PHNG PHP GII NHANH HA HU C

1. Khi t chay hidrocacbon thi cacbon tao ra CO2 v hidro tao ra H2O.Tng khi lng C va H trong CO2 va H2O phai bng khi lng cua hidrocacbon. m hidrocacbon = mC(trong CO2) + m H(H2O) = nCO2.12 + nH2O.2 Thi du1: t chay hoan toan m gam hn hp gm CH4, C3H6 va C4H10thu c 17,6g CO2 va 10,8g H2O. m co gia tri la: A) 2g B) 4g PC) 6g D) 8g. m = nCO2.12 + nH2O.2 = 0,4.12 + 0,6.2 = 6 g

2. Khi t chay ankan thu c nCO2 < nH2O va s mol ankan chay bng hiu s cua s mol H2O va s mol CO2. nAnkan = nH2O nCO2 CnH2n+2 + nCO2 + (n + 1) H2O C th hiu nh sau : ly h s ca H2O h s ca CO2 = n+1 n = 1 =nAnkan Hoc Gi x l s mol ankan => nCO2 = nx , nH2O = (n+1)x Thy nH2O nCO2 = x =nAnkan CT ankan = CnH2n+2 => n = nCO2/nankan = nCO2/(nH2O nCO2)

Hoc khi 2 ankan th = nCO2/(nH2O nCO2) Thi du 2: t chay hoan toan 0,15 mol hn hp 2 ankan thu c 9,45g H2O. Cho san phm chay vao dung dich Ca(OH)2 d thi khi lng kt tua thu c la: A. 37,5g B. 52,5g C. 15g D. 42,5g nCO2 = nH2O nankan = 0,525 0,15 = 0,375 = nCaCO3 m CaCO3 = 0,375.100 = 37,5 g Thi du 3: t chay hoan toan hn hp 2 hidrocacbon lim tip trong day ng ng thu c 22,4 lit CO2(ktc) va 25,2g H2O. Hai hidrocacbon o la: PA. C2H6 va C3H8 B. C3H8 va C4H10 C. C4H10 va C5H12 D. C5H12 va C6H14 p dng CT : = nCO2/(nH2O nCO2 ) = 1/(1,4 -1) = 2,5 => C2H6 v C3H8

3. Da vao phan ng chay cua anken mach h cho nCO2 = nH2O PT CnH2n + 3n/2 O2 => nCO2 + nH2O Gii thch nh phn trn: Gi x l s mol CnH2n => nCO2 = nH2O = nx Thi du 4: t chay hoan toan 0,1 mol hn hp gm CH4, C4H10 va C2H4 thu c 0,14 mol CO2 va 0,23 mol H2O. S mol ankan va anken

co trong hn hp ln lt la: PA. 0,09 va 0,01 B. 0,01 va 0,09 C. 0,08 va 0,02 D. 0,02 va 0,08 Do nCO2 = nH2O(khi anken t chy ) => nankan = nH2O nCO2 =0,23 0,14 = 0,09 ; nanken = 0,1 0,09 mol Thi du 5: t chay hoan toan hn hp 2 hidrocacbon mach h trong cung day ng ngthu c 11,2 lit CO2 (ktc) va 9g H2O. Hai hidrocacbon o thuc day ng ng nao? A. Ankan PB. Anken C. Ankin D, Aren

Suy lun: nH2O = nCO2 => Vy 2 hidrocacbon thuc day anken. (Do cng dy ng ng)

4.Da vao phan ng cng cua anken vi Br2 (hoc H2) co ti l mol 1: 1., Ankin t l 1:2 Thi du7: Cho hn hp 2 anken i qua binh ng nc Br2 thy lam mt mau va u dung dich cha 8g Br2. Tng s mol 2 anken la: A. 0,1 PB. 0,05 C. 0,025 D. 0,005 nanken = nBr2 = 0,05 mol

