Boi duong hoc sinh gioi hoa hoc 8+9 dao huu vinh p3

mjd sau phan ling = rrijjCa(OH)2 + "^002 = ~ "^CaTO^ 1000 1,025 + 44 0,075 - 100 (0,05 : 2) = 1025,8 g ^ n n B,2=0,l 162.M5 N o n g m u o i CaCHCOj) dung dich la: Cdch 2: CO2 X +Ca(0H)2 mol 2CO2 X > CaCOjto + H2O mol > y mol nco2 = x + 2y = 0,075 100 = 0,4% v , „ _ , , , x mol + Ca(OH)2 2y m o l ^^^^^^ V a y A l a C2H4 + _ ,^ 0,05 m o l 0,05 mol 12 8 mc = - ~ 44 H2O + 0,025 m o l CaCOj + H2O ' = 2,4 g; mn = 18 ' *• = 0,6 g; mo = - 2,4 - 0,6 b) Dat cong thiJc A la C,Hy ta c6: x : y = mc 12 mH_ • Ta CO ( + ) n < 40 0,025 m o l (chii y ncaco3 = 0-0-'' - 0-025 = 0,025 mol) Nghiem thich h d p n = 12 ' Cong thtfc p h a n tur A : C2H6 d) D o t chay 12,6 gam hon hdp ba k h i CH4, C2H4 va C2H2, ta thu difdc 39,6 gam CO2 N e u cho 12,6 gam hon hdp tren qua binh diTng dung dich brom thi c6 80 gam b r o m tham gia phan tfng Xac dinh phan phan t r a m k h o i Wdng m o i hidrocacbon hon hdp D o t chay 11,2 l i t (do d dktc) hon hdp metan va etilen r o i cho san pham tac dung v d i dung dich N a O H sau phan tfng ngi/di ta thu difdc 250 m l dung dicb Na2C03 2,6M Xac dinh phan phan t r a m theo the tich m o i k h i hon hdp T" • C H - C H + CI2 c n g thQc p h a n ttf ' ' CaCOjio - c) Chat A CO l a m mat mau dung dich brom khong ? Ca(HC03)2 y mol Ca(OH)2 ,o.u 2, D o t chay gam chat hffu cd, thu duTdc 8,8 gam k h i CO2 va 5,4 gam H2O T i n h TUM sau phan tfng va nong Ca(HC03)2 dung dich nhiTcach CO2 : ng^^ = 0,01 : 0,01 = : b) B i e t phan tuf k h o i cua A nho hdn 40 T i m cong thiJc phan tuT cija A '5 nca(0H)2 = x + y = 0,05 Cdch 3: = , mol; ' Tii phu'dng t r i n h p h a n vtng do't c h a y , v d i so' n g u y e n tuT c u a m o i n g u y e n to' h a i ve' c u a phu'dng t r i n h p h a i b a n g n h a u , ta t i m diTdc g i a t r i c u a x , y I Bdl tap CO Idfl giSl B i e t 0,01 m o l hidrocacbon A l a m mat mau vCfa dii 100 m l dung dich brom 0,1M V a y A la hidrocacbon nao so cac chat sau : a) CH4 b)C2H2 c)C2H4 d) CsHs CxHy + 602 >4C02 + y/2H20 So n g u y e n tijT C, O d h a i ve' c u a phiTdng t r i n h : nc = x = ; n o = = 4.2 + y / s u y r a y = ' • ' - ' Cong thiJc p h a n tuT c u a h i d r o c a c b o n A l a C4H8 205 Cong thtfc cau tao ciia A c6 the la: C H - C H - C H = C H hoac la C H - C H = C H - C H Dot chay hoan toan mot liTctng hidrocacbon X, ngu-di ta thu dxidc 22 gam COj va 13,5 gam HjO Biet khoi Itfdng phan tur ciia X la 30 dvC Tim cong thiJe phan tuf cua X va viet cong thiJc cau tao cua no Hifdng d i n giai Tuf khoi lu'dng cua CO2, H2O ta suy so mol Neu goi cong tMc cua X 1^ C,Hy tiJf phU'dng trinh phan tfng do't chay X se tim diTdc moi quan he x vdi y va tiJf Mx suy gia tri bang so' cua chiing Cdch 1: nco2 = 22 : 44 = 0,5 mol; = 13,5 : 18 = 0,75 mol CMy + (x + y/4)02 > X C O + y/2H20 mol mol 0,5y mol 0,5 mol 0,75 mol Ta c6: x : 0,5 = 0,5y : 0,75 => 0,75x = 0,25y => 3x = y (1) Mx = 12x + y = 30 Giai phming trinh (1), (2) ta diTdc x = va y = Cong thu-c phan tuf cua X la C2H6 Cong thtfc cau tao CH3 - CH3 ^ ^^ '' Cdch : nc = n^oj = 0-5 mol; 2nH20 = = 1,5 mol U H J O X U H = CxHy => X : y = Uc : HH = 0,5 : 1,5 = : Cong thiJc dc(n gian nhat cua X la CH3 Cong thtfc phan tuf la (CH3)„ = M = 30 => 15n = 30 ^ n = Cong thiJc phan tuf cua X (CH3)2 hay CsHfi, Cong thufc cau tao CH3-CH3 ''' Dot chay gam mot chat hffu cd A thu dxidc 6,6 gam C O va 3,6 gam H2O a) Xac dinh cong thtfc cua A, bie't khoi lu'dng phan tuT ciia A la 60 dvC ' b) Viet cong thiJc cau tao c6 the c6 cua A Hrfdng d i n giai ' va m^^Q tim du'dc mc va mn; dxia vao djnh luat bao toan khoi Ixidng sc suy A c6 nguyen to' oxi hay khong (vi dot A > CO2 + H2O ncn A CO the c6 oxi) mco2 Cdch 1: nc = TICQ^ = 6,6 : 44 = 0,15 mol mc = 0,15 12= 1,8 g n H = 2.3,6 : 18 = , m o l = > m H = 0,4 g Theo dinh luat bao toan kho'i lU'cJng ta c6: m A = mc + m H + rno => mo = - 1,8 - 0,4 = 0,8 g 206 UH = thtfc phan tuf cua A : C x H y O , 12x y 16z 60 mH mc mo 12x y _ 16z 60 1,8 0,4 0,8 ||ai c&c phuWng trinh tren ta diTCfc x = 3, y = 8, z = -' : n g thi?c phan tuf cua A la C^U^O Zung CO the tim cong thtfc dcJn gian nhS't cua A roi suy cong thi?c phan tuf: m o _ 1,8 0,4 0,8 z = 12 16 12 16 = 0,15 : 0,4 : 0,05= : : :6ng thtfc ddn gian nhat cua A la CsHjO ; Cong thufc phan tuf la (C3H80)„ = 60 (12 + + 16)„ = 60 n = f^y cong thiJc ddn gian nhat cua A cung la cong thtfc phan tuf ich 2: v6i gam A tao 6,6 gam C O va 3,6 gam H O Neu do't chay (1 mol) 60 g A tao x g CO2 va y g H2O 60.6,6 ^ , = 132 g ;^6ng X = y= 60.3,6 ' - • nco2 = 132 = 72g nH = n H = = mol 'H2O : 44 =3 mol = nc = : 18 = mol Vay mol A c6 mol C vk mol H, cong thtfc phan tur cua A 1^: CsHsO, = 60 •.rami 12 + + 16z = 60=> z = Cong thiJc phan tuf ciia A la CiW C H - C H - C H O H (1) CH3-CH(OH)-CH3 (2) C H - O - C H - C H (3) Hdp chat \\\ixx cd A c6 phan khoi lifdng cac nguyen to' nhiT sau: C = 53,33%, H = 15,55%, N = 31,12% , Xdc dinh cong thtfc phan tiJf cua A, bie't rkng A chi c6 mot nguySn tijf N phan tuf Vie't cong thufc cau tao rut gpn ciia A, bie't N c6 hod tri III H\i6ng d i n g i a i Cdch I : a.a - KiNeu gpi cong thtfc phan tuf cua A la C x H y N , Tuf phan phan tram kho'i IWng nguyen to ta tim dufdc ti 16 so' nguyen tiJf cac nguyen to x : y : z , Sau d6 thay z = ta tim diTdc cong thufc phan tuf 207 X ^ XT %C — XT a) C H y N , : x : y : z = %H : — Mc : MH %N — • ^ = 12 ' 4,444 ^ 14 CxH, „: : 15.55 : 2.222 = : : ^ b) Cong thiJc cau tao rut gpn cua A la: C H - C H - N H hoSc la C H - N H - C H "CO2 (DCjH, '^^^ 12x 53.33 y 31,12.12x= 14.53,33 14 31,12 _ 14 15,55 ~ 31,12 (2) C H , (3) C2H4 C2H2 C2H2 C6ng thtfc cau tao: Tu'dng tiT phan tuf CjHyN;, c6 xC cacbon; y H g hidro va N g nitd 53.33g yg 15.55g (1) 14g CH3-CH3 y = 14.15,55 b) Xet c a p ( l ) : =>x«2 C2Hfi + 7/2O2 =>y«7 C2H4 D o t chay 1,12 l i t hon hdp hai hidrocacbon (the k h i ) c6 ciing so nguyen [\i cacbon, dan san pham phan uTng ch^y Ian liTcJt qua binh (1) diTng P O , binh + 41 +302- >2C02 Xet c a p (2) c moi a) Xac dinh cong thtfc phan tijT cua hai hidrocacbon h o n hcJp V i e t cong > C O + H2O c moi a + c = 0,05 thiJc cau tao cua chiing " H J: N e u g p i cong thiJc cua hai hidrocacbon hon hpp la CxHy ->3a + c = 0,l a = 0,025 va c = 0,025 M o i k h i c h i c m % the tich hon hdp X e t c a p (3) DiTa vao phan tfug (2) va (3) 'hh (HjO 'H20 hap t h u , C O b i K O H hap t h u dp tang k h o i liTdng cua cac binh ciing c + b = 0,05 ->c + 2b = , l b = 0,05 ::> c = v6 l i chinh la k h o i liTdng cua san pham phan iJng chay) ta t i m dirpc x (so nguyc" Let luan: H o n hdp gom hai k h i C2H6 va C2H2 tijr cacbon) pong Ihu-c cau tao: C H - C H va C H ^ C H , ^ a) nhh= 1,12:22,4 = 0,05 m o i 208 if':' v6 l i •3a + 2b = , l (2) tang t h e m 4.4 gam Hifdng d i n giai m 2b m o i C2H2 + 5/2O2 b) Tinh phan phan tram the tich moi hon hpp The tich k h i d dkic (1) + 2H2O a + b-0,05 "HJO 3H2O 3a m o i (2) diTng K O H Sau t h i nghiem k h o i IiTdng binh (1) tang t h e m 1.8 gam, binh P2O5 ->2C02 b moi Cong thurc cau tao (rut gpn) la: C H - C H - N H hoSc la C H - N H - C H C x H , , Tir phircJng trinh phan tfng dot chdy va k h o i lurpng san pham CH=CH CH^CH a myl > (3) C H = C H (2) C H - C H CH2=CH2 31,12g Cong thtfc phan tijr cua A ^ C H N Cdch x = = x a + xb = , l C2H4 Cvt lOOg A t h i C O 53.33 g C ; 15,55g H va 31.12g N xb m o i Hon hpp hai hidrocacbon c6 the la : Cdch 2: Cong thtfc phan tur cua A la C^HyN 12xg XCO2 + Z / H O (x + z/4)02 = a + b = 0,05 'hh tur N (z = 1) nen cong thtfc D G N cung la cong thiJc phan tuf C H N -m|afeli§«0 + b moi • , „, Cong Ihtfc ddn gian nhat ( D G N ) cua A \h C H N V i phan tiJf chi c6 nguyf> '>^-' - xa moi a moi MN 53,33 15.55 31,12 - -> XCO2 + y / H + ( X + y/4)02 C,Hy I i k h i chiem 50% the tich hon hdp * ; ] ,5, ; r mH20= ^0 tang k h o i IiTpng cua binh (1) = l , g => nH^Q = 1,8 : 18 = 0,1 mol tach m c o = ^9 tang k h o i liTpng cua binh (2) = 4,4g =0 Ucoj = 4,4 : 44 = 0,1 rong hon hdp, cong thiJc chung cua chiing la C x H - 2: N e u g p i y la so nguyen tuT hidro trung binh cua hai hidrocacbon , 209 C,H- + (x+ y/4)02 > mol XCO2 mol 0,5 y m o l ' 0,1 m o l 0,1 m o l '* X 0,05 m o l Tir etan: C H - C H + CI2 y =4 (mot hidrocacbon phai c6 so' nguyen lu' CaC03 , > CH3 - CH2 - OH + NaCl Xac dinh cong thiJc phan tuf va viet cong thiJc ca'u tao rut gon cua X > CaO + C02 p ; '""^ (c) > v ^.rh Ca(0H)2 + C2H2 > 3C2H2 k h o i liTdng la 44 : 27 Phan tiJ k h o i cua X la 30 dvC C6H5 ' ' ,v•,,^.M • ^ H d p chat hffu cd A c6 phan k h o i