Khám phá kỹ thuật giải quyết nhanh gọn đề thi Đại học môn Hóa học: Phần 2

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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.

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Nội dung Text: Khám phá kỹ thuật giải quyết nhanh gọn đề thi Đại học môn Hóa học: Phần 2

O CHl/dNG VI CfiC DE THi t>6l HOC THfiM KH^O DE SO 1 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 1 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 do cao trong dieu kien khong c6 khong k h i , thu diioc h6n hop chat ran Y. Chia Y thanh hai phan bang nhau. 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 ra 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 + 2 F e ^ Fe2(S04)3 + 3SO2 + 6H2O Trong cac phan ufng t r e n , phan ufng xay ra 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 8. Sue X vao li/gng diT dung dich AgNOs trong N H 3 den phan ufng hoan toan, t h u dtfgc h6n hgp k h i Y va 24 gam ket tua. H 5 n hgp k h i Y phan ufng vCra du vdi bao nhieu mol Br2 trong 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: (a) 2C + Ca CaC2 (b) C + 2H2 CH4 (c) C + CO2 -> 2 C 0 (d) 3C + 4A1 -> A I 4 C 3 128
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 thanh ancol etyhc (hieu suat phan ufng bang 90%). Hap thu hoan toan liiong k h i CO2 sinh ra 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 6 n hop X gom Ba va A l . Cho m gam X vao nvidc dii, sau k h i cac phan ijfng xay ra 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 3 C H O phan ufng hoan toan vdi li/ong dtf dung dich AgNOa trong 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 ra 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 trong X va hai k i m loai trong 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 do 0,4M tac dung viTa dij vdi 80 m l dung dich NaOH 0,5M, thu ducfc dung dich chufa 5 gam muoi. Cong thufc cua X la A. NH2C3H6COOH B. NH2C3H5(COOH)2 C. (NH2)2C4H7COOH D. N H 2 C 2 H 4 C O O H Cau 10: Cho 1,37 gam Ba vao 1 l i t dung dich CUSO4 0,01 M . Sau k h i cac phan ufng xay ra 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 1 : 1 , 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 trong 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 ra hoan to^n, thu di/gc dung dich X. Khoi liiong muo'i trong 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 Gly- Ala-Gly-Glu. Thijy phan hoan toan m gam h 6 n hop gom X va Y t h u difoc 4 amino axit, trong do 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 2 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? A . N a , N a C l , CuO B. N a , CuO, H C l C. N a O H , N a , CaCOg D . N a O H , Cu, N a C l 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 A. 2,2,4-trimxetylpentan B. 2 , 2 , 4 , 4 - t e t r a m e t y l b u t a n C. 2 , 4 , 4 , 4 - t e t r a m e t y l b u t a n D. 2,4,4-trimetylpentan C a u 18: T o n i l o n - 6 , 6 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 do 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 2 0 % 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 do 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 do 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 trong CCI4. (d) Cho dung dich glucozo vao dung dich AgNOa, trong N H 3 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 ra phan ufng oxi hoa - khijf la A. 5. B. 2 C. 3 D. 4 Cau 23: Cho so do cac phan ijfng: X + NaOH - ^ - ^ Y + Z X + NaOH (ran) > T + P T ) Q + H2 Q + H2O — > Z Trong so do tren, X va Z Ian liicft la A. HCOOCH-CHa va HCHO B. CH3COOC2H5 va C H 3 C H O C. C H 3 C O O C H - C H 2 va C H 3 C H O D. C H 3 C O O C H = C H 2 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. 