Nối tiếp nội dung phần 1 tài liệu Cẩm nang ôn luyện thi đại học 18 chuyên đề Hóa học, phần 2 giới thiệu các nội dung: Ăn mòn kim loại - Điều chế kim loại, kim loại kiềm - Kim loại kiềm thổ - Nhôm, các lý thuyết cơ bản của hóa học,... Mời các bạn cùng tham khảo nội dung chi tiết.
Ca'm nang 6n luyCn thi dai hpc 18 chuy6n dS H6a hpc - Nguygn Van HSi Cty TNHH MTV DWH C h u y e n de + AN M(m V i d v , de bao v? v tau bien bang thep, n g u d i ta gan vao v6 tau (phan c h i m ^-^^ g r m nsffq oyt qsii 6v lorn S0;0 = t i h • ' Phuang phdp dien hoa: No'i k i m loai can bao ve v o i tarn k i m loai khac c6 t i n h k h u m a n h h o n K I M LOAI - IHltU €Hlt K I M LOAI A N M O N K I M L O A I a K h a i n i ? m -' Khang Vijt t r o n g n u o c bien) n h i i n g tarn kem K h i tau hoat dong, tarn k e m b i an m o n dan, v tau d u p e bao v^ Sau m p t t h o i gian n g u a i ta thay cac tarn k e m khac Su pha h u y k i m loai hoac h o p k i m tac d u n g hoa hoc cua m o i truorig x u n g q u a n h goi la s u an m o n k i m loai b Phan loai + VI D U ^-''("OVi'v-r^ -.vrHf rtsrfr: MAU V i dvi 1: K h i d i e u che h i d r o t u Z n va d u n g d i c h H2SO4 loang, neu t h e m vao An m o n k i m loai d u g c chia t h a n h loai chinh: A n m o n hoa hoc va an m6n m p t v a i gipt d u n g d i c h CuS04 t h i tha'y k h i H2 thoat m g n h h o n Ban chat d i ^ n hoa cua h i ^ n t u p n g t r e n la •„-•!•> • ^ i s-: • - f/.'^ - >5 ^ ff'^v*^ 4'^'^ An mon hoa hoc: A A n m o n dien hoa A n m o n hoa hpc la sy pha h u y k i m loai k i m loai p h a n u n g hoa hoc v6i C A n m o n k i m loai ^^"t"' ^' • ' ' ' ^' B A n m o n hoa hpc 'i'i - - D A n m o n hpp k i m axit hoac cac chat k h i (hoi) nhiet dp cao Dac d i e m ciia an m o n hoa hoc la qua t r i n h oxi hoa-khu, electron dupe chuyen Vidu\ , ,.s;Vt«nf Fe + H2S04(io3ng) 3Fe + + 2O2 ^'-S'iffW • +€^q - 'v ,gA^ ^ " ^ ^ '" > Fe304 H2 £0.0 • FeS04 + H2 —> T i n h o x i hoa Cu^* > H * Cu nen k h i t h e m CuS04 se xay qua t r i n h : Zn ' ' " + CuS04 > * ^ x i ii^ ZnS04 + C u - nkqkQ ^ C u thoat b a m tren hat Z n nen thoa m a n d i e u kien ve qua t r i n h an m o n An mon dim hod: d i f n hoa: Co h a i k i m loai khac n h a u ( Z n - Cu), tie'p xuc true tie'p v o i A n m o n d i ^ n hoa la s u pha h u y k i m loai k i m loai tie'p xiic v o i d u n g dich (Cu b a m vao Z n ) va c u n g n h u n g t r o n g d u n g dich chat dien l i (H2SO4 loang) chat d i e n l i tao nen d o n g electron chuyen d a i t u cue a m sang cue d u o n g va D o xay qua t r i n h an m o n dien hoa —> D a p an A phat sinh d o n g d i f n Co che an m o n d i e n hoa: Vi •.^_.,f^y.y :M _y.^;_,^^ v \ Zn^* D V tau bien bSng thep n g a m t r o n g nuoc b i e n ^ Tai cue d u o n g (eatot), i o n H"^ chuyen den be m a t C u de n h a n electron t u cue a m chuyen t o i va b i o x i hoa tao k h i H2: H * + 2e > H2 Ba d i e u k i ^ n de xay an m o n d i f n hoa: 8""*^ ' ^mi>nkdaoBoo.r Laigidi: Loai A, C: T h e p de t r o n g k h o n g k h i kho n o n g hoac dot chay s^t t r o n g khong k h i - » (1) Co hai k i m loai c6 ban chat ^- ' , Tai cue a m (anot), Z n ( k i m loai m a n h hon) b i o x i hoa: , "'^"^ A Thep de t r o n g k h o n g k h i kho, nong V i d u CO tarn k i m loai Z n va C u dup-c n h u n g vao d u n g d i c h H2SO4 Z n - 2e , > ——— Nhqn xet: T r o n g day d i e n hoa, cap t r y c tie'p t u chat oxi hoa deh chat k h u len k h o n g phat sinh d o n g di§n D e u bj an m o n hoa hpc SOB^E L o a i B: K e m de t r o n g axit H2SO4 loang EJJ9>1£ K e m bj an m o n hoa hpc • • i khac n h a u , (2) tie'p xiic t r y c tie'p v a i n h a u (hoac gian tie'p qua day dan) va (3) —> D a p an D dupe n h i i n g vao c i m g m p t d u n g dich chua chat d i ^ n l i Luu y: V tau bang thep la h p p k i m Fe - C (chua d i | n cue c6 ban chat khac c C a c h chong an mon k i m loai: va tie'p xiic t r y e tie'p v o i nhau) va ciing tie'p xiic v a i d u n g d j c h d i ^ n l i + (nuae bien) nen xay an m o n dien hoa Phuomg phdp bdo ve be mat: D i m g n h u n g chat ben v o i m o i t r u o n g p h u len be m a t k i m loai D o la: :,; V i d v 3: Cho cac h p p k i m sau: Cu-Fe (I); Zn-Fe (II); Fe-C (III); Sn-Fe (IV) K h i D u n g son cho'ng gi, vecni, dau m o , trang men, p h u h p p chat poiime tie'p xiic v o i d u n g djeh chat dien l i t h i cac h p p k i m m a t r o n g d o Fe b i an M a m p t so' k i m loai ben n h u c r o m , niken, dong, kem, thie'c len be m a t kif^ m o n t r u o c la: A I , I I va I V loai can bao ve B.I, II vain C.I,IIIvaIV D I I , I I I va IV 1Q1 Cty TNHH MTV DVVH Khang Vi$t C^m nang 6n luy^n thi dgi hpc 18 chuySn 66 Hoa hpc - Nguygn Van HSi Lai gidi: ^ Nhan xet: Trong cac qua trinh an mon dien hoa, k i m loai m ^ h han se bi an mon —> Fe bi an mon truoc no la chat k h u manh hon (dong vai tro la anot - cue am) Trong cac cap (I), (III), (IV) thi Fe deu manh han, rieng cap (II) thi Zn manh hon •^DapanC , , • , :>;;.f:,,,, , ,,• V i di^ 4: Gang bi an mon dien hoa khong am Qua trinh kliir xay tren ãn- be mat aia gang la: ,fe^,^|,.,B,M>i.#ôi: A O2 + 2H2O + 4e C Fe > 40H- ' -)1»ft»l^s*&i^.»v:t Fe2* +2e D Fe > O2 + H - > Fe^* +3e Lai gidi: ' Thep la hop k i m Fe-C (vdi mot so nguyen to'khac) K h i tie'p xiic voi khong am (c6 chiia chat di^n li) se xuat hi^n qua trinh an mon di?n hoa Fe la cue am (bi an mon), C la cue duong se xay su k h u O2 + 2H2O + 4e > 40H- ^nfi^YGx6ôtm-ã > Dap an A Luu y: Cac em de chon riham B v i bo qua vai tro cua chat oxi hoa la oxi khong V i du 5: Co dung dich: a) H C l , b) CuCh, c) FeCb, d) H C l c6 Ian CuCh Nhiing vao moi dung dich mpt Fe nguyen chat So truong hop xuat hien an mon dien hoa la: A.O , B C.2 D.3 yu, Fe203 -> D a p an D _ A C u , FeO, A I O , M g O Mol: + O x i t sat la Fe203 t: t h a i gian d i # n p h a n (giay) F: H a n g so Faraday (F = 96.500) ivi i V L ^ v v i i ach 1: Giai theo p h u o n g t r i n h hoa hpc nP A : Kho'i Itfong m o l n g u y e n t u (mol.phan t u ) ciia chat t h u dug'c a d i ^ n cue I : C u o n g d p d o n g d i e n (ampe) MViiii , , -s^ ,,, B Fe304 v a 4,48 - m »• EOMS D Fe203 va 3,36 ; c u o n g d p d o n g di?n k h o n g d o i 1,34A (hi^u sua't qua t r i n h d i ^ n p h a n la 100%), t h u d u p e chat ran X, d u n g d i c h Y v a k h i Z Cho m o t t h a n h Fe vao Y, sau k h i cac p h a n u n g ke't thiic thay k h o i l u p n g t h a n h Fe tang t h e m 2,36 gam va c k h i N O (san p h a m k h u d u y nhat ciia N"^"^) bay Gia t r i eiia t la A 0,8 B 1,2 .