Nối tiếp nội dung phần 1 tài liệu Kỹ thuật giải quyết nhanh gọn đề thi Đại học môn Hóa học, phần 2 giới thiệu một số đề thi Đại học - Cao đẳng tham khảo các năm 2012 và 2013. Mời các bạn cùng tham khảo nội dung chi tiết.
O CHl/dNG VI CfiC DE THi t>6l HOC THfiM KH^O DE SO D E T H I T U Y E N S I N H D A I H O C N A M 2013 M o n t h i : H O A , k h o i A - M a de : 374 Cho biet nguyen tii khoi cua cac nguyen t o ' : H = 1; C = 12; N = 14; O = 16; Na = 23; M g = 24; A l = 27; S = 32; CI = 35,5; P = 31; Ca = 40; Cr = 52, Fe - 56; Cu ^ 64; Zn = 65; Br = 80; Ag=108; Ba = 137 I P H A N C H U N G C H O T A T C A T H I S I N H (40 cdu, tic cdu den cdu 40) C a u 1: H o n hcfp X gom 3,92 gam Fe, 16 gam Fe203 va m gam A l Nung X d nhiet cao dieu kien khong c6 khong k h i , thu diioc h6n hop chat ran Y Chia Y hai phan bang Phan mot tac dung vcri dung dich H2SO4 loang (dii), t h u diicfc 4a mol k h i H2 Phan hai phan ufng vdri dung dich NaOH duf, t h u diicfc a mol k h i H2 Biet cac phan ufng deu xay r a hoan toan Gia t r i cua m la A 5,40 B 3,51 C 7,02 D 4,05 C a u 2: Trong dieu kien thich hop, xay cac phan ufng sau (a) 2H,S04 + C ^ 2SO2 + CO, + 2H2O (b) H2SO4 + Fe(0H)2 ^ FeS04 + 2H2O (c) 4H0SO4 + 2FeO ^ Feo(SO.,)a + SO2 + 4H2O (d) 6H2SO4 + F e ^ Fe2(S04)3 + 3SO2 + 6H2O Trong cac phan ufng t r e n , phan ufng xay vdi dung dich H2SO4 loang la A (a) _ B (c) C (b) D (d) C a u 3: Trong mot b i n h k i n chufa 0,35 mol C2H2; 0,65 mol H2 va mot i t bot N i Nung nong b i n h mot thcri gian, t h u difdc h6n hop k h i X c6 t i khoi so vdri H2 bang Sue X vao li/gng diT dung dich AgNOs N H den phan ufng hoan toan, t h u dtfgc h6n hgp k h i Y va 24 gam ket tua H n hgp k h i Y phan ufng vCra du vdi bao nhieu mol Br2 dung dich? A 0,10 mol B 0,20 mol C 0,25 mol D 0,15 mol C a u 4: O dieu kien thich hgp xay r a cac phan ufng sau: (d) 3C + 4A1 -> A I C (c) C + CO2 -> C (b) C + 2H2 (a) 2C + Ca CaC2 CH4 128 Trong cac phan iJng tren, t i n h khuf cua cacbon the hien d phan ufng A (c) B (b) C (a) D (d) Cau 5: Len men m gam glucozo de tao ancol etyhc (hieu suat phan ufng bang 90%) Hap thu hoan toan liiong k h i CO2 sinh vao dung dich Ca(0H)2 dif, thu duoc 15 gam ket tiia Gia t r i ciia m la A 15,0 B 18,5 C 45,0 D 7,5 Cau 6: H n hop X gom Ba va A l Cho m gam X vao nvidc dii, sau k h i cac phan ijfng xay hoan toan, t h u ducfc 8,96 Ht k h i H2 (dktc) M a t khac, hoa t a n hoan toan m gam X bang dung dich NaOH, thu duoc 15,68 l i t k h i H2 (dktc) Gia t r i cua m la A 29,9 B 24,5 C 19,1 D 16,4 Cau 7: Khol Itfcmg Ag thu dugfc k h i cho 0,1 mol C H C H O phan ufng hoan toan vdi li/ong dtf dung dich AgNOa NH3, dun nong la A 10,8 gam B 43,2 gam C 16,2 gam ' D 21,6 gam Cau 8: Cho hot Fe vao dung dich gom AgNOs va Cu(N03)2 Sau k h i cac phan ufng xay hoan toan, thu duoc dung dich X gom hai muo'i va chat ran Y gom hai k i m loai H a i muo'i X va hai k i m loai Y Ian iLfgt la: A Cu(N03)2; Fe(N03)2 va Cu; Fe B Cu(N03)2; Fe(N03)2 va Ag; Cu C Fe(N03)2; Fe(N03)3 va Cu; Ag D Cu{N03)2; AgNOg va Cu; Ag Cau 9: Cho 100 m l dung dich amino axit X nong 0,4M tac dung viTa dij vdi 80 m l dung dich NaOH 0,5M, thu ducfc dung dich chufa gam muoi Cong thufc cua X la A NH2C3H6COOH B NH2C3H5(COOH)2 C (NH2)2C4H7COOH D N H C H C O O H Cau 10: Cho 1,37 gam Ba vao l i t dung dich CUSO4 0,01 M Sau k h i cac phan ufng xay hoan toan, khoi liicfng ket tua thu di/oc la A 3,31 gam B 2,33 gam C 1,71 gam D 0,98 gam Cau 11: K h i di/cfc chieu sang, hidrocacbon nao sau day t h a m gia phan ufng the vdi clo theo t i le mol : , thu diioc ba dan xuat monoclo la dong phan cau tao cua nhau? A isopentan B pentan C