3 B. 5 C. 4 D. 2 Cau 25: Cho m gam Fe vao b i n h chiJa dung dich gom H2SO4 va H N O 3 , 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 trong ca hai tri/cfng hop NO la san phan khijf duy nhat, do d dieu kien tieu chuan. Dung dich Y hoa tan vifa het 2,08 gam Cu (khong tao thanh san pham khiif ciia N*^). Biet cac phan uTng deu xay ra 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 trong 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 ra phan ijfng la A. 6 B. 5 C. 4 D. 3 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 3 cation k i m loai. Cho Z phan ufng vdri dung dich NaOH d\i trong dieu kien khong c6 khong k h i , t h u di/oc 1,97 gam ket tua T. Nung T trong khong k h i den khol Itfcfng khong doi, thu difOc 1,6 gam chat ran chi chufa mot chat duy 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 3 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) < > 2 H I (k). (b) 2NO2 (k) < > N2O4 (k). (c) 3H2 (k) + N2 (k) < > 2NH3 (k). id) 2SO2 r^;+02 (k) < > 2SO3 (k). 0 nhiet do 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 ra muoi saUH)? A. CUSO4. B. HNO3 dac, nong, di/. C. MgSOn. _ D. H2SO4 dac, nong, dii. C a u 34: Hoa t a n hoan toan 1,805 gam hon hop gom Fe va k i m loai X vao bkng dung dich H C l , thu diioc 1,064 l i t k h i H2. Mat khac, hoa tan hoan toan 1,805 gam h5n hop tren bang dung dich H N O 3 loang (dif), thu dLfOc 0,896 lit k h i NO (san pham khtjf duy nhat). Biet cac the tich k h i deu do d dieu kien tieu chuan. K i m loai X la A. A l . B.Cr. C. Mg. D. Zn. C a u 35: Cac chat trong day nao sau day deu tao ket tua k h i cho tac dung vdi dung dich AgNOs trong N H 3 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 4 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 0 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 khi (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. 5 1 , 1 . 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. CH3-COO-CH=CH-Cir3. C. CH2=CH-COO-CH2-CH3. D . C H3- C O O - C H2 - C H= C H 2. 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 ( 0 H ) 2 v a K N O 3 . 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 trong hai phan (Phan A hoac Phan B) A. T h e o chifofng t r i n h C h u a n (10 cdu, tii cau 41 den cau 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 do X d o n chufc, Y h a i chufc. C h i a h 5 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. 8 5 , 7 1 % 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 ra 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 ra 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 3 thi/dng sinh ra k h i doc NO2. De han che k h i NO2 thoat ra t i / ong nghiem, ngifcfi ta nut dng nghiem bkng: (a) bong kho. (b) bong c6 tarn nifdrc. (c) bong CO t a m niidrc voi. (d) bong c6 t a m giam an. Trong 4 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 6 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. 2 D.3 C a u 49: Cho phucfng t r i n h phan ilng aAl +bHN03 > cAl(N03)3 + dNO + eHzO. T i le a : b la A. 1 : 3 B. 2 : 3 C. 2 : 5 D. 1 : 4 C a u 50: Cho 25,5 gam h6n hop X gom CuO va AI2O3 tan hoan toan trong dung dich H2SO4 loang, t h u diicfc dung dich chufa 57,9 gam muoi. Phan trSm khoi lifgfng cua AI2O3 trong 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 B 2:3 C. 1:6 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 do CO2 vifot qua tieu chuan cho phep gay ra hieu ufng nha k i n h . (d) Trong k h i quyen , nong do NO2 va SO2 viiOt qud tieu chuan cho phep gay ra hien ti/gng miTa axit Trong cac phat bieu tren , so phdt bieu dung la: A.2 B. 3 C. 4 D. 1 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 trong co the ngifdi c6 sinh ra 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. 4 C. 2 D. 1 Cau 54: Cho 13,6 gam mot chat hCfu ccf X (c6 thanh phan nguyen to C, H , O) tac dung viTa dii vdi dung dich chijfa 0,6 mol AgNOs trong 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 do 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 ra phan ufng? (a) CH, = C H - C H 2 - C 1 + H , 0 ^ ^ (b) C H ^ - C H ^ - C H ^ - C l + H p > 135