- C.1,0 D.0,3 " 195 dm nang On luy^n thi dgi hpc 18 chuy6n di H6a hpc - NguyJn Van HSi Cty TNHH MTV DWH Khang Vigt Lot gidi: Phan ling dien phan dung dich AgNOs: ' " + H2O —^Edd_) 2Ag + ^ 2AgN03 | + 2HN03 Mol: 2a "'^ ' ' ' ' 2a ' 2a ' '^^' Nhqn xet: Vi khoi lug-ng Fe tang them nen Fe dupe A g bam vao-^ y chiia A g N - > AgNOs chua bi di^n phan het* l«l^/;v.' Cac phan ung cho Fe vao Y: Fe Mol: + 4HN03 0,5a > Fe(N03)3 + N O + H , o y a n i ^ M ' 2a -> 0,5a f' Can luu y Fe tac dung voi Fe(N03)3: Fe + 2Fe(N03)3 Mol: 0,25a •(- 0,5a Fe + 2AgN03 Mol: b D e u sinh cap di|n c\fc Fe - C u , tiep xiic tryctiep vai va ciing B 1,3 gam C 1,95 gam D 2,6 gam nhung mgt dung dich di|n li pai 15: Di^n phan 200ml dung dich gom AgNOs 0,5M va Cu(N03)2 I M bang Lieu y: thi nghi?m v a 3, Fe b} an mon hoa hgc theo cac phan ling: li dong dif n mgt chieu vol cuong dg dong dif n 2,68A thai gian gia |,, Kho'i lugng kim loai thoat catot la: A 20,4 gam B 23,6 gam C 10,2 gam D 11,8 gam/' Xuat hi|n an mon di|n hoa -> D a p an B -fy?^ vt;;/''^ Fe + 2FeCla > 3FeCl2 C u + 2FeCla > C u C k + 2FeCh f + Xo^»^ • pai 16: Di#n phan dung dich hon hgp chua 0,15 mol CuS04 va 0,1 mol H C l thai gian gio v a i dong difn c6 cuong dg la 1,34A Biet h i f u suat ciia J?/ = 0,5 mol; n^^^o, = ^ nAl= ^ = ^'^ "H2 ~ = 0'^ qua trinh di|n phan la 100% Khoi lugng kim loai thoat tai catot la: A 6,4 gam B 3,2 gam C 12,8 gam 2A1 D 9,6 gam Bai 17: D i ? n phan dung dich X gom 0,2 mol N a C l va 0,2 mol CuS04 (di^n afc Mol: 2a + 2NaA102 + 2H2O _^ Mian xe^: n H , = as ^ - (a5-2a) = 0,3-> a = 0,15 • Bai6: H2 ' ^ ' ' : ' - ; '^^^ •••1 /Till j^nrv^' H = ^ ^ 0 % = 75% ^ D a p a n C , 0,2 ^ * m = 5,4 ^ nA,= ^ » = X a2 mol ^ " ^ ° > AI3^, pe^^ H M ^ T ? > "CUSO4 -> C u S p h a n u n g h e t fr^ ; , r Nhan thdy: t u tr^ng thai dau tien den trang thai cuoi ciing thi: So'oxi hoa ciia A p dyng bao toan electron: ng=3nAi =3nNO CuS04 ^ < ,, ,^ C u C k het i > , > MSO4 + C u 0,02 1^.1 0,02 , , M = 65(Zn) ^ D a p an D ' A , ; Nhan xet: Y gom kim l o ^ —> Y chiia Mg hoac Fe , Fe khong doi (luon la +3), so oxi hoa ciia A l tang tit len +3 + , n; i" - > MMSOA = - ^ = 161 ^ 0,02 Bai 10: , ^ M Mol: 0,02 „ :h-: = (3,2 - 64x) + 108.(0,04 + 2x) = 12,08 " ^'^ ^'^^ " ^'-^ - > a = 0,50 m o l -)• D a p an D - > 0,02 + x = 0,la mv - > Fe = 56 da p h a n u n g m p t p h a n de tao t h a n h C u = 64 —> Z gom ZnS04 v a FeS04 ^ - _ G o i so Z r ban d a u = a m o l ; nFe(p.) = b N h a n tha'y: 'tidrt > a m o l C u -> K h o i l u g n g g i a m (65-44)a = a b m o l Fe > h mol Cu -> gnv m- r^yi^^:^^^ + a mol Zn Catot: C u - + 2e Mol: 0,05 — ^ j^.,^^, 0,05 r.- ^ Bai 17: , Phan u n g , ^ t' tang (64-56)b = 8b ~^^Zn'^ ^'^Fe(pu) = a + b ^ a = 0,02; b = 0,02 J- 0,1 0,2 T-/-.-,-r « , j , CuCh + -> 0,1 r>i.,tjr K2SO4 mzn = 0,02.65 = 1,3 gam g " ""^ V - > D a p an B Bails: • • Nhan xet: D a y la bai toan d i | n phan d u n g djch chua n h i e u chat, ca.c e m nen ap d u n g cong thuc t i n h so m o l electron trao d o i : m It 2,68.4.3600 ^ , , n„=—.n=—= = 0,4 m o l ^ A F 96500 Ta c6: So'mol electron A g * nh^n vao: 0,1.1 = 0,1 m o l - > So'mol electron Cu^* nh|n vao = 0,3 m o l „ „ 4- " 0,1 " '' T h u t u d i e n p h a n n h u sau: Mol: CuCl2 Cu 0,1 0,1 + CI2 "-"•••"•^ ^-^^^^"'^ 0,1 mgiam = 0,1.(64+ 71)= 13,5 g a m -> K h o i l u g n g d u n g d i c h g i a m tie'p 17,5 - , = g a m cuso4 + Mol: a H20 - i p ^ Cu + a io2 0,5a + H2SO4 , * ''•TDfeMS ^ cO:dr 64a + 32.0,5a = - > a = 0,05 m o l ,.^14 r;(HO)!/ Vc,2 + V02 = (ai + 0,025).22,4 = 2,8 l i t L a o c J i o ioinv -» Dap anD.wi , , q u y d o i gia d m h : CuS04 + 2KC1 Mol: gj./j = , - , =0,14 g a m b - a = 0,14 Bao toan electron: n ^ 2+ -> a04 - > X g o m : C u C h = 0,1 m o l ; CuS04 = 0,1 m o l va K2SO4 = 0,1 m o l Nhan xet: T i n h k h u Z n > Fe ->• Z n p h a n u n g t r u o c r o i m a i d e n Fe D o mx < mt5ng MLM ^, anlAnyH 1,34.2.3600 , ^^^^^ =0,1 m o l • - » D a p an D Theobai: umUMYl , , ^ It molelecti-on: n g = - = ;^a;j i » « v i ,4 m M a t khac: mx = 64(0,05 - x) + 108.0,2a = 12,08 - > x = 0,03; a = 0,5 ? (BH • , • • -""'-^ - » m = 0,05.64 = 3,2 g a m - > D a p an B , , Cach 2: G p i : ncu (pu) = x m o l Bao toan electron: 2n2n + 2ncu(ph4n ijng) = "^g+ 2.0,02 + 2x = 0,2a ^' ' X chua: n^^2+ = 0,15mol; n^^ = O ,lm ol; n^+ = O ,lm ol; ng^2- = 0,15mol 2x n^gNOa " ' IsJhanxet: > Cu(N03)2 + A g mcudu+ mAg - > X = 0,03 m o l - » - 0,04 ^ Bail6: p a y la p h a n l i n g dien phan h o n h g p —> N e n d u n g cong thuc t f n h n h a n h so' > Zn(N03)2 + A g Cu + A g N Mol: X gom Ag , Zn + 2AgN03 = 0,1.108 + 0,15.64 = 20,4 gam D a p an A c6 chua Z n hoac C u —> A g N O s he't D o t i n h k h u Z n > C u —>• Z n phan u n g truac r o i m a i den C u ::.,,f:-./: J.!A.f./'-:'.,-' malBlHqlonyrtq [dmx • , \nnU laix w Cty T N H H M T V DVVH Khang Vi$t dm nang On luygn thi dgi hpe 18 chuyfin dg H6a hgc - NguySn Van Hi\ phan img nhiet phan Chuyen de Na2C03 + CO2T + H2O - 2NaHC03 KIM LOAI KIIIM - KIltM THO - NHOM K I M L O A I K I E M a KimloaiMem + ^rv.'/-., Tac dung vai nude Kim N e u N a d u : 2Na + 2H2O Tac dung vai dung dich muoi ;, > 2NaOH + H t it f + + + 2NaOH > K2SO4 + Cu(OH)2i M " " + H2O Mol: M2O + H2O Ta c6: nMOH NaHCOs + NaOH > NaCl + C02t + H2O ^,,, C.Cs.,jy^|j;^XcmP'lfe)e(*Oe):: 0,02 col' + H2O NaHC03 + HCl j-j ^^_^>„c,,^.a MM.' > M O H + ^ H t ' ^ i l ^ ' H '+ 0,02 x NaHC03 + H C l > NaCl + C t + H2O d Muoi hidrocacbonat + Tinh chat ludng tinh • Mol: > NaHC03 + NaCl , ' ^,^^1^ Lcrigidi: Phan ieng thuy phan (quy tim ~^ xanh; phenolphtalein ~^ hong) >0 Na2C03 + H C l , , J _^ >'•' B.Li Tac dung vdi hap chat luang tinh Cho tu tu dung dich axit HCl vao dung djch Na2C03 j,^^,, 2NOt+4H20 va 0,224 lit H2 (dktc) K i m loai M la > Fe(OH)3i +3NaN03 Tac dung v&i dung dich axit ^ no vao nuoc, thu dugc 300ml dung dich chiia mQt chat tan c6 nong dp 0,2M Fe(N03)3 + 3NaOH HCO3 + OH- , • Cac phuang trinh hoa hpc: > NaAI02 + 2H2O 1,j Vi dii 1: Hoa tan hoan toan 1,7 gam hon hop gom k i m loai kiem M va oxit ciia A.Na Al(OH)3 + NaOH c Muoi cacbonat ^ ^^^^^^^^^ ^ ^ 3Cu2- + CO2 + NaOH > NaHCOa Tac dung vdi dung dich muoi > 2NaA102 + H2O + / VIDUMAU > NaiCQh + H2O AI2O3 + 2NaOH of ^ NaHS04 + H C l t 3Cu +2NO3 + 8H* b Hidroxit,, „„,„.