neopentan D butan Cau 12: Oxi hoa hoan toan 3,1 gam photpho k h i oxi duf Cho toan bo san pham vao 200 m l dung dich NaOH I M den k h i phan ufng xay hoan to^n, thu di/gc dung dich X Khoi liiong muo'i X la A 14,2 gam B 11,1 gam C 16,4 gam D 12,0 gam Cau 13: Cho X la hexapeptit Ala-Gly-Ala-Val-Gly-Val va Y la tetrapeptit GlyAla-Gly-Glu Thijy phan hoan toan m gam h n hop gom X va Y t h u difoc amino axit, c6 30 gam glyxin va 28,48 gam alanin Gia t r i cua m la A 77,6 B 83,2 C 87,4 D 73,4 129 C a u 14: O t r a n g t h a i ccf b a n , cau h i n h electron cua n g u y e n tuf N a ( Z = 11) l a A Is22s22p'^3s- B Is22s22p''3s' C ls^2s"2p''3s^ D Is^2s22p'=3s^ C a u 15: H6n h a p X chufa ba a x i t cacboxylic deu don chufc, m a c h h d , gom m o t a x i t no va h a i a x i t k h o n g no deu c6 m o t l i e n k e t doi (C=C) Cho m gam X tac d u n g viTa du vori 150 m l d u n g d i c h N a O H M , t h u dirge 25,56 gam h o n h o p muo'i D o t chay h o a n t o a n m g a m X , h a p t h u t o a n bo s a n p h a m chay b k n g d u n g d i c h N a O H duf, kho'i liJgng d u n g d i c h t a n g t h e m 40,08 gam Tong k h o i lu'gng cua h a i a x i t cacboxylic k h o n g no t r o n g m g a m X l a A 15,36 g a m B 9,96 g a m C 18,96 g a m D 12,06 g a m C a u 16: D u n g d i c h a x i t axetic p h a n ufng diigc vdri t a t ca cac c h a t t r o n g day nao sau day? D N a O H , Cu, N a C l C N a O H , N a , CaCOg B N a , CuO, H C l A N a , N a C l , CuO C a u 17: T e n t h a y t h e (theo l U P A C ) cua (CH3)3C-CH2-CH(CH3)2 la D 2,4,4-trimetylpentan C , , , - t e t r a m e t y l b u t a n B , , , - t e t r a m e t y l b u t a n A 2,2,4-trimxetylpentan C a u 18: T o n i l o n - , l a san p h a m t r u n g ngUng cua A etylen glicol va hexametylendiamin B a x i t a d i p i c v a g l i x e r o l C a x i t a d i p i c v a e t y l e n g l i c o l D a x i t adipic v a h e x a m e t y l e n d i a m i n C a u 19 : HQn hop X gom N a , B a , Na20 va BaO Hoa t a n hoan toan 21,9 gam X vao nxidc, t h u diigc 1,12 l i t k h i H2 (dktc) va dung dich Y , t r o n g c6 20,52 gam Ba(0H)2 H a p t h u h o a n t o a n 6,72 l i t k h i CO2 (dktc) vao Y , t h u di/gc m gam ket tiia G i a t r i ciia m l a A 23,64 B 15,76 C 21,92 D 39,40 C a u 20: H g p c h a t X c6 t h a n h p h a n g o m C, H , O chufa v o n g benzen Cho 6,9 g a m X vao 360 m l d u n g dich N a O H 0,5 M (diT % so vdti li/gng can phan ufng) d e n p h a n ufng h o a n t o a n , t h u difgc d u n g dich Y Co can Y t h u dugc m g a m chat r a n k h a n M a t k h a c , dot chay h o a n t o a n 6,9 gam X can vi^a du 7,84 l i t O2 ( d k t c ) , t h u dirge 15,4 g a m CO2 B i e t X c6 cong thufc p h a n t i f t r i m g v d i cong thufc don g i a n n h a t G i a t r i ciia m l a A 13,2 B 12,3 C 11,1 D 11,4 C a u 21: B i e t X l a a x i t cacboxylic don chufc, Y l a ancol no, ca h a i chat deu mach h o , CO Cling so n g u y e n t i r cacbon D o t chay h o a n t o a n 0,4 m o l h o n hgp gom X v a Y ( t r o n g so m o l cua X Idm h o n so' m o l cua Y ) can vira du 30,24 l i t k h i O2, t h u dirge 26,88 l i t k h i CO2 va 19,8 g a m H-.O B i e t t h e t i c h cac k h i d dieu k i e n t i e u c h u a n Kho'i liTgng Y t r o n g 0,4 m o l h6n h g p t r e n l a A 17,7 g a m B 9,0 g a m C 11,4 g a m D 19,0 g a m 130 Cau 22: Tien hanh ca!c t h i nghiem sau (a )Suc k h l etilen vao dung dich KMn04 loang (b) Cho hcfi ancol etylic di qua hot CuO nung nong (c) Sue k h i etilen vao dung dich Br2 CCI4 (d) Cho dung dich glucozo vao dung dich AgNOa, N H di/, dun nong (e) Cho FeoOs vao dung dich H2SO4 dac, nong Trong cac t h i nghiem