lifdng cac nguyen to nhU' sau: C = Bai Hay sap xep cac hidrocacbon cho du'di day theo thuT tu" tang dan nhiet soi: 54,5%; H = , % ; O = 36,4% Xac dinh cong thiJc phan tuT cua A biet rang 0,88 gam hdi chat A chie'm the tich 224 c m \ C4Hi(), CH4, C3HX, C2H6 va Ci{)H22, Cho biet d 20"C benzen c6 k h o i lifdng rieng la 0,879 g/ml N e u hoa long 7,8 kg hdi benzen xuong 20"C thi thu duTdc bao nhieu l i t benzen THI B a i giai V i ^ ' t c n g thufc c a u t a o , V i e t phu'orng t r i n h p h a n ufng, Diiu So nguyen tur cacbon cang nhieu, nhiet soi cang tang, do thu* M tang che nhiet soi la C H < C2H6 < CjHx < C4HU, < C,oH22 B a i Ngifdi ta c6 the dieu che metan tiJf C va H2 (co mat N i , t"), tif n h o m cacbua (AI3C4) tac dung v d i nU'dc hoSc nung nong natri axetat v d i N a O H CO mat CaO xiic tac, bie't rang tru'dng hdp ngoai metan chi c6 mgi san pham m u o i v6 cd V i e t ta't ca cac PTPL/ xay Hoa long 7,8kg hdi benzen xuong 20"C thu diWc: I •r ' x ^ ' g = 8873 m l hay 8,873 l i t 0,879g/ml Bai B a i giai c + > C H - C H - CI + H C l ^"^'""^ > C a C + CO C a C + 2H2O Do't chay mot lu'cfng hidrocacbon X, ngU'di ta thu diTdc C O va H2O theo li if a) '" CaO + C I I Bai tap \\f g l i i Chii de '"^''"^ TH than da, da v o i d i e u che' axetilen, benzen nho hdn la C2H2 va mot hidrocacbon c6 so' nguyen tuf H Idn hdn la C2Hf,) C BAI T A P L U Y E N CH2 = CH2 CH3 - CH2CI + NaOH (1 : 0,05) = ( x : , ) = ^ x = (1 : 0,05) = (0,5 y : 0,1) C H ^ C H + H2 TCr axetilen: y/2H20 + < \ liifh^l :^/.;Hf«:W-r- l Cho hon hdp cac k h i C H , C2H4, C2H2, SO2, C O di qua nu-dc brom V i e t c^c phu'dng trinh phan tfng xay H - ^ C H Cho l i t benzen (d = 0,879 g/ml) tac dung v d i 112 l i t CI2 (ct dktc) k h i c6 mSt b) AI4C3 + I2H2O c) CH3 - COONa + NaOH > 4A1(0H)3 '" + 3CH4 > CH4 xiic tac la bot s^t thu dirdc 450g clobenzen T i n h hieu sua't phan iJng B a i giai + Na2C03 1- K h i cho h o n hdp k h i qua nU'dc brom chi CO cac phan iJng sau: B a i Co the dieu che etilen tiJf ru'du etyliq, tiJf axetilen va tijr etan V i e t cac phv[iM trinh phan iJng xay TiJf than d a , da v o i , viet cac phu'dng trinh phan tfng d i e u che' axetilen, benzen (c6 ghi dieu k i e n phan tfng) B a i giai ' S Cic phuWng trinh dieu che etilen: a) Tur riTdu elylic: , ,' C2H5 - O H > C2H4 C2H4 + Br2 > C2H4Br2 C2H2 + 2Br2 > C2H2Br4 ui l« H2SO4 + H B r 2- Phan iJng giiJa benzen va clo k h i CO bot s^t xuc tac ' CfiHe + CI2 — CO2 + Na2C03 > CO2 + H2O + CaO E + O2 niTdc brom: the sue k h i CO2 vao ket tua) khong la > CO2 + H2O + CI2 + H2O > H B r + H2S04 Cho k h i l a i tac dung v d i > CO2 ' • - ^ ''-'^ i,! : Chat Chat B + O2 ' CO2 thi khong: (CO > CO2 + H2O ' ^ >CaS03i+H20 cho d\idi day la hdp cha't hffu cd dtfdc khong: A + O2 ' ' v6i CaCOji Ca(OH)2 + S02 4CO2 + 5H2O + N2 B a i Co the diTa vao san pham dot chay de suy luan cac chat dem dot chay O2 > Ca(CH3COO)2 + H2O + CO21 chat tao ket tua v d i niidc v o i la CO2 va SO2; ' COOH) niTdc la benzen C6H6 Nhan bie't cac chat k h i cd ((NH4)2C03) ho|c hiiru cd (NH2 - CH, - - chat tan nvldc la C2H5OH va C H C O O H , dung da v o i de nhan bie't Con C2H5OH khong tac dung v d i da ' V a y chat C c6 the la hdp chat v6 C T C T cua hidrocacbon X cd - Chat A la hdp cha't hSu cd chiJa C, H hoSc C, H , O - Chat B chi c6 the la C hoSc CO (v6 cd) - Chat C chi c6 the la liTu huynh S (v6 cd) - Chat D la hdp chat hihi cd chuTa C, H, C I hoSc C, H , C I , O - Chat E la hdp chat huTu cd chiJa C, O, Na - ^ 2CH3COOH + CaCOj nguyen to C, H, N va c6 the c6 hoac khong co o x i • CH3COOH hUucd K h i dot chay chat C tao CO2, H2O va N j , dieu chuTng to cha't C chiJa cac Thi du: B a i giai Chat khong tan nu'dc, n o i tren mat , > CO2 + 2SO2 V a y B la hdp cha't ^ J Nhan biet cac chat long: K h i dot chay k h i B tao CO2 va SO2, dieu chtfng to chat B phai chuTa C, s Thi du: Iren niidc v o i trong, nu'dc brom, da v o i , hay cho bie't each nhan biet chat ) I2CQ2+ IIH2O '" Baigi^i jPhan ti-ng dot chay: CJiy + (x + ^ ) ••J^(it^mnm > xCOj + ^ H j O (1) • ' '*'' Chat F la hdp chat hffu cd chtfa C, H , O, Na B a i Trong phong thi nghiem c6 binh thuy tinh khong mau b i mat nhan, m"J binh diTng mot chat long hoac mot chat k h i sau day: etan, etilen, benzen, ^t^' cacbonic, k h i sunfurd, riTdu etylic va axit axetic Chi dU'dc diing them ni/'^'^' k h o i lu'dng binh tang H2O bi hap thu b d i P2O5 P2O5 + 3H2O I D o do: ny^^Q=— , > 2H3PO4 f'j " ^ , ' = 0,5 m o l K h o i lu'dng binh tang CO2 bi hap thu b d i K O H : CO2 + K H — > K2CO3 + H2O ' ' • ' ' 215 17,6 44 = 0,4 mol 2,24 = 0,1 mol nen theo phan uTng (1) ta c6 he phufcfng trinh; Vi so mol X = 22,4 Dodo n C02 '0,lx = , = > x = 0,l.^ = 0,5=>y = 10 Vay cong thtfc phan tur cua X la C4Hi() Cac cong thufc cau tao c6 the c6 CH3 - CH2 - CH2 - CH3 hoac CH3 - CH - CH3 CH3 Bai De dot chay the tich hidrocacbon Y (d dktc, so nguyen tuf C nho hdn 5) can dung 6,5 the tich O2 (d dktc) Tim CTPT cua Y Bai giai Phan u-ng dot chay Y: C,H„ + (x + ^ )02 > XCO2 + - H2O (2) cm the tich Y can 6,5 the tich O2, tuTc cur mol Y can 6,5 mol O2, nen theo phan uTng (2) ta c6: x + - = 6,5 hay 4x + y = 26 4 X -'fv 22 18 14 10 y LapKetbang: C4Hi() luan loai loai loai Bai Dot chay hoan toan gam cha't A chiJa cac nguyen to' C, H, O ta thu diTdc 4,48 lit CO2 (d dktc) va 3,6 gam niTdc Biet lit hdi chat A (tinh theo dktc) nang 2,679 gam Tinh CTPT cua chat A Bai giai Goi cong thtfc cua A la CxHyO, , Phan u-ng dot chay A: C.HyO, + (x + ^ - - )02 > XCO2 + - H O (1) '2/ Tim qua cong thiJc dcfn gian nhat (ttfc ti 16 ddn gian nha't cua so' nguyen ti' cac nguyen to') TriTdc het can tinh khoi lifcJng cua cac nguyen to 4,48x l 2,^ = ^2,4g; , „0,4g; mc= — - 3,6x2 = mo = - 2,4 - 0,4 = 3,2g r^- 2,4 0,4 3,2 , - , Ta CO X : y : z = : :— =1:2:1 12 16 Cong thu-c ddn gian nhat la (CH20)„ (trong n la boi so' cua CTDGN de 216cong thiJc phan tur) Tinh khoi lifdng mol cua A = 2,679 x 22,4 - 60g Vay cong thiJc phan tur cua A la C2H4O2 p^j De dot chay hoan toan 4,6 gam chat B chiJa cac nguyen to' C, H, O can dung 6,72 lit O2, thu diTdc CO2 va H2O theo ti le the tich Vco^ : VH20= : Tim cong thtfc phan tur, vie't cong thdrc cau tao cua B Biet gam chat B d dktc chiem the tich 0,487 lit "" ' " ' * Bai giai Gpi cong thtfc phan tur cua B la C,HyO,: '' Phan u-ng do't chay: + (x + - - - )02 - -> XCO2+ -2H O (1) (a) X Theo dieu kien cho ta c6 ti le: — = — => y =^ 3x x + y x22,4 Theo phiTdng trinh (1) ta c6' ti le: 12x + y + 16z ^ (b) 6,72 4,6 The (a) vao (b), rut ra: x = 2z Vay CO ti le X : y : z = : : Cong thi?c ddn gian nha't cua B la (C2H60)„ Tinh KL mol PT cua B b^ng 22,4 «46 0,487 Tim CTPT cua B: (CzHfiOn = 46n = 46 do n = , Vay CTPT cua B la CjHfiO B&i Tim cong thtfc phan tijr cua mot chat hiJu cd A chiJa 25% hidro va 75% cacbon lit cha't A (d dktc) nhe hdn Ian so vdi lit O2 (c( dktc) Bai giai Vi hidro chic'm 25%) va cacbon chiem 75*% chuTng to A la mot hidrocacbon C,Hy Vi lit A nhc hdn Ian lit O2 chu'ng to khoi li/dng mol phan tur cua A 32 nho hdn khoi lifdng mol tur cija O2 ,2 Ian: MA = 16x75 , phan 16x25 =4 Vay: x = —rr = l; y = 100x12 100x1 Vay CTPT cua A la CH4 Hidrocacbon B chita 20% hidro lit B (d dktc) nSng 1,34 gam Tim cong thiJc phan tiJ cua B Ne'u khong biet khoi liTdng cua lit B co tim difdc cong thiJc phan tur hay khong? 