„:„.„ rfe:)|'iWomr.O.':^ + Tac dung vdi oxit axit ; CO2 f • + Tinh oxi hod (khi c6 mat axit) Cho Cu vao dung dich hon h(?p KNO3 va H2SO4 loang, Cu tan dan tao dung dich mau xanh theo phuong trinh: ^ > 2KOH + H2t 2KOH + CuS04 + H2t 2KNO2 + O2 2KN03 'I.XI Cho K k i m loai vao dung dich CuS04 xay cac phuong trinh: 2K + 2H2O NaCl (ran) + H2S04(^flc) f Muoi nitrat ; Phan ung nhiet phan > 2NaCl + H t * 2Na + CI2T ^ 2NaCl + 2H2O "P""''"" > 2NaOH + C h t + TflcdwM^^ 2NaOH + H t 2Na + 2HC1 + vv,;:,r loai kiem khir mxac de dang nhi?t dp thvrong, giai phong hidro: 2Na + 2H2O + >, iff *• 0,01 ^^^^^^^^^^^^^^ ^ 2MOH f^iapi^^^ 2x = 0,3.0,2 = 0,06 * " 0,02 + 2x = 0,06 - > x = 0,02 Mat khac: 0,02M + (2M + 16).0,02 = 1,7 M = 23 (Na) - > Dap an A ^1 du 2: Hoa tan hoan toan 1,794 gam kim loai kiem M vao 200 m l dung dich H2SO4 0,1M Co can dung dich sau phan ung thu dugc 4,36 chat ran khan Y Kim loai kiem M la A.Rb , ifiui B.K C.Na Lcrigidi: , |oe >• D.Li f^Q Cac phan ung hoa hoc: 2M + H2SO4 ^ M2SO4 +H2 va M +H2O ^ MOH + ^H2 ^ NaiCO^ + H2O 205 Hoa TiQc - NQuyen van Hai Nhdn xet: Bao toan kho'i lugng -> m y = ITIM + Cty TNHH MTV D W H Khang Vijt "^coZ- + ^ Al(OH)3 O H Mol: -> m _ , = 4,36 - 1,794 - 0,02.96 = 0,646 g a m - > n ^ „ = 0,038 moi u O H O H = 0,078 OH 0,04 + 2H2O r 0,04 _^ a = 0,08.78 + 0,1.233 = 29,54 gam 794 -> nM = 2nH,so M H2t>U4 > AIO2 + OH- f-,,, r „•.*„ , , Dap an A yf du 5: C h o 100ml dung dich H3PO4 a mol/1 vao 100ml dung dich K O H M M= ^ = 23 (Na) 0,078 ,i, thu dugc dung dich Y c6 chua 16,42 gam hon hgp muo'i G i a tri ciia a la ~^ D a p an C A 0,9 B 1,0 V i d u 3: H o a tan hoan toan 4,6 gam N a vao SOOjnl H2SO4 0,1M, thu dugc C 0,5 Lai gidi: D.0,8 i,ôTr rr, rằ ,ã - r r H2 va dung dich Y Co can Y thu dugc m gam chat ran khan G i a tri ciia m la Hhan xet: V i Y chua hon hgp muo'i —>^ chac chan c6 chua muoi axit —> A 11,1 K O H phan ung he't Hfga B.7,1 C.14,2 D 18,2 Laigidi: = 0-2 mol; nH2 = 0/1 mol; nH2S04 = 0,05 mol 0,05 Na + 0,05 H2O ^ NaOH + ^ V i d ^ 4: C h o m gam hon hgp gom N a va Ba (ti 1# mol 2:1) tac d y n g voi nuoc (du), thu dugc dung dich X v a 4,48 lit H2 (dktc) C h o 300ml dung dich Hon hgp X gom: ji^^ n.3i>l A.LivaNa B N a v a K C 24,86 _ •••M^^^' Na + H2O NaOH + i H 2a a > Mol: 2a Ba Mol: + 2H2O a I': Trong X: A13^ Mol: 0,12 ' M ^ ,,, 0,06 = 28,33 0,1 + 30H0,36 ' fef >BaS04^ 0,1 ' •*>:• 0,12 , , iOaMyi'' -H2t t iO'ABVi < 0,03 H a i kim loai la N a v a K - > Dap an B • ^ ,^^,,| dugctaorala A 9,40 gam B 8,69 gam ^ ^ d>I nH, = ^ g r m nB/!q Di' :'}H.% Na* + OH" + > + OH- + - H 2 ^J^i^J > Al(OH)3 , ' iOnmA 0,2M Trung hoa dung dich X boi dung dich Y , tong khoi lugng cac muoi Q^^, ^ D.RbvaCs OiM Ba + 2H2O 206 Ba(OH)2 + H2 a • M + H2O D 9,34 -tikAn \M i i i b Ggi kim loai lien tiep nhom l A la M Al2(S04)3 0,2M vao X, thu dugc m gam ket tua Gia tri cua m la Voigidi: , y^,,, , nhom l A ) tac dung voi nuoc d u , thu dugc 0,672 lit hidro (dktc) m = mNa2S04 + rnNaOH= 142.0,1 + 40.0,1 = 11,1 gam —> D a p an A B 6,24 ,• > Muoi + H2O Vi dv 6: C h o 1,7 gam hon hgp X gom hai kim loai (thupc hai chu ki ke tiep -Hi '•^•'•'••al A 29,54 0,2 —> Dap an A ^^^^ ^ Mol: 0,1 nH20 = H K O H = Bao toan khoi lugng: 0,la.98 + 0,2.56 = 16,42 + 0,2.18 - > a = 0,9 Luu y: N a d u se tiep tuc phan ung voi nuoc: - "H-iH3P04) = ^'^^ So phan ling: H3PO4 + K O H > Na2S04 + H t 2Na + H2SO4 Mol: 0,1 " O H - ( K O H ) = ^'2' ^ i f B fiu> r '' > Ba2* + H - + H2 207 Cty TNhH Ca'm nang fin luy$n thi dgi hpc 18 chuy§n de 116a hqc - Nguygn Van H i ! Nhdn xet: n ^ „ = n H , = 0,1 m o l OH Cac p h u o n g t r i n h p h a n l i n g k h i pha t r p n : ^ M a t khac, n o n g d o H C l gap Ian H2SO4 -> T r o n g c i i n g m p t the tich thi n H c i = 3nH2S04- G o i n H c i = 3a -> nH2S04 = a- -> T r o n g Y: n , = a n i o l ; n ' i_ = aniol;n " , \ =3aniol/ + Ivlol: 0,02 - > a = 0,l > H2O + OH" ^ a = 0,02mol K h o i Ivtqmg m u o i t h u d u g c = 5,35 + m^,- + ^cr,2- SO^ ,j ->DapanA AP* * 0,02 '' 'i- •iA.A Uv M + - H2 0,02 - m + mHj - > mx = 1,62 + 0,02.2 - 18.0,04 = 0,9i g a m CuO + 2N02t + :>fd Iv D 1,26 , Laigidi: + H2O M o l : 0,04 NaN02 + - t CU(N03)2 0,02 -> Gpi cong thue c h u n g ciia cac k i m loai la M Cac p h a n ving nhiet phan: NaN03 > Al(OH)34 + O H ' A 0,94 n h i | t ciia cac muo'i t r o n g c h u o n g t r i n h t u lop 10 den I d p 12 ^ 0,03 —> D a p an C Nhan xet: D a y l a cau h o i n h a m k i e m tra kien t h i i c cua hpc sinh v e d p ben 2KMn04 BaS04i dupe 0,448 l i t k h i h i d r o (dktc) v a d u n g d i c h Y Co can Y t h u d u p e 1,62 g a m m o l m u o i p h a n l i n g la: (••A:Ab) cp.tid , u, , V l y : a = 0,01.78 + 0,03.233 = 7,77 g a m ' , V i d\ 8: Cho day cac m u o i : K M n , NaNCh, Cu(N03)2, CaCOs, KCIO3, AgNOs A > + 30H" M o l : 0,01 1, i •, 0,03 Al(OH)3 '•'1 H2O i.^.,,*-i M o l : 0,02 - > 0,06 ''' =5,35 + 0,06.35,5 +0,02.96 = 9,40 gam "I'" > + SO4" Ba2- ' OH" M o l : 0,03 TrunghoaXboiY: 1TV DWH Khang Vigt ••' ;.(,{:••'• vlS + — DM ^ EWS + < ) , • • - f H + rf^^ 0~ ,.