tren, so' t h i nghiem c6 xay phan ufng oxi hoa - khijf la A B C D Cau 23: Cho so cac phan ijfng: X + NaOH - ^ - ^ Y + Z X + NaOH (ran) T Q + H2O — > T + P ) Q + H2 > Z Trong so tren, X va Z Ian liicft la A HCOOCH-CHa va HCHO B CH3COOC2H5 va C H C H O C C H C O O C H - C H va C H C H O D C H C O O C H = C H va H C H O Cau 24: t f n g vdri cong thiifc phan tuf C4H10O c6 bao nhieu ancol la dong phan cau tao cua nhau? A B C D Cau 25: Cho m gam Fe vao b i n h chiJa dung dich gom H2SO4 va H N O , thu difgc dung dich X va 1,12 l i t k h i NO Them tiep dung dich H2SO4 dif vao binh thu dtfctc 0,448 l i t k h i NO va dung dich Y Biet ca hai tri/cfng hop NO la san phan khijf nhat, d dieu kien tieu chuan Dung dich Y hoa tan vifa het 2,08 gam Cu (khong tao san pham khiif ciia N*^) Biet cac phan uTng deu xay hoan toan Gia t r i cua m la A 2,40 B 4,20 C 4,06 D 3,92 Cau 26: Lien ket hoa hoc gifla cac nguyen tijf phan tii H C l thuoc loai lien ket A cong hoa t r i khong cifc B ion C cong hoa t r i c6 cifc D hidro Cau 27: Thiic hien cac t h i nghiem sau (a) Cho dung dich H C l vao dung dich Fe(N03)2 (b) Cho FeS vao dung dich H C l (c) Cho Si vao dung dich NaOH dac (d) Cho dung dich AgN03 vao dung dich NaF (e) Cho Si vao b i n h chijfa k h i F2 131 (f) Sue k h i S02vao dung dich H2S Trong cac t h i nghiem tren, so t h i nghiem xay phan ijfng la A B C D C a u 28: Hoa tan hoan toan m gam A l bang dung dich HNO3 loang, thu di/oc 5,376 l i t (dktc) hon hop k h i X gom N2, N2O va dung dich chOfa 8m gam muoi T i khol ciia X so vdi H2 bang 18 Gia t r i cua m la A 17,28 B 19,44 C 18,90 D 21,60 C a u 29: Cho h6n hop X gom 0,01 mol A l va a mol Fe vao dung dich AgNOa den k h i phan ufng hoan toan, thu diiOc m gam chat ran Y va dung dich Z chufa cation k i m loai Cho Z phan ufng vdri dung dich NaOH d\i dieu kien khong c6 khong k h i , t h u di/oc 1,97 gam ket tua T Nung T khong k h i den khol Itfcfng khong doi, thu difOc 1,6 gam chat ran chi chufa mot chat nhat Gia t r i cua m la A 8,64 B 3,24 C 6,48 D 9,72 C a u 30: Chat nao sau day k h o n g tao ket tua k h i cho vao dung dich AgNOs? A H C l B K3PO4 C K B r D H N O C a u 31: Phenol phan Ofng dufoc vdri dung dich nao sau day? A NaCl B K O H C NaHCOg D H C l C a u 32: Cho cac can bang hoa hoc sau: (a) H2 (k) + h (k) < > H I (k) (c) 3H2 (k) + N2 (k) < (b) 2NO2 (k) < > N2O4 (k) > 2NH3 (k) id) 2SO2 r^;+02 (k) < > 2SO3 (k) nhiet khong ddi, k h i thay doi ap suat chung cua moi he can bang, can bang hoa hoc nao cr tren k h o n g bi chuyen dich? A (a) B (c) C (b) D (d) C a u 33: K i m loai sat tac dung vdri dung dich nao sau day tao muoi saUH)? A B HNO3 dac, nong, di/ CUSO4 C MgSOn _ D H2SO4 dac, nong, dii C a u 34: Hoa t a n hoan toan 1,805 gam hon dung dich H C l , thu diioc 1,064 l i t k h i 1,805 gam h5n hop tren bang dung dich k h i NO (san pham khtjf nhat) Biet tieu chuan K i m loai X la A A l B.Cr hop gom Fe va k i m loai X vao bkng H2 Mat khac, hoa tan hoan toan H N O loang (dif), thu dLfOc 0,896 lit cac the tich k h i deu d dieu kien C Mg D Zn C a u 35: Cac chat day nao sau day deu tao ket tua k h i cho tac dung vdi dung dich AgNOs N H dii, dun nong? A vinylaxetilen, glucozo, andehit axetic 132 B glucozcf, d i m e t y l a x e t i l e n , a n d e h i t axetic C v i n y l a x e t i l e n , glucozcf, d i m e t y l a x e t i l e n D v i n y l a x e t i l e n , glucozo, a x i t pr opion ic Cau 36: T i e n h a n h d i e n p h a n dung d i c h chijfa m g a m h o n hap C U S O v a N a C l (hieu suat 100%, dien ciic trcf, m a n g ngan xop), den k h i nifcfc b ^ t dau h i d i e n phan ca h a i dien ciic t h i ngCtog