i cao.p cao (c) C ^ H , - C ! + N a O H ( d a c ) - ^ v d i (CeHs- l a go'c p h e n y l ) (d) C ^ H j - C l + N a O H ^ ^ A. (a) 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 6 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 tram khoi iLfgrng cua bac t r o n g m a u h o p k i m l a A. 65% B . 30% C. 55% D . 45% +Cl.dif ^V +dungj|chNa()H.dir C a u 59: Cho so do p h a n ufng C r - C h a t Y t r o n g so do t r e n l a A. NaaCraOv B . Cr(OH)2 C. C r ( 0 H ) 3 D. Na[Cr(0H)4] C a u 60: H 6 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% D . 8% DAP AN IC 2C 3D 4A 5A 6B YD 8B 9A lOA IIB 12A 13B 14D 15D 16C 17A 18D 19B 20B 21C 22D 23C 24C 25C 26C 27B 28D 29A 30D 31B 32A 33A 34A 35A 36C 37B 38B 39D 40A 41D 42A 43B 44B 45B 46B 47C 48C 49D 50C 5 ID 52C-- 53A 54C 55A 56D 57B 58D 59D 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 2 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 + Fe203 > AI2O3 + 2Fe 0,2mol 0,1 m o l 0,2mol 3 92 V a y h 6 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) = 3 a m o l H2 n e n — = 3a o a = 0 , 0 4 5 (mol) 2 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 26.0,3.^ + 2.0,65 Cau 3: T a c6 n x = 1 m o l v a ny = = 0,65 m o l Do do so' m o l H2 da p h a n ufng = 1 - 0,65 = 0,35 m o l Chu y r a n g do C2H2 con d i i 0 , 1 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 con 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 2 , 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 > 2 C 0 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 2 0 0 g a m ke't t u a , do do 15.180.100 m = = 15,0. V a y c h o n A 200.90 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
T a p h a i c6 n,nu6i - n^aon = 0,04 m o l n e n Mn = 125. 0,04 V a y M x = 125 - 22 = 103, tufc X l a NHaCgHsCOOH. Vay chon A C a u 10: T h e o de t a c6: n , =0,01mol n^^^ =0,02mol n -0,01mol n^^., = 0 , 0 1 m o l Vay nikettua = m,,__^„^ +mcu,ou,, = 233.0,01 + 98.0,01 = 3,31 (gam). V a y chon A C a u 11: P e n t a n CH3CH2CH2CH2CH3 tao difcrc 3 dfin x u a t monoclo l a : CH3CH2CH2CH2CH2CI; CH3CH2CH2CHCICH3 v a CH3CH2CHCICH2CH3 Vay chon B C a u 12: T a c6 so m o l P - so m o l H3PO4 = 0, 1 m o l ; So m o l N a O H = 0,2 m o l . V a y t h u ducfc d u y n h a t 0 ,1 m o l Na2HP04 tufc 14,2 g a m m u o i Vay c h o n A C a u 13: G o i x , y I a n lifcft l a so m o l X , Y 2x + 2y = ^ = 0,4 x = 0,12 Ta CO h e : y = 0,08 2 x . y = ^ . 0 , 3 2 89 V a y m = m x + m y = 472.0,12 + 332.0,08 = 83,2 (gam). V a y c h o n B C a u 14: (J t r a n g t h a i co b a n , cau h i n h e l e c t r o n ciia n g u y e n tuf N a ( Z = 11) la ls22s'2p^3s\y c h o n D C a u 15: G o i cong thijfc a x i t n o l a CnH2n02; cong thufc t r u n g b i n h 2 a x i t chUa no l a C „ H 2 „ , 2O2. G o i a l a so m o l a x i t n o ; b l a so m o l 2 a x i t chi/a no. a+b=a3 a = 0,15 T a c o he: i a ( 1 4 n + 54) + b(14m + 52) = 25,56 « < b = 0,15 4 4 ( n a + b m ) + 18(an + b m - b ) = 4 0 , 0 8 an + b m = 0 , 6 9 R u t r a n + m = 4,6. C h u y r a n g t a p h a i c6 m ^ 3 n e n c h i c6 n = 1; m = 3,6gam l a p h u h o p . Vay m2axUkh6ngno = b ( 1 4 m + 30) = 0,15(14.3,6 + 30) = 12,06 g a m . Vay chon D 138
Cau 16: D u n g d i c h C H 3 C O O H tac d u n g diidc vdri N a O H ; N a v a CaCOa. Vay chon C Cau 17: T e n t h a y t h e (theo l U P A C ) cua (CH3)3C-CH2-CH(CH3)2 l a 2,2,4-trimetylpentan Vay chon A Cau 18: T o n i l o n - 6 , 6 l a san p h a m t r u n g n g t f n g cua a x i t a d i p i c va hexametylendiamin. Vay chon D Cau 19: G o i a, b , c, d I a n lifot l a so' m o l ciia N a ; N a 2 0 ; B a v a B a O . 23a + 6 2 b + 137c + 153d = 21,9 20 52 Theo de t a c6 he: < 171 a 1,12 - + c = = 0,05 2 22,4 23(a + 2c) + 91(c + d ) + 6 2 ( b + d ) = 21,9 a + 2c = 0 , l 23(a + 2 b ) + 137(c + d ) + 16(b + d ) = 21,9 c + d = 0,12 c + d = 0,12 b + d = 0,14 a + 2c = 0 , l a + 2 b = 0,14 Vay nMaOH = a + 2b = 0,14 mol Suy r a d u n g d i c h sau hoa t a n r a n X c6 chufa 0,12 m o l Ba^"^ v a (0,14 + 2.0,12) = 0,38 m o l O H ' Theo de n,,„ = 0 , 3 m o l n e n n , =n - n „ , = 0,38 - 0,3 = 0,08 m o l .