5.t chay ankin: nCO2 > nH2O va nankin (chay) = nCO2 nH2O

Gii thch: CnH2n 2 + (3n 1)/2 O2 => nCO2 + n-1 H2O Gi x l s mol ankin => nCO2 = nx mol , nH2O = (n-1)x mol Ta thy nCO2 nH2O = nx (n-1)x = x = nankin => CnH2n-2 th n = nCO2 / nankin = nCO2/(nCO2-nH2O) Khi t chy 2 ankin th : = nCO2 / nankin = nCO2/(nCO2 nH2O) Thi du 8: t chay hoan toan V lit (ktc) mt ankin th khi thu c CO2 va H2O co tng khi lng 25,2g. Nu cho san phm chay i qua dd Ca(OH)2 d thu c 45g kt tua. a. V co gia tri la: A. 6,72 lit B. 2,24 lit C. 4,48 lit PB. 3,36 lit b. Cng thc phn t cua ankin la: A. C2H2 PB. C3H4 C. C4H6 D. C5H8 a) nCO2 = nCaCO3 = 0,45 mol => mCO2 = 19,8 g => nH2O = (25,2 19, /18 = 0,3 mol => nankin = nCO2 nH2O = 0,15 mol => V = 3,36 lt b. p dng CT = nCO2/(nCO2 nH2O) = 0,45 / (0,45 0,3) = 3 =>C3H4

Thi du 9: t chay hoan toan V lit (ktc) 1 ankin thu c 10,8g H2O. Nu cho tt ca san phm chay hp thu ht vao binh ng nc vi trong thi khi lng binh tng 50,4g. a) V co gia tri la: A. 3,36 lit B. 2,24 lit PC. 6,72 lit D. 4,48 lit b) Tm CT ankin: A.C2H2 PB.C3H4 C.C4H6 D.C5H10 a)V H2O v CO2 u b kim hp th => m tng = mCO2 + mH2O nH2O = 0,6 mol , nCO2 = (50,4 10, /44 = 0,9 mol nankin = nCO2 nH2O = 0,3 mol => V = 0,3.22,4 = 6,72 lt b) p dng CT : n =nCO2 / nankin = 0,9/0,3 = 3 =>C3H4

6.t chay hn hp cac hidrocacbon khng no c bao nhiu mol CO2 . Mt # nu hidro hoa hoan toan ri t chay hn hp cac hidrocacbon khng no o se thu c by nhiu mol CO2. o la do khi hidro hoa thi s nguyn t C khng thay i va s mol hidrocacbon no thu c lun bng s mol hidrocacbon khng no. VD : t chy : C2H4 + O2 => 2CO2 => nCO2 = 2nC2H4 Hidro ha(phn ng cng H2) C2H4 + H2 => C2H6 C2H6 + O2 => 2CO2 => nCO2 = 2nC2H6 m nC2H6 = nC2H4 C hidrocacbon khng no (Nh anken hoc ankin)

Thi du10: Chia hn hp gm C3H6, C2H4, C2H2, thanh 2 phn u nhau: - t chay phn 1 thu c 2,24 lit CO2 (ktc). - Hidro hoa phn 2 ri t chay ht san phm thith tichCO2thu c la: PA. 2,24 lit B. 1,12 lit C. 3,36 lit D. 4,48 lit

7. Sau khi hidro hoa hoan toan hidrocacbon khng no ri t chaythi thu c s mol H2O nhiu hn so vi khi t luc cha hidro hoa. S mol H2O tri hn chinh bng s mol H2 a tham gia phan ng hidro hoa. Nh Anken + H2 t l 1: 1 (do anken c lin kt i) Ankin + H2 t l 1: 2 (do ankin c lin kt ba) Nu xc tc l Ni nung nng . nH2O(khi hidro ha ri t chy) = nH2O(khi t chy) +nH2 (tham gia hidro ha) Thi du11: t chay hoan toan 0,1 mol ankin thu c 0,2 mol H2O. Nu hidro hoa hon toan 0,1 mol ankin nay ri t chay thi s mol H2O thu c la: A. 0,3 PB. 0,4 C. 0,5 D. 0,6 Suy lun: Do hidro ha hon ton => t l ankin v H2 l 1:2

=> nH2 = 2nankin = 0,2 mol =>nH2O(khi hidro ha v t chy)=nH2O(khi t chy)+nH2 = 0,4 mol

