HYDROCACBONII.1 BI TP GIO KHOAI.1.1 BI TP V CNG THC CU TO NG NG NG PHN DANH PHPI.1.1.1 Bi tp v ng ngv Phng php : C 2 cch xc nh dy ng ng ca cc hydrocacbon : Da vo nh ngha ng ng Da vo electron ha tr xc nh Lu : C lun c ha tr IV tc l c 4e ha tr nC s c 4ne ha tr H lun c ha tr I tc l c 1e ha tr - Parafin chnh l ankan, dy ng ng parafin chnh l dy ng ng ca CH4. - Olefin chnh l anken, dy ng ng olefin chnh l dy ng ng ca C2H4 - Ankadien cn c gi l ivinyl - Aren : dy ng ng ca benzen. - Hydrocacbon : CxHy : y chn, y 2x + 2 v V d 1: Bi tp v d : Vit CTPT mt vi ng ng ca CH4. Chng minh cng thc chung ca dy ng ng ca CH4 l CnH2n+2. GII : Da vo nh ngha ng ng, CTPT cc ng ng ca CH4 l C2H6, C3H8, C4H10,, C1+kH4+2k Chng minh CTTQ dy ng ng metan CH4 l CnH2n+2 :

21

Cch 1:

Da vo nh ngha ng ng th dy ng ng ca metan phi l: CH4 + kCH2 = C1+kH4+2k Tm mi lin h gia s nguyn t C v s nguyn t H t SnC = 1 + k = n SnH = 4 + 2k = 2(k + 1) + 2 = 2n + 2 Vy dy ng ng farafin l CnH2n+2 (n 1)

Cch 2:

Da vo s electron ha tr : - S e ha tr ca nC l 4n - S e ha tr ca 1C dng lin kt vi cc C khc l 2 S e ha tr ca nC dng lin kt vi cc C khc l [2(n-2)+2] = 2n2 (v trong phn t ch tn ti lin kt n) (S d +2 v 1C u mch ch lin kt vi 1C nn dng 1e ha tr, 2C u mch dng 2e ha tr. - S e ha tr dng lin kt vi H: 4n2n-2 = 2n + 2 - V mi nguyn t H ch c 1 e ha tr nn s e ha tr ca (2n +2)nguyn t H trong phn t l 2n + 2. Cng thc chung ca ankan l CnH2n+2 (n 1) Metan c CTPT CH4 dng CnH2n+2 dy ng ng ca ankan l CnH2n+2 CT n gin nht ca 1 ankan l (C2H5)n. Hy bin lun tm CTPT ca cht trn. GII : CT n gin ca ankan l (C2H5)n. Bin lun tm CTPT ankan :

Cch 3:

V d 2:

Cch 1: Cch 2:

Nhn xt: CT n gin trn l 1 gc ankan ha tr 1 tc c kh nng kt hp thm vi 1 gc nh vy na n = 2 CTPT ankan C4H10 CTPT ca ankan trn : (C2H5)n = CxH2x+2 2n = x v 5n = 2x + 2 5n = 2.2n + 2 n = 2. CTPT ankan : C4H10 Ankan trn phi tha iu kin s H 2.s C + 2 5n 2.2n + 2 n 2 n =1 th s H l loi n= 2 CTPT ankan l C4H10 (nhn) Vy CTPT ankan l C4H1022

Cch 3:

V d 3 : Phn bit ng phn vi ng ng. Trong s nhng CTCT thu gn di y, nhng cht no l ng ng ca nhau? Nhng cht no l ng phn ca nhau.?CH3CH2CH3 (1) CH3CHClCH 3 (4) CH3CH=CH 2 (7) CH3CH2CH2Cl (2) (CH3)2CHCH 3 (5) CH2 CH2 CH2 CH2 (8) CH3CH2CH2CH3 (3) CH3CH2CH=CH 2 (6) CH3 CH3 (9) C=CH2

GII : Phn bit ng phn vi ng ng : xem I.2.2/12 Nhng cht l ng ng ca nhau l : 1 v 5 hoc 1 v 3(ankan); 6 v 7 hoc 6 v 9 (anken). Nhng cht l ng phn ca nhau : 2 v 4; 3 v 5; 6 v 9 v 8. v Bi tp tng t : 1) Vit CTPT mt vi ng ng ca C2H4. Chng minh CTTQ ca dy ng ng ca etilen l CnH2n , n 2 nguyn 2) Vit CTPT mt vi ng ng ca C2H2 . Chng minh CTTQ ca dy ng ng ca axetilen l CnH2n-2, n 2 nguyn 3) Vit CTPT mt vi ng ng ca C6H6. Chng minh CTTQ ca cc aren l CnH2n-6, n 6 nguyn

II.1.1.2 Bi tp v ng phn danh php :v Phng php vit ng phn : Bc 1: - T CTPT suy ra cht thuc loi hydrocacbon hc no. - Vit cc khung cacbon Bc 2 :- ng vi mi khung cacbon, di chuyn v tr lin kt bi (nu c), di chuyn v tr cc nhm th (nu c). - Nu c ni i hoc vng trong CTCT ca cht th xt xem c ng phn hnh hc khng. Bc 3 : - in Hidro. Lu : lm xong phi kim tra li xem cc nguyn t ng ha tr cha.

23

v Bi tp v d : V d 1 : a) Nu iu kin mt phn t c ng phn hnh hc? b) Vit tt c cc CTCT cc ng phn ca C5H10; Trong cc ng phn , ng phn no c ng phn hnh hc? c tn cc ng phn . GII : a) iu kin mt phn t c ng phn hnh hc (ng phn cis-trans) : Xt ng phn :a C=C d b f

iu kin : a d v b f - Nu a > d v b>f (v kch thc phn t trong khng gian hoc v phn t lng M)* ta c ng phn cis. - Nu a > d v b C=C > C C. * Gii thch v kh nng tham gia phn ng : - S xen ph trc xy ra vi mt ln lm cho lin kt s bn vng. - S xen ph bn xy ra vi mt nh nn lin kt p km bn vng d b t khi c tc nhn tn cng kh nng tham gia phn ng ca ankan< anken, ankin. - y do lin kt 3 lm cho khong cch 2 nhn C rt gn nhau nn lin kt 3 hi bn hn lin kt i nn kh nng tham gia phn ng ca ankin hi km hn anken. - V cng do khong cch gia hai nhn C b m mt in tch tp trung hu ht nhn nn cc ankin-1 c H linh ng tham gia c phn ng th vi ion kim loi v Bi tp tng t : 1) Gii thch quy tc cng Maccopnhicop? Minh ha bng v d c th. 2) Gii thch ti sao di lin kt n C-C trong butadien-1,3 ch bng 1,46Ao ngn hn lin kt n C-C bnh thng? 3) Ti sao khi nhit phn mui axetat vi xt iu ch ankan tng ng li phi dng xc tc CaO,to? 4) So snh nhit si ca cc hydrocacbon a) Khi khi lng phn t tng dn? b) C cng CTPT nhng khc nhau dng khung Cacbon? 5) Khi thc hin phn ng phn hy ankan bi nhit li c tin hnh nhit trn 1000oC ti sao li nhn mnh trong iu kin khng c khng kh? 6) So snh kh nng tham gia phn ng th ca cc halogen Flo, Clo, Brom, Iod vi cc ankan? 7) Ti sao cao su khi chy li c nhiu khi en? Lm th no khi en t li? 8) Trong phn ng iu ch axetilen t metan c tin hnh nhit 1500oC cn ghi km iu kin lm lnh nhanh? 9) So snh cao su thng v cao su lu ha v thnh phn, bn, ng dng? 10) Gii thch v sao cao su tng hp c tnh n hi km cao su thin nhin? 11) Phn bit cc khi nim: a) CTN, CTG, CTPT v CTCT b) Lin kt s , p . Ly propen lm v d c) ng ng, ng phn l g? Nu cc loi ng phn, cho v d? d) C th coi nguyn t Br trong phn t CnH2n+1Br l mt nhm chc c khng? Ti sao?

