ni = 2 < n = 2 , 4< n 2 = 3 C2H4 6 I C 3 H , l a ) PhiTdng t r i n h p h a n ỉng : 211
pran^anri vS phirn.! pnap giai Hoa ni;ic TI nmrw^TJaTroarr-nimg-CH2=CH2 + H2O CH3-CH2-OH (1) CH2=CH2 + H2O CH3-CH2-OH (1) C H 3 - C H 2 - C H 2 - O H bac 1 (2) CH3-CH=CH2 + H2O ~ CH3-CH-CH3 bac 2 (3) OH
Dat a la s6' mol C2H4 trong 1 mol hon hdp => 1 - a la so mol C3H6
Ta c6 : n = = 2,4 ^ a = 0.6 (mol)
=> nC2H4 =0.6mol; nc3H6=0,4mol
Theo phiTdng trinh (1), (2), (3) thi n^^o = n^nken = 1 mol
Ap dung dinh luat bao toan khdi lifdng ta c6 : nihh ankcn + mnjo - ^hh nrdu Y
<=> 0,6.28 + 0,4.42 + 1.18 = nihh r./<?u Y o mhh n/du Y = 51,6 (g)
Theodg: "'"'^"^'^^^=21
mnrdubac2 15
Th^nh phan phin tram khdi liTdng moi nTdu : %mc3H70H (b|c2) = ^ ^ x l 0 0 % = 34,88% %mc3H70H (b|c2) = ^ ^ x l 0 0 % = 34,88%
%rnc2H50H = X 100% = 53,49%
=> %mc3H70H (He 1) = 11.63%.
b) Phtfdng trinh phan iJng :
CH3CH2CH2OH + CuO C H 3 - C H 2 C H O + Cu + H2O CH3CH2OH + CuO CH3-CHO + Cu + H2Ọ
Bai 4. Dot chdy hoan to^n mpt hon hdp gdm hai ancol kg' tiep nhau trong day dong
ding cua metanol, ngiTcfi ta thu difdc 3,584 lit CO2 (dktc) 3,96g H2Ọ
a) Xdc dinh cong thiJc phSn tuf cua hai ancol va th^nh phSn phan trSm cvia chdng
trong hon hdp.
b) Hai ancol nay c6 the c6 cong thiJc cau tao nhif thd njkỏ
Giai
a) Dat CTTQ cua hai ancol ke tiep nhau la : C-Hj-^jOH C - H j - ^ i O H + ^ 0 2 nC02 + (n + l)H20 C - H j - ^ i O H + ^ 0 2 nC02 + (n + l)H20
0,16 mol 0,22 mol 3 584 3 584
So mol CO2: nco-, = = 0,16 (mol)
2 22.4
212
Sd mol H2O : H H J O = ^ = 0.22 (mol) n + 1
Ta c6 ti le :
V$y CTPT cua hai ancol:
^ C2H5OH + 3O2 X mol X mol n = 2.67 0,16 0.22 ni = 2< n =2.67<n2 = 3 C2H5OH C3H7OH > 2CO2 + 3H2O 2x mol 3x mol I C3H7OH + - O 2 -> 3CO2 + 4H2O
y mol I Ta c6 h? phuWng trinh : I Ta c6 h? phuWng trinh : 3y mol 4y mol '2x + 3y = 0,16 3x + 4y = 0,22 mhh = 46.0,02 + 60.0,04 = 3,32 (g) 46.0,02 x = 0,02 (mol) y = 0,04 (mol) => %mc2H50H = - Y ^ x l 0 0 % = 27,71% ^ %mc3H70H = 72,29%. b) HS tif Viet CTCT. . , . , Bkis. • • " (adsbygoogle = window.adsbygoogle || []).push({});
a) Mpt hdp chat hiJu cd X c6 cong thilc tdng quat CxHyÔ;. Tim dieu kỉn giffa X va
Y de X la ancol nọ Cho x = 3. Viét c6ng thdrc cau tao va gpi ten tat ca cdc ancol
' c6 th^ c6.
b) Ddt chay hoan toan hai ancol X, Y dong ding ke tid'p nhau, ngifdi ta thay ti 1? sd
mol CO2 va H2O tang dan. Cho X, Y la ancol nọ chiTa no hay thdm ?
