C&u 15. C h o m(g ) hidrocacbon A thuoc day d o n g d i n g cua m e t a n tac d u n g v d i clo
C O c h i e u sang, c h i thu dufdc m o t dan x u a t clo d u y nhát B v d i k h o i liTdng 8,52g. D e t r u n g h6a het k h i H C l sinh ra, can vuTa du 80 m l d u n g d i c h K O H I M . C T P T
cua A , B Ian lufdt l a : ::^,i^t:^' j^btdmâi^^ I • ] kmr'^ỷ •)! 1) cMi)
A . C s H n va C s H n C l ' ' , ^j: D ' " B . C s H ^ v^ CsHjoClj C. C 4 H , , , va C4HyCl D . C 4 H H , va C 4 H X C I 2 C. C 4 H , , , va C4HyCl D . C 4 H H , va C 4 H X C I 2
c a u 16. D o t chay hoan toan hon hdp 2 hidrocacbon dong d i n g lien tiep can 1,35 m o l O 2
tao t h a n h 0,8 m o l C O 2 . C T P T cua 2 hidrocacbon l a :
^ • ^ ^ " - ^ ^ " ^ B . C 2 H 2 . C 3 H 4
C. C H 4 , C 2 H 6 , , D . C 2 H 6 , C j H s
C a u 17. D o t chay h o a n toan hon h d p 2 hidrocacbon (phan tuT k h o i h d n k e m nhau 14 d v C ) , t h u diTdc 5 m (g) C O 2 va 3 m (g) H 2 O . C T P T cua 2 hidrocacbon t r e n l a :
A . C 3 H x, C 3 H 6 B. C 2 H 6, C 3 H s , ,, ,
C. C 2 H 2 , C 3 H 4 D . C 3 H 6, C 4 H 6
CSu 18. T r o n a m o l h o n h d p A ( g o m C 2 H 6 , CsHs) va b m o l h o n h d p B ( g o m C3H6, C 4 H S ) t h u difdc 0,35 m o l h o n h d p C r o i d e m dot chdy, thu diTdc h i f u so m o l H 2 O
va C O 2 l a 0 , 2 m o l . G i d t r i c u a a v a b l a n l i W t l a : , , ^ . ' ^ ' ^
Ạ 0,1 va 0,25 B. 0,15 va 0,2 C. 0,2 v^ 0,15 D . 0,25 va 0 , 1 ' '**' c a u 19. H o n h d p X g o m etan va m e t a n , c6 t i k h o i h d i so v d i h i d r o la 9,4. D o t chdy
1 m o l X can the t i c h O 2 (dktc) l a :
A . 44,8 l i t B . 33,6 l i t C. 51,52 l i t D . 35,84 l i t
C&u 20. H o n h d p X g o m 2 hidrocacbon d the k h i c6 so n g u y e n tuf C t r u n g b i n h b^n g
3 va m c o , = 3 m x . D a y d 6 n g d i n g cua chdng l a : » ' " ^ " ^
A . ankan , B. anken C. a n k i n D . aren , C a u 21. K h i dót chdy h o a n toan cdc ankan t h i ta dUdc T = UQQ^ : U H J O • j C a u 21. K h i dót chdy h o a n toan cdc ankan t h i ta dUdc T = UQQ^ : U H J O • j
T b i e n d o i t r o n g k h o a n g : !' i<i s i ^ 1 1 • - r
A . 1 < 'T< 1 , 5 B. 0 , 7 5 <T< 1 C. 0 , 5 <T< 1 D. 1 , 5 <T< 2 „;> c a u 22. TrOn 2 the tich b^ng n h a u ciia CsHx va O 2 r o i d e m dot. Sau k h i cho bay h d i c a u 22. TrOn 2 the tich b^ng n h a u ciia CsHx va O 2 r o i d e m dot. Sau k h i cho bay h d i
nufdc ngufng t u , dufa t r d v.e d i e u k i ? n ban d a u . Mnihfu iF
N h a n x6t d u n g ve Vh<i„ h,.,p UM. pa ( V j ) va Vh<',„
h(.(p sau p\
A . V d : V , = 3 : l B . V d < V s C . V d > V , D . V . , : V , = 10 : 7
K i i A i g f l i n i g r
(^au 23. K h i dot c h a y hoan toan cijng m p t lUOng hidrocacbon X b a n g m p t liTdng vtfa
O 2 hoac C I 2 , n g i r d i ta thay t i 1§ the Úch O 2 va C I 2 b^ng 1,25 ( x e t d c i i n g d i e u
k i $ n ) . X l a : - - r , . , .