(y-)n'^ m.im.n V i d\ 1 : Cho 5,65 g a m h o n h p p X g o m L i , N a va K v a o 100 m l H2SO4 I M , •:>rl nofi, cn.fijjf ?L,c o d " } :\ > CaO + C02t CaC03 KCIO3 KCI + - t •it'.M ^ A g + N02t_+-at_^^^^^,,^^ AgN03 d u n g d i c h Y g o m Al2(S04)3 0,1M va H2SO4 0,1M, t h u d u p e a g a m ket tua ' ptj^si fMsnij! B.4,66 , C 7,77 miiq-.: D.8,55., 2+ = 0,03 m o l ; n + = 0,03 m o l ; n C 22,45 D 17,80 j i ?;[ ni c n III 1., • Bao =n2nH2 0,35 -> n x == 2n= 0,35 A p dtoan u n g electron: d i n h luat tru g hoa=dien: 1.0,35 2- + l^ir^u- ^ • = 0,35 - 0,2 = 0,15 m o l OH = 0,09 m o l n + T r o n g Y: n ^ ,3+ = 0,02 m o l ; n ^ ^ - = 0,04 m o l ; n „ + = 0,02mol A' H jd-iS + 208 B 15,25 n H , = - ^ ^ — = 0,175 , Laigidi: T r o n g X: n A 19,80 Laigidi: V i d\ 9: T r p n 300ml d u n g d i c h X g o m Ba(OH)2 0,1M v a N a O H 0,1M v i lOOml A 6,99 ran k h a n Gia t r i ciia m la 92 —> D a p an B Giatricuaala thoat 3,92 l i t k h i H2 (dktc) v a d u n g d i c h Y Co can Y t h u d u p e m g a m chat Bao toan k h o i l u g n g : =m m xx +1^502+ mgQ + n ^ o H - " ^'^^ in = ^'^"^^ 0,15.17 = 17,8 g a m D a p an D 209 Cjiii (lai lipc 18 chiiyen de Wod tua: [i:in(] ! Cty TNHH MTV DVVH Khang Vi?t fjquyln VSn Hii C a u 49 Tron 0,30 mol H2 voi 0,3 mol brom dun nong voi xiic tac, sau mpt thoj , gian thu dvtgc hon hop san pham Y , hoa tan Y vao 287,85 gam nude thu dugc dung dich Z , lay 100 gam dung dich Z tac dung voi dung dich A g N O , (du), sau phan ling thu dugc 9,4 gam ket tiia Hieu suat ciia phan ung to'ng ' hgpHBrla A.60% / B.30% C.50% < ; D 40% C a u 50 C h o 5,2 gam bot Z n vao dung dich chua 0,06 mol F e C h v a 0,02 mo] C u C h Sau phan ling ket thiic thu dugc chat ran X v a dung dich Y Co ; can dung d}ch Y thu dugc bao nhieu gam chat ran khan? ^, A 10,88 gam B 12,36 gam C 14,69 gam (3) CH3COOCH=CH2 + N a O H — - — > CH3COOH + CH3CHO (4) 2CH2=CH2 + 02 — ^ 2CH3CH0 (5) CH3CHCI2 + 2NaOH D 11,96 gam ' , ' ' Theo phan ung: 2O3 ^ f,j„ > 3O2 Sau 40 giay, so mol O3 bi phan huy la 0,04 mol Vay nong oxi giam sau 40 giay la 0,04 mol/1 Vay toe dg phan ung Cau C h g n B • > K C l + CO2 + H2O (2) Cl2 + H2O ^ '"Vi-ni •'"ir;"'!.' rr.p > HCl + HCIO IB 2C 3B 4C 5B 6A 7D 8D 9B lOB lie H C l + KHCO3 12A 13D 14A 15D 16D 17C 18B 19C 20c 22A 23B 24C 25B 26C 27A 28C 29C (3) O3 + 2KI + H2O 21B 30A 31D 32B 33C 34B 35B 36B 37C 38B 39B 40B 41D 42D 43A 44A 45B 46D 47D 48C 49C 50C > p'M^iJ'' m K i g '(Bfi-X' - ' GQI Y GIAI DE so > K C l + CO2 + H2O > O2 + 2KOH + I2 (4) K H C + KHSO4 > K2SO4 + CO2 + H2O Hiiiih j i#> Cau Chgn A ; 1' u , Gpi so mol Y : A1(N03)3 = a; Fe(N03)3 = b Bao toan nguyen to N ta c6: 3(a+b) + 0,055 = 0,34 Cho Y + N a O H du, thu dugc ket tiia Fe(OH)3, A l ( O H ) luong tinh, tan N a O H d u Vay 3,6 gam cha't ran la Fe203, suy b = 0,045; a = 0,05 - aiv-iS|,4:,iLlKf:M Z la CH2=CH2 nen T la CH2Br-CH2Br Theo so do, X la CH3-CH3, Y la dong ;;, >,,) *• Loai A vi cho Z + Br2 (dung dich) se tao T la dan xuat dibrom Loai C vi X (CH3-CH3) dugc the H bang 2Br thi se u u tien the tai ciing mgt vi tri cacbon Loai D vi Y va T la dong phan, nen cung so lugng nguyen tu brom C a u Chgn C Khi X + H N O : so mol electron A l nhuong (-0,05x3) + so'mol electron FexOy nhuang = 0,165 - > Khoi lugng dung dich H2S04 = 0,1.98.100/10 = 98gam i.r • ;• Vay so mol electron FexOy nhuong = 0,015 = so mol FexOy Bao toan Fe ta c6: 0,015xx = 0,045 x = 3->Fe304 yujtu 'M tmt:m - H toraV& VIJ ' m = 27.0,05 + 232.0,015 = 4,83 gam , i u Chgn D , ; fn m k'ft: ; So mol H2O = 0,65 mol > so mol CO2 = 0,40 mol nen ancol deu no Tong so mol ancol = 0,65 - 0,40 = 0,25 mol So mol H2 = 0,1 ^ So mol H2SO4 = 0,1 f CH3CHO + 2NaCl + H2O trung binh tinh theo ozon la 0,04/40 = 1,0.10-^ mol/(l.s) fit" DAP AN DE SO phan cvia T nen Y la CHBr2-CH3 '^''•%::ff;:^^ Cau Chgn C (1) H C l + KHCO3 C a u l ChgnB ' J f; i),.:i, C.H2„.iOH + ^ - J ^ nC02 + („^i)H20^^''"'"^-^^'"^^ ' Khoi lugng dung dich sau phan ung = 98 + 2,3 - 0,1.2 = 100,1 gam C a u C h g n B - ^'un-m65 (1) CH3CH2OH + C u O (2) H C ^ C H + H2O 386 i"::?-' • CH3CHO + C u + H2O 80"C •> C H C H O n = 1/6 Khoi lugng phan tu trung binh ciia ancol = 40,4 0,40 /.ot - > Tong khoi lugng ancol = 10,1 gam r'",li'it: Trong phan ung ete hoa: m.mcoi = mete+ mniroc —> 10,1 = mete + 0,25.18 - > mete = 5,60 gam ; Cam naiiy on iuyen tin dai hyc 18 chuyen di H6a hgc - NIguySii V i n liai C dung v i kim loai kiem nhe, kiem thuong dung de che tao hop kim dimg Cau8 ChQnD Cac hop chat deu c6 nguyen tu C —> So'mol CO2 = 0,40 mol Id thuat hang khong D khong dung vi muoi an thuong dupe san xua't t u nuoc bien hoac khai Bao toan nguyen to'O: - thae t u cac mo muoi Trong cac phuong an A, C, D (cac chat c6 chua oxi): 0,1.1 + 0,5.2 = 0,4.2 + so mol H2O - > so mol H2O = 0,3 mol - > X chua 6H - > Dap an D I'l Cong thuc phan t u ciia X dang C3nHsn02n hay C2nH4n(COOH)n Trong cac phuong an B (chat c6 chua oxi) : 0,1.2 + 0, 5.2 = 0.4.2 + so'mol H2O - > so mol H2O = 0,4 mol - > X chua 8H: loai ^ Tac6:4n C4H8(COOH)2: thich hpp > C4H8(CC)ONa)2 + 2CO2 + 2H2O C4H8(COOH)2 + 2NaHC03 ; ^ > n C2H4-COOH (loai); n = n =l Cau Chon B chay giam dan theo chieu tang ban kinh nguyen tu u • Cau 14 Chpn A Vay, the tichC02 = 0,02.22,2 = 0,448 lit = 448ml , Cau 15 Chpn D Truoc het ta tim z: Ba2* + SO^" )• BaS04 So mol BaS04 = 3z.0,4 = (3) Sai v i cac kim loai kiem tho ngoai c6 lop electron ngoai ciing la ns^, can 0,144 CO lop v6 electrron ben la ciia hiem (4) Dung vi cac kim loai kiem tho khong c6 cung mot cau true niang tinh the So mol Al(OH)3 = 0,108; so mol NaOH da dung = 0,612 > 3.0,108 nen xay phuong trinh: dang lap phuong tam khoi Cau 10 Chon B z = 0,12 • /-J- •* j.; • '->.• „-•;., A : Cac polime c6 phan phan tram cac nguyen to' bang cac 3' monome c6 cung cong thuc don gian nha't O dap an B, cac polime deu c6 phan don gian nha't la CH2 Cau 11 Chon C Truoc het, can nhan thay rang ca axit tren deu c6 18 nguyen t i i C ?• ? phan tir Vay, tong so mol axit = (1/18) so mol CO2 = 0,9/18 = 0,05 mol Cac axit tren deu don chuc nen so mol = so mol axit = 0,10 mol Vay so mol H = (14,12 - 0,9.12 - 0,10.16) = 1,72 mol Suy so mol H2O = 0,86 mol va m = 0,86.18 = 15,48 gam Cau 12 Chpn A + 30H- > Al(OH)3 va Al(OH)3 + OH" Vay: so'mol OH- = (0,4y + 0,8z)x + (0,4y + 0,8z - 0,108) x = 0,612 - > y = 0,021 U',piS^ ' -aul6.ChpnD So mol X: Na* = 0,16; HCO ^ = 0,12; CO ] ' = 0,02 Hco: + OH- -> col' + va Ba2*+ CO;' — > BaC03 0,02 + 0,5y = 0,04 B dung vi hop kim Na - K c6 nhiet dp nong chay thap nen dung lam chat trao doi nhif t 16 phan ung hat nhan H2O So mol BaCOs = 0,04 mol So mol CO ' =0,04 lam te bao quang di^n |+ Trpn Y vao X xay phan ung: Loai B vi 2-cIobutan tach HCl tao anken khac A diing vi Cs c6 electron hoa tri lien ket yeu voi hat nhan neii dupe dung So mol Y: Ba^* = 0,08 + 0,25y; CI" = 0,16; OH" = 0,5y _ Ta loai C va D vi khong cung khung mach cacbon Cau 13 Chpn D > A l O j + 2H2O , y = 0,04 •o:dM w -1.:.