Giai
a) De X la ancol no thi y = 2x + 2
Vdix = 3=>y = 8 => C6ng thdc X c6 dang : CsHxO, (dk z < 3) * Neu z = 1 => CTPT X : CjHsO * Neu z = 1 => CTPT X : CjHsO CTCT: CH3-CH2-CH2-OH:propan-l-ol CH3-CH-CH3 : propan-2-ol OH * N^u z = 2 => CTPT X : C3H8O2 C T C T : C H 3 - C H - C H 2 : propan- 1.2-diol . O H O H 9- C H o - C H o - C H jI I : propan-1.3-diol O H CH3 J' I
Phan d j n g vi phudnp ph^p giSi H6a hpc 11 HQu co - D8 Xuan Hung
* Ncu z = 3 => CTPT X : C3HHO3
CTCT : CH2-CH-CH2 : propan- 1,2,3-triol (glixerol) OH OH OH
b) Ncu X. Y Ik ancol no : C-H2;j^,OH 3n
PhiTdng trlnh phan rfng : C-Hj-^jOH + y O j > nC02 + (n + DHzO
Ta C O ti 1$ : " ^ " ^ = se tang dan khi n tang dan. Neu thoa man de bai thi hai ancol nay la no; con neu X, Y la chifa no hoac thdm thi khi dot chdy ta c6
•J^ hoac • = ^ , neu thay n v^o ta nhan thay ti 1$ giam dan. n - i n - 3
Bai 6. Cho 2,84g mpt hon hdp hai nTdu đn chdTc la d6ng ding lien tiep nhau tac dung vdi mpt liTdng Na vCfa dii, tao ra 4,6g chat r^n va V lit khi H2 d dktc.
a) TinhV.
b) Viet sd do dieu che moi nTdu tCir CH4.
(Trich TSDH Nong nghi^p, khoi A)
Giai
a) Dat CTTQ hai ri/(?u đn chiJc d6ng d k g ke tiep Ik R-OH va R-CH2OH
PhiTdng trlnh phan líng : R-OH+ Na > R-ONa + - H z t i ; . - / 2 X mol — mol R-CH2OH + Na > R-CH20Na + ^H2t (1) (2) y mol ^ mol 2
Thco phifdng trinh (1) v^ (2), khoi lifdng tang : 23 - 1 = 22 (g) => So mol riTdu : nnf,,u = '^'^22'^'^ ~ ^'^^
X V 0 08
S o m o l H j : n^^. = _ + i = ^ - ^ = 0,04 mol 2 2 2 2
=> The tich H2: = 0,04 X 22,4 = 0,896 (lit)
b) M hai rtfifu = ' = 35,5
0, Oo
CTTQ hai n/du diTdc viet lai C-H-OH
=^ 12x + y + 17 = 35,5 => 12x + y = 18,5 => y = 18,5-12x Chi c<3 X = 1 hoac x = 2 mdi phi) hdp (adsbygoogle = window.adsbygoogle || []).push({});
X = 1 => CTPT : CH3OH metanol , !
X = 2 => CTPT : C2H5OH etanol.
'ami J)
(,) phi^dng trlnh phan iJng dieu che :
* CH3OH : C H 4 + CI2 CH3CI + HCl
CH3CI + NaOH CH3OH + NaCl
* C2H5OH : 2CH4 ^°°°°^> CH^CH + 3H2
CH^CH + H2 CH2=CH2
CH2=CH2 + H2O > CH3-CH2OH.
BJki 7. Biet ring d 20"C, khoi li/dng rieng cua etanol hlng 0,789g/ml, cija nÚdc coi nhi/bing Ig/ml, ciia dung dich etanol 90% trong niTdc bkng 0,818g/ml. Hoi khi pha dung djch etanol 90% thl the tich dung dich thu difdc bing, Idn hdn hay nh6 hdn tdng the tich cua etanol va cua niTdc da diing ?