A. C 3 H 6 * B . C J H H C. C 4 H f i H D. C 4 H S
c a u 24. Thifc h i e n p h a n tfng c r a c k i n h 11,2 l i t h d i isopentan (dktc), t h u dúdc hon hdp A c h i g o m cac a n k a n va a n k e n . T r o n g h o n h d p A c6 chiJa 7,2 g a m m o t chát X (adsbygoogle = window.adsbygoogle || []).push({});
tnh k h i d o t chay t h i thu dufdc 11,2 li t C O 2 (dktc) va 10,8 g a m H 2 O . H i e u suat phan iJng c r a c k i n h isopentan l a :
Ạ 8 0 % B . 8 5 % C. 9 0 % ' ' ' D . 9 5 % ' c a u 25. D o t c h d y h o a n toan hon h d p etan va propan t h u diTdc h o n h d p k h i C O 2 va c a u 25. D o t c h d y h o a n toan hon h d p etan va propan t h u diTdc h o n h d p k h i C O 2 va
hdi H 2 O theo t i le the tich 1 1 : 1 5 . T h a n h phan % theo k h o i lufdng cua etan trong hon hdp l a :
Ạ 1 8 , 5 2 % B . 18,63% C. 16,18% D . 18,16%
c a u 26. N u n g m p t lifdng butan trong b i n h k i n (c6 x u c tac t h i c h hdp) t h u diTdc hon hdp k h i X g o m a n k a n va a n k e n . T i k h o i cua X so v d i k h i h i d r o la 21,75. Phan U-am the tich cua b u t a n trong X la " i : A . 3 3 , 3 3 % B . 5 0 , 0 0 % C. 6 6 , 6 7 % D . 25,00%.
" Trich de thi tuyen sinh Cao đn^ nam 2012"
c a u 27. C h o butan qua x u c tac ( d n h i ^ t dp cao) thu diTdc h o n h d p X g o m C4H|(i, C4Hg, C 4 H 6 , H 2 . T i k h o i cua X so v d i butan la 0,4. N e u cho 0,6 m o l X v a o dung
djch b r o m (dxi) t h i so m o l b r o m t o i da phan uTng la
A . 0,48 m o l B . 0,36 m o l C. 0,60 m o l D . 0.24 m o l
i f c i f i "fit!, o r " ^"y^" '^'"^ ^'-'^ ^ "'•'"^ 2011"
c a u 28. K h i c r a c k i n h h o i n toan m p t the tich ankan X t h u difdc ba the tich h o n hdp Y (cdc the t i c h k h i do d c i l n g d i e u ki§n n h i e t do va ap suat); t i k h o i cua Y so v d i
H2 b^ng 12. C o n g thu-c p h a n tuf cua X l a :
A . C 6 H, 4 . B . C 3 H 8 C. C 4 H, o D . C H u .
c a u 29. N h i ^ t p h a n m e t a n t h u diTdc hon h d p X g o m C 2 H 2 , C H 4 va H 2 . T i k h o i hdi cOia X so v d i H 2 b k n g 5. H i ? u suát qud trinh nhiQt phan l a : • A . 5 0 % V B. 6 0 % C. 7 0 % , , 5 D . 80% ,
Cfiu 30. C r a c k i n h 0,1 m o l pentan diTdc h o n h d p X . D o t c h a y het X r o i h a p thu san p h a m c h a y b d i nufdc v o i trong dÚ. H d i k h o i li/dng d u n g d i c h c u o i c u n g t h u difdc tSng hay g i a m bao n h i e u gam?