8.Da va cach tinh s nguyn t C va s nguyn t C trung binh hoc khi lng mol trung binh + Khi lng mol trung binh cua hn hp:

+ S nguyn t C: + S nguyn t C trung binh: ; Trong o: n1, n2 la s nguyn t C cua cht 1, cht 2 a, b la s mol cua cht 1, cht 2 CT trn ly t phn ng t chy hidrocacbon Trng hp c bit : Khi s nguyn t C trung binh bng trung binh cng cua 2 s nguyn t C thi 2 cht co s mol bng nhau. VD : = 1,5 ; 2,5 ; 3,5 th mol n1 = mol n2 Tng t c 1 s trng hp khc : = . , 67 ; hoc .... , 33 = ,2 ; = , 8 . Ch .... l mt s bt k nh 1 ; 2 ; 3;4 Thay vo trn th tm c t l mol ca 2 cht. VD = 1,67 th n1 = 1 , n2 = 2

=>1,67 = (a + 2b)/(a+b) 0,67a = 0,33b 2a = b Tc l s mol cht 2 = 2 ln s mol cht 1 Ch : Cch tm % theo th tch nhanh ca bi 2 cht lin tip nhau. VD . Khi tm c = 1,67 => % Vcht c C ln nht (Tc l n =2) = 67% => %V Cht c C nh = 100 67 = 33% Nhn thy % V cht c C ln nht l s ... ,67 . cn nh nht th tr i l c VD . = ,3 (Ch c th l 1 hoc 2 hoc 3 .) %V c C ln nht = 30% => %V nh hn = 70% Nu bi bo tnh % theo khi lng th da vo t l s mol VD: Cho 2 ankan lin tip tm c = 1,67 => t l mol 2a = b(va lm trn) n1 = CH4 , n2 = C2H6 => %CH4 = 16 a /(16a + 28b) = 16a /(16a + 28.2a) = 22,22% => % C2H6 = 77.78%

Th d 12: Hn hp 2 ankan la ng ng lin tip co khi lng la 24,8g. Th tich tng ng cua hn hp la 11,2 lit (ktc). Cng thc phn t ankan la: A. CH4, C2H6 B. C2H6, C3H8 P B. C3H8, C4H10 D. C4H10, C5H12

Suy lun: Gi CT CnH2n + 2 (n trung bnh) . => 14 + 2 = 49,6 => = 3,4 => 2 hidrocacbon la C3H8 va C4H10.

Th d 14: Cho 14g hn hp 2 anken la ng ng lin tip i qua dung dich nc Br2 thy lam mt mau va u dd cha 64g Br2. 1.Cng thc phn t cua cac anken la: PA. C2H4, C3H6 B. C3H8, C4H10 C. C4H10, C5H12 D. C5H10, C6H12 2. Ty l s mol 2 anken trong hn hp la: A. 1:2 B. 2:1 C. 2:3 PD. 1:1 Suy lun: T l anken : Br = 1 : 1 ,, CT CnH2n ( n trung bnh) => nanken = nBr = 0,4 mol ; => =>o la : C2H4 va C3H6 2. Thy n = 2,5 thy (2 + 3)/2 = 2,5 (Trung bnh cng) =>nC2H4 = nC3H6 => t l 1:1 Thi du 15: t chay 2 hidrocacbon th khi k tip nhau trong day ng ng thu c 48,4g CO2 va 28,8g H2O. Phn trm th tich mi hidrocacbon la: A. 90%, 10% B. 85%. 15% PC. 80%, 20% D. 75%. 25%

nCO2 = 1,1mol , nH2O = 1,6 mol => nH2O > nCO2 : ankan( c th khng ni cng c) => = nCO2/(nH2O nCO2) = 1,1/(1,6-1,1) =2,2 => %V cht c C ln hn (Tc l 3) = 20% =>%VNh hn = 80%