dien p h a n , t h u dJcfc dung d i c h X v a 6,72 l i t (dktc) a anot D u n g dich X hoa t a n t o i da 20,4 g a m AI2O3 G i a t r i cua m l a A 25,6 B 23,5 C , D 50,4 Cau 37: C h a t nao sau day k h i d u n n o n g vdri d u n g d i c h N a O H t h u di/oc san pham CO a n d e h i t ? A CH3-COO-C(CH3)=CH2 B C CH2=CH-COO-CH2-CH3 D C H3- C O O - C H2 - C H= C H CH3-COO-CH=CH-Cir3 Cau 38: D u n g d i c h nao sau day l a m p h e n o l p h t a l e i n d o i mau? A g l y x i n B m e t y l a m i n C a x i t axetic D alanin Cau 39: Cho 0,1 m o l t r i s t e a r i n ((Ci7H35COO)3C3H5) tac d u n g h o a n t o a n v d i dung d i c h N a O H d i i , d u n n o n g , t h u diTcfc m g a m g l i x e r o l G i a t r i cua m l a A 27,6 B 4,6 C 14,4 D 9,2 Cau 40: Day cac chat deu tac d u n g difoc vdri d u n g d i c h Ba(HC03)2 l a : A HNO3, Ca(0H)2 v a Na2S04 B HNO3, C a ( H ) v a K N O C HNO3, N a C l v a Na2S04 D N a C l , NagSO^ v a Ca(OH)2 II P H A N R I E N G (10 c a u ) Thi sink chi diic/c lam mot hai phan (Phan A hoac A T h e o chifofng t r i n h C h u a n (10 cdu, tii cau 41 den cau Phan B) 50) Cau 41: Cho X v a Y l a h a i a x i t cacboxylic m a c h h o , c6 cCing so' n g u y e n tii cacbon, t r o n g X d o n chufc, Y h a i chufc C h i a h n h o p X v a Y t h a n h h a i phan b a n g n h a u P h a n m o t tac d u n g h e t vdri N a , t h u di/gc 4,48 l i t k h i H2 (dktc) D o t chay h o a n t o a n p h a n h a i , t h u diigc 13,44 l i t k h i CO2 ( d k t c ) P h a n t r a m kho'i lufcfng cua Y t r o n g hSn h a p l a A 28,57% B 57,14% C , % D 42,86% Cau 42: D o t chay h o a n t o a n h o n h a p X g o m 0,07 m o l m o t ancol da chufc va 0,03 m o l m o t ancol k h o n g no, c6 m o t l i e n k e t d o i , m a c h h d , t h u di/gc 0,23 mol k h i CO2 v a m g a m H2O G i a t r i ciia m l a A 5,40 B 2,34 C 8,40 D 2,70 Cau 43: D a y cac chat deu c6 M f e n a n g t h a m g i a p h a n ufng t h i j y p h a n t r o n g dung d i c h H2SO4 d u n n o n g l a : A fructoza, saccarozo v a t i n h hot B saccaroza, t i n h b o t v a x e n l u l o z a C glucoza, saccaroza v a fructoza D glucoza, t i n h hot v a x e n l u l o z a 133 C a u 44: Cho cac cap oxi hoa - khuf di/cJc sSp xep theo thiif t i i tSng dan t i n h oxi hoa ciia cac ion kirn loai: AI^VAl; Fe^^/Fe, Sn^VSn; Cu^VCu Tien hanh cac t h i nghiem sau: (a) Cho sat vao dung dich dongdl) sunfat (b) Cho dong vao dung dich nhom sunfat (c) Cho thiec vao dung dich dongdl) sunfat (d) Cho thiec vao dung dich sat(II) sunfat Trong cac t h i nghiem tren, nhijfng t h i nghiem c6 xay phan I'tag la: A (b) va (c) B (a) va (c) C (a) va (b) D (b) va (d) C a u 45: Cho cac phat bieu sau: (a) Trong bang tuan hoan cac nguyen to hoa hoc, crom thuoc chu k i 4, nhom VIB (b) Cac oxit cija crom deu la oxit bazcf (c) Trong cac hop chat, so' oxi hoa cao nhat ciia crom \k +6 fd) Trong cac phan Ofng hoa hoc, hop chat crom(III) chi dong vai tro chat oxi hoa (e) K h i phan iJfng vdi k h i CI2 dii, crom tao hcrp chat crom(III) Trong cac phat bieu tren, nhuTng phat bieu diing la: A (a), (b) va (e) B (a), (c) va (e) C (b), (d) va (e) D (b), (c) va (e) C a u 46: T h i nghiem vdi dung dich H N O thi/dng sinh k h i doc NO2 De han che k h i NO2 thoat t i / ong nghiem, ngifcfi ta nut dng nghiem bkng: (d) bong c6 t a m giam an (c) bong CO t a m niidrc voi (b) bong c6 tarn nifdrc (a) bong kho Trong bien phap t r e n , bien phap c6 hieu qua nhat la A (d) B (c) C.