A . g i d m 17,2g B . tang 32,8g C. tSng 10,8g D . tang 22g
Cau 31. C r a c k i n h V l i t butan thu difdc 35 l i t hon hdp A g o m H 2 , C H 4 , C 2 H 4 , e 3 H 6 , C 4 H 8 va 1 p h a n butan chtfa b i c r a c k i n h . Cho hon h d p A l o i rat tiT tiT qua b i n h dUng b r o m duT t h a y the t i c h c o n l a i 20 l i t . % butan da p h a n iJng l a :
A . 2 5 % B . 6 0 % C. 7 5 % D . 8 5 %
Phan djina va phuang pMp g i i i H6a hgc 11 HOu co - Pi Xuan Hung
C a u 32. Nhi^t phan 8,8g CjHs ta thu diTdc hon hop khi A gom C H 4 , C 2 H 4 , C3H6, H ,
va C3H8 chUa bi nhi#t phan. Biet c6 90% C s H g bi nhi^t phan. T i khoi hOi cua hoti
h d p A s o v d i H z l a : ; Ạ 11,58 , B. 15,58 C. 11,85 D. 18,55 ;
C a u 33. Hon hcJp khi gom etan propan c6 t i khoi so vdi hidro 1^ 20,25 diTdc nung trong binh vdi chát xiic tac de thifc hỉn phan ufng de hidro hdạ Sau mgt thdi ; gian thu difdc hon hdp khi c6 ti khói so vdi hidro la 16,2 gom cd cac ankan, anken v^ hidrọ Tinh hieu suát phan iJng de hidro hda biet rang toe dp phan iJng ciia etan va propan Ih nhiT nhau^ , j
Ạ 30% B 5 0 % C 25% D 4 0 % HUOfNG D A N G I A I
Cflu 1. X6v. n ^ j o • "CO2 = 1.2 => X la hidrocacbon no
J ' X 1^ ankan: CnH2n+2 ' - n" 2 n + 2 + — — ^ 2 = l , 2 = > n = 5 ,.^ii£y>tu,''i:ii: C H 3 CH3 C •nCOj H-ln + D H j O It H ' . I >i - C H , CH A> ,U 1 I,
Ta thay 2,2-dimetylpropan chi cho 1 dan xuat mono duy nhat.
't => Chpn Ạ • " 1 " J , M l «i
C S u 2. M x = 2,625.32 = 84 dvC
Cong thiJc cua X la: CnHjn = 84 => n = 6 (adsbygoogle = window.adsbygoogle || []).push({});
=> Chi cd xiclohexan cho 1 dan xuat. => Chon C. 29,55 I C S u 3 . T a c d : m BaCO3 = - j ^ = 0 . 1 5 m o l ^ n c o 2 = 0,15 Matkhac:Amdjgia„, = m i- ( m c o 2 + m H 2 o ) ' ' ' I => m c o j + m H 2 0 = - Amđgiam = 29,55 - 19,35 = 10,2 g mol m c o 2 = 0,15.44 = 6,6 g mn^o = 3,6 g n H 2 0 = 0.2 mol
Ta thay: n^^^Q > n^Q^ => hidrocacbon X la ankan va:
" x = n H 2 0 - n c o 2 = 0,2 - 0,15 = 0,05 mol
so nguyen tuf C trong X = = = 3 => X la CjHs => D a p a n D.