217 rr., ^ , CxHyO, / N Bai giai De do't chay m gam chat A chtfa cac nguyen to C, H , O can 0,3 m o l O2 a) K h o i lifting m o l phan lit cua B b^ng: 1,34 x 22,4 = 30 g/mol thu diTdc 0,2 mol C O va 0,3 mol H2O So nguyen lu* H = ^^^^'^ = ; So nguyen tu" C = ^ ^ ^ ^ ^ = 100x1 ^ ' 100x12 < ' g) T i n h kho'i liTOng m J,) T i m C T P T cua A , vie't cong thiJc ca'u tao cua A V a y CTPT cua B la C2H6 Bai giai b) N c u khong biet k h o i liTOng cua lit B, nghla la khong biet K L P T , luc ta Phan u-ng dot chay A : A + O2 — » chi mcti biet t i le so nguyen^uT C va H cua C^Hy x : y = ^ : ^ = l:3 ^ 12 V a y cong thufc dcfn gian nhat cua B la (CH3)n V i so nguyen tu" H phai la so chan nen n chi c6 cac gia t r i 2, 4, K h i n = la CO cha't C:Hf„ diing hoa trj cua oacbon K h i n = ta c6 cong ihiJc C4H12, cong thiJc khong dung v i cacbon c6 h o a tri l(^n h(Jn m A + nio2 = mco2 + niA rnuxo = 0,2 x44 + 0,3 X 18 - 0,3 x 32 = 4,6 gam b) Goi cong thiJc cua A la CxHyO,: So'mol H = Ian so m o l H2O = x 0,3 = 0,6 m o l S6' m o l O A = to'ng so' m o l O C O va H2O triT s6' m o l O O2 = x 0,2 + , - X 0,3 = 0,1 m o l V a y t i 16 x : y : z = 0,2 : 0,6 : 0,1 = : : K h i n = 6, thi cong thiJc thu diTdc cang sai C6 the vie't C „ H „ r o i b i e n luan theo tifng loai hidrocacbon Bai T i le khoi lu'cing cua cacbon va hidro hidrocacbon X la mc : m H = 12 T i m cong thufc phan tu" cua X biet k h o i liTcfng phan tur cua X Idn gap 1,3 I a n khoi liTdng phan tuT cua axit axctic Cong thiJc ddn gian nhat cua A la ( C j H f i O n Ta thay n c h i c6 the bang v i neu n bang t h i d\i hoa tri cua cacbon C4H12O2 (so nguyen tuf H t o i da bang 10, nghTa la C4H|,)02) V a y cong thtfc phan tur cua A la C2H6O 12.1 x :y= —: 12'l bkng thu diTdc 3,52 gam C O va 1,62 gam H2O T i m CTPT, vie't C T C T 1:1 cua cac hidrocacbon Bai giai V a y cong thuTc ddn gian nhat cua X la (CH)„ K L P T ciJa C H - C O O H bling Trirdc het can tinh so m o l C O va H2O 60 K L P T cua X = 60 x 1,3 = 78 V i (12 + l ) n = 78 «St* ii < ' ••• Bai 10 Do't chay hoan toan hon hcJp hidrocacbon CxHy va CxH,, c6 so' m o l B a i giai Goi cong thiJc cua X la C^Hy ta c6 t i le: C O + H2O 2) Theo djnh luat bao toan khoi lifdng: ' ^^*^' Co the t i m cong thuTc phan tuT b^ng each bien luan nhiT sau: ' ' n = Do C T P T cua X la CfiH^ nco2 = ' 52 62 ^ = 0,08mol; U H ^ O = ^ = 0,09mol ; : Ji-i-'* 'Iv • B a i T i m t i le so nguyen tijT C , H , O hdp chat Y chiJa , % H , 18,67'7rN va 42,67% O B i e t rSng dot chay hoan toan m o l Y thu duTdc 11,2 lit N: V i so m o l H2O nhieu hcfn so m o l C O nen phai c6 ankan (trU'dng hdp (d dktc) T i m C T P T cua Y ankan can l o a i v i luc y = z = 2x + nghTa la chi c6 chat chtf khong phai r hon hdp) B a i giai ke't d o i - (c6 the n o i tdng so l i e n ke't n) va so' vong = 0,5 m o l N2, dieu chtfng 16 m o i phan tuf c6 nguyen tu" N , do cong thuTc phan tuT cua m C2H5O2N 218 • a to'ng so lien ke't d o i , l i e n ke't ba - lien ke't ba tiTdng diTdng l i e n ^ ^ ^ , =2:5:2:1 V i k h i dot chay m o l Y thu diTdc 11,21 N2 ttjTc •• Goi cong thiJc cua ankan la CxH2x+2 va cua hidrocacbon thi? hai la CxH2x+2-2a Gpi cong thuTc cua hdp chat la C x H y O , N , ta c6 t i le: 32 6,67 42,67 18,67 X : y : /.: t = — : : :— — • 12 16 14 - M «'» Cac phan iifng dot chay CxH2x.2 + ^ ^ — ^ CxH2x.2-2a + ^ ^ " " ^ " ' XCO2 + ( X + 1)H20 XCO2 + ( X + - ,07 a)H20 219 9ni V V i dot m gam A thu diTcJc — gam H2O nen ta c6: - ( x + y) = M S = 9y V i so' mol ciia hidrocacbon b^ng nen ta c6 ti le so' mol C O va HjCj nhiT sau: ^ "CO2 HHJO 2x 2x Hay y = 2x, ttfc CTDGN Ih (C^UiA- ^ 0,08 + 2-a Rut X = - 4a hay a = 2,3 nr.;- 0,09 8-x X KLPT cua A n^m khodng 29 = 66,7 va 2,5 x 29 = 72,5 Tu-c 66,7 < 14x < 72,5, gia tri x nhai t^ng Vay CTPT cua A la CsH,,, pai 13 Hon hcfp X gom hidrocacbon A, B thupc loai ankan (no), hoSc anken (CO lien ket doi) hoSc ankin (c6 lien ket ba) T i le KLPT cua chung la 22: V i a la so nguyen nen chi co nghiem x = va a = I nhat 13 Dot chay hoan toan 0,3 mol X v^ cho san pham chay hap thu vao binh CTPT cua hai hidrocacbon la dung dich Ba(OH)2 dvl thay khoi liTdng binh dung dich tang 46,5 gam va c6 C4H1,, va C4HX CHi - (pH - C H CTCT: C H - CH2 - CH2 - C H 147,75 gam ket tua Tim CTPT cua A, B CH2 = C H - CH2 - CH3 C H - C = CH2 CH3 - CH = CH - CH3 CH2-CH2 CH3 Cho 0,3 mol X di tijf tvt qua 0,5 lit dung dich nMdc Br2 0,2M thay nifdc brom CH2 CH2-CH2 mat mau hoan toan va c6 5,04 lit bay (dktc) •',1 CH3-CH-CH3 Phan urng tao ket tua: Ba(0H)2 + CO2 5rt Viet CTCT cua tat ca cic dan xuat chiJa brom c6 the c6 Goi X la dan xuat chiJa nhieu brom nhat c6 CTPT C^HyBr,, KLPT cija X bang: x 202 ti?c z toi da Ih Do c6 the viet lai CTPT cua X la - 12x + y + 80x = 202 Hay 12x + y = 42 v.^: A TriTcfng hdp 1: anken „ Taco: 80 X y 30 18 C,H2, 14x = CyHjy Lap bang: (1) Dodo so'molnirdc nH20 = ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ = 0,75moL , 18 Vi 0^02 = " H20 CO triTcfng hdp: = 202 dvC Tac6: 12x + y + SOz = 202 Vay z < > B a C O j i + H2O S ' m o l C = nB,co3 = ^ ^ ^ - , m o l Bai giai 101 t.'": Bai giai Bai 11 Cho hidrocacbon A tdc dung v6i BT2 thu diTcJc mot so dan xuaft chij-a brom, dan xua't chuTa brom nhieu nhat c6 ti khoi so vdi H2 bhng 101 I iiJ- -U Hoi thu diTcJc san pham gi, bao nhieu gam? 22 X 22 ^ 44 = — t t f c - = — h c a c — v v loai 14y 13 y 13 • 26 Trirdng hdp 2: anken va ankin • CxH2x+2 + O2 > XCO2 + (x + 1)H20 C,H2,.2 + ^ O2 > yC02 + (y - )H20 (2) Vay CTPT cua X la C3H6Br2 Bai 12 A la mot h d p chat hHu ccJ chiJa nguyen to Dot chay hoan toan m gam 9m •> ' A thu di^rtc -y gam niTdc T i khoi cua A so vdi khong nam khoang 2,3 den 2,5 Tim CTPT, viet CTCT cua tat ca cac dong phan cua A , f Bai giai ^* (3) 5oi a, b la so mol cua ankan va ankin, ta c6 a + b = 0,3 nco2 =ax + by = 0,75 ' |,^ n^^^o = a(x +1) + b(y - ) = ax + by + a - b = 0,75 , Hdp chat A chiJa nguyen to cacbon va hidro, c6 the bieu dien A Ik C^Hy, CO phan i^ng chay: Giai he phu'cJng trinh tim di/dc a = b = • CxHy + (X + ^ )02 > XCO2 + ^ H2O = 0,15mol Xet ti le KLPT 221 220 a) Mankan Mankin _ 14x + 14y - 22 13 V i a = b = 0,15 nen x + y = hay s6 m o l H2O nhieu hdn so' m o l CO2 nen A la ankan Thee phan tirng do't 3x + = 2 y 0.75 0,15 chiy A: =5 C„H2„.2 + Ke't hdp phi/dng trmh cuoi cung ta c6 x = va y = Tac6tyl$: Vay CTPT cua cac hidrocacbon la C^Hx va C2H2 b) M ankin ankan 14X-2 22 14y + 13 hay 22x = I y - > nC02 + (n + D H j O = Rdt n = V a y cong thuTc phan tuf cua A la: CeHu Suy do't chiy loai - 1,5 = 1,5 m o l CO2 va 2,5 - 1,75 = 0,75 m o l H2O C H = C H + Br2 CHBr = CHBr (4) C H B r = C H B r + Br2 CHBr2 - C H B r j (5) 5,04 So m o l k h i d i k h o i bmh midc brom bang '-^^ = 0,225mol 22,4 vi X khong l a m mat m a u dung dich nifdc brom n e n B phai thupc loai hidrocacbon thdm (aren), tiJc CnHnph^i thoa m a n : n = 2n - 6, tuTc n = va c6ng thtfc phan tuf cua B la CeHe (benzen) G p i X , y la so m o l cua A, B Theo phdn tfng chay: Nhir vay so m o l C2H2 da tham gia phan uTng b^ng: 0,3 - 0,225 = 0,075 m o l CfiHu + , — ^ G o i p, q la s o m o l C2H2 tham gia cac phan tfng (4, 5) ta c6: CfiHfi + 7,502 — ^ ,; C O + 3H2O T a c 6 x = l , v a y = l , , tij'cx = y p + 2q = n g ^ = 0,5 x 0,2 = , p = 0,05; q = 0,025 M o i chat A , B c h i e m % so m o l 143 — - , = 0,25 m o l CO2 v 44 K h o i li/dng C H B r ^ - C H B r j b^ng 0,025 x 346 = 8,65 gam B a i 14 Dot chay hoan toan 41 gam hon h d p X gom hidrocacbon A, B thu dUdc 132 gam CO2 va 45 gam H2O Neu them vao X mot nuTa liTdng A c6 X roi So nguyen tuf cacbon = 0,05 dem dot chay hoan toan thi thu dU'dc 165 gam CO2 va 60,75 gam H2O T i m C T P T cua A , B biet hon h d p X khong l a m mat m a u dung dich niTctc brom A, B thuoc cac loai hidrocacbon c h i T d n g trinh da hoc T h e m 0,05 m o l hidrocacbon D vao hon hdp X d tren r o i d e m dot chdy hoan toan t h i thu d i T d c 143 g CO2 va 49,5 gam H2O T i m CTPT, viet tat ca dong ^ I^TPT 49 a 18 2,5 = 0,25 m o l H2O = ; So nguyen tuf hidro = x = 10 ^ ' 0,05 cua D : C5H10 B i i 15 H o n hdp k h i X (dktc) g o m ankan (C„H2„ + 2) anken ( C ^ H z J Cho Nhif vay do't chay A se thu dU'dc: = 1,5 mol CO2 va (3,375 - 2,5) X = 1,75 m o l H2O 6,72 l i t hon hdp X nSng 13 gam 1- T i m C T P T cua ankan va anken, biet so nguyen tijf cacbon m o i phan tuf kh6ngqud4 v^o dung dich N a O H diT, sau 66 them BaCl2 diT t h i thu diTdc bao nhieu gam = m o l CO2 va ^ = 2,5 m o l H2O 18 di/dc — = 3,75 m o l CO2 va • 44 X J 2- D o t chay hoan toan 3,36 l i t h o n hdp X va cho tat ca san p h a m chdy hap thu B a i giai (3,75 - 3) ^ ' : 3,36 l i t h o n hdp X qua binh nirdc brom dU' tha'y c6 g a m b o m phan tfng B i e t T i n h % so m o l cua A, B t r o n g X D o t chay X diTdc ^ 44 ; ,, Dot chay hoan toan 0,05 m o l D diTdc: K h o i l i T d n g C H B r = C H B r bang 0,05 x 186 = 9,3 gam phan ciia D « C O + 7H2O p + q = 0,075 Do't chay X+-k ^ B diTdc: V i so m o l CO2 ga'p d o i so' m o l H2O nen cong thiJc ddn gian cua B la (CH)2 va Cac phan ufng: rtRn^i,,^a O2 18 = 3,375 m o l H2O ke't tua B a i giai \jUid:jj^, j f^^r^khP^ A = 0,4mol > mol CO2 + mol H2O ^^.f Suy mol X c6 mol C, mol H va mol O Vaycongthurcphantijrcua A , BlaC4Hx02(M = 88) J Ung vdi cong thtfc phan tuf C4HSO2 ta thay c6 cac este nhiT sau: ' CH3 - CH2 - C00CH3 (I) CH3 - COO - CH2 - CH3 (II) , ; ^^ ; / H - COO - CH2 - CH2 - CH3 hoac HCOO - CH - CH3 (III) ^'rWii CH3 275 phan uTng vdi N a O H : C H - C H - C O O C H + NaOH C H O H + C H - C H - COONa (M = 9^) CH3-COOCH2 - C H + NaOH CH3 - CH2OH + CH3 - COONa (M = 82) HCOO-CH2-CH2-CH3 + NaOH -> CH3CH2 - CH2OH + HCOONa (M = 6;^^ Theo cac phan iJng tren thi: n^stc = n^mt, = = 0,02mol 88 V i KLPTTB cua m u o i bang ^15, do c6 hai triTcJng hdp xay