9. CT tng qut nht ca cht c cha C,H,O ( nh nh ) Phn ny b tr cho 12 phn este. CnH2n +2 - 2a m (Chc)m Trong a l s lin kt pi c tnh nh sau a = (2.s C + 2 s H)/2 (p dng cho c hidrocacbon nhng phn hidrocacbon da vo CT TQ l c .VDC3H4 th l ankin c2lk pi . Nu p dng CT cng tmc = 2) m l s nhm chc ( VD : 1 ,2 ,3 nhm OH) (CHc: OH , -O-,COOH,COO) VD:chtC4H6O2 hi c mi lk pi? p dng CT s pi =( 2.4 +2 6)/ = 2 VD: CT tng qut ca ru : Ru c nhm chc OH CnH2n + 2 -2a m(OH)m Nu l ru no th khng c lk pi nh hidrocacbon no: => a= 0 => CT : CnH2n + 2 m(OH)m hoc CnH2n+2Om

Nu l ru khng no c 1lk pi (nh anken) hoc 2 lin kt pi(nh ankin) th thay vo trn ta c CT .

i vi hidrocacbon th khng c nhm chc : CT tng qu l CnH2n+2 2a VD1:Cng thc ca ru no, 3 nhm OH l: A.CnH2n-3(OH)2 B.CnH2n+1(OH)3 C.CnH2n-1(OH)3 D.CnH2n+2(OH)3

p dng. 3 nhm OH => m =3, ru no => a = 0, => CT : CnH2n + 2 -3(OH)3 => C VD2:Cng thc phn t tng qut ca ru 2 nhm OH c 1 lin kt i trong gc hidrocacbon? A.CnH2n+2O2 B.CnH2n-2O2 C.CnH2nO2 D.CnH2n-1O2 L ru 2 nhm OH => m = 2, c 1 lin kt i tc l 1 lk pi => a = 1 CT : CnH2n+ 2 2.1 2 (OH)2 = CnH2nO2 => C VD3: Chtsinh bi axit n chc,c 1 lk i? (Gi axit c gc COOH) A.CnH2n-1COOH B.CnH2n+1COOH C.CnH2nCOOH D. CnH2n2COOH

Axit n chc => m =1, 1 lk i => a =1 => CT CnH2n+2 -2 -1(COOH) = CnH2n-1(COOH) => A Bi tp vn dng: Bi 1: Ru 2 chc c 2 lk pi .Tm CT tng qut? Bi 2: Ru 3 chc c 1 lk pi.Tm CT tng qut? Bi 3.Tm s lk pi trong cc cht sau: C6H10O2;C8H12O4; C9H10O(C6H5COCH2CH3)(Vng benzen c 4 v CO c 1) Bi 4:Ru n chc 2 lk pi(Ging ankin). Tm CT tng qut? (CnH2n-2Om) Bi 5: X la ancol mach h co cha mt lin kt i trong phn t. Khi lng lng phn t cua X nho hn 58 vC. Cng thc phn t cua X la: A. C2H4O B. C2H4(OH)2 C. C3H6O D. C3H6(OH)2

10. Phn ng t chy ca Ru. T phn 10 Ta tm c CT sau: - Ru no : CnH2n + 2 m (OH)m hoc CnH2n + 2Om ( m l s chc) ng khung ging ht CT ca ankan Bi tp ging ankan

n Ru = nH2O nCO2 , n = nCO2/(nH2O nCO2) (C th l) (Khng tin th Vit PT ri lm nh phn ankan) -Ru khng no,c 1lk pi (Ging anken):CnH2n+ 2 -2-m(OH)m = CnH2n Om Ging anken => nCO2 =nH2O -Ru khng no, c 2lk pi(Ging ankin): CnH2n+2 -4-m(OH)m = CnH2n-2Om Ging ankin =>n Ru = nCO2 nH2O, n=nCO2/(nCO2 nH2O) VD4:t chy hn hp 2 ru ng ng c cng s mol nhau, ta thu c kh CO2 v hi nc H2O c t l mol nCO2:nH2O = 3:4. Bit khi lng phn t 1 trong 2 cht l 62. Cng thc 2 ru l ? A.CH4O v C3H8O B,C2H6O v C3H8O C.C2H6O2 v C4H10O2 D.CH4O v C2H6O2 p dng CT: nH2O>nCO2 => ru no = nCO2/(nH2O nCO2) = 3/(4-3) = 3 => C VD 5: Khi t chay mt ancol a chc thu c nc va khi CO2 theo ti l khi lng . Cng thc phn t cua ancol la: A. C2H6O2 B. C4H8O2 C. C3H8O2 D. C5H10O2 mCO2:mH2O = 44:27 => nCO2/nH2O = 2/3 => ( Ru no v nH2O > nCO2) => n = nCO2/(nH2O nCO2) = 2 /(3-2) =2 => A