(a) D (b) C a u 47: H n hop X gom H2, C2H4 va CgHe c6 t i khoi so vdri H2 la 9,25 Cho 22,4 l i t X (dktc) vao b i n h k i n c6 sSn mot i t hot N i Dun nong binh mot thcfi gian, t h u dtfcfc hon hgp k h i Y c6 t i khoi so vdri H2 b&ng 10 Tong so' mol H2 da phan ijfng la A 0,070 mol B 0,015 mol C 0,075 mol D 0,050 mol C a u 48: Trong cap dung dich CH3-CH2-NH2, H2N-CH2-COOH, HzN-CHz-CHCNHg)COOH, HOOC-CH2-CH2-CH(NH2)-COOH, so dung dich lam xanh quy t i m la A.4 B.l C C a u 49: Cho phucfng t r i n h phan ilng aAl +bHN03 D.3 > cAl(N03)3 + dNO + eHzO T i le a : b la A : B : C : D : C a u 50: Cho 25,5 gam h6n hop X gom CuO va AI2O3 tan hoan toan dung dich H2SO4 loang, t h u diicfc dung dich chufa 57,9 gam muoi Phan trSm khoi lifgfng cua AI2O3 X la A 40% B 60% C 20% D 80% 134 B Theo chitofng t r i n h N a n g cao (20 cdu, ti£ cdu 51 den cdu 60) Cau 51: Cho phtfcfng t r i n h phan ufng: aFeS04 + bKzCraOv + CH2SO4 > dFe2(S04)3 +6X2804 + {Cr2(S04)3 + gHgO Ty le a:b la A.3:2 C 1:6 B 2:3 D 6:1 Cau 52: Cho cac phat bieu sau: (a) De xuf ly thiiy ngan rai vai, ngifcfi ta c6 the dung bot liAi huynh (b) K h i thoat vao k h i quyen , freon pha huy tan ozon (c) Trong k h i quyen, n6ng CO2 vifot qua tieu chuan cho phep gay hieu ufng nha k i n h (d) Trong k h i quyen , nong NO2 va SO2 viiOt qud tieu chuan cho phep gay hien ti/gng miTa axit Trong cac phat bieu tren , so phdt bieu dung la: A.2 B C D Cau 53: Cho cac phat bieu sau: (a) Glucozo CO kha nang tham gia phan uTng t r ^ n g bac (b) Sir chuyen hoa t i n h bot co the ngifdi c6 sinh mantozO (c) Mantorazo c6 kha nang tham gia phan ufng trang bac (d) Saccarozo difgc cau tao iii hai go'c p-glucozo va a-fructozcf Trong cac phat bieu t r e n , so phat bieu dung la: A.3 B C D Cau 54: Cho 13,6 gam mot chat hCfu ccf X (c6 phan nguyen to C, H , O) tac dung viTa dii vdi dung dich chijfa 0,6 mol AgNOs NH3, dun nong , thu diioc 43,2 gam Ag Cong thijfc cau tao cua X la : A C H j - C ^ C - C H O B CH, = C = C H - C H O C CH = C - C H , - C H O D C H S C - [ C H J -CHO Cau 55: Peptit X bi thiiy phan theo phiiomg t r i n h phan ilng X + 2H2O -> 2Y + Z (trong Y va Z la cac amino axit) Thuy phan hoan toan 4,06 gam X thu diicfc m gam Z Dot chay hoan toan m gam Z can vi/a dii 1,68 l i t k h i O2 (dktc), thu difcfc 2,64 gam CO2; 1,26 gam H2O va 224 m l k h i N2 (dktc) Biet Z c6 cong thufc phan tijf trung vdi cong thufc don gian nhat Ten goi ciia Y la A glyxin B lysin C axit glutamic D alanin Cau 58: Trifdng hop nao sau day k h o n g xay phan ufng? (a) CH, = C H - C H - C + H , ^ ^ (b) C H ^ - C H ^ - C H ^ - C l + H p > 135 (c) C ^ H , - C ! + N a O H ( d a c ) (d) C ^ H j - C l + A (a) i cao.p cao ^ v d i (CeHs- l a go'c p h e n y l ) N a O H ^ ^ B (c) C (d) D (b) C a u 57: Trtfcfng h o p nao sau day, k i m l o a i b i a n m n d i e n h o a hoc? A D o t day s&t t r o n g k h i o x i k h o B T h e p cacbon de t r o n g k h o n g k h i a m C K i m l o a i k e m t r o n g d u n g d i c h H C l D K i m l o a i sSt t r o n g d u n g d i c h HNO3 l o a n g Cau 58: C h o 12 g a m h g p k i m ciia bac v a o d u n g d i c h HNO3 l o a n g (dii), dun n o n g d e n p h a n ufng h o a n t o a n , t h u di/oc d u n g d i c h c6 8,5 g a m AgNOa P h a n B 30% A 65% cua bac t r o n g m a u h o p k i m l a tram khoi iLfgrng C a u 59: Cho so p h a n ufng C r - C 55% +Cl.dif ^V D 45% +dungj|chNa()H.dir C h a t Y t r o n g so t r e n l a A NaaCraOv C C r ( H ) B Cr(OH)2 D Na[Cr(0H)4] C a u 60: H n h o p X g o m ancol m e t y l i c , ancol e t y l i c v a g l i x e r o l D o t chay h o a n t o a n m g a m X, t h u dLfgc 15,68 l i t k h i CO2 ( d k t c ) v a 18 g a m H2O M a t khac, 80 g a m X h o a t a n duac to'i d a 29,4 g a m Cu(0H)2 P h a n t r a m k h o i lifgng cua ancol e t y l i c t r o n g X l a A 46% B 16% C 