Trirdng hdp 1: m u o i HCOONa va CHjCOONa ttfc cong thiJc cau tao cua cac este iJng v d i III va II Trurdng hdp 2: m u o i HCOONa va CH3CH2 - COONa, ttfc cong thtfc cau tao cua cac este iJng v d i III va I Bai Cdc De dot chay m o l chat X can 6,5 m o l O2, thu dtfdc m o l CO2 va m o l HjO Hay xac dinh cong thiJc phan tuT cua X j Dot chay hoan toan gam chaft Y chtfa cac nguyen to' C, H, O thu diTdc 2,24 lit CO2 (d dktc) va 1,8 gam H2O Biet gam chat Y c h i e m the tich 0,3732 lit (tinh theo dktc) Xac dinh cong thuTc phan tvt, viet cong thtfc ca'u tao cua Y, biet r^ng Y la m o t este Ta c6 phan tfng: X + 6,502 Theo dinh luat bao toan k h o i Bai giai > 4C02 + 5H2O liTdng thi m o l cha't X phai c6 m o l C 10 mol H va khong chuTa oxi Vay cong thiJc phan tur cua X la C4H1,, Khoi li/dng mol phan tu" cua Y b i n g : ^^'^ = 60.02 0,3732 M Y = 60g/mol • lay chinh xdc: So nguyen tijT C - ^'^'^^^^ = (suy tijr gia thiet dot chay m o l Y) X 22,4 , c-^' n 1,8x60x2 So nguyen tiT H = =4 3x18 o-^' v ^ - x - x = ^2 So nguyen tur O = Vay CTPT cua Y la C2H4O2 Vi Y la este nen CTCT la: H - COO - CH3 Bai Chat beo B cd cong thtfc (C„H2„+ ,COO)3C3H5 Dun ndng 16,12 g chat B v d i 250 m l dung djch NaOH 0,4M tdi k h i phan iJng xa phong hoi xay hoan toan ta thu du-dc dung dich X De trung hoa NaOH dU" c6 1/1'dung dich X can 200 m l dung dich HCl 0,02M 276 Hoi x^ ph6ng hoa kg cha't b6o B tidu ton bao nhieu gam NaOH va thu di/dc bao nhifiu gam glixerin? X^c dinh CTPT cua axit tao chat beo B y ,t ? Bai giai !r ^ v Phan tfng xa phong hoa cha't beo B: (C„H2„+,COO)3C3H5 + N a O H > 3C„H2„+,COONa + C3H5(OH)3 (1) Phan ifng trung hoa NaOH diT: HCl + NaOH > NaCl + H2O (2) Tong so mol NaOH diT = 10 x UHCI = 10 X 0,2 X 0,02 = 0,04 mol 56'mol NaOH ban dau = 0,25 x 0,4 = 0,1 mol So' mol NaOH tham gia phan uTng xk phbng hoa (1) b^ng:0,l - 0,04 = 0,06 mol Kho'i Itfdng NaOH can de xa phong kg chat beo B bing: 0,06x1000 16,12 x40 = 148,8gam Khoi lufdng glixerin thu diTdc bang: ^'^^^/^^^ x 92 = 114, Igam X 16,12 Theo phan ufng (1) so mol chat beo B bang - so' mol NaOH = = 0,02mol 3 Do KLPT cua B = = 806 dvC 0,02 Va KLPT cua axit beo tao B b^ng: ^ ^ ^ ^ ^ + = 256 fta c6 14n + 46 = 256 Rut n = 15 I Vay cong thtfc cua axit beo la C15H31COOH |li Hdp chat X chtfa cac nguyen to C, H, O Ctf 0,37 gam hdi chat X chiem the tich b^ng the tich cua 0,16 gam oxi d ciing dieu kien nhiet do, ap sua't Mat khac cho 2,22 gam X vao 100ml dung dich NaOH IM (d = 1,0262 g/ml); sau phan iJng lam bay hdi dung dich tdi kho thi thu duTdc 100 gam chat long, phan kho lai nSng Y gam Tim CTPT cua X Bai giai Vi the tich Vx = V02 do nx = n^^ = ^ = 0,005mol r- > VaMx=-^^-74z:>nx- —-0,03mol ' 0,005 ^ 74 :>ci;i Kho'i li/dng dung dich NaOH = 1,0262 x 100 = 102,62 gam So mol NaOH = 0,1 X = 0,1 mol; Khoi liTdng H2O dung dich NaOH = 102,62 - 0,1 x 40 = 98,62g 277 - Neu X khong phan tfng vdi NaOH, nghia Ik bay hdi cilng nxidc thi khoi liTdng chat long m x + n i H j o = 2.22 + 98,62 = 100,84 gam dieu khac vdi so Ueu cho Vay X tac dung vdi NaOH - VI X tac dung vdi NaOH nen X Ih axit hoSc este v^ vl Mx = 74 nen chi c6 nhdm axit ( - COOH) hoac este (-COO - ) - Goi cong thu-c cua X la R - COOR' (neu axit thi R' la H) R-COOR'+NaOH ' > R - COONa + R'OH (1) KLPTcua Y=^ 12x + y+16z = 88 (5) W T I J T cac phtfdng trinh (2, 3, 4, 5) ta c6 x = 4; y = 8; z = Vay CTPT cua Y la C4HSO2 I Phan iJng vdi NaOH: Vi Y(C4Hx02) + NaOH => rtfdu (mi gam) + muoi (m2 gam) nen Y phai la mot este; vi so nguyen tijT cacbon ru'du = so nguyen tur cacbon axit = - = nguyen tur C K;? So mol NaOH d\i = UNaOH diu - nNaOH phan ifng = 0,1 - 0,03 = 0,07 mol Theo phan tfng (1) khoi lifdng cha't ran Y bang Do cong thtfc cua riTdu la C2H5OH vdi mi = 0,05 x 46 = 2,3g c6ng thiJc cua axit la CH3 - COOH vdi m2 = 0,05 x 82 = 4,lg CHjCOONa A la mot este ddn chiJc cua mot axit no R - COOH va ru'du R'OH Khi thuy phan hoan toan 7,4 gam A ngiTdi ta dijng 45,5 ml dung dich NaOH 10% (d= 1,1 g/ml) LiTdngxutda dungdir25% , Xac dinh CTPT va viet CTCT cua A ' g i * ; s : my = mx + mjd NaOH - 100 = 2,22 + 102,62 - 100 = 4,84 gam Khoi lurdng R - COONa = 4,84 - 0,07 x 40 = 2,04 gam M 04 MRcooNa = = 68 , d o d o R = 1, tu-c H Yh R' = 74 - 45 = 29 suy d o la C2H5 - , Dot chay hoan toan 1,48 gam A va cho tat ca san pham chdy hap thu \ho midc voi cMa 2,96gam Ca(0H)2 thi c6 bao nhieu gam ket tua tao thanh? Bai giai ' Vay CTPT cua X la HCOOC2H5 Bai Dot chay hoan toan 4,4 gam hdp chat hi?u cd Y chiJa C, H, O can vuTa 5,6 lit oxi (dktc), thu diTdc CO2 va hdi n\idc vdi the tich b^ng Xac dinh CTPT cua Y, biet rang KLPT cua Y la 88 dvC du , Dat cong tMc cua este la R - COOR' Phan tfng thuy phan R - C O O R ' + NaOH Cho 4,4 gam Y tac dung hoan toan vdi mot lifdng vufa du dung dich NaOH sau dc lam bay hdi hon hdp thu diTdc mj gam hdi cua mot rifdu ddn chiJc va m2 gam muoi cua mot axit hi?u cd ddn chiJc So' nguyen tu" cacbon d rU"du va axit thu diTdc bkng Hay xac dinh cong thtfc cau tao va -? ten goi cua Y Tinh khoi lifdng mi va m2 >xC02+^H20 Theo phan tog (l)ta CO tile: 7,4 Khi R = (tto H) thi R' = 29 (tto C2H5 - ) Cong thto cua este la: HCOO- C2H5 j.,, > iCOj + 3H2O ^ > CaC03 i + H2O CO2 + Ca(OH)2 CO2 + H2O + CaC03 tinh: nc,(0H)2 =^ = 0,04mol; • ^ ' I Cac phan tog: C3H6O2 + ^ Vi the tich CO2 = the tich hdi niTdc, do ta c6: , > Ca(HC03)2 n c o = 3nA (2) - (3) ' ' (4) ' ' y = 2x = — = > R + R' = 74-44 = 30 0,1 j ^1 K h i R = ( C H ) t h i R ' = 15(CH3-) (1) Tinh ny = — = 0,05mol; no, = = O,25mol 88 °2 22,4 nco2=0,05x; > R - COONa + R'OH (1) So'mol NaOH tac dung viifa du vdi este bang x—x = 0,Imol ^ 10 40 125 „ Bai giai Goi cong thtfc ciaa chat Y la CxHyO,, Phan tog dot chiy Y: C,HyO, + ( x + ^ - - ) , (3) (4) n c o > nca(OH)2 CO xay phan tog (1) Theo cac phan tog (3, 4): ncacoa = 0,04 - (0,06 - 0,04) = 0,02 mol IVay khoi lifdng ket tua b^ng 0,02 x 100 = gam ^ •• 279 B a i De dot chay hoan toan 13,275 gam cha't A (chi chtfa c&c nguyen t6' C, H, O ) can 15,12 l i t O (dktc), san pham chay gom nhiJng the tich bang C O va h d i nirdc LiTdng C O thu dUdc k h i dot chay 0,2 m o l A b^ng lifdng C O dot chay 0,25 m o l butan I C the tim theo g o c R , c h i c a n chd ^ go'c R c n h d m - O H tdc dung B dufdc v d i Na mi PHAN GLUXIT - B a i X la cha't long hiJu c d chtfa (% k h o i liTdng) 40,45% cacbon, 7,87% hidro, TimCTPTciiaA 15,74% nitd, l a i la oxi T i k h o i cua X so v d i khong k h i la 3,069 V i e t C T C T ciia A biet r^ng: A tac dung v d i Na theo t i le m o l : va A cung tdc dung v d i N a O H theo t i le so m o l : 1, k h i cho 5,9 gam A tac dung vdi T i m C T P T c u a X 150 m l dung dich N a O H , M , sau c6 can dung dich thu diTdc 9,6 gam Y la mot dong phan cua X T i m C T P T cua chung, bie't r^ng k h i tac dung v d i chat r ^ n khan dung dich N a O H X cho m u o i C H N N a , Y cho m u o i C3H602NNa B a i giai B a i giai r N e u n la so nguyen tijT cacbon A , ta c6 he thtfc 0,2n = 0,25 x (4 la so nguyen tv( cacbon butan C4H|()) ttfc n = D o c6 the b i e u d i e n cong Phan urng dot chay: CsHjoO, + ^^y^Oi > 5CO2 + 5H2O (1) Co the dat C T T Q cua A la RCOOR' (trong do, R, R' c6 the la H ) R - C O O - R' + N a O H > R - COONa + R ' O H n^^ = — ^ = 0,05mol; n^jgOH = ' x l = = , m o l m u o i + ri/du (+ N a O H diT) K h o i liTdng rifdu = 5,9 + 0,15 x 40 - 9,6 = 2,3 g a m D o K L P T c u a riTdu = — •— = So nguyen tuf H = 100x12 ^ ' ' ^NT 89x15,74 , So nguyen tiT N = = 100x14 89-3x12-7x1-1x14 ^ So nguyen tiT O = = I f 89x7,87 ^ — = 100x1 t>;! • )• 16 1- nguyen tijf cacbon cua X nen X phai la este, phan goc axit phai chtfa N, V i Y tac dung v d i dung dich N a O H cho ta m u o i cd so nguyen tuf cacbon Idn hdn cacbon nen Y phai la: aminoaxit N H - CH2 - CH2 - C O O H hoac CH3 - C H - C O O H • , NH2 a i D o t chay hoan toan 0,74 g mot hdp chat hffu c d X mach hd (chtfa cac P2O5 dU" va ong diTng K O H duT Sau thi nghiem thay k h o i li/dng ong tang 0,54 gam v a ong tang 1,32 gam i T i m C T P T cua X biet t i k h o i cua X so v d i He b i n g 18,5 = 46 I V i e t C T C T cua tat ca cac chat mach hd cd C T P T nhiT X va c h i ro chat n^o 0,05 Ta d e d a n g t i m diTdc C T P T cua riTdu la CjHfiO ho5c C H O H = n g u y e n tuT c a c b o n ; - = n g u y e n tijr hidro tic dung dufdc v d i Na, v d i N a O H , v d i NaHC03 Xac djnh C T C T chinh xac ciia X , biet r^ng dun nong 0,74 gam X vdi 100 gam dung dich N a O H % , sau dem c6 can thi thu diTdc 4,42 gam chat r^n khan (, • D o C T C T c u a A la: hoSc CH3 - C H - COOC2H5 