- c:Oi.-' Cau 17 Chpn C Hon hpp gom k i m loai A l het (neu A l d u se thu dupe kim loai) V$y hai kim loai la Zn d u va Fe Mat khac, vi Zn d u nen FeCls bj k h u he't Fe Vay so mol Fe = 0,16 mol Suy so'mol Zn d u = (11,56 - 0,16.56)/65 = 0,04 mol -• ^V.'" • 389 ca'rn nang On luyjn thi dgi hpc 18 chuygn 66 Ho.i iioi, Cty Tl\i, NguyfnVan Hii Bao toan electron: so' mol electron A l va Z n nhuang = so' mol F e C h nhan 3x + 0,06.2 = 0,16.3 - > x = 0,12 CaumChonB " , , =0,24mol/l ' /ô'M^'Ov^^i ã ^ ã Cau 22 C h g n A uc'v-; N2 van a trang thai rang (1) v a (6) dung —> loai C > tach dugc NH3 Vi saccarozo khong k h u dugc AgN03/NH3 va Cu(OH)2 d u n nong nen Loai B v i N H se tac dung voi H2SO4 dac (3) khong dung —> loai A, D Loai C vi H2 v a N H se tac dung voi C u O (3) Khong dung vi saccarozo khong khir dugc AgN03/NH3 dun nong (4) Khong diing v i glucozo la monosaccarit nen khong c6 phan u n g thuy phan > CO2 nen Y chiia muoi cacbonat ' , f nuoc) I Day la mot cau hoi If thuye't tong hgp, c6 lien quan den kien thuc lop 10 (1, 5); lop 11 (2, 7, 8) va lop 12 (3, 6) Trong can l u u y den cac phuong SO2 + Br2 + 2H2O Ag2S + 2KNO3 N2 H- 2H2O > 5KC1 + KCIO3 + 3H2O 0,06 2KNO2 + O2 '" ) Khoi lugng chat beo can diing = 0,7843.890.100/89 = 784,3 kg K C l (Z) + A g N O s > 2Ag + SO2 NH4NO2 Cau 25 Chgn B 1.4FeS2 + 1 _ i ! _ > 2Fe203 + 8SO2 3S + 2H2O Mtom C H C O O C H tac dung duoc voi: H C l , K O H trinh then cho't la va •" ) Vay so kmol chat beo = 0,7843 kmol C6H5NH2 (anilin) tac dung dugc voi: H C l , dung djch Br2 C a u 20 C h g n C 4.2H2S + SO2 nOfi:) „ « ;r, " CH3NH2 tac dung dugc voi: H C l Vay andehit lai la C H = C H - C H O '" ) H g + SO2 •' H N C H C O O H tac dung dugc voi: H C l , K O H , C H O H (xuc tac) Khoi lugng phan tu andehit = (1,770 - 30.0,045)/0,0075 = 56 H g S + O2 ' ' N;>iuih 3Ci7H35COONa + C3H." ' ^ Cau 23 Chgn B Caul9 ChgnC 390 ?, K h i N H phan cue nen nen va lam lanh se hoa long, H2 va Tap trung vao nhan dinh (1) v a (6) xua't hien nhieu nhat D e nhan thay Do Y + H C l TV DVVH Khang Vigt -Cau 27 Chgn A ' ' '•S'"*''''"' V i (1) tac dung voi (3) c6 sui bgt nen (1), (3) la Na2C03, H2SO4 V i Ig (2) tac dung v a i cac Ig (1), (3), (4) lai deu tao ket tiia - > lo (2) la B a C h Vay Ig (4) la Na2S04 ' ^ *'' L'a'n: n, ng On luyen ihi dai ligi; 18 chuyfin de Hoa hpc Cty TNHH M i V [)\/VH Khang Vigt Nguyfit Van HSI au 32 Chpn B Cau 28 Chon C A diing: protein duoc chia logi, 1-protein don gian la loai protein Goi kim loai kiem la M Kho'i lugng dung djch M O H bang 80.1,25 = 100 gam ma thiiy phan chi cho Hon h^p cac a-amino axit, 2-protein phuc teip —> Khoi lugng M O H bang 11,2 gam, khoi lugng dung moi = 88,8 gam loai protein duoc tao tu protein don gian cong vai phan "phj Ta CO phan ung: protein" n h u axit cacboxylic, chat beo, RCOOR' + M O H B diing, v i du long trang trung tan nuoc, dun soi thi long trang trung dong tu lai C sai vi chi cac peptit c6 mat xich tra len moi c6 phan ung mau bioure D diing vi lien ke't peptit de bi thiiy phan moi truong axit hoac bazo Cau 29 Chon C T u ti khoi suy ti I f mol CsHe: H2 = : 0,1 a n-jsO >RCOOM + R'OH 0,1 >i.rj :.tj':.ô i ui/x uiiA l''^\U'i'^id'%dUJ\o:J'^'ããã.? Trong 93,4 gam chat long c6 88,8 gam n u o c lai la ancol d _> Khoi lugng ancol = 4,6 gam ^ ' • ffifta fi^ifD Tf if; Khoi lugng phan tu dia ancol R ' O H = 46 ^ R' la C2H5- ( ' RCOOC2H5 = 88 ^ R la CH3- ,j Cau 33 ^ - > So mol C3H6 phan ung = so'mol H2 phan ung = 3a - 0,09 Mol: • mc-a Bao toan khoi lugng: khoi lugng este = 15,4 + 4,6 - 11,2 = 8,8 gam J Q ^ ^ i ?, Goi so'mol Y : CsHb = 3a; H2 = 2a So'mol brom phan ung = 0,09 ^^ ex M\ ly.^ (.,,.^1,1;, Khi dot chay Y thu dixqic C O = 9a mol; H O = 11a mol Kho'i luong binh nude voi giam = kho'i lup-ng C a C O s - (kho'i lugng C O + khoi lugng H O ) —> a = 0,05 mol Goi so'mol C u = x; A g = y Ta c6: 64x + 108y = 2,68 Bao toan electron ta c6: 2x + y = so mol N O = 0,06 Vay x = 0,025 mol; y = 0,01 mol Khi cho tie'p 0,025 mol C u , xay hai phuong trinh phan ung: Hieu suat (tinh theo H2) = ^'^'^^'^'^ 100% = 60% C u + 8HNO3 > 3Cu(N03)2 + N O + H O Mol: 0,015 Zau 30 Chon A C u + 2AgN03- Do cac este don chuc nen so' mol K O H = tong so mol cac este, v^y truoc -> Cu(N03)2 + 2Ag he't can tinh duoc tong so'mol este , Mol: Nhan tha'y, phan tu cac este tren deu c6 nguyen tu cacbon dot chay Cha't ran gom: C u = 0,005 mol; A g = 0,01 mol mol hon hop este se thu duoc mol C O - > tong so mol este ^ (1/4) so mol C O = 0,03 mol _> the tich dung dich K O H = 0,03/0,15 = 0,2 lit = 200ml :au31 ChgnD 0,005 m = 0,005.64 + 0,01.108 = 1,40 gam Cau 34 Chgn B M C O s + 2HC1 De biet y can biet so mol H N O R2CO3 + 2HC1 Mat khac: so' mol H N O = so' mol NO2 = so' mol N O 0,01 0,01 > MCI2 + CO2 + H2O 2RC1 + CO2 + H2O Ta nhan thay: so mol C O = so mol H O = (1/2) so mol H C l = 0,02 mol So moi N O phy thugc vao so mol H * va N O ban dau va dugc tinh tu ban Bao toan khoi lugng: 2,42 + 0,04.36,5 = m + 0,02.44 + 0,02.18 >f'y \m A phan ung: - > m = 2,64 gam \, NO; (mol): 0,008 + 3e + 4H^ 0,024 0,032 > N O + 2H2O 0,008 Vay [H*] = [HNO3] = 0,008/0,8 = 0,01 ** 192 • p H = y = if) • jr? • ^'iiri'? ; Cau 35 C h o n B liC^H":: Gia s u khoi lugng muoi X la 100 gam ' cfvr! i ' ' —> Khoi lugng muoi Y la 88,875 gam * » «' > giam 11,125 gam ^ u u i n i i 6D-v.; 393 Ill naiiy tin liiyrii llii ilai 1H>C CI2 + 2KBr gam: cluiyfMi (In iloa hqc - IMguygn Van H^i Cty Ti\ 11 [ i M i V DVVH Khang Vigt 149 a : giam 89 gam Do la cac chat chua no'i doi C=C mach ho (stiren, axit metacrylic, etilen, : giam 11,125 gam butadien); bi oxi hoa dun soi vai thuoc ti'm (toluen) v a chua nhom chuc , v V''-^', ' andehit (propanal) * ' "''I a = 29,75 g a m ^ % K B r = 29,75% :au36.ChQnB 4!) |Cau 42 Chon D -» 2KC1 + Br2 238 gam: ^ 18 :au43 C h o n A •'^'"""^ ' ' ^' "" ^ ^ ' T a c : a + b = 0,15 cac chat oxi hoa I :au 37 Chon C So mol H2 = 0,2 ^ tu la 13; Y la 26 1,5a + b = 0,2 Vay: a = , l ; b = 0,05 (1) Khong diing vi X, Y khong ciing so hieu nguyen tii: X co so hieu nguyen - > T i 1^ ve so mol A l : Z n = : " Trong m gam X: A l = 2x mol; Z n = x mol (2) Khong dung vi X, Z khong ciing so hieu nguyen tu 102X-81 x = 18,3 (3) Diing vi X, Z deu ciing so khoi la 26 x = 0,l ••:!olvj ISu 44 C h o n A ; f ,4,^ - = 13; Y la - = 14 AI2O3 = X mol; Z n O = x mol Vay m = 0,2.27 + 0,1.65 = 11,90 gam (4) Khong diing v i X, Y khong cung so hieu nguyen ttr: X c6 so natron la :au38 C h o n B I Dot chay ancol X thu dugc the tich CO2 < the tich hoi nuac K h i cho X tac dung voi AgNOs/NHa, dun nong ta c6 ti le: so mol X : so mol -> X la ancol no - > cong thuc CnH2n+2-m(OH)m AgNOs = 0,2 : 0,6 = : ^ Tu ti le the tich C O : the tich hoi nuoc = 3/4 - > n = X chiia chuc andehit v a I H dinh vao cacbon I -> C3H^m(OH)m mang noi ba —> loai C va D loai A ,^ 3, , m = - > chat: propan-l,2-diol va propan-l,3-diol Tuy nhien, X ton tai the ran dieu kien thuong - > X khong the la este Amino axit: m = —> chat: glixerol i u 45 C h o n B H2N-CH2-CH2-COOH; CH3-CH(NH2)-COOH; CH3-NH-CH2-COOH I De thay (3) la phan ung axit-bazo, khong phai phan u n g oxi hoa-khii' Muoi amoni: C H = C H - C O O N H loai cac dap an A, C ••JJ :au 40 Chon B Cac chat: (1) mocphin; (2) cafein; (3) nicotin; (4) cocain v a (7) heroin lau 41 C h o n D ãã oô rwril 'mdfi f B sai v i nhieu hop chat cpng hoa tri tan tot nuoc n h u H C l , NH3, -J': Loai dap an D vi (6) khong phai la phan ung oxi hoa-khu: C u O + H N O T CU(N03)2 + H2O * A sai v i trang thai ran thi cac hop chat ion each dien : C sai v i hgp chat ion thuong c6 nhiet nong chay, nhi^t soi cac D diing v i trang thai nong chay, cac hop chat ion phan li cac ion trai 194 i I m = —> chat: propan-l-ol va propan-2-ol V i X luong tinh _> X la amino axit, muoi amoni hoac este da'u va c6 tinh dan dien , ,.,:,,,„,, ,;,,,„, „ V i m < n nen ta c6: :au39 C h o n B CH3OH, Trong 0,15 mol X: A l = a; Z n = b Glucozo the hien tinh khir nhom chuc andehit cua glucozo tac dung voj Til-dii kien phan ung chay , -1 • I u 46 Cho.i D I Goi cong thuc cac kim loai kiem la M : I 2M Mol: + 2H2O > 2M0H 0,04 + H2 0,02 I Bao toan kho'i luong: , '' ' I m +0,04.18 = 1,62 + 0,02.2 -> m = 0,94 gam \ A;y.'y 395 dm nang On luy^n thi djii hpc 18 chuypn de Hoa hpe - Nguyjn Van HSi Cty C a u 47 Chon D ;-V Twmi M i v DV\/H Khang Vi^t DE 50 A sai v i phenol la hop chat c6 nhom - O H dinh true tiep vac vong benzen B sai v i phenol khong tae dung duoc voi KHCOa (vi tinh axit cua phenol liCau 1: Nho t u t u tdi d u dung dieh K O H vao dung dieh muo'i X thu dugc ket yeu hon axit eaebonie) tiia, sau ket tiia tan dan Day gom cac muo'i eo tinh chat gio'ng muo'i X la C sai v i hop ehat CsHsO ean phai la ancol bac A ZnCh, AlCb va Cr2(S04)3 B AlCb, CrCh va ZnCh D dung: CaHs-OH + (CH3CO)20 C CuCh, Al2(S04)3 va CrCh ' ' ' ' ' D ZnCh, CrCb va FeCh CHaCOO-CeHs + CHsCOOH Cau 2: Cho cac chat tae dung voi nhau: Cau 48 Chon C ;)ti Amino axit X eo nguon goe thien nhien _> X la a-amino axit -> lo^i phuong an A Nho t u t u toi d u dung djch K O H va dung djch ZnCh Nho tix tu dung dieh K H C O vao dung dieh FeCb * ?i Dua vao khoi luong phan t u de tinh so mol chan - > phuong an C Cau 49 Chon C " Mol:' ik '-'^^ AgN03 + HBr > AgBr + HNOs " 0,05 ' ^ ^ ' ^ ' '' ^ Nho t u t u tdi d u dung djch NH3 vao dung dieh ZnCh Nho t u tir tdi d u dung dieh NH3 vao dung dieh FeCb 0,05 NH3 tdi d u Zn + Cu^* Mol: 0,02 0,02 Z n + Fe^* Mol: 0,03 0,03 c H f-y, B (1), (3), (5), (7) 50.0 mo; C (1), (3), (5), (6) D (1), (3), (6), (7) B AS,C«, C .,, r D I lEf i Cau 4: Tien hanh dien phan 100ml dung djch CuS04 (vdi d i f n eye tra) mot thdi ' 0,02 gian, thu dugc dung djch X, khoi luong dung djch giam 0,8 gam (d eatot chua cd hot khi) Them 1,68 gam hot Fe vao X, den phan ung hoan toan, > Zn2* + Fe thu dugc 1,24 gam kim loai Nong mol cua dung dieh CuS04 ban dau la 0,03 '' ' t l i + i »V, r - A (1), (2), (3), (7) A > Zn^^ + Cu _> Khoi luong chat ran khan = 136.0,08 + 127.0,03 = 14,69 gam c vdi dung dieh NaOH la: ' 0,06 Dung dieh Y chua: ZnCh = 0,08 mol va FeCh = 0,03 mol , \,((m yu:^ :< ; u.A Cau 3: So' dong phan cau tao mach hd cd cong thue C4H8O2 va tac dung duoc So mol Zn = 0,08 mol Cac phuong trinh phan ung 0,06 vfii b :n.n Nhirng thi nghiem nao tao ket tua, sau ket tua tan? -'^r > Zn^* + 2Ve^^ '^ Nho dung djch KCl vao dung djch AgNOs, sau nho tiep dung dieh H i ^ u suat tong hop HBr = ^ - 0 % = 50% Zn + 2Fe^* i Nho t u t u toi d u dung djch NH3 vao dung dieh AlCb Vay 287,85 gam nuae eua Z eo chua: 3.0,05 = 0,15 mol HBr CauSCChpnC Nho t u t u toi d u dung dieh NH3 vao dung dieh CuS04 K _> 100 gam Z CO chua: 4,05 gam HBr (0,05 mol) va 95,95 gam nvroc Mol: 0,03 ' ; , ,- A.0,15M t * B 0,25M C 0,05M D 0,20M !au 5: Di^n phan dung djch mudi MSO4 (vdi dien cue tro) vdi cudng dg dong dien khdng ddi bang 1,5A Sau thdi gian 1351 giay thi dirng d i f n phan, d eatot chua cd bgt va khd'i lugng eatot tang 0,672 gam Cong thiic muo'i la A CuS04 B ZnS04 C FeS04 D NiS04 Cau 6: Cho day eae muo'i: KMn04, NaN03, Cu(N03)2, CaC03, KCIO3, AgNOa So muol day nhiet phan hoan toan thi tao so mol nho hon I so'molmuoithamgiaphanungla: A 396 B.3 K^t'x C.4 D.5 397 Cam iiaiiq on luyen thi dai hoc 18 chuyeri H6a hpc - NguySri Vftn Hai Cau 7: H n h o p X g o m K O , K H C O , BaCla va N H C I , c6 so m o l bang 0,01 mol H o a tan hoan toan X vao m o t l u o n g nu-oc (du), d u n n o n g Sau p h a n ung B 3,850 g a m C 3,725 g a m D 4,225 gam C a u 8: Cho m gam h o n h o p X g o m N a v a A l vao nude t h u d u o c d u n g d i c h X , B 5,60 lit C 8,96 lit C a u 15: Tien h a n h cac t h i n g h i e m sau: (1) N h o iie t u d u n g d i c h H C l vao d u n g d i c h KHCO3 (5) C h o k i m loai bac tac d u n g v o i k h i ozon N h i i n g t h i n g h i e m xay p h a n u n g va tao bot k h i la B cue