- Tir n/du tifdng tfng: ,•^
b) Tinh khoi lifdng stiren thudifdc ti f1 tán benzen neu hi^u suat ciia qui trinh 1^
a) Cic phifdng trinh phan tfng : -^..i&fcijjv, rs& j o ^ /i^:^ j - M Y . .
CfiHe + C 2 H 4 " > C 6 H 5 - C 2 H 5
C6H5-CH2-CH3 C6Hj-CH=CH2 + Hz • ' ^
1 i n "
b) So mol CfiHfi : ncgH^ = (mol) f' # in | ^ nf: ^= i
1 10
Theo sd do : n^j^en = = (mol) imtib 8) s i H p &l A V^V:)
7o
=> Khoi lifdng stiren : m = — . 1 0 4 . — - = 104.10'* =1,04(tlfn). imi nh:.
• ^ 78 100
Bfti 5. Mpt chat hỉu cd X chi chifa hai nguyen to biet 150 < Mx < 170. Dot chdy
hoan loan a gam X thi sinh ra a gam H2Ọ Biet X kh6ng Ihm m^t m^u dung djch
brom, cung khong phan tfng vdi brom khi cd bpt F e xdc tdc, nhUhg lai phan iJng vdi brom khi cd chig'u sdng tao thanh mgt dan xuat monobrom duy nhat. Xdc dinh C T P T v^ C T C T cua X . Biet ring X cd tinh d^i xtfng.
Giai
Vi X \h chat hilu cd, khi d6t chay tao ra H2O => trong X cd hai nguyen to C , H .
Dat C T T Q cua X : C,Hy .^ao-^fn* '" ' H
^JHâJ y i a i n u a i i y ^ • • i luu w— w /\uaii i i u i i y T a c 6 : niH = 2 n H 2 0 = = | ( g ) => mc = a - | = ^ ( g ) Ta CO t i 1? : x : y = — ^ : — ^ = : — = — : 1 = 2 :3 •\: j , 12 1 9.12 9.1 12 ; ; = : > X c 6 d a n g ( C 2 H 3 ) „ . Theo de : 150 < M x < 170 150 < 27n < 170 o 5,55 < n < 6,29 => n = 6 => CTPT X la C,2H,8. . J. ;uiS
* V i X khong lam mat mku dung dich brom => X khong phai la hidrocacbon mach hd
CO noi doi C=C => X 1 ^ day dong d i n g cua benzen.
X khong tdc dung v d i Bra (bpt Fe xuc tdc) => X khong c6n H l i e n ket triTc tiep vao vong benzen v^ X cd tinh doi xuTng. . c i ; H ; ') rH';> ,51
= * V a y C T C T c u a X m : "^"foT^"'. v , r r Phifdng trinh phan uTng: 1 ^ <> < "
3
Bai 6. A la mgt dong d i n g cua benzen c6 cong thiJc (C3H4)„. T i m cong thỉc phan l\i
Ạ Via va g o i ten cdc d6ng phan cd t h ^ cd cua A . dufnwmH.n^
. ' : Oidi : , -iiHft'? íH'P + ^ i - '
A : (C3H4)„ hay C 3 „ H 4 „ . tiỌ-;(H3-.eH,ỵ V I A Ijl dong d i n g cua benzen ndn cd dang CmH2nv-6-
^ T a c d : | ™ " ^ " =>\"'^^ V ' ' ..Hpy^^
l 2 m - 6 = 4n I n = 3
Vay CTPT A Ik C9H12 (8 dong p h a n ) ; ' f n . ^ir-^-jmpn :
B a i 7.
a) Di san xuat cumen, ngiTcli ta cho benzen phan iJng v d i p r o p e a c d xiic tdc axit, hay viet phUdng trinh hda hoc cua phan iJng.
b) De san xuát 1 tan cumen can dung t o i thieu bao nhieu m ' (dktc) hon hdp khi tdch diTdc tir k h i crackinh gom 60% prdpen va 40% propan (v^ the tich) ? Biet rkng hỉu suat phan u^ng dat 80%.
c) Hay viet phi/dng trinh hda hoc cua phan vJng k h i cho 1 mol cumen t^c dung v d i 1 mol brom cd mat bpt Fe, 1 mol brom cd chieu sang.