OH OH ,;v, , nguyen to C , H , O) v^ cho san pham chay Ian liTdt ong dufng d u - - , = 0,lmol Co the tim R' difa vao D L B T K L : A + N a O H ^r,, n i each khac X la este cua mot a m i n o a x i t : N H - C H - C O O C H Phan ufng cua A v d i dung dich N a O H : CH2 - CH2 - COOC2H5 , V i X tac diing v d i N a O H tao muo'i c so nguyen tu* cacbon nho hdn so V a y C T P T cua A la C S H H A Phan g o c R g o m : 5-2 V a n g u y e n tuf o x i 89x40,45 C T P T cua X la C3H7O2N 22,4 "NaOH ,^ So nguyen tuf C = T h e o d i e u k i d n cho vk theo phu'dng trinh (1) ta c6 t i 1?: 15-x _ 70 + 16X RiJt X = 13,275 15,12 Tinh: ; \h M , = 3,069 x 29 = 89 thiJc cua A la C s H i o O , ( v i so m o l C O = so m o l H O ) 280 PROTEIN - POLIME B a i giai •uiW:>-i>' '] ' D o t chay X thu diTdc C O va hdi nifdc; k h i cho san p h a m chay qua ong P2O5 I t h i nirdc b i giff l a i 1^ 281 HzO + P2O5 Do so mol H2O = Bai giai ( hoac V i e t H3PO4 cung dtfdc) = 0,03 mol 18 / Khi qua ong KOH thi CO2 bi hap thu: CO2 + 2KOH K2CO3 + H2O Do so mol CO2 = — = 0,03 mol 44 Khoi lu-dng oxi = 0,74 - 0,03 x x - 0,03 x 12 = 0,32 gam Goi cong thu-c cua X la CxHyO,, ta c6 ti le X : y : z = 0,03 : 0,03 x : — = : : CTDGN cua X la (C3H602)„ Mx16= 18,5 X = 74 Do n = 1, nghia la CTPT cua X la C3H6O2 Cac CTCT c6 the c6 cua X: > 2HPO3 Goi cong thtfc cua aminoaxit la CxHyO,N„ ta c6 ti le: 46,6 8,74 31,07 13,59 X :y :z :t = 12 16 14 = : : : Vay cong thtfc phan tuf cua aminoaxit la C4H9O2N va cong thufc ca'u tao la: CH3 - CH2 - CH - COOH > CH3-CH2-COOH (I) CH2 - CH2 - CHO " OH (IV) CH2-C-CH3 iH O Chu de HCOOCH2-CH3 (II) CH3 - CH - CHO OH (V) (III) CH3 - O (VI) CH2 Nen UNaOH d>, = 0,1 - 0,01 = 0,09 mol hay 0,09 x 40 = 3,6 Do khoi liTdng muoi bkng 4,42 - 3,6 = 0,82 gam V$y KLPTcuamuoibSng: — = 82 0,01 t - CHO rfS^^^' (VII) - : gam • ^ Uti., Xac dinh phan h5n hgrp Bai Chia h6n hdp ruTdu etylic va axit axetic hai phan bkng Cho phan tac dung het vdi Na du'thu drfdc 5,6 lit H2 (d dktc) Phan thu" hai dem dun nong vdi axit sunfuric thu diTdc 8,8 gam este Biet hi^u suat phan uTng este hoa la 100% Tinh % khoi lifdng cua rU'du etylic hon hdp ban dau CH3-COOCH3 Tdc dung du-dc vdi Na c6:1, IV, V, VII Tac dung diTdc vdi NaOH CO I, II, III Tdc dung du-dc vdi NaHCOs chi c6:1 _ _ 0,74 ^ ^, , 100x4 ^ , , Tmh Ox = = 0,01mol; nM.,oH = = 0,lmol 74 ^"^^ 100x40 VI R - COOR' + NaOH > RCOONa + R'OH oiia i !.:•«'> a't ' NH2 K i! Ttfc R + 67 = 82 do R = 15, ta de dang suy luan R la goc CH3 -Vay CTCT chinh xac cua X la CH3 - COOCH3 Bai Mot aminoaxit chiJa 46,6% cacbon, 8,74% hidro, 31,07% oxi va 13,59% nitd a) Tim cong thiJc phan tuT cua aminoaxit, biet rang moi phan tuf chi c6 mol nguyen tuf nitd 282 b) ke't Vie'tvdicong cacbon thiircIan ca'ucantao,nhom biet -r^ng COOH mach cacbon th^ng va nhom - NH2 lien Cic phan iJng: Bai giai - OH + Na > C2H5 - ONa + 0,5H2 (2) CH3 - COOH + Na > CH3 - COONa + 0,5H2 CH3 - COOH + C2H5OH > CH3 - COOC2H5 + H2O (3) C2H5 5,6 TrU'dc he't can tinh so mol cac chat: = 22,4 = 0,25mol Do theo cac phan iJng (1,2) tdng so mol axit va rU'du bang Ian 86' mol H2 = X 0,25 = 0,5 mol 88 S6' mol este: n^sic = — - 0,Imol 88 VI hieu suat phan uTng este hoa la 100%, do xdy triTdng hdp: Trurdng hdp 1: riTdu thieu, ttfc so mol riTdu la 0,1 mol va axit la 0,4 mol TriTdng hdp 2: riTdu diT, tu^c so mol riTdu la 0,4 mol va axit la 0,1 mol , Tinh phan tram khoi lu'dng cija riTdu: 0,1x46x100 Trirdng hdp 1: %C2H50H = 0,1x46 + 0,4x60 = 16,08% 0,4x46x100 = 75.41% Trtfdng hdp 2: %C2H50H = 0,4x46 + 0,1x60 ki Dot chay hoan toan 1,1 gam hon hdp X gom metan, axetilen va propilen (C3H6) thu diTdc 1,792 lit CO2 (dktc) Mat khac cho 448cm^ hon hdp X (dktc) di tuf tuf qua dung dich nMdc brom dU" tha'y c6 gam Br2 tham gia phan tfng Tinh % the tich va % khoi lufdng cua moi chat X Bai giai Cdc phan tfng xdy ra: * 283 Tinh: C3H6 + Br2 > C3HfiBr2 (4) 2Na + 2C2H5OH — ^ 2C2H50Na + H j t (2) Tru'dc het can tinh so' mol cua riTdu va cua nu'dc: 10x92x0,8 , 10x8x1 ^ , nr-,H.nH = = 0,16mor, n u ^ n = — = 0,044mol C2H5OH 100x46 "20 iQQ^jg C2H2 + 2Br2 > C2H2Br4 (5) Theo cac phan urng (1, 2): CH4 + 2O2 > C + 2H20 (1) C3H6 + 4,502 > 3CO2 + 3H2O (2) C2H2 + 2,502 > 2CO2 + H2O (3) 1,792 , 0,448 iicoj = ^^^TT = ^ 0^^^ol; nx = = -i:^ = 0,02mol 0,08mol;nv22,4 ^ 22,4 ' ^"2 = — = 0,025mol ^ 160 nx = 0,020 = 1,25 l l n 100x46 cac "2 100x18 22,4 '""-s- Giai phiTdng trinh tren ta c6: x = 85,7 ti^c dp ri/du ciia rtfdu X la 85,7% ^ phtfcfng trinh: "'• 16x + 42y + 26z= 1,1 S a i x + 3y + 2z = 0,08 ^'1 Lfng vdi cong thufc phan t^ C2H6O c6 cong thuTc cau tao CH3 - CH2 - OH va CH3 - O - CH3 Khi cho 46 gam chat C2H6O tdc dung vdi Na AM chi thu diTdc 11,2 lit H2 (cl dktc) Vay theo em cong thtfc nao tfng vdi riTdu etylic? Neu lay lit rtfdu etylic d trang thai long (d 20"C, d = 0,80 g/ml) cho bay hdi het thi thu difdc bao nhieu lit hdi rifdu (tinh theo dktc) Bai giai 5I Ta nhan thay cho mol chat (46 gam) tac dung vdi Na chi thu di/dc 11,2 112 lit H2 ti3fc — ^ = 0,5 mol H2 Dieu d6 chtfng to chi c6 nguyen tuf H diTdc 22,4 thay the bdi Na, nghia la cong thtfc C2H5 - OH la dung y + z = l , ( x + y + z) Giai he phiTdng tfmh ta dufdc x = 0,01; y = 0,01; z = 0,02 Vay % the tich c u a c^c k h i X l a : % V c H , - % V c H , = " ' Q Q ^ " " = 25%; %Vc2H2 = 100 - 25 - 25 = 50% % khoi lirdng: CH4 = M i ^j jl l ^ = 14,55%;%me,H C3H6 =Ml^il^ = 38,18% Ta thay mot lit riTdu etylic chtfa: ^ ^ " ^ ^ " ' ^ = 17,39 mol riTdu 46 %mc2H2 = 100 - 14,55 - 38.18 = 47,27% Chu de Vay the tich hdi ri/du = 17,39 x 22,4 = 389,565 hdi riTdu (tinh theo dktc) Tinh thco p h i ^ g trinh phan ufng, II PHAN AXIT hi^u suS't phan ufng, n6ng dp dung d|ch |Bai 1, Hoa tan 12 gam axit axetic vao ni^c 500 ml dung dich (dung dich A) I PHAN Rl/OU Tinh nong dp mol va nong dp % cua dung dich A (d = 1,00 g/ml) Bai :2 Can ddng bao nhieu ml dung dich A de hoa tan vtlfa du gam CaCOs Cho 10 m l rirdu e t y l i c 92" tac dung het v d i Na Tinh the tich H2 b a y r a (d dktc) Biet k h o i liTcJng r i e n g c u a riTdu etylic n g u y e n cha't la 0,80 g/ml, c u a nurdc la 1,00 g/ml ^ De x a c d i n h ri/du c u a mot l o a i riTdu etylic (gpi la rifpu X) ngiTcJi ta l a y 10 I ml rircfu X cho t a c d u n g het vdi.Na thay bay r a 2,564 lit H2 (cl dktc) Tinh dp rtfdu c u a X Coi the tich dung d j c h riTpu b^ng tdng the tich c u a riTpu Bai giai 2Na + 2H20 >2NaOH + H t nifdc Can them bao nhieu gam axit axetic vao 100 ml dung dich A de c6 dung dich axit axetic 10% Bai giai So mol axit n = — = 0,2mol Nong dp mol CA = — = , m o l ' 60 0,5 CA%= i ^ ^ ^ 500 Phan tfug hoa tan: Nong dp %: CdcphaniJng: 284 = ^ (nn,* + n„u C a C C H j C O O z + H2O + CO21 d) ^^^75o "^'^"^'^^ Goi V la so lit axit can diing, ta c6: V x 0,4 = 0,1 nit ra: V = 0,25 lit tuTc 250ml Gpi X la so' gam axit axetic can them vao, ta c6 bieu thiJc ve nong phan tram nhiT sau; fx.M^lxlOO l^oj 10=i 100 + x Giii ta c6: x = 8,44g Bai Dung djch X chtfa HCl va CH3 - COOH De trung hoa 100 ml dung dich X can dilng 30 ml dung dich NaOH IM, c6 can dung dich da trung hoa thi thu diTcJc 2,225 gam muo'i khan Tinh nong mol cua cac axit dung dich X '2 Can bao nhieu ml dung dich X de trung hoa 25 ml dung dich hon hcJp BaCOH), 0,02M va NaOH 0,05M Bai giai Cac phan tfng trung hoa: HCl + NaOH > NaCl + H2O (1) CH3 - COOH + NaOH > CH3 - COONa + H2O (2) Goi a va b la so mol cua HCl va CH3 - COOH ta c6 he phiTdng trinh: a + b = nNaOH = 0,03 x = 0,03 mol Tong khoi liTdng muoi: 58,5a + 82b = 2,225 Giai he phu-dng trinh ta c6 a = 0,01 mol va b = 0,02 mol Vay nong mol cua cac axit la: I Ngay trUdc ngufcfi ta thUcJng diing dau lac, dau viTng de th^p sdng Vict phiTdng trinh phan uTng dot chay hoan toan mot chat beo long c6 cong thtfc; (C,7H3,COO)3C3H5 Ne'n la mot loai hidrocacbon c6 cong thiJc phan tu" C25H52 Trong mot hop kin dung tich Im^ chiJa khong (oxi chiem 20% the tich) thap cay ne'n nang 35,2 gam Hoi cay ne'n c6 chay he't khong? (gdi y: lufpng oxi c6 du khong) -^ ' Bai giai " I Phan iJng dot chay dau an: I = 0.1M ; CCH3-C00H = ^ = 0,2M 4^ Cac phan iJng trung hoa: HCl + NaOH >NaCl + H20 (3) , 2HCl + Ba(OH)2 >• BaCl2 + 2H2O (4) „ ,, CH3 - COOH + NaOH > CH3 - COONa + H2O (5) , 2CH3 - COOH + Ba(0H)2 > Ba(CH3COO)2 + 2H2O (6) Ta nhan thay cu" mol Ba(0H)2 can mol HCl hoSc CH3 - COOH, nghia la tong so mol - OH b^ng tong so mol axit, do ta c6 bieu thiJc: 0,025 (0,02 X +djch 0,05) V (0,1 0,2)V = 0,0075 lit tiirc 7,5ml 286Trong Tong sodomol V la- soOH lit=dung Tong so' mol X =Giai axit ta+c6: + 78,502 > 57C02 + 49H2O (1) • i Phan tfng dot chay ne'n: C25H52 + 3802 > 25C02 + 26H2O (2) 35 = O,lmol ' Tinh: So mol ne'n n = —— 352 Theo phan iJng (2), liTcfng O2 can no2 = 0,1 X 38 = 3,8 mol I tu'c3,8 X 22,4 = 85,121 -kmo-i I LiTdng O2 CO hop hlng = 2001 - ! Ket luan: Lufpng oxit dxi, cay ne'n chay he't Bai Dun nong 10 kg chat beo ran ( C | H C 0 ) C H vdi liTdng vira du dung dich NaOH thi thu diTpc bao nhieu kg xa phong, bao nhieu kg ghxerin va tieu ton bao nhieu kg xut? Bai giai (C,7H35COO)3C3H5 + 3NaOH > SCnHjsCOONa + C3H5(OH)3 (1) ' I CHC, = ^ (C,7H3,COO)3C3H5 Mbcc,= 283 X 3+41 M , , ph,ng = 283 + 23 = 890 = 306 MgUxcnn = 41 + 17 x = 92 MNaOH = ^.