" x 0,05
Cflu 4. Ta cd: nhh = 0,3 mol; n^ j O = 0.6 m o l ; ncoj = 0,5 mol
Ta tháy: n H 2 0 ^ " 0 0 2 ^ ^'^^^'^ ^ P^^''^^ 1 chS't la ankan ,; : • . nc = = M = 1,67 => trong hon hdp cd C H 4 " h h 0,3 Va: M y > M x => X la CH4 => D a p a n C . j . , v c a u 5. C 2 H 4 : H - C = C - H H H C 2 H 2: H - C ^ C - H I Ú I ' I 1 I •> I ' l l f' ) ú C 4 H 6 : H - C = C - C = C - H I I I I H H H H
C 2 H 4 : liSn ket giffa C va H la lien ket xich ma (lien ket đn); lidn kd't C va C
g 6 m 1 xich ma va 1 p i
C 2 H 2: lien ket giffa C va H la lien ket xich ma ; lien ket C va C gom 1 xich ma v^ 2 pi
C 4 H 6 : lien ket giffa C va H la lien ket xich ma ; lien ket C v^ C gom 1 xich ma
va 1 pi .JIO ::-lr •
= > D a p a n B . .^„^ \ •uur?Mi]mr:y>--.^^^^^^ :K_
Cflu 6. Goi cong thiJc phan tuf cua ankan X la C„H2„ + 2.
Trong C l i n g dieu kien ve nhi^t dp va ap suat thi ti I9 ve the tich cung chinh la t i 1^ ve so mol 1 mol X c r a c k i n h
•> 3 mol Ỵ m
Ta cd: M y = 12.2 = 24 <=> ^ - 2 4
=> my = 24.nY = 24.3 = 72. «*vr^.=vv:^v,:;;:vfc,%,tri,v «1 "r.)T:;;'
A p dung dinh luat bao toan khoi liWng ta cd: mx = my = 72 ' ' H > ' '
M - — - 7 2 • '^^ O 14n + 2 = 72 => n = 5 => X m CsH^. ?AX •••• , O 14n + 2 = 72 => n = 5 => X m CsH^. ?AX •••• ,
= > D a p a n D .
Cflu 7. CO 4 san pham
CH3 - CH - C H 2 - C H 2 C I CH3 - CH - CHCl - CH3 C H , C H , CH3 - C C l - C H 2 - C H 3 C H 2 C I - C H - C H 2 - C H 3 CH3 D a p a n D . C H , 73
Phan dgng va pnijcing ph^p giai H6a hpc 11 HQu ca - D5 Xuan Hung
C§ku h. X > 6 C O 2 => X CO 6 C .
X m a c h h 5 , c h i chiJa l i e n ke't a = > X la C f i H u . i j , . , , ' ! . . ,
X c d 2 n g u y e n tuf C bac ba = > X l a : 2 , 3 - d i m e t y l b u t a n .
C h o X tac d u n g v d i CI2 (ti l e m o l 1:1) c h i sinh ra t o i da 2 d a n x u a t m o n o c l o :
C H 3- C H - C H- C H 3 ^''^'^''^v'"'^^^
CH3 CH3 (adsbygoogle = window.adsbygoogle || []).push({});
=> Dap an C .
C S u 9. T r o n g hidrocacbon só n g u y e n luf H luon la so c h a n nen:
C T D G N cua h i d r o c a c b o n (CnH2n^i) => C T P T (C„H2„+,)2 hay C2nH2(2„+i) ' '^'v '
D a t 2 n = m => CTPT: Cn,H2m+2 => hidrocacbon d o thuQC d a y d o n g d a n g ankan
=> Dap an Ạ j r i ii r i k i
Cfiu 10. T a c d : n H j o =0,132 m o l > nco2 = 0.11 mol. am ; iHp
=> hidrocacbon X la ankan: C„H2„ + 2. H *«» ^J*** I . c„ H 2 „ . 2 n C 0 2 + ( n + l) H 2 0 ^»:'«*.S«^^'^»''^ir*'rfnaa:iHta 0,11 0,132 = > 0,132.n = 0,1 l . ( n + 1) = > n = 5 = > ankan CsH.z. '' ' ' C T C T cua C5H12. ' C H 3 - C H 2 - C H 2 - C H 2 - C H 3 : p e n t a n , l l H & q M ! < = C H 3 - C H - C H 2 - C H 3 'T^ - .<;. : 2 - m e t y l b u t a n '* '^'h -t^^' Sn^'* •
... -''..CHj i - •,jH'n oj'.' fifi;;^ u y i t ^;;nuy g n o i T
* (pHs ' ' " " ^ ,V ' ' ' ' o m I c:: l o m ^ 7 1?)