, 0 . ^ ' ' ' . V '''~\'miM:um.iiMmr r,l:Kr•
ạ) Pmdng tnnh phan \ing aieu Che cumen :
(C^ + CH2=CH-CH3 ( 0 ) " ? " ' ^ " ^ (isopropyl benzen)
^<:>' CH3 cumen
168
b) So m o l c u m e n : n = (mol) ^^^^^^'> ,
10^ 10^ Theo phifdng trinh: ncgHg =n^umen =T:;7:("IO1) "^c^He = 7 : ^ ^ x 2 2 , 4 (lit)
^.•i>_ m^fiMi h:!'h> -^n'/• ^-^^ i 2 U ;r ,
V l C 3 H 6 c h i ^ m 6 0 % =^ V c 3 H , ( 6 0% ) = ^ x | ^ x 2 2 , 4 ( / ) = ^ . 1 0 3 ( m ^ )
V d i h i ^ u su^t phSn iJng dat 80% => Vhh = ^ ^ . 1 0 ^ — = 388,89 ( m ' ) . 72 80
Bdi 8. T i khoi hdi cua mpt hidrocacbon A vdi khong khi Ik 3,586. Biet 2,08g A phan d-ng t o i da v d i 1,792 lit H2 (dktc) va 3,12g A phan tfng viifa du v d i 100ml dung dich brom 0 , 3 M . T i m cong thiJc cáu tao v^ g p i ten A . , . , : i : . : i i ,
Theo de : M A = 3,586.29 = 104 , . (5) XU): «,H«3 • ' * S^ m o l H 2 : n H , = ^ = 0,08 (mol) S n c T n H K ( « » i l > i d ^Ir^ f,.:n. J^>,r " 2 22,4 ' ^ ^ • • '^^.^^ . , ' . , . , . i , j „ . ; » , 0 2 08 <VI hí••]>• So mol A k h i p h a n i J n g v d i H 2: nA = -— = 0,02 (moD Ta cd t i Ip HA : iiH2 = 0,02 : 0,08 = 1 : 4
=> Trong A cd 4 lien ket n (1) (T
* So m o l Br2: nBr2 =0,3.0,1 = 0,03 (mol) I d «h j ' - ' V f n^' i
So m o l A k h i phan iJng v d i B r 2: n A = — = 0,03 (mol) .0 mifi ( i ; = i > T a c d t i l p n A: n H 2 = 0 . 0 3 : 0 , 0 3 = 1 : 1 \ . e» 0 0 ; O ( k - > => Trong A cd 1 lien ket d o i d msich C=C S<! * <^)f6i< f n ^ IP
Ta l a i cd : M A = 104 '''V'>v,(i() KJ^O i;,£(;iK),0 <k..i-,
CxHy = 104 => 12x + y = 104 . ( 3 ) ^.^
: k e t hdp v d i dieu kipn (1), (2), bipn luan ta difdc x = 8; y = 8 ' => CTPT A : CsHs r.,;,-,,,,.. !....,iw>^.ru ,£lXJO,,n ();,v/:
C T C T A : j g j- C H = C H 2 (stiren). '^^im^ -r ?:CKf,:^^*^ ^^.a : .,0 !f.>;::'
B^i 9. A n k y l benzen X cd phan t r i m khoi liTdng cacbon b^ng 91,31%. " i " a) T i m cong thiJc phan tuT cua X . b) V i e t cong thiJc cau tao, g p i ten chat X .
am Dat C T T Q X : C„H2„-6. " ' ^^^li^-?!felvlV'ifi^^^K-V.. v Dat C T T Q X : C„H2„-6. " ' ^^^li^-?!felvlV'ifi^^^K-V.. v T h e o d e : %C = — 1 0 0 % = 91,31% => n = 7 ^ " v<rj'^ / ^ 1 4 n - 6 ^ . • , - ; „ . V§y C T P T : C 7 H S . " - . ^ ^ 169
b) => C T C T cua X : O (toluen) ' < : i t • ' • • ' ' i ' / ' -
, ,,,,,, '''''^ metylbenzen. , ^ '/'^ ^.,,5,;.,, ,,.,,!,.v-.'. ^
B a i 1 0 . Mot loai xSng c6 thanh phan ve khoi Itfdng nhif sau : hexan 43%; heptan 49,5%; pentan 1,8%, c6n lai 1^ octan. Hay tInh xem can phai hon hdp Ig xSng do t6'i thieu vdi bao nhieu lit khong khi (dktc) de dam bao sir chdy diTdc hôn toan
\h khi do tao ra bao nhieu lit CO2.
i Khoi lildng cdc hidrocacbon trong 1 gam ••:•;u^;..G;:blrl. M4 W'^^,^* 0 , •: '
: * hexan : CfiHu : 0.43 (g) I^K^ st. < W;' {:>y^ti^ iH S,( C?T,i ,st)v i;b h snv
* heptan : CvHis : 0,495 (g) : u}^ t;v vl.; Á-fU %n6o H i l l " ,M£,0 nKnr> ih-i
* pentan i C j H i a : 0.018(g) • }W,y
* octan : Q H i x : 0.57 (g) ' -^i^^M - vL^^ctt^.^ : oaf^" => So mol cdc hidrocacbon trong h6n hdp : IC'US , , , v
" c. „ , 4= ^ - 0 , 0 0 5 ( » o l ) ; „ c , H , = ^ = 0 0 0 4 9 5 ( , „ o l )
" C 5 H , 2 = 0,00025(mol); n ^ H . g = ^ = 0,0005 (mol)
PhiWng trinh phan iJng : ' ' "^""p4 a&H r^K.- A j^iioff'^-'^'
CfiHu + y O j > 6CO2 + 7H2O .0 f , t * . L O - ••••5^^ i«n^"''='
(mol) 0,005 0.0475 0,03 . j t i i f i u ^oti l i i r i q i d. i A k>sa o5
C7H,6 + IIO2 > 7C02 + 8H2O ^ (mol) 0,00495 0,05445 0,03465 . H " ' ? i , j ! o ^ 6 <
C5H,2 + 8O2 > 5CO2 + 6H2O rf3R ( O:J A x;<!(;aT
(mol) 0,00025 0,002 0,00125 ^^Oi « ; 5a if,l »1
^ C s H . -H f O , ^ 8CO2-H 9H2O ^s^-^'''^--.^^'^'^^
(mol) 0,0005 0,00625 0,004 J > ••
* So mol O2: = 0,0475 + 0,05445 + 0,002 + 0,00625 = 0.1102 (moO "'>,,;
=> Vo, = 0.1102 X 22.4 = 2.4685 (lit) , , , , ^ , „ ^ , „ ^ A ^ . ,
T h I tich khong k h i : =^^\i, &f,|«fe|,j^
=> Vku = 5 X Vô = 5 X 2,4685 = 12.3425 (1ft) • ' ^ ' 5
* So mol CO2: Vco2 = 0,03 + 0,03465 + 0,00125 + 0.004 = 0.0699 (mol) ,:>,,( 1 => The tich CO2: V c o j = 0.0699 x 22.4 = 1.56576 (lit). , , . ^ 170
p^i 1 1 . A la dong dang cua benzen c6 9.43%H ve khoi luTdng.