48

II.2 BI TON LP CTPT HYDROCACBONII.2.1 CC PHNG PHP LP CNG THC PHN T CA HYDROCACNONII.2.1.1 Phng php khi lng hay % khi lng.1) Phng php gii :

Bc 1 : Tm MA : ty theo gi thit bi cho m s dng cc cch tnh sau tm MA Tm MA da trn cc khi nim c bn, cc nh lut c bn. C nhiu cch tm khi lng phn t, ty tng gi thit bi cho m dng cch tnh thch hp. 1. Da vo khi lng ring DA (ktc) MA = 22,4 . DA vi DA n v g/l 2. Da vo t khi hi ca cht hu c A MA = MB . dA/B MA = 29 . dA/KK 3. Da vo khi lng (mA ) ca mt th tch VA kh A ktc MA = (22,4 . mA)/ VA mA: khi lng kh A chim th tch VA ktc 4. Da vo biu thc phng trnh Mendeleep Claperon: Cho mA (g) cht hu c A ha hi chim th tch VA (l) nhit T (oK) v p sut P(atm) PV = nRT M =mRT (R = 0,082 atm/ oKmol) pV

5. Da vo nh lut Avogadro: nh lut: cng iu kin nhit v p sut, mi th tch kh bng nhau u cha cng mt s phn t kh. VA = VB => nA = nB => MA = mA M BmB mA m = B MA MB

49

Bc 2 : t CTPT cht A: CxHy Xc nh thnh phn cc nguyn t trong hydrocacbon. Cch 1 :Dng khi bi -Khng cho khi lng hydrocacbon em t chy -Tnh c mC, mH t mCO2, mH2O * Tnh khi lng cc nguyn t c trong A v mA (g) cht A. - Xc nh C:mC (trong A) = mC (trong CO 2 ) = 12. m CO2 V = 12.n CO2 = 12. CO2 44 22,4 m H 2O 18 = 2.n H 2O

- Xc nh HmH(trong A) = mH (trong H2O) = 1.2n H 2O = 2

- Xc nh mA mA = mH + mA * Xc nh CTPT cht hu c A: CxHy Da trn CTTQ cht hu c A: CxHyM .m 12 x M y = = A => x = A C ; 12.m A mC mH mA y= M A .m H mA

Cch 2 : Khi bi cho bit thnh phn % cc nguyn t trong hn hp * Dng cng thc sau: CTPT A.12 x MA M .%C y = = => x = A % C %H 100% 12.100

;

y=

M A .%H 100

Cch 3 : * Tm CTG nht => CTN => CTPT Ax:y= mC mH %C %H : = : hoc x : y = : = : 12 1 12 1

- CTG nht : CaHb => CTTN : (CaHb)n - Xc nh n: bin lun t CTTN suy ra CTPT ng ca A : y 2x + 2; y chn, nguyn dng ; x 1, nguyn dng. T xc nh c CTPT ng ca cht hu c A. Lu : Khi bi tan yu cu xc nh CTG nht ca cht hu c A (hay CTN ca A) hoc khi khng cho d kin tm MA th ta nn lm theo cch trn. 2) V d 1 : Cc v d :

50

Mt hydrocacbon A c thnh phn nguyn t: % C = 84,21; %H = 15,79; T khi hi i vi khng kh bng dA/KK = 3,93. Xc nh CTPT ca A GII Bc 1: Tnh MA: Bit dA/KK => MA = MKK. dA/KK = 29.3,93 = 114 Bc 2 : t A : CxHyM 12x y = = A %C %H 100 M .%C 114.84,21 x= A = =8 12.100 12.100 M .%H 114.15,79 y = A = = 18 1.100 1.100

Suy ra CTPT A: C8H18 V d 2 : Mt hydrocacbon A th kh c th tch gp 4 ln th tch ca lu hunh ioxit c khi lng tng ng trong cng iu kin. Sn phm chy ca A dn qua bnh ng nc vi trong d th c 1g kt ta ng thi khi lng bnh tng 0,8g. Tm CTPT A. GII * Tm MA : 1VA = 4VSO2( cng iu kin ) nA = 4nSO2 mSO2 mA 1 4 =4 = MA M SO2 M A M SO2 M SO 2 4 = 64 = 16 4

(A v SO2 c khi lng tng ng nhau)

MA =

Cch 1 : gii theo phng php khi lng hay % khi lng : t A : CxHy Bnh ng Ca(OH)2 hp th CO2 v H2OCa(OH)2 + CO2 CaCO3 + H2O

m = mCaCO3 = 1g nCO2 = nCaCO3 = 1/100= 0,01mol nC = nCO2 = 0,01mol mC = 12.0,01=0,12g mCO2 = 0,01.44 = 0,44g rmbnh = mCO2 + mH2O mH2O = 0,8-0,44 = 0,36gmH = 2 m H 2O 18 =2 0,36 = 0,04 g 18

LBT khi lng (A) :mA = mC + mH = 0,12 +0,04 = 0,1651

Ta c

M .m 12 x M y 16.0,12 = = A => x = A C = =1 mC mH mA 12.m A 12.0,16 M .m 16.0,04 y= A H = =4 mA 0,16

Vy CTPT A : CH4 Cch 2 : Bin lun da vo iu kin y 2x + 2; y chn, nguyn dng ; x 1, nguyn x =1 v y = 4 CTPT A. V d 3: t chy hon ton 2,64g mt hydrocacbon A thu c 4,032 lt CO2 (ktc). Tm CTPT A? GII * Tm thnh phn cc nguyn t : mC (trong A) = mC (trong CO2) = (4,032/ 22,4)*12 = 2,16g mH = mA mC = 2,64 2,16 = 0,48gx:y=

CTN : C3H8 CTTN : (C3H8)n Bin lun : S H 2 s C +2 8n 6n + 2 n 1 m n nguyn dng n = 1 CTPT A : C3H8

m C m H 2,16 0,48 : = : =3:8 12 1 12 1

II.2.1.2) Phng php da vo phn ng chy:Du hiu nhn bit bi ton dng ny : bi t chy mt cht hu c c cp n khi lng cht em t hoc khi lng cc cht sn phm (CO2, H2O) mt cch trc tip hoc gin tip (tc tm c khi lng CO2, H2O sau mt s phn ng trung gian). 1) Phng php gii:

Bc 1 : Tnh MA ( phn II.2.1.1) Bc 2 : t A : CxHy * Vit phng trnh phn ng chy.0 y y C x H y + x + O 2 t xCO 2 + H 2 O 4 2