^^ 40 ^ Theo phan rfng (1) ta thay ci? 890g (hoSc kg) chat beo tao diTpc 306 x = 918g (hoac kg) xa phong va 92g (hoSc kg) glixerin va liTcJng xiit tieu ton la 40 X = 120g (hoac kg) Vay: liTdng xa phong thu diTcfc b^ng ^^^^^^ = 10,3kg 890 v ' ' •• • * 92x 10 Li/dng glixerin thu dUpc b^ng = 1,03kg LiTcJng NaOH tieu ton bing ^ ^ ^ = 1,35kg , > |]r' f\ • !'>v 287 VI PHAN GLUXIT - PROTEIN - POLIME Cho 36 gam glucozd tac dung vdi AgjO nhieugam Ag, biethieusuatphanu'ngla 100% NH3 (TRICH TLf OE THI HQC SINH GI61 HOA 9) (dii) thi thu diroc b '^^^ a i 1: Cho 6,45 gam hon hOp hai kim loai hoa tri (II) A va B tac dung vdi dung dich H S O loang dU', sau phan uTng xong thu diTdc 1,12 lit (dktc) va 3,2 gam chat ran LiTdng chat ran tac dung vifa du vdi 200 ml dung djch AgNOs 0,5M thu diTdc dung dich D va kim loai E Loc lay E roi c6 can dung djch D thu diTdc muoi khan F , Xac dinh cac kim loai A, B biet rang A di^ng triTdc B "day boat dong Cho 4,5 kg glucozd len men Hoi thu d i T d c bao nhieu lit riTdu nguyen c h a t r bao nhieu lit C O (d dktc), biet hieu suat phan tfng la 80% va khoi la^^^ rieng cua riTcJu la 0,8 g/ml ^ B a i giai ' :fi:.i: Phan iJng trang giTdng cua glucozcJ C6H,206 + A g - i M ^ C6H,207 + A g l Mg,u™„,= ; M A g = (1) hoa hoc cac kim loai" ! Dem liTdng muoi khan F nung d nhiet cao mot thdi gian thu diTdc 6,16 gam chat ran G va V lit hon hdp Tinh the tich V (dktc), biet nhiet phan muoi F tao oxit kim loai NO2 va O2 108 Theo phan iJng (1): 11^^ = x ng,„,„,,, = x ^ = x , = 0,4mol 180 Khoi liTdng Ag bang 0,4 x 108 = 43,2g Phan u-ng len men: C6H,206 ^""/^^ Theo phan uTng (2): n,,,„ = n„,^ = x Nhurng VI > 2C2H5OH + C O ng,u,o„, hieu sua't 80% nen thiTc te" chi thu =2 x diTdc Nhung mot kim loai A vao 400 ml dung dich muoi F co nong mol la C M Sau phan iJng ket thuc, lay kim loai rufa sach, lam kho va can lai tha'y khoi liTdng cua no giam 0,1 gam Tinh nong C M , biet rang ta't ca kim loai sinh sau phan uTng bam len be mat cua kim loai A (2) = 50mol 180 so mol r i / d u cung bkng so B a i giai Kim loai khong tan dung dich on mol CO2, bkng 50 X — = 40 mol 100 40 X 46 Vay V^„u = = 2300 ml = 2,3 lit va the tich C O b^ng: 0,8 40 X cac am TOAIV KH6 ^Chaong VI B a i H2SO4 loang phai la B (dufng sau hidro) Khoi lu'dng kim loai A = 6,45 - 3,2 = 3,25 gam Phan u-ng: Vi HA = nH2 A + H2SO4 > ASO4 + H i t 1.12 = 0,05mol; = 22,4 22,4 = 896 lit Do K L N T cua A = B a i TO nguyen lieu chinh la v6 bao, miin ctfa chtfa 50% xenlulozd (khoi li/dng) ngirdi ta dieu che ri/du etylic vdi hieu suat 75% Viet cdc PTPl/ ciia qua tnnh dieu che va tinh khoi lifdng nguyen lieu can thiet de dieu che 1000 lit 90" Khoi liTdng rieng cua rifdu etylic nguyen cha't la 0,8 g/ml 3,25 0,05 Phan urng: (1) r = 65 Vay A la Zn B + 2AgN03 - — > B(N03)2 + A g i V i n ^ g N O j = 0-2 X 0,5 = 0,1 mol; do n g = 0,1 (2) = 0,05mol B a i giai Cac PTPLf xay ra: (C6H,„05)„ + n H C.H,20 > nCfiH^Ofi (1) ""^"7" ) 2C2H30H + C t (2) 30-32" C Khoi liTdng C H O H CO 1000 lit c6n 90"C = ^""^""^^ x 80 = 720k2 100 * Kho'i liTdng v6 bao miin ciTa c i n thie't = l^^ilZ^x 92 x 75 = 3380,86kg 50 I va K L N T cua B = — = 64 Vay B la Cu 0,05 Dung djch D la dung djch Cu(N03)2, muoi khan F la Cu(N03)2 Theo phan i?ng (2) np = ne = 0,05 mol Phan u-ng nhiet phan F: Cu(N03)2 — n u y = ^ CUO + 2NO2T + ^ ^ ! Neu lifdng Cu(N03)2 bi phan buy het thi liTdng chat r^n CuO bang 0,05 X 80 = gam, mau thuan vdi 6,16 gam 288 (3) • ' 289 Goi n la so' mol Cu(N03)2 da bi nhiet phan, ta c6 phufdng trinh ve khoi li/cJng cha't r^n G: (0,05 - n)188 + 80n = 6,16, rut n = 0,03 mol Vay theo phan iJng (3): V = (2 x 0,03 + ^ x 0,03) x 22,4 = 1,68 lit Phan iJng: Zn + Cu(N03)2 — > Zn(N03)2 + Cu 4^ (4) Gpi a la so' mol Zn tham gia phan tfng (4), ta c6 phu'dng trinh giam khoi lu'dng cua Zn: a - a = 0,l=> Rutraa = 0,lmol Vay nong mol cua Cu(N0,)2 b^ng — = 0,25M 0,4 Bai 2: Hon hdp A gom muo'i clorua cua kim loai hoa tri II Dien phan nong chay 15,05 gam A tdi hoan toan thu du'dc 3,36 lit CI2 d anot (+) va m gam hon hdp kim loai d catot (-) Tinh m Hoa tan hoan toan 3,01 gam A vao ntfdc thu du'dc dung dich A Them 17 gam AgNOs vao dung dich A thay tao p gam ket tua va dung dich B (nUdc loc) Co can B thu difdc q gam muoi khan Tinh p, q Cho biet A so mol cua kim loai gap doi so mol kim loai Hoi chiing la nhffng kim loai nao so cdc kim loai cho dU'oti day: Mg (24); Ca (40); Cu (60); Zn (65); Ba (137) B a i giai '^P"' ) X + C l t YCI2 '^P"' ) (1) Y + Clzt Theo DLBTKL ta c6: m = m„u,ti - mci^ = 15,05 - • ^ ^ x (2) 71 = 4,4gam 22,4 C^cphan^ng: XCI2 + 2AgN03 YCI2 + 2AgN03 > 2AgCl i + X(N03)2 > 2AgCl i + Y(N03)2 Theo cac phan tfng (1, 2, 3, 4): nAgci = B a i giai Gpi Z, N, E va Z', N', E' Ih so hat proton, ndtron va electron cua nguyen tuf A, B; ta c6 phu'dng U-inh: Z + N + E + Z' + N' + E' = 142 Hay (2Z + 2Z') + (N + N') = 142 (1) (2Z + Z ' ) - ( N + N') = 42 (2) 2Z'-2Z=12 Hay Z' - Z = (3) Lay (1) + (2), sau ket hdp vdi (3) ta c6 Z = 20 va Z' = 26 Vay cAc kim loai d6 la Ca va Fe ^ a i 4: m Tinh nhiet lufdng toa dot cMy Im^ (dktc) hon hdp gom (% the tich): 14% H2, 2%CH4, 15,5% CO, 12,5% CO2, 56% N2 Biet nhiet liTdng toa I dot chay mol H2 Ik 241,8 kJ, mol CO la 283,2 kJ va mol CH4 la I 802,4 kJ m Hoa tan 2,22 gam hon hdp Al, Fe b^ng 500 ml dung dich HNO3 0,5M thu diTdc dung dich A va 1,12 lit nhat NO (dktc) a) Tinh % khoi liTdng moi kim loai hon hdp dau, biet ring Fe bj tan I Goi CTPT cua cac muoi la XCI2 va YCI2 Cac phan iJng dien phan nong chay XCI2 0,05Mx + 0,1MY = 4,4 hay X + 2Y = 88 Nhu-vay Y < 88/2 = 44 Khi Y = thi X = 70 loai Y = 24 thi X = 40 dung, la Ca va Y la Mg i Y = 40 thi X = loai ihi 3: Cho biet tong so hat proton, ndtron, electron nguyen tijf kim loai WL A, B la 142, so' hat mang dien nhieu hdn so' hat khong mang dien la 42 hat, so hat mang dien cua B nhieu hdn so hat mang dien cua A la 12 hat Hoi A, B la kim loai gi? Cho dien tich hat nhan cua mot so kim loai sau: ZNa = 11, ZMS = 12, ZA, = 13, ZK = 19, Zca = 20, ZFC = 26, Zcu = 29 ^'^^^^'^^ x Fe(N03)3 b) Cho dung dich A tac dung vdi 210 ml dung dich NaOH IM, roi lay ket tua nung d nhi$t dp cao thi thu dUdc bao nhieu gam chat r^n? B a i giai (3) Cac phan ufng chay: (4) = 0,06mol 15,05 Vay p = 0,06 X 143,5 = 8,61 gam Theo DLBTKL q = 3,01 + 17 - 8,61 = 11,4 gam Goi X gam la =soXA,mol cua XCI2 ngiTdc lai) thi so mol cua 15,05 nci2 + ta2xc6 = c^c 0,15 phiTdng => (hoSc X = trinh: 0,05 YCI2 H2O H2 la 2x CH4 + 202 -> CO2 + 2H2O CO + -> CO2 -02 (l)i (2) (3) 291 Trxidc he't can linh so' m o l cua cac k h i cMy: r o i lay k e t tua nung d nhiet cao thi thu du'dc 0,40 gam cha't rin 1000x14 , -, 1000x2 , HH, = = 6,25mol; n r u = = 0,893mol; "2 100x22,4 ^ * 100x22,4 Hrn = 1000x15,5 T i n h k h o i lu'dng m o i k i m loai hdp k i m ban dau , = 6,92mol 100x22,4 Cho k h i B tac diang v d i 0,672 l i t clo (d dktc) r o i la'y san pham hoa tan vao 19,72 gam nuTdc, ta thu duTdc dung dich D Lay gam dung dich D cho tac N h i e t li/dng tea ra: dung v d i AgNOs diT thay tao 0,7175 gam ket tua T i n h hieu sua't nhan Q = 6,25 x 241,8 + 0,893 x 802,4 + 6,92 x 283,2 = 4187,5 kJ Xing giifa k h i B va clo B a i giai a) Cac phan tfng ho^ tan A l , Fe: A l + 4HNO3 — > Fe + H N O A1(N03)3 + N O T + H O (I) + 2H2O (2) > Fe(N03)3 + N O t '27x + 56y = 2,22 ' ' \ 12 x + y = n N o = ^ = 0,05 G i a i he phiTcfng trinh ta c6: x = 0,02mol; y = 0,03 m o l III, + NaOH > N a N + H2O (3) Fe(N03)3 + N a O H > Fe(0H)3i +3NaN03 (4) A1(N03)3 + N a O H > A l ( H ) i +3NaN03 (5) A1(0H)3 + N a O H > NaAlOa + 2H2O (6) 2Fe(OH)3 — ! ^ Fe203 + 3H20T (7) 2A1(0H)3 — ^ Al203 + H t (8) ^ MgCl2 + 2NaOH -> M g ( H ) i + N a C l (3) AICI3 + 3NaOH -> A I ( H ) ; (4) (1) 2AlCl3 + 3H2t (2)1 +3NaCl N a A l O j + 2H2O (5) - > M g O + H20t ——> 2Cu + O2 •(is- (6) z2CuO uuu (7) /) K h o i liTdng M g = 0,01 x 24 = 0,24 gam : 0,80 = 0,01mol Theo phan ufng (7): ncu = ncuo = 80 K h o i liTdng Cu = 0,01 x 64 = 0,64 gam K h o i liTdng A l = 1,42 - 0,24 - 0,64 = 0,54 gam Cac phan lirng: I H2 + CI2 = 0,21 X - 0,05 - X 0,02 - X 0,03 = 0,01 mol NhU" vay c6 0,01 m o l A1(0H)3 b i tan theo phan tfng (6) va l a i 0,02 - 0,01 = 0,01 m o l A1(0H)3 > 2HC1 Theo de b a i : n ^ ^ = 0,672 22,4 143,5 VI 0,01 0,03 — — X + - : — x l = 2,91gam 2 (8) j ).AgCli+HN03 (9) ' = 0,005moi = 0,03niol Theo cac phan ifng ( , ) : n H = n M g,+-n., - K h d i lu-dng chat r ^ n g o m Fe203 va AI2O3 b^ng: =0,01 + - x ^ = 0,04mol > nci2 n M g C b + H z t Mg(0H)2 b) Cac phan ufng: HNOsdir M g + 2HC1 A1(0H)3 + NaOH ' % A = ' x x 0 ^ , % ; %Fe = 100 - 24,32 = 75,68% 2,22 'tj -^.a^ift ^ Cac phan ufng: t G p i X , y la so m o l A l , Fe ta c6 he phufdng trinh: ; M a t khac, dot nong cha't rin C khong k h i thi thu du'dc 0,80 gam mot oxit mau den • Cach : G o i x la to'ng so' m o l H C l thu di/dc Difa theo so m o l H C l c6 gam D suy tdng so' m o l D ; v •f • •: T 293 (kho'i liTdng D = 19,27 + 36,5x) ta c6 b i ^ u thtfc: x = Q-005xa9^27 + 36,5x) B a i giai Cac phan iJng nhiet phan: G i a i ta dtfdc: x = 0,02 mo l 2CH4 > C2H2 + 3H2 V a y hieu suat phan iJng: h % = ^'^^"^^^^ = 33,33% • ^ ^ 0,06 CH4 > (1) C + 2H2 (2) Cach 1: G o i x, y la so m o l C2H2 va C tao d cac phan iJng (1) va (2), U A Cach 2: Kho'i lifdng H O gam dung dich: D = - 0,005 X 36,5 = - 0,1825 = 4,8175 g D i e u CO nghia la cxS 4,8175 g H2O c6 0,005 m o l H C l la so' m o l hon hdp A Ta c6 he phu'dng U-inh ^^.^ D ' i v d i C H : x = 0,12nA j j,/,; D o i v d i H2: 3x + 2y = 0,78 V a y tdng so' m o l H C l = ^ ' 0 ^ ^ ^ , ^ Q Q2mol ^ 4,8175 nA => y = 0,21 ^ UA Theo so'mol tang d phan iJng ( , 2) ta c6: B a i 6: Hoa tan m gam tinh the Na2C03 IOH2O vao V m l dung dich Na2C03 b% ^> U A = n„ (so m o l C H ban dau) + 2x + y = n,, + x 0,12nA + 0,21nA (kho'i li/dng r i e n g d g/ml) thu dtfdc dung dich X V, = 909,0W L a p bieu thiJc tinh nong C% cua dung dich X theo m , b, V, d RiJt ra: n ^ =•0,55 Cho m = 28,6 gam, b = 5,3%, V = 500ml, d = 1,2 g/ml Cach 2: DiTa theo phan hon hdp A suy lu'dng C H ban dau, theo T i n h gia t r i cu the cua C% nMa2C03.1()H20 -"Na2C03 m -mol 286 106 + 10x18 Kho'i liTcfng dung dich N a C = V x d, do so' m o l N a C c6 dung dich 100x106 mol (n + n ' ) x 106x100 m _ V 286 Vxdxb - + +Vxd 10600 m x 106x100 28,6^500x1,2x5,3 286 10600 -6,78% — ^ C H + 3H2 (1) — C + 2H2 ' C2H2 = x 0 = 43,63% 55 %CH4 > C = 21x100 (2) H o n hdp k h i thu diTdc sau phan iJng (hon hdp A ) chiJa 12% C2H2, 10% CH4 va % H2 (ve the tich) T i n h the tich ciJa hon hdp A (dktc) 55 Cac phan liTng dieu che nCH2 = C H • Cl = 38,18% \, PVC: CH = CH + HCl ,0 CH4 ' ' Difa theo each ta de dang tinh % C H b i phan huy: X 106x100 B a i 7: Cho 5(X)m^ metan (dktc) di qua ho quang Gia su" luc chi xay phan uCng: 2CH4 Dod6VA=^5^=909,09m^ %CH4 28,6 + 500x1,2 • 55 +Vxd Thay cac gia t r i vao: C%adx = ' Ngoai c6 10 m o l C H khong bi phan huy m o l C H = 100 m o l hon hdp A m D o C%ddx = Theo phan uTng (2) ci? 21 mol C H tao 21 m o l C va 42 m o l H2 Nhir vay cur 24 + 21 + 10 = 55 m o l C H tao 78 m o l H2 + 12 m o l C2H2 + 10 Vxdxb n'Na2C03 0,55 Suy n ^ j cl phan tfng (2) bang 78 - 36 = 42 m o l m TaCO: 0,55 phan iJng (1) ciJ 24 m o l C H tao 12 m o l C2H2 + 36 m o l H2 B a i giai " - 1^' (xt) -> CH2 = C H - C l ^^,^>(-CH2-CH-)„ i '•• (3) (4) Cl Theo cac phan tfng (3, 4) so' m o l m^t xich PVC = so m o l C2H2 T i n h % C H b i chuyen hoa C2H2 va cacbon N e u lay tat ca C2H2 c6 hon hdp A de dieu che PVC t h i thu dtfdc bao nhieu k g PVC, bid't hieu suat dieu che la % 294 V a y kho'i Itfdng PVC bkng: mpvc = ^^^1 ^ ^ x 62,5 x 100 100x22,4 =213,06kg 295 Bai 8: Hon hdp B chiJa metan va axetilen Cho biet 44,8 lit hon hdp B nang 47 gam Tinh % the tich m6i B Dot chay hoan toan 8,96 lit hon hdp B va cho tat ca san pham chay hap thy vao 200 ml dung dich NaOH 20% (d = 1,2 g/ml) Tinh nong % cua m6i chat tan dung dich N a O H sau hap thu san pham chay D nang 271 gam; tron V lit hon hdp B vdi V lit hidrocacbon X ta thu diTdc hon hdp E nang 206 gam Biet V - V = 44,8 lit Hay xac dinh CTPT cua hidrocacbon X C a c the tich deu dktc >- Goi n la so mol C2H2 mol hon hdp B ta c6 phiTdng trinh ve khoi lifdng (1 - n) = — = 23,5 , trinh (b) va thay V - V = 44,8 ta c6: 2M - 47 = 65 => M = 56 • C a c phan u'ng: cua X la C4HX 9: Dinh nghla phan uTng hoa hdp, phan u'ng phan tich (phan huy), cho cac thi du Hay lay thi du vc sir oxi hoa, si/ khuT, chat oxi hoa, cha't khijf Dinh nghla phan u'ng oxi hoa - khuf • — ^ C + H20 (1) CH4 + 2O2 — (2) ^ CO2 + 2H2O Dinh nghla phan u'ng chay Cho mot thi du ve phan ijTng dot chdy hdp chat v6 cd va mot thi du ve phan iJng dot chay hdp cha't hdu cd + CO2 > Theo cac phan u'ng (1,2): phan tach nhieu chat mdi: A E"co2 = 0,3X + 0,1 X = 0,7mol; ^,^1^2^ = 0,3x + 0,1 x = 0,5mol Thi du: CO2 + N a O H > Na2C03 + H2O > NaHCOj (3) (4) Gpi a, b la so mol Na2C03 va NaHCOj ta c6 he phiTdng trinh: nco2 = a + b = 0,7; nNaOH = 2a + b = 1,2 Rut a = 0,5 mol Na2C03 va b = 0,2 mol NaHCOj Khoi li/dng dung djch NaOH sau hap thu CO2 va HjO bang: 200 X 1,2 + 0,7 X 44 + 0,5 x 18 = 279,8 gam =1^^^^ 279,8 = 18.94%; %NaHCO, x , x 0 ^ ^ 279,8 Ta CO cac phiTdng trinh ve khoi lu-dng hon hdp D va E : V 22,4 x23,5 ' + 22,4 x M = 271 (a) ' '* >X ' CaCOj Thi dii: CaO CO2 + NaOH * Phan uTng hoa hdp la phan u'ng hoa hoc c6 mot chat mdi dtfdc tao Tinh: U g = T T — = 0,4 mol c6 0,3 mol C2H2 va 0,1 mol CH4 NT ^ r r 200x1,2x20 , ^ , So mol NaOH = • = l,2mol 100x40 x m - ^ ^ i , ^ nco2 < " N a O H < X n^Oj do tao th^nh hon hdp muoi ' Bai giai tijrhaihaynhieucha'tbandau: A + B + 96 V ' Sy oxi hoa cham la gi? Cho thi du minh hoa C2H2 + , Vay: %Na2C0, > The nao la phan ufng toa nhiet, phan tfng thu nhiet Giai C O n = 0,75 tuTc axetilen chiem 75% va metan chiem 25% 296 Trong M la KLPT cua hidrocacbon X Lay phiTdng trinh (a) trijr phiTdng minh hoa Bai giai X (b) Goi cong thuTc cua X la C^Hy ta c6: 12x + y = 56 va de dang tim cong thi?c Tron V lit hon hdp B vdi V lit hidrocacbon X (chat khi) ta thu diTdc h6n hdp mol B = 26n + 16 23,5+ - ^ x M = 206 22,4 22,4 Phan u'ng phan tich (phan huy) la phan uTng hoa hoc tilf mot cha't bi 2Fe(OH)3 — ^ > X + Y + Fe203 + 3H2O 4- Sir oxi hoa mot chat la siT tac dung cua chat vdi oxi i Thi du: qua trinh oxi hoa s^t sit (III) oxit la sir oxi hoa sat: 2Fe+ - O — ^ FcjOs ' Sir khur la sir tach oxi khoi hdp chat va chat la'y oxi goi la chat khuf Thi du: khuT oxit dong dong kim loai bang hidro: CuO + H2 — ^ Cu + H2O Phan u'ng oxi hoa - khuf la phan ufng hoa hoc xay dong thcfi sir oxi hoa va sir khijf: Trong thi du trcn thi qua trinh: CuO H2 > Cu la sir khi3f > H201a sir oxi hoa Chat oxi hoa la chat nhudng oxi cho chat khac, chat ' khvt ' * ' la chat lay oxi cua chat khac Trong hai thi du tren thi: Fe, H j la chat khuf; O2, CuO la chat oxi hoa 297 aumt 2H2O H2 + CI2 > 2HCI C2H2 + , > 2CO2 + H2O Sy oxi hoa cham la su" oxi hoa c6 toa nhiet nhtfng khong phat sang Thi du: Nhffng vat bang sat de troflg khong bi hoen ri dan Su" oxi hoa cham cac cha't huU cd cd the (Thi du: Glucozd, mot phan aminoaxit, v.v ) tao nang lifdng (nhiet) can cho sy song Bai 10: Hdp chat hi?u cd X chiJa cac nguyen to C, H, O va c6 cong thiJc phan tu" trung vdi CTDGN Cho 2,85 gam X thuy phan hoan toan (c6 mat H2SO4 xuc tac) thu du'dc chat hffu cd Y, Z Dot chay hoan toan Y thu diTdc 3,96 gam CO2 va 1,62 gam nufdc Dot chay het Z thu du'dc 1,32 gam CO2 va 0,81 gam H2O Tong khoi lu'dng oxi can cho phan iJug dot chay Y va Z dung bang Itfdng oxi thu diTdc nhiet phan hoan toan 42,66 gam KMn04 Tim cong thufc phan tu" cua X Biet khoi lu'dng phan tu* cua Y bang 90 dvC va X tac dung du'dc vdi Na giai phong H2, hay tim cong thufc cau tao cua Y, Z, X Bai giai Co the tom tat sd phan iJng nhiT sau: X + H2O > Y + Z (1) ' - Va * Y + O2 > C + H20 (2) Z + O2 > C O + H2O (3) 2KMn04 — ^ K2Mn04 + MnOz + O21 (4) Tac6theo(4) no2 = ^ n K M n = ^ x ^ ^ = 0,135mol Do khoi lu'dng O2 dung dot chay Y, Z bang 0,135 x 32 = 4,32g Theo DLBTKL, theo phan iJng (2, 3) ta c6: Khoi liTdng Y + Z = khoi lu'dng CO2 + H2O trir kho'i lu'dng O2 = 0,12 X 44 + 0,135 X 18 - 4,32 = 3,39 gam 298 Ap dung DLBTKL cho phan tfng (1) ta c6: 54 3,39 - 2,85 = 0,54 gam hay -j^ = 0,03mol ^ Trong 2,85 gam X c6: mc = m c CO2 = 12 X 0,12 = 1,44 gam mH = mH phan iJng chay tao trir mn H2O thuy phan = X 0,135 - 0,03 X = 0,21 gam mo = m z - m c - mH = 2,85 - 1,44 - 0,21 = 1,2 gam Goi cong thuTc cua X la CxHyO, ta c6: x : y : z = 1,44 0,21 1,2 = 8:14:5 12 16 Vay CTPT cua X la C^U.