' . n 0 38 O day dieu kien nay thoa vi " " = —— = 1,26. CO, 0,3 C a u 20: Goi cong thufc X la CxHyO^ (a mol), ta c6 phan ufng chay: CxHyO, + (x + y^ ~ - ) 0 , ^ XCO2 + ^ H , 0 a(x + ^ - - ) = 0,35 4 2 ax = 0.35 Ta CO he: a(12x + y + 16z) = 6,9 ay = 0,3 ax = 0,35 a/. = 0 , i 5 Rut ra X : y : z = 7 : 6 : 3. Vay X c6 cong thufc C7H6O3. 6,9 So mol X = = 0,05 mol. 138 Goi b la so mol NaOH da phan ufng vdi X, ta c6 b + 0,2b = 0,18 b = 0,15. Do nx : nxaOH = 1 : 3 nen X phai c6 cau tao H C O O C 6 H 4 O H . Phan ufng xay ra: HCOOC6H4OH + 3NaOH > HCOONa + C6H4(ONa)2 + 2H2O 0,05mol 0,lmol Bao toan khoi liTdng cho m^n = 6,9 + 40.0,18 - 0,1.18 = 12,3 (gam). Vay chon B C a u 21: Dat cong thuc X la CxHt02; Y la CxHgx + 20k Goi cong thufc trung binh cua h6n hgrp la CxHyOj, (a mol), ta c6 phan uftig chay: CxHyO, + (x + ^ - - ) 0 XCO2+ ^ H , 0 0 , 4 ( x + ^ - - ) = l,35 4 2 x =3 Ta CO he: 0,4x = l , 2 y = 5,5 z=2 0 , 4 ^ = 0,55 2 Do so O t r u n g b i n h = 2 ma X c6 2 0 nen Y cung c6 2 0 . Vay Y 1^ C3H8O2 Do so H trung b i n h = 5,5 ma Y c6 8 H nen ta phai c6 t < 5,5. Vay Y la C3H4O2 hoac C3H2O2 Goi a, b la so mol X, Y trong 0,4 mol hon hop. 140
a + b = 0,4 a = Neu chon Y la C 3 H 2 O 2 , ta c6 he: 30 2a + 8b = 5,5 7_ 0,4 b = 30 (loai v i naxit < Hancol) a + b = 0,4 a = 0,25 Vay Y la C3H4O2 va ta c6 he 4a + 8b _ g ^ « < b = 0,15 0,4 Do do m = 76.0,15 = l l , 4 g a m . Vay chon C Cau 22: Cac t h i nghiem c6 xay ra phan ufng oxi hoa - khuf la: — - Sue k h i etilen vao dung dich KMn04 loang. - Cho hoi ancol etylic d i qua hot CuO nung nong. - Sue k h i etilen vao dung dich Br2 trong CCI4. - Cho dung dich glucoza vao dung dich AgN03, trong N H 3 duT, dun nong. Vay chon D Gau 23: X la CHsCOOCH^CHa. That vay: CH3COOCH=CH2 + NaOH - ^ CH3C00Na + C H 3 C H O CHsCOONa + NaOH -> CH4 + Na2C03 I5(m"t 2CH4 ^ C2H2 + 3H2 C2H2 + H2O - ^ ^ ^ CH3CHO Vay chon C Cau 24: Co 4 dong phan ancol C4H10O la dong phan cau tao eua nhau: CH3CH2CH2CH2OH; C H 3 C H ( C H 3 ) C H 2 0 H ; CH3CH2CH(CH3)OH va C(CH3)30H Vay chon C. Cau 25: De y rang Fe cuoi bai da b i oxi hoa toan bo thanh Fe^* Tong so mol NO = 0,05 + 0,02 = 0,07; So mol Cu = 0,0325 CHO NHAN Fe — -> Fe"^ + 2e N^^ + 3e N^ T+2 a mol 2a mol 0,21mol 0,07mol Cu — Cu^^ + 2e 0,0325mol 0,065mol Vay 2a + 0,065 = 0,21 mol < » a = 0,0725 mol. Do do m = 56.0,0725 = 4,06 (gam) Vay chon C 141