* Lp t l tnh x,y

MA(g) mA(g)

44x mCO2

9y mH2O

y y x+ MA 1 x 44x 9y 4= hoc = = 2 = = n A n O2pu n CO2 n H2O m A m CO 2 m H 2O

52

x=

M A .m CO 2 44m A

,y =

M A .m H 2O 9m A

* T suy ra CTPT A Mt s lu : 1) Nu bi cho: oxi ha han tan mt cht hu c A th c ngha l t chy han tan cht hu c A thnh CO2 v H2O 2) Oxi ha cht hu c A bng CuO th khi lng oxy tham gia phn ng ng bng gim khi lng a(g)ca bnh ng CuO sau phn ng oxi ha. Thng thng trong bi ton cho lng oxi tham gia phn ng chy, tm khi lng cht hu c A nn ch n nh lut bo ton khi lng mA + a = mCO2 + mH2O 3) Sn phm chy (CO2, H2O) thng c cho qua cc bnh cc cht hp th chng. 4) Bnh ng CaCl2 (khan), CuSO4 (khan), H2SO4 c, P2O5, dung dch kim, hp th nc. Bnh ng cc dung dch kimhp th CO2. Bnh ng P trng hp th O2. 5) tng khi lng cc bnh chnh l khi lng cc cht m bnh hp th. 6) Nu bi ton cho CO2 phn ng vi dung dch kim th nn ch n mui to thnh xc nh chnh xc lng CO2. 7) Vit phng trnh phn ng chy ca hp cht hu c vi oxy nn oxy li cn bng sau t v sau n v trc. Cc nguyn t cn li nn cn bng trc, t v trc ra v sau phng trnh phn ng. 2) Bi tp v d : V d 1 : t hon ton 0,58g mt hydrocacbon A c 1,76g CO2 v 0,9g H2O. Bit A c khi lng ring DA @ 2,59g/l. Tm CTPT A Tm tt : 0,58g X + O2 (1,76g CO2; 0,9 g H2O) DA @ 2,59g/l. Tm CTPT A? GII : * Tm MA : Bit DA => MA = 22,4.2,59 @ 58 * Vit phng trnh phn ng chy, lp t l tm x,y0 y y C x H y + x + O 2 t xCO 2 + H 2 O 4 2

MA(g) mA(g)

44x mCO2

9y mH2O

53

MA 58 44x 9y 44x 9y = = = = = m A m CO 2 m H 2O 0,58 1,76 0,9

x=4 y =10 Vy CTPT A : C4H10 V d 2 : Khi t chy han tan 0,42 g mt Hydrocacbon X thu tan b sn phm qua bnh 1 ng H2SO4 c, bnh 2 ng KOH d. Kt qu, bnh 1 tng 0,54 g; bnh 2 tng 1,32 g. Bit rng khi ha hi 0,42 g X chim th tch bng th tch ca 1,192 g O2 cng iu kin. Tm CTPT ca X Tm tt :0,42g X (CxHy) +O2 CO2 Bnh 1ng ddH2SO4 Bnh 2 ng KOHd CO2 -CO , H2O -H2O, 2 m2=1,32g m =0,54g1

Tm CTPT X? GII * Tnh MX : 0,42g X c VX = VO2 ca 0,192g O2 (cng iu kin) => nX = nO2 => => M X =m mX = O2 M X M O2 = 0,42.32 = 70 0,192

m X .M O 2 m O2

* Gi X : CxHy0 y y C x H y + x + O 2 t xCO 2 + H 2 O 4 2

MX 0,42 Ta c :MX 44x 9y = = m X m CO2 m H2O

44x 9y (g) mCO2 mH2O (g) (1)

bi cho khi lng CO2, H2O gin tip qua cc phn ng trung gian ta phi tm khi lng CO2, H2O * Tm mCO2, mH2O : - Bnh 1 ng dd H2SO4 s hp th H2O do tng khi lng bnh 1 chnh l khi lng ca H2O : rm1 = mH2O=0,54g (2) - Bnh 2 ng dd KOH d s hp th CO2 do tng khi lng bnh 2 chnh l khi lng ca CO2 : rm2 = mCO2 =1,32g (3)

54

(1), (2), (3)

70 44x 9y = = 0,42 1,32 0,54

x=5 y = 10 Vy CTPT X : C5H10 (M = 70vC)

II.2.1.3 Phng php th tch (phng php kh nhin k):v Phm vi ng dng : Dng xc nh CTPT ca cc cht hu c th kh hay th lng d bay hi. v C s khoa hc ca phng php : Trong mt phng trnh phn ng c cc cht kh tham gia v to thnh ( cng iu kin nhit , p sut) h s t trc cng thc ca cc cht khng nhng cho bit t l s mol m cn cho bit t l th tch ca chng. 1) Phng php gii Bc 1 : Tnh th tch cc kh VA, VO2, VCO2, VH2O (hi) Bc 2 : Vit v cn bng cc phng trnh phn ng chy ca hydrocacbon A di dng CTTQ CxHy Bc 3 : Lp cc t l th tch tnh x,y0 y y C x H y + x + O 2 t xCO 2 + H 2 O 4 2 y y 1(l) x(l) (l) x + (l) 4 2

y y y y x+ x+ 1 x x 1 4 = 4 = = = 2 hay = = 2 VA V 2 VCO 2 VH 2O nA nO2 nCO 2 n H 2O x= y= VCO 2 VA 2VH 2O VA = n CO 2 ; nA 2n H 2O nA

VA(l)

VO2 (l)

VCO2 (l) VH2O (hi)(l)

=

Cch khc : Sau khi thc hin bc 1 c th lm theo cch khc: - Lp t l th tch VA: VB : VCO2 : VH2O ri a v t l s nguyn ti gin m:n:p:q. - Vit phng trnh phn ng chy ca hp cht hu c A di dng: mCxHy + nO2 t pCO2 + qH2O - Dng nh lut bo ton nguyn t cn bng phng trnh phn ng chy s tm c x v y =>CTPT A * Mt s lu : - Nu VCO2 : VH2O = 1:1 => C : H = nC : nH = 1: 2o

55

- Nu tan cho oxy ban u d th sau khi bt tia la in v lm lnh (ngng t hi nc) th trong kh nhin k c CO2 v O2 cn d. Bi tan l lun theo CxHy - Nu tan cho VCxHy = VO2 th sau khi bt tia la in v lm lnh th trong kh nhin k c CO2 v CxHy d. Bi tan l lun theo oxy. - Khi t chy hay oxi ha han ton mt hydrocacbon m gi thit khng xc nh r sn phm, th cc nguyn t trong hydrocacbon s chuyn thnh oxit bn tng ng tr: N2 kh N2 Halogen kh X2 hay HX (ty bi) 2. Bi tp v d V d 1: Trn 0,5 l hn hp C gm hydrocacbon A v CO2 vi 2,5 l O2 ri cho vo kh nhin k t chy th thu c 3,4 l kh, lm lnh ch cn 1,8 l. Cho hn hp qua tip dung dch KOH (c) ch cn 0,5 l kh. Cc V kh o cng iu kin. Tm CTPT ca hydrocacbon A. Tm tt : CxHy : a (l) Gi 0,5 l hn hp CO2 : b (l) o t CO ,O d,H O ll(- H2O) CO ,O d 0,5l hn hp + 2,5l OKOH(- CO2)2 2 2 2 2 2