11.Da trn phan ng tach nc cua ru no n chc thanh anken n andehit = n ru (v s nguyn t C khng thay i. Vi vy t ru va t anken tng ng cho s mol CO2 nh nhau.) VD6: Chia a gam ancol etylic thanh 2 phn u nhau. Phn 1: mang t chay hoan toan 2,24 lit CO2 (ktc) Phn 2: mang tach nc hoan toan thanh etylen, t chay hoan toan lng etylen m gam H2O. m co gia tri la: A. 1,6g PB. 1,8g C. 1,4g D. 1,5g Suy lun: nCO2(khi t chy ru) = nanken(khi t chy ru) = 0,1 mol M khi t chy anken th nCO2 =nH2O = 0,1 mol => mH2O = 1,8g

12. t 2 cht hu c, phn t co cung s nguyn t C, c cung s mol CO2 thi 2 cht hu c mang t chay cung s mol. V s mol CO2 lun = sC(trong cht hu c) . Mol hu c VD: C2H5OH => 2CO2 v C2H6 => 2CO2 VD7: t chay a gam C2H5OH c 0,2 mol CO2. t chay 6g CH3COOH c 0,2 mol CO2. Cho a gam C2H5OH tac dung vi 6g CH3COOH (co H2SO4 xt, t0 Gia s H = 100%) c c gam este. C co gia tri la: A. 4,4g PB. 8,8g C 13,2g D. 17,6g

Suy lun: nC2H5OH = nCH3COOH = 1/2nCO2 = 0,1 mol. PT: Hc bi axit ( Ni sau tng qut hn phn este) =>nC2H5OH=

13. Da trn phan ng t chay anehit no, n chc cho s mol CO2 = s mol H2O. Anehit ru cung cho s mol CO2 bng s mol CO2 khi t anehit con s mol H2O cua ru thi nhiu hn. S mol H2O tri hn bng s mol H2 a cng vao andehit.(Phn ny ging phn 7) nH2O(Khi t chy ru) = nH2O(hoc n CO2 khi t chy andehit) + nH2 (khi phn ng vi andehit)

VD8: t chay hn hp 2 anehit no, n chc thu c 0,4 mol CO2. Hidro hoa hoan toan 2 anehit nay cn 0,2 mol H2 thu c hn hp 2 ru no, n chc. t chay hoan toan hn hp 2 ru thi s mol H2O thu c la: A. 0,4 mol PB. 0,6mol C. 0,8 mol D. 0,3 mol Suy lun: p dng CT trn nH2O = nCO2 + nH2 = 0,4 + 0,2 = 0,6 mol

14. Da va phan ng trang gng (Nhng cht c gc CHO) Phn ng ca andehit: T l gia andehit vi Ag = 1:2n ( vi n l s

gc CHO VD C2H5(CHO )2 => c 2 gc CHO) =>nHCHO : nAg = 1 : 4 , H C - H c 2 gc CHO (2 th hin 2 gc)

O

nR-CHO : nAg = 1 : 2(trng hp c 1 nhm CHO)

VD9: Cho hn hp HCHO va H2 i qua ng ng bt nung nong. Dn toan b hn hp thu u sau phan ng vao binh nc lanh ngng tu hi cht long va hoa tan cac cht co th tan c , thy khi lng binh tng 11,8g. Ly dd trong binh cho tac dung vi dd AgNO3/NH3 thu c 21,6g Ag. Khi lng CH3OH tao ra trong phan ng hp H2 cua HCHO la: A. 8,3g B. 9,3g PC. 10,3g D. 1,03g Suy lun: H-CHO + H2 CH3OH ( ) cha phan ng la 11,8g. T l mol gia : HCHO vi H2 = 1 : 4 . mHCHO = 0,05.30 = 1,5g ; VD10: Cho hn hp gm 0,1 mol HCOOH va 0,2 mol HCHO tac dung ht vi dd AgNO3/NH3 thi khi lng Ag thu c la:

PA. 108g B. 10,8g C. 216g D. 21,6g Suy lun: 0,1 mol HCO-OH 0,2 mol Ag 0,2 mol HCHO 0,8 mol Ag nAg = 1 mol => ap an A.

