
-An
mA= 4 )
11)(214(
yxyynx
+
+ = 4
)(
)214(yxy
xynx
+
+ = 4 y
x
yxy
ynx.
)(
214
+
+
= 4 y
x.12 = 4
3
1=
y
x y = 3x.
MA =yx
ynx
+
+214= 12 14nx = 12x + 10y = 42x 14n = 42 n = 3
CTPT ca anken X l C3H6 .Trong A ( % Vanken= 25 , % VH2 = 75 )
Hoc :
MB =xyx
xyxn
+
++ )(2)214(=
x
ynx 214 += 16 14nx = 14y = 42x 14n= 42 n = 3
CTPT ca anken X l C3H6 .
( Ngoi ra bn cng c th tm ra y = 3x vi phng php tng 2 t khi bng cch tng t )...
Cch 4: Hiu 2 t khi + Bo ton khi lng + S ng cho :
dB/H2 - dA/H2 = 2 1/2 ( MA - MB ) = 2 Bn
mB
-An
mA = 4 )11
)(214(yxy
ynx+
+ = 4
)(
)214(yxy
xynx
+
+ = 4 y
x
yxy
ynx.
)(
214
+
+
= 4 y
x.12 = 4
3
1=
y
x y = 3x.
S ng cho :CnH2n 14n 10
MA = 12
H2 2 14n - 12
3
1
1214
10
2
2
==
=
y
x
nn
n
H
HC nn n = 3
CTPT ca anken X l C3H6 .Trong A ( % Vanken= 25 , % VH2 = 75 )
Hoc :
CnH2n+2 14n + 2 14
MB = 16H2 2 14n - 14
2
1
1414
14
2
22
=
=
=+
xy
x
nn
n
H
HC nn n = 3
CTPT ca anken X l C3H6 .Trong A ( % Vanken= 25 , % VH2 = 75 )
Cch 5 : Bo ton khi lng + S ng cho :mA= mB 6.2 ( x + y ) = 8.2 ( x+y - x ) 6 ( x + y ) = 8y y = 3x
S ng cho :CnH2n 14n 10
MA = 12
H2 2 14n - 12
3
1
1214
10
2
2
==
=
y
x
nn
n
H
HC nn n = 3
CTPT ca anken X l C3H6 .Trong A ( % Vanken= 25 , % VH2 = 75 )
Hoc :
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CnH2n+2 14n + 2 14
MB = 16H2 2 14n - 14
2
1
1414
14
2
22
=
=
=+
xy
x
nn
n
H
HC nn n = 3
CTPT ca anken X l C3H6 .Trong A ( % Vanken= 25 , % VH2 = 75 )
Cch 6 : Khi lng mol trung bnh + S ng cho :Bn c th chn 1 trong 2 cch nh sau :
MA =yx
ynx
+
+214= 12 14nx + 2y = 12x +
12y 14nx = 12x + 10y ( 1 )
S ng cho :CnH2n+2 14n + 2 14
MB = 16H2 2 14n - 14
xy
x
nn
n
H
HC nn
=
=+
1414
14
2
22
( 2 )
( 1 ) & ( 2 ) 14nx = 12x + 10y = 14y y = 3x n = 3. CTPT ca anken X l C3H6 .
Trong A ( % Vanken= 25 , % VH2 = 75 )
MB =xyx
xyxn
+
++ )(2)214(=
x
ynx 214 += 16
14nx = 14y ( 3 )
S ng cho :CnH2n 14n 10
MA = 12
H2 2 14n - 12
y
x
nn
n
H
HC nn=
=
1214
10
2
2
( 4 )
( 3 ) & ( 4 ) 14y = 12x + 10y y = 3x n = 3. CTPT ca anken X l C3H6 .
Trong A ( % Vanken= 25 , % VH2 = 75 )
Cch 7 : S ng cho ( qu tuyt ) :CnH2n 14n 10
MA = 12
H2 2 14n - 12
y
x
nn
n
H
HC nn=
=
1214
10
2
2
( 1 )
CnH2n+2 14n + 2 14
MB = 16H2 2 14n - 14
xy
x
nn
n
H
HC nn
=
=+
1414
14
2
22
( 2 )
( 1) & ( 2 ) 110
1214
14
1414
=
nn n = 3
CTPT ca anken X l C3H6 .Trong A ( % Vanken= 25 , % VH2 = 75 )
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