O2 d GII :

* O2 d , bi tan l lun theo Hydrocacbon A0 y y C x H y + x + O 2 t xCO 2 + H 2 O 4 2 y y a ax + (lt) ax a 2 4

CO2 CO2 b b Ta c Vhh = a + b = 0,5 (1) VCO2 = ax + b = 1,8 0,5 = 1,3 (2)y = 3,4 1,8 = 1,6 2 y VO2 d = 2,5 - a x + = 0,5 4 y ax + a = 2 4

(lt)

VH2O = a

(3)

(4) (5)

ax + 3,2/4 = 2 ax = 1,2 (2), (3) VCO2 = b = 0,1

56

Vhh = a + b = 0,5 a = 0,4 x = ax /a = 3 y = ay/a = 8 Vy CTPT ca A l C3H8 V d 2 : Trn 12 cm3 mt hydrocacbon A th kh vi 60 cm3 oxi (ly d) ri t chy. Sau khi lm lnh nc ngng t ri a v iu kin ban u th th tch kh cn li l 48 cm3, trong c 24cm3 b hp th bi KOH, phn cn li b hp th bi P. Tm CTPT ca A (cc th tch kh o trong cng iu kin nhit v p sut) Tm tt :12cm CxHy 60cm O2 (d)3 3

ot

CO2 lam lanh H2O -H2O O2d

CO2 O2 d

24cm kh b hap thu bi KOH (- CO2)3

3

kh con lai b hap thu bi P (-O2) (V=48cm )

GII : * Tnh cc V: VCO2 = 24cm3 VO2 d = 48 24 = 24cm3 VO2 p = 60 24 = 36 cm3 * Tm CTPT : Cch 1: Tnh trc tip t phng trnh phn ng t chy:0 y y C x H y + x + O2 t xCO2 + H 2 O 4 2 y 12 x + 12 12x (cm3) 4

VCO2 =12x = 24 => x = 2 VO2 d = 60 12 x + = 24 => y = 4 CTPT ca A: C2H4 Cch 2: Lp t l th tchx+ y 4 = y 2 y 4

1 = VA VO 2

x = VCO 2 VH 2O

0 y y C x H y + x + O 2 t xCO 2 + H 2 O 2 4 y y 1 x (cm3) x + 2 4

12

36

24

(cm3)

57

1 = VA VO 2

x+

y 4 =

y x+ x 1 x 4 VCO 2 12 = 36 = 24 => x = 2 v y = 4

CTPT ca A: C2H4 Cch 3: Nhn xt: t 12 cm3 A dng 36 cm3 oxy v to ra 24 cm3 CO2 Suy ra 12C x H y + 36O 2 t 24CO 2 + ?H 2 O 0

LBT (O): => 12C x H y + 36O 2 t 24CO 2 + 24H 2 O LBT (C): 12x = 24 => x = 2 LBT (H) :12y = 48 => y = 4 Vy CTPT ca A l C2H4 V d 3 : Trong mt bnh kn th tch 1dm3 c mt hn hp ng th tch gm hydrocacbon A v O2 133,5 oC, 1 atm. Sau khi bt tia la in v a v nhit ban u (133,5 oC) th p sut trong bnh tng ln 10% so vi ban u v khi lng nc to ra l 0,216 g. Tm CTPT A Tm tt :CxHy(A) ot O2 3 V = 1dm o t=133,5 C,P1=1atm sp chay3

0

(lng H2O tao ra la 0,216g)

V=1dm o t=133,5 C, P2 tang 10%

GII : Tm CTPT A?n1 = PV 1.1 = = 0,03(mol) RT 0,082.(273 + 133,5)

V hn hp ng th tch nn nA = nO2 = 0,03/2 = 0,015 mol => CxHy d, bin lun theo O2 Sau khi a v nhit ban u, cc kh to p sut c trong bnh gm H2O, CO2, CxHy d c s mol l : n2 = n1 . P2/P1 = 0,03.110/100 = 0,033 mol nH2O = 0,216/18 = 0,012 mol LBT khi lng (O) : nO2 = n CO2 + 1/2n H2O => n CO2 = nO2 1/2nH2O = 0,015-0,012/2 = 0,009mol nCxHyd = n2 - nCO2 - nH2O = 0,033-0,012-0,009 =0,012mol =>nCxHyphn ng = 0,015-0,012 = 0,003 mol0 y y C x H y + x + O 2 t xCO 2 + H 2 O 4 2

58

1 0,003 Ta c :x+

y x + 4

x 0,009

y 2

(mol) (mol)

0,015

0,012

y y 1 4 = x = 2 = 0,003 0,015 0,009 0,012

=> x = 3 y=8 Vy CTPT A : C3H8

II.2.1.4 Phng php gi tr trung bnh (xc nh CTPT ca hai hay nhiu cht hu c trong hn hp):L phng php chuyn hn hp nhiu gi tr v mt gi tr tng ng, nhiu cht v mt cht tng ng v c im Phng php gi tr trung bnh c dng nhiu trong ha hu c khi gii bi tan v cc cht cng dy ng ng. Mt phn bn cht ca gi tr trung bnh c cp n vic tnh phn trm n v v khi lng hn hp kh trong bi tan t khi hi chng u lp 10. Do , hc sinh d dng lnh hi phng php ny xc nh CTPT ca hai hay nhiu cht hu c trong hn hp. II.1.4.1 Phng php khi lng phn t trung bnh ca hn hp ( M hh ) Cht tng ng c khi lng mol phn t M hh l khi lng mol phn t trung bnh ca hn hp. Cc bc gii : Bc c bn : Xc nh CTTQ ca hai cht hu c A,B Bc 1 : Xc nh CTTB ca hai cht hu c A, B trong hn hp Bc 2 : Tm M hh qua cc cng thc sau :M hh = m hh n A .M A + n B .M B %A.M A + %B.M B %A.M A + (100 - %A )M B = = = n hh nA + nB 100 100 VA .M A + VB .M B VA .M A + VB .M B %A.M A + (100 - %A )M B = = VA + VB V 100

HocM hh = d hh/X .M X =

Gi s MA< MB => MA< M hh < MB Bc 3 : Bin lun tm MA, MB hp l => CTPT ng ca A v B Phm vi ng dng: s dng c li nhiu i vi hn hp cc cht cng dy ng ng 1) Phng php CTPT trung bnh ca hn hp: v Phm vi p dng : Khi c hn hp gm nhiu cht, cng tc dng vi mt cht khc m phng trnh phn ng tng t nhau (sn phm, t l mol gia nguyn liu v59

sn phm, hiu sut, phn ng tng t nhau), c th thay th hn hp bng mt cht tng ng, c s mol bng tng s mol ca hn hp. Cng thc ca cht tng ng gi l CTPT trung bnh. v Phng php gii : Bc 1 : t CTPT ca hai cht hu c cn tm ri suy ra CTPT trung bnh ca chng : t A : CxHy ; B : CxHy CTPTTB : C x H y Bc 2 : Vit phng trnh phn ng tng qut v d liu bi cho tnh x, y Bc 3 : bin lun Nu x n hh S Khi lng hn hp trc v sau phn ng bng nhau (LBTKL). mhh T = mhhS78