Giai
Gia sijf trong 100ml dung dich C2H5OH trong nifdc 90% c6 khoi liTdng : 'f'
mđ = V . D = 100.0,818 = 81,8 (g)
C%.m^,, 90%.81,8 „ , ,
=> inciH<:OH = = = 73,62 (g)
^2"5"n 100
=> Khoi liTdng nuTdc : m^^o = 81,8 - 73,62 = 8,18 (g) => T h l tich C2H5OH nguyen chat can diing Ik :
VcHsOH = - - = 93.31 (ml)
t. 2 H 5 0 H D 0,789
The tich nufdc can diing: V H 2 0 = ^ = ^ = 8,18 (ml) T a c 6 : V H ^ O + V C 2 H 5 0 H = 8.18 + 93,31 = 101,49ml > 100ml
=> The tich dung dich etanol nho hdn tdng the tich cua etanol va nufdc da diing. ^ ^ i 8. Mpt ancol no da chiJc A mach hcl c6 n nguyen tuf C va x nh6m OH trong cau
tsio phan tuf. Cho 7,6g ancol tren phan iJng vdi liTdng dif Na thu diTdc 2,24 lit khi
Wktc).
^) U p bieu thỉc lien hp giiTa n va x. " '
Cho n = X + 1. Xac dinh cong thiJc phan tilf cua A, tuf d6 viet cong thiJc cau tao
ciia cac dong phan Ạ '
Giai v.: .V:
Phan dging va phuong phAp giai H6a hoc 11 HOu co - B8 Xuan Hung CnH2„^2-x(OH), + xNa- 0,1.2 (mol) X S o m o l H 2 : n^^^ = 0,2 T a CO : 2,24 22,4 7,6 -^CnH2„+2-x(ONa)x+ - H 2 0,1 mol » , = 0,1 (mol) <=> 2,8n + 0,4 = 4,4x o 7n + 1 = 1 Ix. X 14n + 2 + 16x b) Cho n = X + 1 => 7(x + 1) + 1 = 1 Ix => x = 2 => n = 3 V3y CTPT A : C3H6(OH)2 CTCT A : CHg -CHg -CHg;I C H 3 -CH-CHg OH OH OH
Bai 9. Mpt hon hdp hai nfdu dufdc chia thanh hai phan b^ng nhaụ Cho phan I tac
dung vdi dung dich H2SO4 dac ndng thu dUdc hon hdp hai olefin. Dem phan II dot chay hoan toSn thu dtfdc 313,6 lit khi CO2 (d 546"k, latm) va 171 gam hdi dot chay hoan toSn thu dtfdc 313,6 lit khi CO2 (d 546"k, latm) va 171 gam hdi nifdc. Biet rkng ti so giffa khoi lUdng phan tuT nTdu thu" nhat va rifdu thu" hai (ct (adsbygoogle = window.adsbygoogle || []).push({});
23
Cling dieu ki?n 546"K, latm) la 37
a) Viét cic phUdng trlnh phSn iJng xay rạ
b) Tim cong thuTc phan tuT moi ri/dụ
Giai (Trich TSCD Kinh téKT thucit CN H)
a) VI hai rifdu khi dun ndng vdi dung dich H2SO4 dac -> hai olefin => hai nTdu phai
Ih no đn ehiJc. ,^
Dat CTTQ hai nfdu : C-Hj-^jOH
Phifdng trinh phan iJng: ' ' Phin I: C-H,- ,OH "2^°^*^ ) C-H,- + HjO Phin I: C-H,- ,OH "2^°^*^ ) C-H,- + HjO
n 2n+l iirPr n ' illi
Phan II: C-H,-,,OH +
170"C 3n 3n ' n 2n+l X mol O, b) S o m o l C 0 2 : n^Q^ = pV 1.313.6 RT 0.082.546 171 —)• n C 0 2 + nx mol = 7 (mol) (n + 1)H20 (n + l)x mol So mol H 2 O : n H 2 0 = - ^ = 9,5 (mol) 18 j n = 2,8 Ta CO : nx = 7 nx + x = 9,5 x = 2,5 Mn/du thrf hai = ^ '^^ ~
Vi tao ra olefin nen phai c6 mOt rifdu tfif 20 trd len ^ C T P T rUdu thiJ nhS't
C2H5OH (M = 46).