23% DAP 54C 53A 52C ID 45B 44B 43B 42A 41D 35A 34A 33A 32A 31B 25C 24C 23C 22D 21C 15D 14D 13B 12A IIB 5A 4A 3D 2C IC D 8% AN 6B 55A 58D 57B 56D 49D 48C 47C 46B 40A 39D 38B 37B 36C 30D 29A 28D 27B 26C 20B 19B 18D 17A 16C lOA 9A 8B YD 59D 50C 60C BAI GIAI C H I T I E T C a u 1: T a c6 so m o l Fe203 = 0,1 m o l V i p h a n t a c d u n g vdri d u n g d i c h N a O H dii g i a i p h o n g H2 chufng t o Fe203 da p h a n ufng h e t t h e o p h a n ufng: 2A1 0,2mol > AI2O3 + 2Fe + Fe203 0,1 m o l 0,2mol 92 V a y h n h o p sau n h i e t n h o m g o m A l d i i v a (— 1-0,2) = 0,27 m o l F e 56 136 Theo de, F e d m o i p h a n g i a i p h o n g (4a - a) = a m o l H2 n e n — = 3a o a = , (mol) Do a = 0,045 ( m o l ) n e n A l d i i a m o i p h a u = ^ ' - 0,03 mol Vay m = 27(0,2 + 2.0,03) - 7,02 g a m Vay chon C Cau 2: L o a i A , C, D v i H2SO4 l o a n g k h o n g t h e o x i h o a F e ( I I ) t h a n h F e ( I I I ) Vay chon C Cau 3: T a c6 n x = m o l v a ny = 26.0,3.^ + 2.0,65 = 0,65 m o l Do so' m o l H2 da p h a n ufng = - 0,65 = 0,35 m o l Chu y r a n g C2H2 d i i , m o l n e n x e m n h i / c h i c6 (0,35 - 0,10) = 0,25 mol C2H2 tac d u n g vdi H2 Vi 0,25 m o l C2H2 c6 k h a n a n g t a c d u n g to'i d a vdri 0,5 m o l H2 n e n h o n h o p Y CO k h a n a n g t a c d u n g t o i d a v(5i (0,5 - 0,35) = 0,15 m o l H , tufc cung 0,15 mol Br2 V a y c h o n D Cau 4: T r o n g p h a n ijfng C + CO2 > C t h i C c6 so' o x i h o a t a n g sau p h a n iJng n e n C d o n g v a i t r o c h a t khijf V a y c h o n A Cau 5: TCf p h u o n g t r i n h CeHizOe > 2C2H5OH + 2CO2 Ta tha'y cuf 180 g a m glucoza se t a o 0 g a m ke't t u a , do m = 15.180.100 200.90 = 15,0 V a y c h o n A Cau 6: G o i a, b l a so' m o l B a v a A l t r o n g X C h i i y r k n g cr t h i n g h i e m v 2HC00H + HOCH2CHO V a y chon C C a u 32: T a c6 n , , „ = 14,4 = 0,09 m o l n e n n , „ = 60 '"^'"^ 0,09.2 = 0,06 m o l C h i i y r a n g p h a n tuf Fe304 d a cho le, p h a n t t f K2Cr207 d a n h a n 6e, ta CO 0,06 n K,Cr,0- = 0,01 m o l Vay V = 0,01 l i t V a y chon D C a u 33: Theo de, A l a CsHgCOOCH^CHs v a B l a CH3CHO V a y c h o n B C a u 34: Goi a, b I a n liiot la so m o l CH3OH b i oxi hoa t h a n h HCHO v a HCOOH Chu y r a n g nHciio = a; nHcooH = b ; n , ^ ^ = a + b ; n^,^^,OH(cli/) = c , t a 4a + 2b: b = ^ 22,4 he: 12,96 108 b a+ b c —+ +— CO = u, 1/ 0,56 22,4 a = 0,025 = 0,025 « ^ b = 0,01 c = 0,005 = 0,01 247 Vay % C H H b i oxi h o a = (^ + b)^00 =87,5(%) Vay c h o n C a+b+ c C a u 35: Do am dien cua C < N < O < F nen muTc phan ciTc hen ket giOra cac nguyen tuT phan tijT CH4 < NH3 < H2O < H F Vay chon B C a u 36: Ham lifcfng H2S khong k h i = ^^^^^^-^^ 3.239 C a u 37: V i X chay cho : n^^ : 0,017 mg/1 Vay chon D ^ = 1,5 : : nen X c6 dang (CHO)n 230 Ta phai c6 n chSn va n < = 1,9 29 Nhitag n = hoac n = deu khong c6 chat X thoa de b ^ i Vay n = 6, tuTc X l a CeHeOe ufng \6i cong thijfc cau tao CeCOIDe Vay chon D C a u 38: Theo de a, b, c cung Ian lutft la so m o l k h i tren c6 m o l h o n hcfp Suy X = z = 44a , X 44 z va t = — = c 44a + 28b + 2c 44a + 28b + 2c va k = — = a 44a + 28b + 2c 44a + 28b + 2c 2c N h i m g < 44a + 28b + 2c < 44 nen k > 1, t < Vay chon C C a u 39: De y r k n g hSn hcfp X c6: nc = 0,25 mol; na - 0,4 mol; no = , - ( , + 0,4.1) ^ ^^^^^^ 16 Vay nn = no tufc h6n hop X phai c6 tong so H = tong so O Do X gom axit la HCOOH va CH2(COOH)2 Vay chon D C a u 40: Cong hoa t r i va so' oxi hoa cua N cac hcfp chat NH3; NH4CI va HNO3 Ian luot la 3; 4; va -3; -3; +5 Vay chon D C a u 41: Dung nxS&t NH3 cho vao mau MSu tao ket tua t r i n g ben la