M T < M S dT < dS + Phn ng vi dd brm v thuc tm d, tng khi lng ca dd chnh l khi lng ca hydrocacbon cha no. CnH2n+2-2k + kBr2 CnH2n+2-2kBr2k + Phn ng c trng ca ankin-1 : 2R(C CH)n + nAg2O 2R(C CAg)n + nH2O Khi lm ton hn hp do s mol cc cht lun thay i qua mi th nghim do khi qua th nghim mi ta nn lit k s mol ca hn hp sau v trc mi th nghim. Lu : trong cng thc tnh PV = nRT th V l Vbnh. V d : Mt bnh kn c dung tch 17,92 lt ng hn hp gm kh hidro v axetilen ( OoC v 1 atm) v mt t bt Ni xc tc. Nung nng bnh mt thi gian sau lm lnh n 0oC. a) Nu cho lng kh trong bnh qua dd AgNO3/NH3 s sinh ra 1,2 gam kt ta vng nht. Tm s gam axetilen cn li trong bnh. b) Cho lng kh cn li qua dd Brom thy khi lng dung dch tng ln 0,41 gam. Tnh s gam etilen to thnh trong bnh. c) Tnh th tch etan sinh ra v th tch H2 cn li sau phn ng. Bit t khi hn hp u (H2 + C2H2 trc phn ng) so vi H2 = 4. Bt Ni c th tch khng ng k. GII a) Tnh lng axetilen cn d : v Phn 1 : Sn phm chy to kt ta vng nht vi ddAgNO3/NH3 chng t hn hp cn axetilen d Cc ptp : C2H2 + Ag2O ddAgNO C2Ag2 + H2O /NH 3 3

nC2Ag2 =

1,2 = 0,005 (mol) 240

Lng axetilen cn li trong bnh : nC2H2 d = 2nC2H2 p = 2nC2Ag2 = 2.0,005 = 0,01 (mol) b) Tnh s gam etilen to thnh trong bnh : v Phn 2 : Cc ptp : C2H4 + Br2 C2H4Br2 b b b (mol) C2H2 + 2Br2 C2H2Br4 0,005 2.0,005 (mol) p dng LBT khi lng : mbnh tng = mC2H4 + mC2H2 mC2H4 = mbnh tng mC2H2 = 2(0,41- 0,005.26) = 0,56 (g)

79

nC2H4 =

0,56 = 0,02 (mol) 28

c) Th tch etan sinh ra v th tch H2 cn li : v Phn 3 : nhh =17,92 = 0,8 (mol) 22,4

Gi x (mol) l s mol H2 trong 0,8 mol hn hp ban u. M hh = 4.2 = 8M hh = 2.x + 26(0,8 - x) =8 0,8

x = 0,6 (mol) nC2H2 b = 0,2 (mol) Cc ptp :t C2H2 + H2 Ni, C2H4 C 0,02 0,02 (mol) Ni, t C C2H2 + 2H2 C2H6 y 2y y (mol) Gi y l s mol etan to thnh. nC2H2 p to etan = y = 0,2 (0,01 + 0,02) = 0,17 (mol) nEtan = 0,17 (mol) nH2 cn li = 0,6 (0,02 + 2.0,17) = 0,24 molo o

Ghi ch : ta nn t cc n s ngay t u v phi cng n v. Qua mi th nghim s gip ta tm mt n s. Lu lng hn hp mang phn ng trong mi th nghim c th khc nhau nhng t l thnh phn cc cht trong hn hp khng i.

II.3.2 PHNG PHP GII MT S BI TON XC NH THNH PHN HN HP CC HYDROCAC BIT CTPTBi 1 : t chy hon ton 100cm3 hn hp A gm : C2H6, C2H4, C2H2 v H2 th thu c 90cm3 CO2. Nung nng 100cm3 A c s hin din ca Pd th thu c 80cm3 hn hp kh B. Nu cho B tip tc qua Ni, to th thu c cht duy nht. Tm % cc cht trong hn hp. GII : t 100cm3 hh A gm : C2H6 : a C2H4 : b80

a + b + c + d = 100 (cm3) v TN1 : Cc ptp :

C2H2 : c H2 : d (cm3)

C2H6 + 7/2O2 2CO2 + 3H2O a 2a (mol) C2H4 + 3O2 2CO2 + 2H2O b 2b (mol) C2H2 + 5/2O2 2CO2 + H2O c 2c (mol) H2 + 1/2O2 H2O d d (mol) Lu : H2 cng chy trong Oxi, sn phm l H2O. VCO2 = 2(a + b + c) = 90 (cm3) a + b + c = 45 (cm3) d = 100 45 = 55(cm3) (1) v TN2 : xc tc Pd,toC th mt lin kt p b t, sn phm cng l anken. t C2H2 + H2 Pd, C2H4 C c c (cm3) Th tch hn hp gim : Vkh gim = 2c c = c = 100 80 = 20 (cm3) (2)o

Hn hp kh B gm : C2H6 : a C2H4 : b + c (cm3) H2 : d c = 55 20 = 35 v TN3 : t C2H4 + H2 Ni, C2H6 C b+c b+c (cm3) V ch thu c mt kh duy nht C2H4 v H2 u ht. b + c = 35 b = 35 c = 35 20 = 15 (cm3) a = 100 (b + c + d) = 100 (15 + 20 + 55) = 10 (cm3) % th tch cc cht trong hn hp :o

a 10 .100% = .100% =10% 100 100 b 15 %VC2H4 = .100% = .100% = 15% 100 100

%VC2H6 =

81

20 c .100% = .100% = 20% 100 100 d 55 .100% = .100% = 55% %VH2 = 100 100

%VC2H2 =

Bi 2 : Cho 11 gam hn hp gm 6,72 lt hydrocacbon mch h A v 2,24 lt mt ankin. t chy hn hp ny th tiu th 25,76 lt Oxi. Cc th tch o ktc. a) Xc nh loi hydrocacbon. b) Cho 5,5 gam hn hp trn cng 1,5 gam hidro vo mt bnh kn cha sn mt t bt Ni ( ktc) un nng bnh phn ng xy ra hon ton ri a v OoC. Tnh thnh phn % hn hp cui cng v p sut trong bnh. GII : Da vo ptp chy, t s mol cc cht v gii h phng trnh tm cc gi tr x, n. a) Xc nh loi hydrocacbon : S mol cc cht : nA = nankin =6,72 = 0,3 (mol) 22,4

2,24 = 0,1 (mol) 22,4 25,76 nO2 = = 1,15 (mol) 22,4 A : C x H y : 0,3 (mol) Gi 11g hn hp Ankin : C n H 2n - 2 : 0,1

Cc ptp :0 y y C x H y + x + O 2 t xCO 2 + H 2 O 4 2

0,3 0,3(x + y/4)C n H 2n -2 +

(mol)

0,1 0,1(3n-1)/2y 3n - 1 nO2 = 0,3(x + ) + 0,1( ) = 1,15 4 2

3n - 1 O 2 nCO 2 + (n - 1)H 2 O 2

(mol)

mhh = (12x + y)0,3 + (14n - 2)0,1 = 11 36x + 3y + 14n = 112 (1) 4x + y + 2n = 16 (2) (1) 7.(2) y = 2x Thay y = 2x vo (1), (2) :82