am, i o n N a * b j k h u A C.4 g a m chat ran E Gia t r j eua m la -«C 5,17 m o t l u o n g d u N a O H , d u n n o n g t h u duoc 0,672 l i t k h i ( dktc) va m gam ket }i> t i i a N u n g k e t tua ngoai k h o n g k h i den k h o i l u o n g k h o n g d o i t h u duoc 0,8 j i , g a m chat ran M a t khae, ne'u cho X tac d u n g v o i d u n g d i c h BaCh d u , thu , d u o c 4,66 g a m k e t tua T o n g kho'i l u o n g chat tan t r o n g X la C 6,75 g a m D 7,65 gam C a u 13: Cho so d o bien hoa: X 4-NaOH Y + CO2 + H2O , k e m p h a n c h i m d u o i nuoc (4) K h i cho Fe tac d u n g v o i l u o n g d u d u n g djch A g N O s t h u duge A g ket tua va d u n g d i c h chi ehiia A g N O s va Fe(N03)2 A B.2 C.3 D.4 C a u 17: Phat bieu nao sau day sai? A Glueozo va f r u c t o z o tac d u n g v o i H2 deu sinh sorbitol B Glueozo v a f r u c t o z o deu c6 kha nang t h a m gia p h a n u n g trang bae C Glueozo va f r u c t o z o la d o n g p h a n cua n h a u D T i n h b o t va x e n l u l o z o la d o n g phan eua n h a u C a u 18: T r p i i 5,4 g a m bot A l v o i 17,4 g a m bot Fe304, tien h a n h phan u n g nhiet n h o m t r o n g d i e u k i e n k h o n g eo k h o n g k h i H o a tan san p h a m ran t h u d u g e X •Al sau p h a n u n g b a n g d u n g djch H2SO4 loang (du), t h u d u g e 5,376 l i t H2 Biet X, Y, Z la eac h o p chat ciia n h o m X, Y, Z c6 the la A AICI3, NaA102, A l ( O H ) B AI2O3, NaA102, A l ( O H ) (3) D e bao v§ t a u bien l a m bang thep n g u o i ta gan vao v tau n h i r n g t a m T r o n g cac phat bieu tren, so phat bieu d u n g la D.5,98 'i C a u 12: D u n g d i c h X chiia cac i o n : Fe^*, SO42- NH4% C h C h o X tac d u n g vol B 3,375 gam d u o i tac d u n g cua d o n g dien m o t chieu hoa va phat s i n h d o n g d i ^ n D.5 d u n g djch T Loc ket t u a Z, n u n g d e n k h o i l u o n g k h o n g d o i t h u duoc m A 3,73 g a m D 5.^^^^ (2) Ban chat cua s u an m o n hoa hoe la phan u n g oxi hoa k h u , k i m loai b i o x i g o m 0,05 m o l N a O H v a 0,02 m o l Ba(OH)2 T r o n Y v a o X t h u ket t u a Z v a B.5,79 C (1) Ban chat eua s u d i e n p h a n la p h a n l i n g o x i hoa k h u xay d i e n cue C a u 1 : D u n g d i c h X g o m 0,02 m o l H C l va 0,01 m o l Al2(S04)3 D u n g dich Y A 5,58 B C a u 16: C h o cac phat bieu sau : C cue d u o n g , n g u y e n t u N a b i oxi hoa B.3 , ^ ,^ ^ A cue dirong, i o n Na+ bj oxi hoa A (dktc) H i e u suat eua p h a n u n g nhiet n h o m la i-r'hl ,^ , ' A 60% B.70% C.80% D 90% • Eau 19: Phat bieu nao sau day sai? C A1(N03)3, NaA102, A l ( O H ) A D u n g d i c h H F c6 the hoa tan Si02 D AI2O3, A l ( O H ) , A l C h B K h i cho t h e m n u o c vao o l e u m t h i n o n g d o cua axit sunfurie dan dat den 100% 398 ; (4) Sue k h i o z o n vao d u n g djch K I D 7,84 l i t So chat t r o n g day k h i tac d u n g d u n g d i c h Ba(OH)2 d u tao ket t i i a la ' D M g , A l , Fe (3) Sue k h i clo t o i d u vao d u n g d i c h KBr C a u 10: C h o d a y cac cha't: NH4CI, (NH4)2S04, Al2(S04)3, M g C h , ZnS04, AlCh, f> C K, Ba, Ca bao nhieu l i t k h i CI2 (dktc)? D cue am, i o n C I " b i o x i hoa , B L i , Be, A l (2) D u n g d j c h H 2 v a o d u n g d i c h K M n (c6 mat axit H2SO4 loang) C a u 9: K h i d i e n p h a n N a C l n o n g chay ; d i e u k i e n thuong? 4,48 l i t H (dktc) va 2,7 g a m chat ran k h o n g tan Ne'u o x i hoa m g a m X can A 6,72 lit ' Cau 14: D a y nao sau day g o m cac k i m loai c6 cau true lap p h u o n g t a m kho'i A L i , N a , K ; l o c ket tua, c6 can d u n g dich t h u duoc k h o i l u o n g muo'i k h a n la A 2,235 g a m If' Cty TNHH MTV DVVH Khang Vi§t , , „ ^ ••vr,' „ , : : ' ' a ^ ' 9 Cty TNHH MTV DVVH Khang Vigt Ca'm nang an luygn thi dai hgc 18 chuy6n 6i H6a hgc - Nguygn van.Hai C Phan ung cua Ch voi nude, thi CI2, la chat oxi hoa v a nuoc la chat k h u D K h i nho dung dich H2O2 vao dung dich K M n (co mat H2SO4 loang) thi CO C a u 20: C o bao nhieu dong phan cau tao, mach ho, co cong thiic phan tit va CO kha nang phan ung voi dung dich N a O H giai phong k h i X (X lam qui tim doi mau xanh)? A ^'5'^ B.4 ' C.3 ' D.2 " B K i m loai nhom la kim loai luong tinh Z^'' ' '^*' '^^ '"^J ' ' *' ' ' (1) Andehit bi k h u boi AgNOs moi truong N H tao A g (2) Phan ung este hoa moi truong axit sunfuric dac la phan ung mot chieu (3) Phan ung cua CI2 voi dung dich kiem thi CI2 vira la chat oxi hoa vira la ' •j'y (4) Phan ung thiiy phan este moi truong kiem la phan ling thuan = ^ B.2 ,j ^ C dung dich thu duoc thuc hien phan ung trang guong thi dugc 10,8 gam bac kim loai Biet hieu suat toan bo qua trinh la 50% Gia trj ciia m la -B 10,25 D fJ''f-' *'' nhom nao sau day? I.M A Mg, AI2O3, A l ' - ''''^ ^ nt^mgn trtj (Hi.J:'\t t, B.Mg,K,Na ^ K ^ p D Fe, AI2O3, Mg ' cao su buna; nilon-7 So'chat co the dugc che bMng phuong phap trung hgp la A B.3 C.4 D.5 au 30: Thanh phan chinh to nilon-6 la MM A poli(hexametylen adipamit) B poliacrilonitrin C policaproamit D poli(etylenterephtalat) J: au 31: C h o day gom cac chat: natri phenolat, anilin, alanin, kali axetat, day tac dung dugc voi dung dich H C l la D.4 C a u 23: Tien hanh thuy phan m gam bpt gao chua 80% tinh bot, roi lay toan bp A 16,20 C etylamin, metylamoni clorua, triolein, phenylamoni clorua So' chat Trong cac phat bieu tren, so phat bieu diing la AA B Cau 29: Cho cac chat: to capron; to nilon-6,6; to nitron; to triaxetat xenlulozo, C a u 22: Cho cac phat bieu sau: nghich fructozo, glixerol, axit axetic, saccarozo, natri axetat, mantozo C o bao nhieu C Z n , AI2O3, A l D N h o m bi pha huy moi t r u o n g kiem chat khu D Cau 28: C h i diing dung dich K O H de phan biet dugc cac chat rieng biet C Na3AlF6c6 phan chinh ciia quang criolit ' C.5 Cau 27: C h o cac chat sau day: propin, vinylaxetilen, andehit axetic, glucozo, A ' A H o p chat AI2O3 la oxit luong tinh B chat tac dung dugc voi dung dich AgNOa/NHs va tao ket tiia? C a u 21: Phat bieu nao sau day sai? ' So dung dich day co phan ung voi dung dich etylamin la A bay C4H9O2N Cau 26: Cho day cac