C H 3 - C - C H 3 ^ ' '^'^ ' ' : 2.2 - d i m e t y l p r o p a n /J^^., ^ ^^j.
C H 3
D i f a v a o C T C T ta thay c h i c d 2,2 - d i m e t y l p r o p a n k h i t a c d u n g v d i GI2 ( t i 1§
m o l 1 : 1) t h u d i M c san p h a m hCTu cd duy nhat. â^,, o i d fMb jjnufa
= > D a p a n C . . ,,,„ i j r v,m .. C a u l l . C„H2„+2 + B r 2 ^ C n H 2„ . , B r + H B r c •i>.;,:>i:>: ^^a,,
T a c d : Mc^H2„„Br = 75,5.2 = 1 5 1 ^ « ? « n < = + o <=> 14n + 81 = 151 n = 5 = > a n k a n C5H12 • -^^ '^^'K
C , H, 2 ) 1 dan x u a t m o n o b r o m duy n h a t ' ' '^^'^^^ ^^'^ ^' => C T C T d u n g cua ankan la: , , J- i
C H 3 ''-^S'.,;.:..:-'; C H 3 - C - C H 3 : 2,2 - d i m e t y l p r o p a n •-. jlf:>~ HO - jHD C H 3 - C - C H 3 : 2,2 - d i m e t y l p r o p a n •-. jlf:>~ HO - jHD C^H3 , , ^ r r - = > D a p a n B . .f ÁA%.f:ry ,;. KiiAMfiirnnr c a u i 2 . * each 1: ' H o n h d p k h i g 6 m C H 4 , C 2 H 6 v a C 3 H 8 la cdc ankan. ' , i . .' : • G o i c o n g thiJc c h u n g cua h o n h d p X la C-H2-^2 ' ' ' • , ' ' 7 , 8 4 ^ „ , ' i o a i ^ O i ; . " * ± : ' ' . . ,: T a c d : nco2 = = 0 , 3 5 m o l nH2o = ^ = 0 , 5 5 m o l -^.^ .-^^ 'UM.):.^^ 3n + l ' - ' - ' d';iOA)':". j i c - ( S ) liv : ' C , H 2 , , 2 - — - nC02 . ( n . l ) H 2 0 . ^^^^ ^ 0.35 0,55 , ( f « i a . . , ; = ^ 0 . 5 5 n = 0 . 3 5 . ( n + l ) = > n = 1 , 7 5 sO ,0 ^ 8 0 , 0 . ..HỌ.^^ : O . « . T h e o phi/dng t n n h p h a n u-ng: ' • . - - j ^ j ^ 0,35 3n + l 0,35 3 . 1 , 7 5 + 1 „ n o , phan tfng = - = - . — — = 7 — • " = 0,625 m o l i ! J X + ; : ' " 2 1,75 2 ^^^^^ 100 Vkhong khi = (0,625.22,4). — = 7 0 l i t ,x H o n h d p k h i g o m C H 4 , C 2 H 6 v a C 3 H 8 l a cdc hidrocacbon.
M a k h i dot chay 1 hidrocacbon baft k i ta l u o n c d : f * " ' s a L r t H t q ;
1 „ 1
phantfng = n c o 2 + - n H 2 0 = 0'35 + - . 0 , 5 5 = 0,625 m o l - , , f^^v
=> Vkhongkhi = ( 0 , 6 2 5 . 2 2 , 4 ) . — = 7 0 l i t r: > D a p a n A . w ) f . = i i f
CSu 13. G o i c o n g thiJc p h a n tuT cua ankan X \h CnH2„ + 2.