^O, Vi Y chay tao ncoj ^ nH20 nen CTPT cua Y Ik C^HzxC, ta CO MY = 14x + 16z = 90, chi c6 x = 3, z = thoa man Vay cong thtfc cua Y la C3H6O3 mHjo = C3H6O3 + 3O2 > 3CO2 + 3H2O So mol Y = 1/3 so mol C O = 0,09 = 0,03mol Nhtf vay phan iJng thuy phan X (phan tjfng 1) c6 ti le so mol la 2,85 0,54 , - nx : n u - n '• H20 = : : 0,03 =1:2:2 Y do phan iJng (1) c6 the viet lai CsHuOs + H O > 2C3H6O3 + Z Suy CTPT cua Z la CjHfiO CTCTciaa chatla: Z: CH3 - CH2OH Y: CH3 - CH - COOH hoSc (^Hz - CH2 - COOH O OH OH p X: C H - CH - C ^ O - CH2 - CH2 - c OH \ - c H HoSc // O X: C H OH CH2 - O- CH2 - CH2 - C P ^0-C2H5 O X: C H j - C H - C OH O \ O - CH - C \ CH, Hoac O - C2H5 X: C H I OH o CH2-C ^- ^ O - C H - C , I \ O - C2H5 CH3 B a i 11: H o n hdp k h i A (dktc) g o m nhffng the tich k h i bang cua metan v a hidrocacbon X c6 kho'i liTcfng rieng bkng 1,34 g/1 X a c djnh cong thtfc phan tijT cua X D o t chay V lit hon hdp A va cho ta't ca san pham hap thu vao binh diTng dung dich B a ( H ) diT thi thu di/dc 15,76 gam ket tua a) T i n h t h e tich V (1) b) H o i k h o i liTdng b m h dung djch B a ( H ) tang hay g i a m bao nhieu gam." (2) B a i giai Theo cac phan iJng ( , 2) de dot chay m o l xang can: K h o i l u - d n g m o l p h a n t i y c i j a h o n h d p A: = 1,34 x 22,4 = g / m o l 0,6 V i the tich bang n e n m o l A c6 0,5 m o l C H va 0,5 m o l X , do ta C O phiTdng trinh: M X = 30 = 0,5 X 16 + 0,5 X Mx X + 0,4 G o i cong thiJc phan tuf cua X la C,Hy ta c6 12x + y = 44 9,5 = 8,6 m o l O2 V a y the tich khong k h i can de do't chay gam xang bSng: 8,6x5x22,4 11,6 M x = 44 X • = 12,411it N h i e t l i f d n g t o a r a k h i do't c h a y m o l x a n g b ^ n g : Xet: 0,6 X 3534 + 0,4 X 4196 = 3798,8 k J X y 32 20 Kctluan loai loai C3HK V a y n h i e t l i f d n g t o a r a k h i d o t c h a y 100 g a m x a n g b ^ n g : H9M!ll»»=4895.4kJ 77.6 VayCTPTcuaXlaCjHs -> C O + H O C3HS+502 3CO2 + 4H2O C + Ba(OH)2 Theo phan iJng (3): > BaCOjl +H2O n c o j ^nBacoj = ^ ^ ^ = 0,08mol ^ ^ Tijf 100 k g g a o chtfa % t i n h b o t c6 t h e d i e u c h e diTdc b a o n h i e u l i t riTdu n g u y e n c h a t ( d = 0,80 g/ml) b i e t h i e u suat d i e u c h e l a % (1) TO ri/du n g u y e n c h a t d o d i e u c h e diTdc b a o n h i e u l i t ru-du 46"? (2) L a y 10 m l ruTdu " c h o t a c d u n g h e t v d i Na T i n h t h e t i c h H b a y ( d k t c ) % ^ (3) n + 3n = 0,08 => n = 0,02 m o l Vay: V = x 0,02 x 22,4 = 0,896 l i t B a i giai Cac p h a n iJng d i e u c h e riTdu e t y l i c : (CfiHuAOn + n H G o i n la so m o l C H , C H X ; theo cac phan iJng ( , 2) ta c6: 300 ^ B a i 13: L c n m e n t i n h b o t d e d i e u c h e ' r i T d u e t y l i c J- Cac phan ufng: CH4 + 2O2 _ CftHuOfi > nC6H,206 2C2H5OH + 2CO2 ' , \h • (1) (2) T h e o c a c p h a n liTng ( , ) , so m o l ruTdu b a n g Ian so" m ^ t x i c h C H O ; 301 , , w , 1o 100x10^x81 75 , vay tong so mol riTdu etylic: =2x x = 750mol ^ • ' 100x162 100 The tich rirdu nguyen chat: Z^2iii^ = 43125ml = 43,125 lit ^ ^ 0,8 CiJ 46 lit rxidu nguyen chaft dieu che diTdc 100 lit rifdu 46", do the tich n.du46"b^.ng^^:i^^.93.751it 46 Cac phan uTng vdi Na: H20 + Na >NaOH+^H2t C H , H + Na > CjHsONa + Can tinh: nf^^u^nH = 10x46x0,8 C2H5OH , 10x54x1 , = 0,08mol; nu^o = = 0,3mol 100x46 Theo phan u-ng (3, 4): Vay t 100x18 =i(nc2H50H+nH2o) = ^(0.08 + 0,3) = 0,19mol =0,19x22,4 = 4,2561it Bai 14: Phan iJng tong hdp glucozcf cay xanh can du'dc cung cap 2816 kJ nang lUcmg mat trdi de tao mol glucozd: 6CO2 + 6H2O ^^^^ ) CfiHuOfi + 6O2 Hay tinh nhiet lUdng can thiet de cay xanh tao 900 gam glucozd Neu cay xanh c6 tong dien tich la 100 dm^ thi can thcJi gian bao lau de san sinh du'dc 900 gam glucozd, bie't r^ng mol cm^ la cur phut nhan dtfdc 20J nang li/dng mat trcJi nhifng chi c6 10% nang li/dng difdc su" dung cho phan iJng long hdp glucozd, Tinh the tich C O (dktc) cay da hap thu va khoi Itfdng O2 difdc giai phong B^lgiai ' Ta CO phan uTng quang hdp: C O + H O So' mol glucozd du'dc tao bkng ^".^"1^^ 5mol Nang liTdng can cung cap = x 2816 = 14.080kJ E = 10.000 X = 20.000 J = 20 kJ ^ 7040 phut ttfc 117,3 gicJ The tich C O can = x x 22,4 = 672 lit Kho'i liTdng O2 diTdc giai phong = x x 32 = 960 gam 302 (1) 900 Tinh nang lu'dng mol phut cay xanh hap thu difdc: Thdi gian can thiet bang: ) CfiHijOs + 6O2 6% [...]... H 3 - C H 2 O H + O2 CH3COOH +CH3-CH2OH > C H 3 C O O H + H2O HgS04c6c t" >CH3COO-CH2-CH3+ H2O Boi dwdng hoc sink gidi Hod hoc 8, 9 - Dao HOU Vinh m H 2 0 = 1 8 0,1 = 1,8 g => V H 2 0 = 1,8 II Bdi tSp tif g i i l 1 V i e t phiTdng trinh phan tfng (ghi ro dieu k i e n neu c6) di thifc h i e n scf d6 chuyen hoa sau : - ^ C z H s O H — ^ CH2 = CH2 CH3COOH—^ (CHjCOOzZn e5r.du= • I) CaCOa +H2O >CaO 2000"c... d§ tan trong n\Xdc CHjCOONa CH3COOC2H5 C6H,206 Axit gluconic + Ag CI2H220II >• Khong cho phan tfng trang giTcJng +H2O, t", H * Glucozd + Fructozd , , Cty TNHH MTV DWH Khang Viet Boi dudng hoc sink gidi Hod hoc 8, 9 - DAo HOu Vinh CizHzzOn axit.to +H2O GlucozcJ Fructozd VI Tinh bOt va xenlulozcf (-CgHioOs -)„ * Tinh bot la chat r^n mau tr^ng, khong tan trong nifdc lanh, tan trong nirg n6ng tao thanh... saccarozd, tinh bot, xenlulozd HuTdng d§n giai a) Thuoc nhom nhien lieu (cha't dot) 6 Vie't cac phifdng trinh phan iJng thtfc hien cac bie'n hoa sau: b) Thuoc nhom gluxit Etilen — r i T d u etylic 3 Vie't cac phiTdng trinh phan uTng thtfc hien cac bien hoa hoa hoc sau : ^'^ > Glucozd — R i T d u etylic — A x i t Etyl a x e t a t — R i T d u etylic — E t i l e n (-C6H,„05-) + nH20 ) CfiH.zOfi Mennf^»u... phong hod xay ra hoan toan, ta thu dffdc dung dich X De trung hoa NaOH con dff trong 1/10 dung dich X can 20 ml dung dich HCl 0,2M a) Hoi khi xa phong hoa 1 kg chat b6o A can bao nhieu gam NaOH vk thu dffdc bao nhieu gam glixerol b) Lap cong thffc phan tff cua axit b6o tao th^nh chat beo A va viet cong thffc cS'u tao A 4 J Hifdng d§n giai * Tff phffdng trinh phan ffng v^ Iffdng HCl trung hoa ta tim dffdc... V a y thanh phan % H bien thien trong khoang: f* |v K h i qua niTdc brom chi c6 C2H4 phan uTng: 1 +1 K h i n v6 ciing Idn thi = 19,64gam ^ Ket luan: H o n hdp k h i A nSng hdn hon hdp k h i B 2 Thanh phan % cua hidro trong ankan CnH2n+2 diTdc tinh theo bieu thiJc: (2n + 2 ) x l 0 0 22,4 ml dung dich B r j 0,04M H o i % the tich cua CH4 bien doi trong khoang nao Hoac ( % 0 = 100 - 32 - 18,67 - 6,67 =... trung hoa ta tim dffdc mwaOH dilng cho x^ phong hoa 1 6 , 1 2 g A, suy ra mNaOH can cho xa phong hoa 1 kg hay Hirdng dSn giai lOOOg A, va suy ra mgnxcroi tdi afi/jdq ix Sf imM Tff UNaOH suy ra U A va mA suy ra M A tff day tim dffdc cong thffc cua A va suy A la protein, vi chi c6 protein mdi chffa nguyen to N ra cong thffc cua axit tao nen A j^^, , 3 D6 trung hoa 15 ml dung dich axit hffu ccf c6 dang C^Hy-COOH... |De xa phong hoa 16,12 gam A can : 0,06 4 0 = 2,4 gam NaOH ' CxHy-COOH + NaOH > C^Hy-COONa + H 2 O De xa phong hoa 1 kg hay lOOOg A can: ^'"^'^^^^ = 148,89 g NaOH 0,03 mol 0,03 mol 16,12 Trong 15ml dung djch c6 0,03 mol axit Khoi Iffdng glixerol thu dffdc khi xa phong hod 16,12 g A la: 0,02 92 g Vay trong 250 ml dung djch c6 "'"^^^^^ = 0,5 mol Kho'i Iffdng glixerol thu dffdc khi \k phong hoa 1 kg hay... Dun n6ng 4,03 kg este tten vdi li/dng AM dung dich NaOH Tinh khoi liTdng glixerol sinh ra va khoi liTcfng xa ph6ng thu diTcJc, biet vlng xa phong c6 chiJa 72% muoi sinh ra tiif phan tfng noi tren C BAI TAP LUYfN THI 1 Viet c6ng thtfc cau tao cua cac riTcJu c6 cong thtfc phan tur CjWfP, C^n^O 2 Viet phan tfng dot chay hoan toan rifcJu C 2 H 5 O H va C„H2„ + , 0 H - PE m ( - C H 2 - C H 2 - ) „ - PVC... h bay phu'dng phap hoa hoc de phan biet cha't beo long (dau lac, dau Bai giai CH3 r>-te ;^ OH > Cu(CH3COO)2 + H2O + vCfng) v d i dau nhdn (la hon hdp nhieu hidrocacbon) > u^^n^u CH3 - C = CH2 H2SO4 (lac CH3 Bai CUCO3 + 2CH3 - C O O H I CH3 CH3 CH3 > Cu(CH3COO)2 + H2O + H2O - H' > •n.^-'vG£,, p i phan biet dau an (dau lac, dau vuTng ) v d i dau nhdn, ta dUa tren phan ijng xa phong hoa dau an tao thanh... di/dc m u o i an tinh khie't can ke't tinh l a i : hoa tan m u o i v^o niTdc soi t d i bao hoa, sau dd lam lanh ta diTdc m u o i an tr^ng tach ra) His ^' Hai Ip dau nhdn (ban chat la hidrocacbon) va dau an (thi du dau lac, dau vtfng (C|7H33COO)3C3H5) nhin be ngoai rat giong nhau (trong suot, mau vang nhat) Hay phan biet hai Ip dau dd bang phufdng phap hoa hpc ^' Co 7 g d i bpt tr^ng giong nhau: v o i