36x + 6x +14n = 112 4x + 2x + 2n = 16 3x + n = 8 x < x=2 n=2 8 = 2,66 3

C2 H 4 C2H2

b) Tnh thnh phn % hn hp cui cng v p sut trong bnh : nH2 = 1,5/2 = 0,75 (mol) Hn hp mi gm C2H4 : 0,015 mol C2H2 : 0,05 mol H2 : 0,75 mol Cc ptp : t C2H4 + H2 Ni, C2H6 C 0,15 0,15 0,15 Ni, t C C2H2 + 2H2 C2H6 0,05 0,1 0,05 Sn phm thu c gm : C2H6 : 0,2 mol H2 d : 0,5 mol T l %th tch chnh l t l % s mol :o o

%VC2H6 =

0,2 .100% = 28,6% 0,2 + 0,5 0,5 %VH2 = .100% = 71,4% 0,2 + 0,5

Tnh p sut : PVbnh = nRT Trc phn ng n1 = nC2H4 + nC2H2 + nH2 = 0,15 + 0,05 + 0,75 = 0,95 (mol) Sau phn ng n2 = nC2H6 + nH2d = 0,7 (mol) cng iu kin Vbnh, T = constP2 n 2 n 0,7 = P2 = 2 = @ 0,73 (atm) (P1 = 1atm ) P1 n 1 n 1 0,95

Bi 3 : Mt hn hp kh A gm C2H2 v H2 c t khi hi so vi khng kh bng 0,4. un nng A vi xc tc Ni mt thi gian thu c hn hp kh B, t khi ca B so vi khng kh bng4 . Nu cho ton b lng B qua dung dch KMnO4 d th cn li kh D thot ra 7

ngoi, t khi ca D so vi H2 bng 4,5. Cc th tch o ktc. a) Tnh thnh phn % th tch ca hn hp A. b) Tnh t s th tch ca A so vi th tch B. Gii thch s thay i th tch . c) Tnh thnh phn % th tch ca hn hp kh D83

d) Bit VB = 3,136 lt, hi nu hp thu ht lng B ny trong dd Brom d th khi lng cc sn phm thu c l bao nhiu? Tm tt :hh A C2H2 Ni,toC dd KMnO4 d hh B kh D thoat ra H2 dA/KK=0,4 dD/KK= 4,5 dB/kk= 4/7 dd Br2 d m sp? 3,136 lt hh B

GII : a. % th tch ca hn hp A : Trong A, t C2H2 : a (mol) H2 : b (mol) M hh A = 0,4.29 = 11,6

26a + 2b = 11,6 b = 1,5a a+b a a %C2H2 = .100% = .100% = 40% a+b a + 1,5a

%H2 = 100% - 40% = 60% b. T l th tch ca A so vi B : Phn ng c th xy ra : t C2H2 + H2 Ni, C2H4 C t C2H2 + 2H2 Ni, C2H6 C Phn ng cng H2 lm gim s mol kh nhng khng lm thay i khi lng kh nB < nA VB < VA. Ta c :o o

mA = mB M hh A .nA = M hh B .nB

n A M hhB = n B M hhA

116 nA 4 116 10 Vi M hh B = 29. = (gam/mol) v M hh A = 11,6 = 7 = 7 7 n B 11,6 7

V t l th tch bng t l s mol kh nn VA : VB = 10 : 7 c. % th tch hn hp kh D : Hn hp B gm C2H4, C2H6, C2H2 d v H2 d khi cho qua dung dch KMnO4 th C2H4 v C2H2 d b oxi ha v gi li trong dung dch : C2H4 + [O] + H2O C2H4(OH)2 C2H2 + 4[O] HOOCCOOH

84

Hn hp kh D thot ra gm C2H6 v H2 d. Trong hn hp D gi C2H6 : x(mol) v H2 d : y (mol). Ta c M hh D = 4,5.2 = 9 %C2H6 =30x + 2y = 9 y = 3x x+y

x x .100% = .100% = 25% x+y x + 3x

%H2 = 100% - 25% = 75% d. Khi lng cc sn phm : t C2H2 + H2 Ni, C2H4 C u u u (mol) t C2H2 + 2H2 Ni, C2H6 C x 2x x (mol) Gi u (mol) l s mol C2H4 thu c B gm : C2H4 : u (mol) C2H6 : x (mol) C2H2 d : a (u + x) (mol) H2 d : 3x (mol)o o

nB =

3,136 = 0,14 (mol) u + x + a (u + x) + 3x = 0,14 a + 3x = 0,14 (1) 22,4

nH2 ban u = 1,5a(mol) u + 2x +3x = 1,5 1,5a = u + 5xn A 10 2,5a 10 = a = 0,8 (mol) = nB 7 0,14 7

(2)

T (1) x = 0,02 (mol) T (2) u = 0,02 (mol) Trong B ch c C2H4 : 0,02 (mol) v C2H2 : 0,04 (mol) cho phn ng cng vi dung dch Br2 : C2H4 + Br2 C2H4Br2 0,02 0,02 (mol) C2H2 + 2Br2 C2H2Br4 0,04 0,04 (mol) mC2H4Br2 = 0,02.188 = 3,76 (gam) mC2H2Br4 = 0,04.346 = 13,84 (gam) Bi 4 : Mt bnh kn 2 lt 27,3oC cha 0,03 mol C2H2 ; 0,015 mol C2H4 v 0,04 mol H2 c p sut P1. Tnh P1 - Nu trong bnh c mt t bt Ni lm xc tc (th tch khng ng k) nung bnh n nhit cao phn ng xy ra hon ton, sau a v nhit ban u c hn hp kh A c p sut P2. - Cho hn hp A tc dng vi lng d dd AgNO3/NH3 thu c 3,6 gam kt ta. Tnh P2. Tnh s mol mi cht trong A..85

Tnh p sut P1 : Tng s mol cc cht trc phn ng : n1 = nC2H2 + nC2H4 + nH2 = 0,03 + 0,015 + 0,04 = 0,085 (mol) p dng phng trnh trng thi kh l tng :PV = nRT P1 =

GII :

Tnh p sut P2 v s mol cc cht : Cc ptp : t C2H2 + H2 Ni, C2H4 C a a a (mol) Ni, t C C2H4 + H2 C2H6 b b b (mol) V s mol H2 = 0,04 < nC2H2 + nC2H4 = 0,045 (mol) nn phn ng ht H2 t a, b l s mol H2 tham gia hai phn ng trn a + b = 0,04 (1)(mol) C2H2 cn d sau phn ng trn tc dng vi lng d dd AgNO3/NH3 : HC CH + Ag2O ddAgNO C2Ag2 + H2O /NH 0,015 0,015 (mol)o o 3 3

n1 RT1 0,085.0,082.300,3 = = 1,05 atm V 2

nC2Ag2 =

nC2H2 d = 0,015 mol nC2H2 phn ng = a = 0,03 0,015 = 0,015 (mol) b = 0,04 a = 0,04 0,015 = 0,025 (mol) nC2H4 = nC2H4 b + a = 0,015 + 0,015 = 0,03 (mol) nC2H4 d = 0,03 b = 0,03 0,025 = 0,005 (mol) p sut P2 : n2 = nC2H2 d + nC2H4 d + nC2H6 = 0,015 + 0,005 + 0,025 = 0,045 (mol)P2 = n2 RT 0,045.0,082.300,3 = = 0,554 (atm) V 2