^ * M C 2 H 5 O H 23 „ 37
Theode—-^-^ = J^nrdu thtf hai ^ ' J^nrdu thtf hai ^ '
D$t C T T Q rifdu thu" hai: CmHzm+iOH T a cd : 14m + 18 = 74 => m = 4 T a cd : 14m + 18 = 74 => m = 4 Vay C T P T nTdu thiJ hai: C4H9OH.
B^i 10. Cho ttf tuf niTdc brom vao mpt hon hdp g6m phenol va stiren den khi ngtfng
mat mau thi het 300g dung dich ntfdc brom nong dp 3,2%. De trung hoa h5n hdp thu dUdc can dDng 14,4ml dung dich NaOH 10% (D = 1,1 Ig/cm^). Hay tinh thanh thu dUdc can dDng 14,4ml dung dich NaOH 10% (D = 1,1 Ig/cm^). Hay tinh thanh phin phan trSm cua hon hdp ban daụ
Giai OH OH O H 6 . (mol) 0,04 3Br2 0,04 C H = C H o + 3 H B r 0.04 I Br Br + Br2 (mol) 0,02 0,02
NaOH + HBr > NaBr + H2O
(mol) 0.04 0,04 , , , ,
mđNaOH= 1,11.14,4= 16(g) ^ n N a O H = ^ x - ; ^ = 0,04 (mol) 40 100 40 100
300 3 2
So mol B r 2: ne- = x = 0,06 (mol)
^•2 160 100
Khoi Irfdng phenol: mc^ H s O H = - J - >^ 94 = (g)
S6 mol Br2 d phSn tfng (2): n^^ (2) = 0,06 - 0,04 = 0,02 (mol)
=> Khoi Itfdng stiren : mcgH5CH=CH2 = ^'^'^ x 104 = 2,08 (g)
Khoi lifdng hon hdp : mhh = — + 2,08 = — (g) (adsbygoogle = window.adsbygoogle || []).push({});
3 . 3 ( j
Thanh phan phan tram cua hon hdp ban dau :
3,76 %m C 6 H 5 O H %m C 6 H 5 O H 10 3 •X100% = 37,6% %mC6H5CH=CH2 =62,4%. (1) (2) 217
Phan d jng va phuang phap giSi H6a hpc 11 HOu co - 08 Xufln Hang
Bai 11. Dót chdy hoan tôn 3 lit hon hdp X gom 2 anken két tiep nhau trong day
dong d i n g can viifa du 10,5 lit O 2 (cdc the tich khi do trong cClng dieu kien nhi^t
dp, i p suS't). Hidrat h6a hôn tokn X trong dieu kien thich hdp thu dUdc hon hdp
ancol Y, trong d6 khoi li/dng ancol bac hai b^ng 6/13 Ian tong khoi liTdng c^c ancol bac mpt. Phan trSm khoi iiTdng cua ancol bac m^t (c6 so nguyen tiJ cacbon
Idn hdn) trong Y Ih
Ạ 46,43% B. 31,58%. C. 10,88%. D. 7,89%.
. , r - i • " Trkh de thi tuyen sink Dai hoc khoi A nam 2012 •
Dot anken thu dUdc V c o 2 = V H J O (cung dk). t„5fej^ t n w
Bao toan nguyen to Oxi: 2. Vco2 + "^HJO = 2. = 2 1 lit => V c o j = V H J O = 7 lit Ta c6: so C trung binh = = - ^ 2 anken la: C 2 H 4 va C 3 H 6 .
C 2 H 4 2 ^ • ^ ^ ^
' C 3 H , 3 ^ 3 ^ 1 / 3 ^^,^,3^,,.^,
=> = — = 2 = > , = 2 lit; Vc,H^ = 1 lit. V i t i le ve the tich cung la ti VcjHf, 1/3 ^2H4 C3H6
1? ve so mol nen: Hidrat h6a X thu diTdc C2H5OH (bac I) 2 lit; C3H7OH (bac I) X lit; C H 3- C H ( O H) - C H 3 (bac II) (1 - x) lit
mancoi = niankcn + mHjO = 2-28+ 1.42 + 3.18 = 152
% nii_c3H70H = 6/(13+ 6) = 31,58%
% mcjHuO = 60/152 = 39,47% ?