T a c d : % C = . 1 0 0 % = 8 3 , 7 2 % n = 6 = > a n k a n X Ih C e H u .
C S H M — " ^ " 2 > 2 d a n xu2ft m o n o c l o d o n g p h a n cua n h a u = > X c d 2 hifdng thẹ => C T C T d u n g cua X : ' " ' ^ ^ftsaSnfi « l nodaipo-'^^c; :
C H 3 - C H - C H - C H 3 : 2,3 - d i m e t y l b u t a n : * r ! 5
C H 3 C H 3 . j t >. i
=> Dap an B. • ;# .^eSCkf ^5;S;,>.ị*t'?l :i , ' ,.,:.'
Cau 14. T a c d : ncjHx bandJu = ^ = 0,05 m o l ; nci2 = ^ = 0,05 m o l
44
C j H s + C I 2 C 3 H 7 C I + H C l a a .;v a (adsbygoogle = window.adsbygoogle || []).push({});
Phan dgng va phuong ph^p giSi H6a h(?c 11 HQu CO - Bfi Xufln HUng
CjHs + 2Cl2 C 3 H 6 C I 2 + 2HC1 , i i
b 2b b 2b
T h e o d e b ^ i : mx = l,3894mY <=> 78,5a = 1,3894.113b = 157b hay a = 2b (1) Khi con lai di ra khoi dung dich NaOH dif la C 3 H 8 Knilx'sa i f pỉ 7 ; • - :
riQMn dir = = 0,02 mol '
^ "C3H}j phan ilng = "C3H){ bandSu - "C3Hj{ dir = 0,05 - 0,02 = 0,03 Hiol
=> a + b=:0,03 (2) , ,
' ' T t f ( l ) v a ( 2 ) = > a = 0 , 0 2 ; b = 0,01.
=> mc,H,ci =0,02.78,5 = 1,57 g va nic,H„ci, = 0-01.113 = l,13g
Dap an D.
CSu 15. T a c o : n N a O H = 0,08.1 =0,08 mol ' PhUdng trinh phan itng clo hda: ' ' ' ' ' \
C n H 2 „ ^ 2 + XCI2 C„ H 2 „ ^ 2- x C l x + XHCl (1) ' ' ,„
5a, , COS
HCl + NaOH -> NaCl + ' ' ' ' V ' ' ' " ' ' ' ' ' ' " ' ' " ' ^ •
0,08 0,08 ' . • Theo phiMng trinh phan iJng (1): ne = - n H c i = — :,r.ml I -{mMmM
X X ^ J .
^(I4n + 34.5X + 2) = 8,52 o + 2,76 + ^ = 8,52
o 7n + 1 = 36x =i> nghi$m phQ hdp m x = 1 va n = 5. ^ mr'xiýf" ™'
o CT cua A B Ian lifdt la: CsHu C 5 H 1 1 C I
^ Dap an Ạ siifaA*?^*;i»#4op:^^
nA %^ T ' 'ý a. > s ' '-Mil
C a u l 6 . T a c o : nojphantfng = nco2 + - n H 2 0 ; T; i f. I , ,
• ^ n H 2 0 = 2(no2 phan J n g - nco2 ) = 2(1,35 - 0,8) = 1,1 > nco2 = 0.8
=> 2 hidrocacbon la ankan. -•-'^••^ ' ':l !>>kyr:^&rTSr:)
Goi cong thut long qudt cua A m C - H 2 - ^ 2 ^ ' ' " ' ' ^ '
^ S " 2 n + 2 + ^ Y^ 0 2 ^ • i C 0 2 + ( n + l ) H 2 0 ffliUfi^ffcv
=> 1.1 n = 0 , 8 . ( n + 1) n =2,67 ^ 2 ankan la C 2 H 6 . C J H R
=> D a p a n D . ,
Cfiu 17. Goi cong thiJc chung cua 2 hidrocacbon la C - H - (vdi x , y Ian luTdt la so nguyfin tu: C trung binh va s6' nguyen tur H trung binh).