3,6 = 0,015 (mol) 240

Bi 5 : Cho a gam CaC2 cha b% tp cht tr tc dng vi nc th thu c V lt C2H2 (ktc) 1) Lp biu thc tnh b theo a v V 2) Nu cho V lt trn vo bnh kn c than hot tnh nung nng lm xc tc,to trong bnh toC p sut P1. Sau phn ng thu c hn hp kh, trong sn phm phn ng chim 60%V, nhit khng i, p sut P2 Tnh hiu sut ca phn ng. 3) Gi s dung tch bnh khng i, th tch cht rn khng ng k. Hy86

a) Lp biu thc tnh p sut P2 theo P1 v hiu sut h. b) Tnh khong gi tr ca P2 theo P1 GI I : 1) Lp biu thc tnh b theo a v V : Cch 1 : CaC2 + 2H2O Ca(OH)2 + C2H2V 22,4 V 22,4 V 22,4 V 22,4

(mol)

Gi V l th tch C2H2 sinh ra. nC2H2 =

mCaC2 = 64.a-

mtp cht = (a-

64V ) 22,4

b% =

Cch 2 : CaC2 + 2H2O Ca(OH)2 + 64(g) a 0,01b(g) b(%) Ta lp c t l :

64V 7a - 20V 22,4 .100% = .100% a 7a

C 2 H2 22,4(lt) V(lt)

64 22,4 = 64V = 22,4a - 0,224b a - 0,01b V 22,4a - 64V 7 a - 20V 7a - 20V b = = b%= .100% 0,224 7 7a

2) Tnh hiu sut phn ng : Xt phn ng : 3C2H2 C hoat tinh ,600 C6H6 C x x/3 (lt) Gi x l th tch C2H2 tham gia phn ng trn Tng s mol cc cht sau phn ng :o

C2 H 2 : V - x (lt) x C6 H 6 : 3 n = V x + x/3 = V 2/3x (l)

VC6H6 = 60%Vhh

87

Cch 1 : Tnh theo cht tham gia : 2 9 x = 0,6(V - x) x = V 3 3 11 9 V x 9 Hiu sut phn ng h = .100% = 11 .100% = .100% = 81,81% V V 11

Cch 2 : Tnh hiu sut phn ng theo sn phm :x 2 11 = 0,6(V - x) V = x 3 3 9

Theo phn ng : 3C2H2 C hoat tinh ,600 C6H6 C V V/3 (theo l thuyt)o

h% =

V(thuc t' ) x/3 x/3 9 .100% = .100% = .100% = .100% = 81,81% 11x/9 V(l thuy' t) V/3 11 3

3) a) Lp biu thc tnh p sut P2 theo P1 v hiu sut h : Ta c pt TTKLT : PVbnh = nRT Vbnh, T = const P2 n 2 V2 = = P1 n 1 V1

V2, V1 : S mol cc cht trong bnh. V V2 = V 2/3x V1 = V P2 V2 = = P1 V1 P 300 - 2h 300 - 2h 2 = P2 = P1 P1 300 300 V2 x 3 = 1 - 2 x = 1 - 2h V 3V 3.100

b) Tnh khong gi tr ca P2 theo P1 : Ta c 0 < h 100300 - 2h =1 300 300 - 2h 1 h = 100 = 300 3

h=0

1 P1 P2 P1 3

88

Bi 6 : Khi sn xut t n ta thu c hn hp rn gm CaC2, Ca v CaO (hh A). Cho hn hp A tc dng ht vi nc th thu c 2,5 lt hn hp kh kh X 27,0oC v 0,9856atm. T khi ca X so vi metan bng 0,725. a) Tnh % khi lng mi cht trong A b) un nng hn hp kh X vi bt Ni xc tc mt thi gian th thu c hn hp kh Y, chia Y lm hai phn bng nhau. - Phn th nht cho li t t qua bnh nc Brom d thy cn li 448 ml hn hp kh X (ktc) v t khi so vi Hidro l 4,5. Hi khi lng bnh nuc Brom tng bao nhiu gam? - Phn th hai em trn vi 1,68 lt O2 (ktc) trong bnh kn dung tch 4 lt. Sau khi bt tia la in t chy, gi nhit bnh 109,2oC. Tnh p sut bnh nhit bit dung tch bnh khng i. GII : a) Tnh % khi lng mi cht trong A : Gi 5,52g hh A CaC2 : a Ca : b CaO : c (mol) mhh X = 64a + 40b + 56c = 5,52 (1) Lu : hn hp A tc dng vi nc, c Ca v CaO cng c phn ng. CaC2 + 2H2O Ca(OH)2 + C2H2 a a (mol) Ca + H2O Ca(OH)2 + H2 b b (mol) CaO + H2O Ca(OH)2 c (mol) 2,5 lt kh thu c gm : C2H2 : a H2 : b M X = 0,725.16 = 11,6 nhhX = a + b =PV 2,5.0,9856 = = 0,1 (mol) (2) RT 0,082.300

M X = 0,725.16 = 11,6 (gam/mol) mX = M X .nX = 11,6.0,1 = 1,16 (gam)

26a + 2b = 1,16 (3) (2), (3)

a = 0,04 (mol) b = 0,06 64.0,04 .100% = 46,38% 5,52

Theo cc phn ng trn : nCaC2 = nC2H2 = 0,04 mol %CaC2 =

89

nCa = nH2 = 0,06 mol %Ca =

40.0,06 .100% = 43,48% 5,52

% CaO = 100% - (46,38 + 43,48)% = 10,14% b) tng khi lng bnh Brom : * Khi nung nng hn hp X vi xc tc Ni, c th xy ra 2 phn ng : C2H2 + H2 C2H4 C2H2 + 2H2 C2H6 hn hp kh Y c th gm C2H4, C2H6, C2H2 d, H2 d. * Khi cho hn hp Y qua bnh ng dd Br2 d th C2H2, C2H4 b hp thu : C2H2 + 2Br2 C2H2Br4 (lng) C2H4 + Br2 C2H4Br2 (lng) hn hp kh Z thot ra gm C2H6 v H2. nZ = 0,02 mol v Mz = 4,5.2 = 9 mZ = 9.0,02 = 0,18 gam * p dng LBT khi lng, ta c : m1/2hh Y = m1/2hh X = 1,16/2 = 0,58gam So snh hn hp Y v Z, ta thy tng khi lng bnh ng dung dch Br2 l tng khi lng ca C2H2 v C2H4 tc l mY - mZ Vy tng khi lng bnh Brom = 0,58 0,18 = 0,4 gam v Tnh p sut bnh sau phn ng chy : So snh hn hp X vi Y v p dng LBT nguyn t, ta c : n C trong hn hp Y = n C trong hn hp X = 2.0,02 = 0,04 (mol) trong hn hp Y = n H trong hn hp X = 2.0,02 + 2.0,03 = 0,1 (mol) * Sn phm chy gm : nCO2 = nC = 0,04 (mol); nH2O = nH = 0,05 (mol) Mt khc, n O trong CO2 v trong H2O = 0,04.2 + 0,05 = 0,13 (mol) nO ban u l 0,075.2 = 0,15 (mol) nO d = 0,15 0,13 = 0,02 (mol) nO2 d = 0,02/2 = 0,01 (mol) n cc kh trong bnh sau khi t = 0,04 + 0,05 + 0,01 = 0,1 (mol) Vy p sut bnh l :H