• =>% mcjH^oH = 39,47% - 31.58% = 7,89% => Dap an D.
Bai 12. Cho hon hdp X gom ancol metylic, etylen glicol va glixerol. Dot chay hoan
toan m gam X thu dUdc 6,72 lit khi C O 2 (dktc). Cung m gam X tren cho tac dung vdi Na dif thu dufdc toi da V lit khi H2 (dktc). Gia tri cua V la
Ạ 3,36 B. 11.20 C.5,60 D. 6.72
. , " Trich de thi tuyen sinh Dai hoc khoi B nam 2012"
Giai
Sd do phan iJng:
C H 3 Q H ) C Q 2 ; C H 3 O H > O.5H2
C2H4(OH)2 ) 2CO2 ; C2H4(OH)2 > H 2 C3H5(OH)3 ) 3 C O 2 ; C3H5(OH)3 ^^'^ > 1.5 H 2
Ta thay so mol H2 thu diTdc luon b i n g ^ so mol C O 2 = 0.15 mol -> V = 3.36 lit (adsbygoogle = window.adsbygoogle || []).push({});
C a c h k h a c : V
Ta thay trong moi chat trong X d^u c6 s6 C \h sd nh6m O H b^ng nhaụ n^^Q^ = 0,3 => S6'mol nh6m O H = 0,3
Sđ6: R O H - > - H 2
2
0,3 0,15 ' ' • " •
=> = 0,15.22,4 = 3,36 lit => Dap an A .
Biki 13. Dot chay hôn toan hon hdp X gom hai ancol no, hai chiJc, mach hd can
vOra du V i lit khi O 2 , thu di/dc V 2 lit khi C O 2 v^ a mol H 2 O . C^c khi d^u do d dieu k i ^ n tieu chuan. Bieu thiJc lien h? giffa cdc gid tri V|, V 2 , a 1^
Ạ V| = 2V2-11,2a B. V i = V 2+ 2 2 , 4 a C. V, = V 2 - 22,4a D. V, = 2 V 2 + 11,2a
" Tn'ch de thi tuyen sinh Cao đnf; nam 2012"
Giai
Gpi cong thỉc cua 2 ancol no 2 chiJc mach hd Ian lúdt la : , i C„H2„(OH)2 vaC„H2„,(OH)2 r ; A
C„H2„(OH)2 + ^ ^ — ^ 0 2 n C 0 2 + ( n + l) H 2 0 x(mol) ( 3 n ^ 1) ^ ( n + l ) x C„H2„(OH)2 + ^ ^ ^ ^ 0 2 m C 0 2 + (m+DHzO y(mol) ^^ILlily my ( m + l ) y • " f t o d Theode : nr.- = " 0 2 l - ^ x + 2 ^ ^ 2 2 ^ " ^ 3(nx + m y ) - ( x + y( 3 n - l ) ( 3 m - 1 ) V, . , ^ V, ) = y j ^ (1) n c o 2= n x + m y = ^ (2) T i r ( l ) v a ( 2 ) = : > x + y = 22,4 nH20 = ( n + l)x + ( m + l)y = a V2 3 V 2- 2 V , O n x + my + (x + y) = a o ^ + ^ ' = a =>V| = 2 V 2 - l l , 2 a = > D a p a n A . • ' 'C\'. '
Phan d^np va phuong phAp gi^i H6a hpc 11 HOu c o - D8 XuSn Hung
Bai 14: Khi dót chdy hôn loan m gam hon hdp hai ancol no, dOn chiJc, mach hd thu
dMdc V lit khi CO2 (cl dktc) a gam H2Ọ Bieu thuTc lien h§ giffa m, a V IS:
V V V V Ạ m = a B. m = 2a C. m = 2a D . m = a + 5.6. 11,2 22.4 5.6 Giai GoiCTTQcua2ancolnođnchỉcla: C- H 2 - ^ 2 ^ ? _ ;V \, Jit's ' !'
Phtfdng trtnh phan urng ch^y: C-Hj-^j^"*" y " ^ ^ ^ nCOi + ( n + l) H 2 0
Difa vao pt piJ ta c6: noj ptf = - nco2 = 2 • ^ ^'""'^
Ap dung DL bao tôn khoi liTdng ta c6: mhhancoi + = mcoj + ' " H 2 0
3 V V 4V V *
= > m + —. .32 = .44 + a=> m = a = &-•
2 22,4 22.4 22,4 5,6
i^ D d p a n A .