15. Da vao cng thc tinh s ete tao ra t hn hp ru hoc da vao LBTKL. VD11: un hn hp 5 ru no n chc vi H2SO4 , 1400C thi s ete thu c la: A. 10 B. 12 PC. 15 D. 17

Suy lun: Ap dung cng thc : ete thu c 15 ete. VD12: un 132,8 hn hp gm 3 ru n chc vi H2SO4 c, 1400C hn hp cac ete co s mol bng nhau va co khi lng la 111,2g. S mol mi ete la: A. 0,1 mol PB. 0,2 mol C. 0,3 mol D. 0,4 mol Suy lun: un hn hp 3 ru tao ra 6 ete. Theo LBTKL: mru = mete + = 132,8 111,2 = 21,6g

Do nmi ete = .

16. Da vao phng phap tng giam khi lng: Nguyn tc: Da vao s tng giam khi lng khi chuyn t cht nay sang cht khac xac inh khi lng 1 hn hp hay 1 cht. Cu th: Da vao pt tim s thay i v khi lng cua 1 mol A 1mol B hoc chuyn t x mol A y mol B (vi x, y la ti l cn bng phan ng). Tim s thay oi khi lng (AB) theo bai z mol cac cht tham gia phan ng chuyn thanh san phm. T o tinh c s mol cac cht tham gia phan ng va ngc lai. P i vi ru: Xet phan ng cua ru vi K:

Hoc ROH + K ROK + H2 Theo pt ta thy: c 1 mol ru tac dung vi K tao ra 1 mol mui ancolat thi khi lng tng: 39 1 = 38g. Vy nu cho khi lng cua ru va khi lng cua mui ancolat thi ta co th tinh c s mol cua ru, H2 va t o xac inh CTPT ru. P i vi anehit: xet phan ng trang gng cua anehit R CHO + Ag2O R COOH + 2Ag Theo pt ta thy: c 1mol anehit em trang gng 1 mol axit

m = 45 29 = 16g. Vy nu cho manehit, maxit nanehit, nAg CTPT anehit. P i vi axit: Xet phan ng vi kim R(COOH)x + xNaOH R(COONa)x + xH2O Hoc RCOOH + NaOH RCOONa + H2O 1 mol 1 mol m= 22g P i vi este: xet phan ng xa phong hoa RCOOR + NaOH RCOONa + ROH 1 mol 1 mol m= 23 MR P i vi aminoaxit: xet phan ng vi HCl HOOC-R-NH2 + HCl HOOC-R-NH3Cl 1 mol 1mol m= 36,5g VD13: Cho 20,15g hn hp 2 axit no n chc tac dung va u vi dd Na2CO3 thi thu c V lit CO2 (ktc) va dd mui.C can dd thi thu c 28,96g mui. Gia tri cua V la: A. 4,84 lit PB. 4,48 lit C. 2,24 lit D. 2,42 lit E. Kt qua khac. Suy lun: Goi cng thc trung binh cua 2 axit la: Ptpu: 2 + Na2CO3 2 + CO2 + H2O Theo pt: 2 mol 2 mol 1 mol m = 2.(23 - 11) = 44g

Theo bai: Khi lng tng: 28,96 20,15 = 8,81g. S mol CO2 = Th tich CO2: V = 0,2.22,4 = 4,48 lit VD14: Cho 10g hn hp 2 ru no n chc k tip nhau trong day ng ng tac dung va u vi Na kim loai tao ra 14,4g cht rn va V lit khi H2 (ktc). V co gia tri la: A. 1,12 lit PB. 2,24 lit C. 3,36 lit D. 4,48 lit Suy lun:2 R-OH +2Na => 2RONa + H2 PT 2mol 2mol 1mol m Tng = 22.2 = 44g Theo u bi khi lng tng = 14,4 10 = 4,4 g nH2 = 4,4 /44 = 0,1 mol => V = 2,24 lt