n c 0 2_ 2 x ^ 5 m 2 ^ 2x ^ _15 '' ^ ' (adsbygoogle = window.adsbygoogle || []).push({});
^ nH20 ~ y ~ 4 4 " 3 m ' ^ y " 2 2
y = 2,93 X > 2 X => 2 hidrocacbon la ankan., , ,.r,..p , , v a l u e d6: y = 2 x + 2 => 2,93 x = 2 x + 2 => x =2,15
=> 2 hidrocacbon la C 2 H 6 , C 3 H 8 => Dap an B.
Cau 18. Cong thiJc chung cua hon hdp A la CnH2„ + 2: a mol
Cong thiJc chung cua hon hdp B la Cn,H2m: b mol /. ' C„H2n + 2 n C 0 2 + ( n + l ) H 2 0 ,+-0 4. /- ^ a na (n + l)a mCỌ + mH-,0 . • r ^f^% C„H2m n i C 0 2 + mHzO b mb mb ' • T a c d : n c o 2 = na + mb; n H 2 0 = + l)a + mb J x r - 5-
=i> n H j o - n c 0 2 = a = 0,2 mol => b = 0.35 - 0.2 = 0,15 mol=> Dap an C.
Cflu 19. Goi c6ng tht?c chung cua hon hdp X la C - H j - ^ j
=> 14n + 2 = 9.4.2 = 18.8 n = 1.2 . O J ' / . , ,.
C ; ? « 2 ; i . 2 + ^ 0 2 ^ ^ 0 2 + ( n + i ) H 2 0
1 0 . 5 ( 3 n + 1) : • ' . ' (Jf> • "
Vo2 = 0 . 5 ( 3 n + l ) . 2 2 , 4 = 0,5(3.1,2 + l).22.4 = 51.521it = > D a p a n C .
Cflu 20, Gói c6ng thiJc chung ciia 2 hidrocacbon la C-H2-^^2-2k • ^
C-H2-^2-2k > n C O ; + ( n + l - k ) H 2 0 • , tSv?
x m o l • n x m o l . • c-:^n'[ ^ b-fótO « Theo de bai ta c6: mcô = 3mx 4 4 n x = (14n + 2 - 2k).3x Hay: 4 4 n = ( 1 4 n + 2 - 2 k ) . 3
Thay n = 3 => k = 0 = > 2 hidrocacbon nay la ankan => D^p an Ạ
C a u 2 1 . C„H2„+2 n C 0 2 + ( n + l ) H 2 0
T " C O 2 n 1 , , ,. ' , ,.••,„ ,
n H 2 0 n + 1 , ^
11 Hau CO - P5 Xuan Hung
K h i n = 1 =5. T = 0,5 K h i n —> 00 => T —> 1.
=> D a p a n C .
0 , 5 = > T < 1
C a u 2 2 . G o i VcjHs = a l i t =ị VQ^ b a n d 3 u = a l i t • ;4J'3i Ajirj^^' x,i
C S H H H- 5 0 2 - > 3 C 0 2 + 4 H 2 0 , v^, ;J — a — V ró- f*-'* V (>..!••!"<
5 5
=> sau k h i H 2 O ngUhg t u , h 5 n hdp k h i sau phan iJng dot chay g o m C O 2 CsHg dụ
3 a a "'^•^^ ''^ '•
= > V h h khi sau phan tfng = V ^ Q , + H„ dlT = — " + (a - - ) = 1,4a l i t
Vj^u = a + a = 2a l i t .j^^;;,, ... /.js^,,;. ^ f i r j , ; , [ ; • : ; smuiu 'XXtiil :ssn63 J:|,if
= > V s a u : Vj^u = 1.4a : 2a = 0.7 hay V j^ u : V^au = 10 : 7 D a p d n D . gn&O (adsbygoogle = window.adsbygoogle || []).push({});