AICI3; MSu tao ket tiia t r ^ g roi tan la ZnCl2; MSu tao ket tiia xanh roi tan la CUCI2; Mau tao sir phan Idrp la C6H5NH3CI (do xuat hien a n i l i n khong tan); M^u tao dung dich dong nhat la NaCl Vay chon C C a u 42: Chii y rang chifa biet v i t r i ciia M day dien hoa nen can xet hai trifcfng hcfp: + M diing sau Al day dien hoa Goi a la so' mol moi chat da dung Chu y 4,8 gam r d n sau phan ijfng gom a mol Fe va a mol M , ta c6 he: 248 56a + a M = 4,8 2a + 2a = 0,2 i a = 0,05 M = 40 Vay M la Ca Nhifng Ca dufng trLfdrc A l day dien hoa, loai + M diing tricac Al day dien hoa Goi a la so' mol moi chat da dung Chu y 4,8 gam ran sau phan lifng gom a mol Fe va a mol MO, t a c6 he: 56a + a ( M + 16) = 4,8 2a + 2a = 0,2 Vay M la M g , phu hop v i M g duTng trade A l day dien hoa Vay chon B C a u 43: Ta phai chon nhuTng chat la muoi amoni hoac muo'i metylamoni; etylamoni Vay CO chat thoa de bai la: C2H5COONH4 : amoni propionat C H C O O N H C H : metylamoni axetat H C O O N H C H : etylamoni fomat HCOONH2(CH3)2: dimetylamoni fomat Vay chon B C a u 44: ^ 42,65 58,5.0,3 ^ 89(alanin) Vay chon B C a u 45: De dang t i m dtfoc n^^.^ = n^.^, = 0,1 mol Dat cong thufc X la FexOy (a mol) Bao toan electron cho he: a(56x + 16y) = 26 a(3x - 2y) = 1.0,1 + 2.0,1) ax = 0,355 [ay = 0,3825 Bao toan Fe cho npc = ax = 0,355 mol nen m = 0,355.56 = 19,88 gam Vay chon A LUu y Co the dung cong thiic gidi nhanh sau: 249 Litu y Khi ddn qua bot Ni nung lam mat cong mdu hdn hap X goni tiong niiac thiic phdn de phdn brom) anken CnH2n ling xdy hodn c6 phdn tii khoi va H2 cd phdn tii khoi todn dilgc hdn hap Y Id M2 thi anken la Mj (khong CnH2n can tim c6 tii Id: n = (M^-2)M, 14(M -M,) T a se chuTng m i n h cong thijfc t r e n X e t m o l h n h o p Y K h o i Itfofng m o l h o n h o p Y l a M g a m Theo d i n h , , l u a t bao t o a n k h o i l i f o n g t a c6 m x = m y = M g a m n e n so' m o l h n h o p X - M, —M, M Theo q u i tfic h i e u so' m o l t h i so' m o l H da p h a n ufng vori a n k e n = — ^ - M, Do H dii sau p h a n ufng n e n t a p h a i c6 so' m o l a n k e n t r o n g X = so m o l H M- da p h a n ufng = M, C H ; :(—^-l)mol " M, Vely X g o m : H , :1 mol Do kho'i l a g n g hSn h o p X l a M g a m n e n : M, 14n(M - M , ) n ( — ^ - ) + = M —^ M, M, ^ n = (M -2)M, 14(M, - M , ) ^ ^ 95 232 100 100 , , C a u 47: T a co mq^ang manheut = 0 — ——- — — t a n Vay chon B 100 168 80 99 Cau 48: Ta Vay m - c6 U x a o n = 0,4 147(n - n H mol; n^j^ij^^^^ = , mol ) = 147(0,85 - 0,8) = 7,35 g a m V a y c h o n D OH C a u 49: V i H S O dac c6 t i n h o x i hoa m a n h n e n H2SO4 dac k h o n g di/oc dung de l a m k h o cac c h a t c6 t i n h khuf m a n h V a y cac k h i c6 t h e l a m k h o d d a y l a H C l , C I v a C O V a y c h o n A C a u 50: A n k i n C„H2n c6 ( n - 2) n g u y e n tOf I I NhOfng n g u y e n t t f H n a y l i e n k e t triTc t i e p vdri cac n g u y e n t t f C b a n g (2n - 2) l i e n k e t a M a t k h a c giOfa n n g u y e n tuf C c6 ( n - 1) l i e n k e t a 250 V a y t o n g so' l i e n k e t a t r o n g a n k i n CnH2n - l a ( n - + n - l ) = 3n-3 Vay chon B Litu y Co the gidi so lien ket "meo" bdng each viet cong thiic cdu tao mot ankin a cua ankin a va so C cila bat ky roi dem he giiia so lien ket ankin Vi du xet ankin H - C = C - H ta thay n = thi so so lien bieu thiic so lien ket Trong se tim dugc bieu thtic lien a = 3n - Id phii triCang hap c6 phiiang khdc nUa se c6 ket qua can kit a = nen hap an trd lai ciing thoa thi xet them mot ankin tim C a u 51: Cac h n hcfp (cung so' m o l ) c6 t h e t a n h o a n t o a n t r o n g nifdrc (dtf) c h i