T thng lm theo cch # cng tng t nh trn nhng khng vit PT Ta bit R-OH => RONa M tng 22 g M m tng = 4,4 g => nR-OH =nRONa = 0,2 mol M t l gia R-OH vi H2 l 2:1 => nH2 = 1/2nR-OH = 0,1 mol

17. Da vao LBTNT va LBTKL:

- Trong cac phan ng hoa hoc, tng khi lng cac cht tham gia phan ng bng tng khi lng cua cac san phm tao thanh. A+BC+D Thi mA + mB = mC + m D - Goi mT la tng khi lng cac cht trc phan ng MS la tng khi lng cac cht sau phan ng Du phan ng va u hay con cht d ta vn co: mT = mS - S dung bao toan nguyn t trong phan ng chay: Khi t chay 1 hp cht A (C, H) thi Gia s khi t chay hp cht hu c A (C, H, O) A + O2 CO2 + H2O Ta co: Vi mA = mC + mH + mO VD15: t chay hoan toan m gam hn hp Y: C2H6, C3H4, C4H8 thi thu c 12,98g CO2 va 5,76g H2O. Tinh gia tri m? (ap s: 4,18g) VD16: cho 2,83g hn hp 2 ru n chc tac dung va u vi Na thi thoat ra 0,896 lit H2 (ktc) va m gam mui khan. Gia tri cua m la: A. 5,49g B. 4,95g C. 5,94g PD. 4,59g VD17:Cho 4,2g hn hp gm ru etylic, phenol, axit fomic tac dung va u vi Na thy thoat ra 0,672 lit H2 (ktc) va 1dd. C can dd thu c hn hp rn X. Khi lng cua X la:

A. 2,55g PB. 5,52g C. 5,25g D. 5,05g Suy lun: Ca 3 hp cht trn u co 1 nguyn t H linh ng S mol Na = 2nH2 = 2.0,03 = 0.06 mol Ap dung LBTKL: mX = m hn hp + mNa mH2 = 4,2 + 0,06.23 0,03.2 = 5,52g. Hoc dng tng gim khi lng mX = m hn hp + m Tng = 4,2 + 0,06(23-1)=5,52 VD18: Chia hn hp 2 anehit no n chc lam 2 phn bng nhau: P1: em t chay hoan toan thu c 1,08g H2O P2: tac dung vi H2 d (Ni, t0) thi thu hn hp A. em A t chay hoan toan thi th tich CO2 (ktc) thu c la: A. 1,434 lit B. 1,443 lit PC. 1,344 lit D. 1,444 lit Suy lun: Vi anehit no n chc nn s mol CO2 = s mol H2O = 0,06 mol Theo BTNT va BTKL ta co: lit VD19: Tach nc hoan toan t hn hp Y gm 2 ru A, B ta c hn hp X gm cac olefin. Nu t chay hoan toan Y thi thu c 0,66g CO2. Vy khi t chay hoan toan X thi tng khi lng CO2 va H2O la:

A. 0,903g B. 0,39g C. 0,94g PD. 0,93g nCO2(t chy ru) = nCO2(t chy anken) = nH2O(t chy anken) = 0,015 mol m = 0,015(44 + 18)=0,93 g

18. P ca hp cht C,H hoc C,H,O Vi hidrocacbon: CxHy + (x+y/4) O2 => xCO2 + y/2H2O Vi hp cht cha (C,H,O) : CxHyOz + (x +y/4 z/2)O2 => x CO2 + y/2H2O Phn ny ch cn nh h s O2 ViCxHy => x+y/4 ,,,, Vi CxHyOz => (x+y/4 z/2) VD20 : t chy 1mol cht hu c cn 3,5 mol O2 Vy cht hu c c th l : A.C3H6O2 B.C2H5OH C.C3H7OH D.C5H9OH PT CxHyOz => (x + y/4 z/2)O2 TheoPT: 1 mol => (x+y/4 z/2)O2 Theo B: 1 mol => 3,5 mol x+y/4 z/2 = 3,5 => p n A ph hp Nh nO2/n hp cht = x + y/4 z/2 (Vi CxHyOz) cn = x+y/4(vi CxHy)