n

P=

nRT 0,1.22,4.(109,2 + 273) = = 0,784 atm V 273.4

90

II.4 BI TP TNG HP V HYDROCACBONv Mt s lu khi gii bi tp tng hp v hydrocacbon : Bi tp tng hp l mt dng bi tp trong c c phn tnh ton km theo cu hi l thuyt hoc cu hi th nghim. Bi tp hn hp thng c cc dng sau : Tm CTPT ca mt hay nhiu hydrocacbon sau yu cu : Xc nh CTCT ng ca cc cht qua th nghim cho cht tc dng vi mt cht no thu c sn phm c th. Xc nh CTCT ri vit phng trnh phn ng iu ch mt cht hydrocacbon khc hoc iu ch cht t nguyn liu chnh ban u l g. a ra phng php phn bit cc hydrocacbon mi tm c hoc nu cch tch ring, tinh ch tng cht trong hn hp cc cht mi tm c. V phng php lm bi tp loi ny, chng ta vn dng cc phng php hng dn trong phn bi tp l thuyt v bi tp tm CTPT, bi tp hn hp gii. Sau y l mt s bi tp v d : Dng 1 : bi yu cu xc nh CTPT ca sn phm th, t gi thit cho xc nh ng CTCT ca hydrocacbon ban u. Bi 1 : Khi tin hnh phn ng th gia ankan B vi hi Br2 c chiu sng ngi ta thu c hn hp X ch gm 2 sn phm phn ng (mt cht v c v mt cht hu c) th hi. T khi hi ca X so vi khng kh bng 4. a) Lp CTPT ca B v chn cho M mt CTCT thch hp. b) Nu tin hnh phn ng th 3 nguyn t hidro trong phn t B bng Clo th c th thu c my ng phn? GII : bi cho t khi hi ca sn phm th nn ta tm CTPT sn phm ri suy ra CTCT B a. Lp CTPT ca B v chn CTCT ng ca B. Gi k l s nguyn t Brom th vo phn t B : CnH2n+2 + kBr2 CnH2n+2-kBrk + kHBr a a ak (mol) Gi a (mol) l s mol B tham gia phn ng Sn phm phn ng gm : CnH2n+2-kBrk : a mol v HBr : ak molM hh X = 29.4 = 116 14n + 44k = 114 (14n + 2 - k - 80k )a + 8ak = 116 a + ak

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n=

114 - 44k 14

k 1 2 3 n 5 1,8 100oC v p sut 0,8amt. Bt tia la in t chy hon ton hn hp ri a bnh v nhit T1, o li p sut trong bnh vn c tr s 0,8atm. Lm li th nghim vi cc hn hp X c thnh phn A, B, C, D khc nhau vn thu c kt qu nh c. a) Lp CTPT A, B, C, D bit rng MA < MB < MC < MD. b) Vit ptp iu ch D t A v B t C GII : Nhit sau khi t T1 > 100oC H2O th hi cng iu kin nhit , th tch p sut bnh trc v sau khi t khng i s mol kh trong bnh trc v sau phn ng bng nhau. Khi thay i thnh phn ca hn hp X m kt qu khng thay i khi t chy tng cht th tng s mol trc v sau phn ng cng bng nhau. t cng thc ca mt cht trong hn hp l : CxHy0 y y C x H y + x + O 2 t xCO 2 + H 2 O 4 2

a a(x + y/4)

ax ay/2

(mol)

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Ta c : nT = nS a + a(x + 0,25y) = ax + 0,5ay 1 + x + 0,25y = x + 0,5y 0,25y = 1 y = 4 Vy c 4 hydrocacbon trn u c 4 nguyn t H trong phn t. Mt khc do A, B, C, D u th kh nn x 4 Vy 4 hydrocacbon trong X l CH4, C2H4, C3H4, C4H4 Theo th t MA < MB < MC < MD th A : CH4, B: C2H4, C: C3H4, D: C4H4. b. Vit cc ptp iu ch : iu ch D t A : 1500 c ,l ln 2CH4 C2H2 + 3H2 CuCl , HCl ,100 C 2C2H2 C4H4 (vinylaxetilen) iu ch B t C : t C3H4 + 2H2 Ni, C3H8 C t C C3H8 CH4 + C2H4O o o o

Dng 3 : Tm CTPT ca cc hydrocacbon sau nu cch nhn bit hoc tinh ch hoc tch cc cht trong hn hp hydrocacbon . Bi 3 : t chy mt s mol nh nhau ca 3 hydrocacbon L, L, M ta thu c lng CO2 nh nhau v t l s mol H2O v CO2 i vi K., L, M tng ng bng 0,5; 1, 1,5. a) Xc nh CTPT K, L, M b) Nu cch nhn bit 3 kh trn ng trong 3 l mt nhn c) Hy tch ring 3 cht trong hn hp trn. GII t cng thc chung cho 3 hydrocacbon l CnH2n +2-2k vi k l s lin kt p trong phn t cc hydrocacbon trn.C n H 2n + 2-2k +

a an a(n+1-k) (mol) 3 hydrocacbon t vi s mol nh nhau thu c lng CO2 nh nhau nn K, L, M c cng s C trong phn t. T=n H 2O n + 1 - k = nCO 2 n

3n + 1 - k O 2 nCO 2 + (n + 1 - k)H 2 O 2

K th T = 0,5 0,5n = n + 1 k n = 2(k 1) 0 n 4 v k 0 n = 2, k = 2 K : C2H2 L th T = 1 n = 2 v k = 1 CTPT L : C2H4 M th T = 1,5 n = 2 v k = 0 CTPT M : C2H6

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b) Nhn bit 3 kh trn ng trong 3 l mt nhn : - Ly mi kh mt t lm mu th. - Dn ln lt 3 kh vo dd AgNO3/NH3, kh no to c kt ta vng nht l C2H2.C2H2 + Ag2OddAgNO3/NH3 AgC CAg (vang) + H2O

- Hai kh cn li khng c hin tng g c dn tip qua ddBr2 d, kh no lm mt mu nu ca dd Br2 l C2H4, kh cn li khng c hin tng g thot ra ngoi l C2H6 H2C=CH2 + Br2 BrH2CCH2Br c) Cch tch 3 cht trn ra khi hn hp ca chng : - Cng thc hin qua cc th nghim nh trn ta thu c kh C2H6 thot ra ngoi. - Tinh ch li C2H2 bng cch cho dd axt HCl vo kt ta bc axetilua, kh axetilen c hon nguyn s bay ra ngoi : C2Ag2 + 2HCl C2H2- + 2AgCl - Tinh ch li C2H4Br2 bng cch cho thm ddKOHc/ancol vo dd Br2 b mt mu th kh C2H4 c hon nguyn s bay ra ngoi :C2H4Br2 + KOH ancol C2H4 + 2KBr

Hoc :C2H4Br2 Zn/ru C2H4 + ZnBr2

Ghi ch : Trn y ch l mt s bi tp v d nh, nu cc em lm tt bi tp phn II.1& II.2 th s lm c bi tp phn ny.

94