Bai 15. Dot chdy hoSn t o i n hSn hdp M g6m hai nfdu (ancol) X Y d6ng dSng
ke tiep cua nhau, thu di/dc 0,3 mol CO2 va 0,425 mol H2Ọ Mat khdc, cho 0.25
mol hon hdp M tdc dung vdi Na (dtf). thu diTdc chiTa d^n 0,15 mol H2. Cong thiJc phan tiir cua X . Y l i :
A . C 2 H 6 O 2 C J H K O Z B . CzHfiO C H 4 O C. C3H6O va C4H8O ^ D. C2H6O va CjHgO C. C3H6O va C4H8O ^ D. C2H6O va CjHgO
{Trich de thi tuyen sinh cao đng khoi A, B)
Hon hdp 2 ancol X, Y ^^2> = 0.425 mol > n^Q^ = 0.3 mol (adsbygoogle = window.adsbygoogle || []).push({});
hon hdp 2 ancol X. Y la ancol nọ i f o t f l i
Goi cong thuTc chung cho hon hdp 2 ancol X, Y 1^: ^'^2^+2^a
CKH2n+20a " ^ n C O z + (n + l)H20 0.3 0,425 => 0,3(n + 1) = 0,425n => n=2.4. Matkhdctacd: CnH2n.2-ăOH)a+aNa-^ C-H2-^2-ăONa)a+f H21 0.25 0.125a ' => 0,125a < 0,15 a < 1,2 ma anguyen => a= 1.
V$y CTPT cija 2 ancol m C2H6O v^ C 3 H 8 0 = > Dap D.
220
BAI T^P T R I C NGHI|;M
u 1. Dun ndng XxX t\X mpt hon hdp g6m ancol etylic, ancol n-propylic vdi H2SO4
d$m dSc tor nhi#t dp th^p den 180"C. V|y so san pham hffu cd toi da thu difdc
trong q u i trinh tren l£k: ' ' 1 J> 1 - . .
C. l l D. 10
Ạ 6 B.9 Cfiu 2. Cho sd d6 phSn ỉng sau: Cfiu 2. Cho sd d6 phSn ỉng sau:
C l , , V +C1;.H20
Propen X •>Y + NaOH,t" -> z HjSOjdac +HNO
Glixerol trinitrat V $ y X , Y, Z i a :
A . 2-clopropen; l,3-diclopropan-2-ol, glixerol | ; B. 3-clopropen; l,3-diclopropan-2-ol, glixerol B. 3-clopropen; l,3-diclopropan-2-ol, glixerol
C. 3-clopropen; 1,3-diclopropan-l-ol, glixerol D. 2-clopropen; l,2-diclopropan-2-ol, glixerol D. 2-clopropen; l,2-diclopropan-2-ol, glixerol
Cflu 3. D^t chdy hoSn tôn 0.15 mol ancol no đn chiJc thu diTdc 6.72 lit khi CO2 d dktc. V|y C T P T cua ancol la: ,; ^.
A. C H 3 O H B. C 2 H 5 O H C. C 3 H 7 6 H D. C 3 H 5 O H
Cfiu 4. Dot chay 0,2 mol nTdu no đn chiJc mach hd thu diTdc 8,8g CO2 va m(g)
H2Ọ m c 6 gii trila:
A . 4,6 B.5,4 C. 3,6 D. 7,2
Cflu 5. Dot chdy x mol 1 rufdu no, mach hd thu difdc 6,72 lit khi CO2 (dktc) v^ 7,2g
H2Ọ Vay X c6 gid tri la: t > .fs v
A . 0,2 B.0,15 C.0,1 D.0,3
Cflu 6. Cho but-l-en tdc dung vdi H C l ta thu diTdc X. Biet X tdc dung vdi NaOH cho
sin pham Ỵ Dun ndng Y vdi H2SO4 dSc, ndng d 170°C thu diTdc Z. V$y Z1^:
A . but-2-en B. but-l-en