(Dng lm mo nhanh)

19.Dng bi tm CT khi bit % ca Oxi. VD: Cht hu c X thanh phn gm C, H, O trong o %O: 53,3 khi lng. Khi thc hin phan ng trang gng, t 1 mol X 4 mol Ag. CTPT X la: PA. HCHO B. (CHO)2 C. CH2(CHO)2 D. C2H4(CHO)2

Phng php gii PT 3 n. (Tm thy trn mng th xem nh hi kh din t khi no ni sau). Gi CT ca X : CxHyOz % ca 1 cht trong hp cht = M cht / M hp cht VD. Tm % ca Na trong Na2CO3 => %Na = 23.2.100%/(23.2 + 12 + 16.3) = 43,4% Theo nh trn th %O = 16z.100%/(12x + y + 16z ) = 53,3 % Ly 100/53,3 ng ly 53,3 . (12x + y + 16z) lm g 16z.100/53,3 = 12x + y + 16z 14z = 12x + y Gii PT 3 n : ta th z = 1 vo trn => 12x + y = 14

Cch tm x = 14/12 = a,.... th a = x (Tc l khi thay z = 1 vo th x = 14z/12 ly phn nguyn ko ly sau du , hoc lm trn) => x = 1 => y = 2 => CT n gin (CH2O)n hoc ( HCHO)n Theo u bi nAg/nX = 4 = 2.2 => c 2 gc CHO m HCHO c 2 gc CHO => n= 1 CT HCHO

Ni thm v bi ton % O ca hp cht cha C,H,O Gi s bi cho nh sau:A la hp cht hu c mch h cha C,H,O v %O = 43,24 . Bit m oxi trong 1 mol A< m Nito trong 150g NH4NO3.Tm Tng h s ca hp cht ( bi c tnh du p n mi hi tm tng h s) Phng php gii PT ba n: Nh bi trn CT A. CxHyOz %O = 16z.100%/(12x + y + 16z) = 43,24% 16z.100/43,24 = 12x + y + 16z 21z = 12x +y Thay z = 1,2,3 vo tm nghim nguyn.( xem hp l th chn) Nu thay nhiu th u tm ra c 1 cng thc n gin . VD. Thay z = 1 => x = a,.... = 21/12 = 1,... => y = 9 Ta thy v l khng c cht l CH9O

Ta thay tip z = 2 => x = a,.... = 21.2/12 = 3 ( ln nh ly s trc du ,) y = 42-36 = 6 CT: (C3H6O2)n Ta ch cn thay n y thi Nu thay tip z = 4 => (C6H12O4)n khng ti gin bng CT kia Theo bi mO trong 1mol A < m N trong 150gNH4NO3 Nh m ca 1 cht trong hp cht : Cho a mol AxBy => m A = a.x . MA, m B = a.y.MB 32n < 14.2.1,875 n n = 1 => CT : C3H6O2 => tng h s = 3+6+2 = 11 Tng t VD: Cho 1 hp cht hu c c mch h cha 3 nguyn t C,H,O c % O = 50. t chy 0,1 mol A thu c V lt CO2 Tm gi tr V.

Cng nh bi trn ta c c nh sau: %O = 16z.100%/(12x + y + 16z) = 50% 16z = 12x + y Thay z = 1 => x= 1 ,y = 4 => CT : (CH4O)n iu Kin : s H n n = 1 => CT : CH4O

nCO2 = s C(trong hp cht ) . mol hp cht = 1.0,1 mol => V = 2,24 lt Tng t vi % O = 55,17% ; 34,78 % Ta c bng sau ( cc cu lm cho nh cch lm dng bi ny nh c th cng tt )

%O 55,17% 53,33% 50% 43,24% 37,21% 34,78%

CTCT (CHO)n (CH2O)n CH4O (C3H6O2)n (C2H3O)n C2H6O

Vi % = 37,21 => 27z = 12x + y => z = 1 , x = 2 , y = 3 => CT(C2H3O)n [font=Times New Roman]linhkute