dung dich: H C l , FeCb, CH3OH, NaHS04, H N O , Na2C03 C 6,48 D 20,25 C a u 24: C h o cac chat: KBr, S, Si02, P, Na3P04, FeO, C u va Fe203 Trong cac A B.7 C.4 D.5 au 32: H o a tan 6,21 gam mot kim loai M vao dung dich H N O loang d u , thu dugc 1,68 lit hon hgp X (dktc) gom khong mau, khong bj hoa nau khong va ty khoi ciia X so voi hidro bang 17,2 K i m loai M la: A C u B Z n C Fe D A l C a u 33: Cho cac dung dich: axit axetic; glyxerol; etanol; alanylglyxylalanin; chat tren, so chat co the bj oxi hoa boi dung djch axit H2SO4 dac, nong la natri hidroxit; glucozo Thuoc thu dung de phan biet cac dung dich tren la A A Q u y t i m B.6 C D.5 , ; B CaCOs C CuO ^ D CuS04 C a u 25: H o p chat hiiu co don chuc X chua C , H , O L a y 13,6 gam X phan ung au 34: Trgn C u O va R O (R la kim loai hoa tri 2) theo ti 1? so moi la 1:2 thu vua du voi dung dich K O H 11,667 % thu duoc dung dich Y C o can dung dugc hon hgp X Thoi C O d u vao 4,8 gam X nung nong, thu dugc hon dich Y dugc phan hoi chi co 86,6 gam nuoc va 23 gam chat ran Z Cong thiJfC hgp Y H o a tan hoan toan Y can 50ml H N O M , thu dugc N O (san phan tu cua X la ,tt A C H O pham k h u nhat) K i m loai R la B C H O C C4H8O2 D C4H5O4 A.Pb B.Zn C C a Kfi D Mg 401 Ca'm nang an luygn thi dgi hpc 18 chuySn dg H6a hpc - Nguygn VSn HSi Cty TNH!< MTV DWH Khang Vigt C a u 35: Cho m gam C u vao 150ml dung dich A g N O s 0,2 M , sau mot thoi gigj^ C a u 44: H o n hgp X gom andehit oxalic, metyl fomat va etylen glicol Do't chay thu duQ-c 3,840 gam chat ran X va dung dich Y Cho 6,210 gam Pb (du) vao hoan toan 0,2 mol X can toi thieu 0,35 mol O2 Khoi lugng phan tu trung Y, thu dugc 4,745 gam chat ran Z va dung dich T G i a tri cua m la binh cua cac cha't hon hgp X (tinh theo u) la | ^ j | | A 2,24 B.3,36 C 1,68 D 1,12 A 59 C a u 36: C h o cac polime: (1) amilozo; (2) xenlulozo; (3) amilopectin; (4) to nilor,, 6; (5) cao su l u u hoa; (6) n h y a bakelit Cac polime c6 m^ich khong phan nhanh la: ' A.(l),(2),(3) C.(2),(4),(6) dung dich K O H , thu dugc 219,95 gam muo'i Khoi lugng K O H da phan ung D (2) (5), (6) A 43,4 gam glucozo + Cu(OH)2 N h u n g phan ling tao sorbitol la C (3) va (5) B 42,4 gam C 34,4 gam D 46,4 gam C a u 46: N h a n djnh nao sau day dung ? O H - ; (3) dd fructozo + H2 (xt N i , t°); (4) dd saccarozo +H2O (xt H ^ ; (5) dd B ( l ) v a ( ) D 54 la C a u 37: C h o cac thi nghi^m sau: (1) dd glucozo + H2 (xt N i , t°); (2) fructozo + A.(l)va(2) C.62 -it j i C a u 45: C h i so axit cua mgt lipit X bang Lay 200 gam X tac dung vira du voi rB (1), (2), (6) B.56 i A Trong mgt chu ki, nguyen tu c6 dien tich hgt nhan Ion thi ban kinh ,^ nguyen tu Ion D (2) va (4) tin B Cac nguyen to'thugc cimg nhom A thi c6 so electron lop ngoai cung bang C a u 38: H o n hgp X gom axit acrylic, vinyl axetat, metyl acrylat va axit metacrylic C h o 8,32 gam X tac dung vira dii voi 100 m l dung dich K O H C Hang so'can bang cua phan ung thuan nghjch phu thugc vac nong dg 1,0M Dot chay hoan toan 8,32 gam X sau phan ung thu d u ^ c m gam CO2 va D Thuy phan este moi truang kiem la phan ung thuan nghich 5,04 gam H2O G i a trj ciia m la A 14,96 C a u 47: Dot chay he't hon hgp X gom metan, etan va hidro can V lit oxi (dktc) B 13,2 C 17,60 D 16,72 C a u 39: D u n g dich lam qui h'm hoa xanh la A alanin B glyxin " »i'»v' C lysin H a p thu san pham chay vao binh dung nuoc voi du, thay c6 gam 'i'- ket tua va khoi lugng dung dich binh giam 1,92 gam Gia trj cua V la D valin A 2,24 ' " ' B 1,792 C a896 D 1,344 * C a u 40: C h o day gom cac chat: phenol, anilin, glyxin, axit axetic, etylamin, Cau 48: Xenlulozo trinitrat dugc dieu che'tu xenlulozo va axit H N O d^c c6 metylamoni clorua, axit acrylic So chat day tac dung dugc voi dung xiic tac De c6 5,94 kg xenlulozo trinitrat can dung dung djch chiia m kg axit dich K O H la H N O (xenluloza lay d u va hi^u sua't cua phan ung la 75% ) G i a tri cua m A • B ^ C D C a u 41: Dot chay hoan toan m gam hon hgp X gom cac axit beo: oleic, stearic A 5,67 kg v a linoleic bang oxi, thu dugc 12,42 gam H2O va 16,128 lit CO2 (dktc) Gia trj C 15,12 kg D 5,04 kg '" B 14,6 C 12,40 D 11,3 B K h i cho glucozo tac dung vai anhidrit axetic thu dugc C Metyl axetat c6 the lam mat mau dung dich brom ung la D K h i dot chay este no thi thu dugc so mol nuoc bang so mol CO2 A 1,68 lit B 4,48 lit C 1,12 lit D 0,896 lit mol tuong ung la 1:2 T h u y phan hoan toan m gam X bang dung dich H C l fructozo, metanol, dung dich alanylglixin va tripanmitin So' chat tac dung (vua dii) C o can dung dich sau phan ung thu dugc 16,45 i B y.^i C .ti D.3 .u ' ' u 50: H o n hgp X gom tripeptit Ala-Gly-Gly va dipeptit Ala-Gly, c6 ti 1? so C a u 43: C h o cac chat: protein, polipropen, etylen glicol, axit fpmic, dung djch dugc voi Cu(OH)2 la este C6H70(OCOCH3)5 thu dugc 3,02 gam hon hgp oxit The tich oxi (dktc) da tham gia phan A '' A Phenyl axetat dugc tao cho axit axetic tac dung voi phenol C a u 42: Do't chay hoan toan 1,74 gam hon hgp Mg, A l va Fe oxi (du) 402 ^'" au 49: N h a n djnh nao sau day dung noi ve este? cua m la A 13,5 B 1,68 kg gam chat ran khan Gia trj cua m la A 11,13 B.9,90 C.9,45 D 8,49 ' 403 ca'm nang 6n l u y § n thi dai hgc 18 c h u y § n cl6 Hoa lioc - iMguygn van Hii K h i c h o Fe v a o d u n g d i c h , x a y cac p h a n u n g D A P A N D E SO lA 2B 3D 4D 5A 6A 7A 8D 9C IOC n c 12C 13B 14A 15C 16B 17D 18C 19C 20B 21B 22A 23D 24D 25B 26B 27C 28A 29B 30C 31A 32D 33D 34D 35A 36B 37B 38D 39C 40D 41D 42D 43C 44A 45A 46B 47A 48D 49B SOB G O I Y G I A I DE SO dung dich K O H d u Loai B, C v i ke't tiia Cr(OH)2 la baza, k h o n g tan t r o n g K O H d u b b F e + H2SO4 b F e S + H2 Mol: 0,01 0,01 ,„ Q,|5j , > r Cau2:ChQnB ; ' Zn(OH)2 - > K2Zn02 (tan) C u C h Cu(OH)2 V a y b = 0,015 FeCb + 3NH3 + 3H2O Fe(OH)3 i K C l + A g N - > A g C l i + KNO3 A g C l + 2NH3 ' i + 3NmC\ ,',efe4Q:6Moxoluln.x) [Ag(NH3)2]Cl (tan) MSO4 + H2O N h $ n thay cac h o p chat can t i m la axit cacboxylic d a n chiic hoac la este don chiic A x i t cacboxylic: CH3CH2CH2COOH; C H C H ( C H ) C O O H ' ,