tao r a d u n g d i c h l a : Na20 v a A l ; B a O v a Z n O V a y c h o n - B C a u 52: T a c6 so' m o l ancol = I a n so' m o l H2 = 0,1 m o l Vay CO M l + Mh6n hop ancol = 39 V i ancol tao t h a n h c6 so m o l b a n g n h a u n e n t a p h a i M2 = 39.2 = 78 Do ancol l a C H O H v a C2H5OH v a este p h a i l a C3H6O2 V a y c h o n A Cau 53: N e u 0,1 m o l C u ( N ) p h a n ufng h e t se tao m Y > g a m T h e o de t h i m y = 6,4 g a m chu'ng to C u ( N ) d i i V a y A l v a Z n da p h a n ufng h e t Goi a, b , c I a n lufot l a so' m o l A l , Z n v a Cu b a n dau Chu y r a n g r f i n Y c h i la CuO v a r a n Z c h i l a AI2O3, t a c6 h e : 27a + 65b + 64c = 3,76 l,5a + b + c = — = 80 a _ 1,02 ~ 102 a = 0,02 0,08o b = 0,02 c = 0,03 = 0,01 Vay % m z n = 3,76 = , ( % ) Vay chon C C a u 54: D a t cong thufc t r u n g b i n h a n d e h i t l a CxHyO v a a l a so m o l a n d e h i t , t a CO he: a(12x + y + 16) = l l , ax = ay 24,64 44 7,2 18 = 0,56 0,24 chdfng to c6 mot andhit la HCHO 108 Goi b, c Ian Ixiat la so mol H C H O va andehit c6n l a i , ta c6 he: M = 56 4b + 2c = 0,64 c = 0,16 b + c = 0,24 b = 0,08 b + M c = 11,36 Vay h6n hap gom H C H O va C H = C H - C H Vay chon C C a u 55: V i X chay cho so' mol X bkng hieu so' mol CO2 v^ so mol H2O nSn cong thuTc X CO dang CnH2n Vi X trang giXOng 20k cho A g theo t i le mol : n e n X la andehit n h i chufc Vay cong thufc X c6 dang CnH2n 2O2 Do Mx = 100 la phu hgp Vay chon B C a u 56: So' dung dich lam mat mau nirdc brom la fomon; glucozcf; mantozcf; axit fomic va axetandehit Vay chon D Litu y Phenylamoni anilin C6H5NH2 du mot doi electron veto vdng electron benzen, clorua CeH^NHsCl cho dugc phdn khong lam mat iCng nay, vi nguyen chUa tham gia lien nguyen tii N da tham gia phoi mau nUdc tvC N ket nen c6 the tham gia lien phenylamoni clorua khong brom anilin hap doi tri vai ion H* C a u 57: Ta c6 so mol CO2 = so mol SO2 - 0,1 mol Chu y r k n g 24 gam r&n Z la 0,15 mol Fe203, sijf dung bao toan nguyen to' va bao toan kho'i liicfng ta c6: 35,4 + 0,5.98 = 152.0,3 + 160(0,5 - 0,3 - 0,1) + 0,2.54 + 18.0,5 + m O m =r gam Vay chon C C a u 58: Goi a, b Ian Itfot la so' mol C va S c6 mau t h a n tren Ta 12a + 32b = CO NhLftig 12(a + b) < 12a + 32b < 32(a + b) — >a + b> 12 32 Nen 12(a + b) < < 32(a + b) o 0,25 > a + b > 0,09375 252 o 0,25 > 0,75 h6ng * D T : ( 3 ) 9 - 04 L y Thai T o - T P D a N i n g *DT: 0511.3823421 - 259 Le Duan - T P Vinh - D T : 0383.554777 - X V i ^ t N g h T T n h - C a n T h d - D T : ( O) 8 - TTnh 10 - TT.CCi C h i - T P H C M *DT: (08) 37924216 xac ^nh sach chinh phamTX , chiing toi in chim of b'la va chih ) - 51 L y T h U d n g K i e l - T P D o n g H i - Q B * D T : (0523) 8 - 66 L y T h a i T o - T h j x a Q u a n g Trj ISBN: 978-604-934-832-7 - L e T h a i T o - VTnh L o n g - D T : 9 - H a n T h u y e n - T P H u e - 78 B a c h D i n g - Da N i n g - D T : (0511) 3834328 - V T h i S a u , / T o n DiJfc T h i n g - L o n g X u y e n 935092 Gia: 754802 55.000d - y y ... A 24 B 20 C 36 D 18 154 DAP AN lA 2D 3A 4A 5A 6B 7B 8D 9B lOB lie 12B 13A 14A 15D 16B 17B 18D 19B 20 A 21 D 22 B 23 C 24 B 25 C 26 C 27 C 28 D 29 D 30C 31A 32D 33A 34C 35D 36B 37C 38C 39A 40B 41D 42B... 45B 44B 43B 42A 41D 35A 34A 33A 32A 31B 25 C 24 C 23 C 22 D 21 C 15D 14D 13B 12A IIB 5A 4A 3D 2C IC D 8% AN 6B 55A 58D 57B 56D 49D 48C 47C 46B 40A 39D 38B 37B 36C 30D 29 A 28 D 27 B 26 C 20 B 19B 18D 17A... 77,6 B 83 ,2 C 87,4 D 73,4 129 C a u 14: O t r a n g t h a i ccf b a n , cau h i n h electron cua n g u y e n tuf N a ( Z = 11) l a A Is22s22p'^3s- B Is22s22p''3s' C ls^2s"2p''3s^ D Is^2s22p'=3s^