C a u 26: Lien ket hoa hoc giOTa cac nguyen tis trong phan tuf H C l thuoc loai lien ket cong hoa t r i c6 c\ic. Vay chon C C a u 27: Cac t h i nghiem c6 xay ra phan ufng la: - Cho dung dich H C l vac dung dich Fe(N03)2 - Cho FeS vao dung dich H C l . - Cho Si vao dung dich NaOH dac. - Cho Si vao binh chufa k h i F2. - Sue k h i SO2 vao dung dich H2S. Vay chon C C a u 28: Ta c6 so mol N2 = so mol N2O = 0,12 mol. Gia sijf mudi thu diigc gom A1(N03)3 va x mol NH4NO3, ta c6 213 — + 80x = 8m « x= m 27 720 Bao toan electron cho 3 — = 0,12.10 + 0,12.8 + 8- m = 21,6 (gam) 27 720 Vay chon D C a u 29: Theo de, dung dich Z gom 0,01 mol Al''^; X mol Fe^* va y mol Fe3+. Ran Y chi la Ag. De y rang 1,97 gam ket tua gom x mol Fe(0H)2 va y mol Fe(0H)3; 1,6 gam ran la 0,01 mol Fe203, ta c6 he: 90x + 107y = 1,97 f x = 0,01 x + y = 2.0,01 = 0,02 [y = 0,01 NhLf vay hon hop X ban dau da cho (0,01.3 + 2x + 3y) = 0.08 mol electron va Ag^ da nhan 0,08 mol nay tao 0,08 mol Ag. Do do m = 108.0,08 = 8,64. Vay chon A C a u 30: Dung dich AgN03 tao ket tua difcrc vdi HCl; K3PO4 va KBr. Vay chon D C a u 31: Phenol phan ufng di/gc vdi dung dich K O H . Vay chon B C a u 32: Can bang hoa hoc H2 (k) + h (k) < > 2 H I (k) c6 tong so mol k h i a 2 ve b a n g nhau nen k h i thay doi ap sua't chung cua he, can b a n g hoa hoc tren k h o n g b i chuyen dich. Vay chon A C a u 33: Fe tac dung v(5i dung dich CUSO4 tao mudi Fe^"^. Vay chpn A C a u 34: Goi n la hoa t r i cua X va a, b la so mol cua Fe va X, ta c6 he: '56a + bX = 1,805 a = 0,025 2a + bn = 2 ^ ^ = 0,095 « ^ bX = 0,405 22,4 bn = 0,045 22,4 Rut ra X = 9n (Al). Vay chon A 142
Cau 35: Cac chat trong day vinylaxetilen, glucozcf, andehit axetic deu tao ket tua khi cho tac dung vcfi dung dich AgNOs trong NH3 dif, dun nong, trong do vinylaxetilen tao ket tua AgC = C - C H = C H ^ , hai chat con lai tao ket tua Ag. Vay chon A Cau 36: Gpi a, b la so mol CUSO4 va NaCl trong dung dich ban dau Vi dung dich sau dien phan hoa tan diicfc toi da 0,2 mol Al20;j nen dung dich nay chi'/a 1,2 mol hoac 0,4 mol O H + Neu dung dich dung dich sau dien phan chiifa 1.2 mol H ^ Anot (+) Catot (-) so;;cr ;H.O Cu'";Na^;Hp 2 CI > CI2 + 2e Cu^" + 2e > Cu b 0,5b b a 2a 2H2O > 4H" + O2 + 4e l,2mol 0,3mol Loai vi anot thoat ra 22,4(0,5b + 0,3) > 6,72 h t . + Neu dung dich dung dich sau dien phan chifa 0.4 mol OH" Anot (+) Catot {-) so; ;C1 ; H p Cu' ;Na^H,0 2Cr > CI2 + 2e Cu^" + 2e > Cu b 0,5b b a 2a 2H2O + 2e - > 20H + H2 0,4mol 0,4raol 2a + (),4 = b a = 0,l Ta CO he: [0,5b = 0,3 b = 0,6 Vay m = 160a + 58,5b = 51,lgam. Vay chon C Cau 37: Ta CO CH3COOCH=CH-CH2 + NaOH - ^ - > CHgCOONa 1- CH3CH2CHO Vay chon B Cau 38: Dung dich metylamin lam phenolphtalein tii khong mau hoa hong. Vay chon B Cau 39: Ta c6 ngUxeroi = 92.0,1 - 9,2 gam. Vay chon D Cau 40: Cac phan ufng: Ba(HC03)2 + HNO3 > Ba(N03)2 + 2CO2 + 2H2O 143
Ba(HC03)2 + Ca(OH)2 -> BaCOa + CaCOa + 2H2O Ba(HC03)2 + Na2S04 - -» BaS04 + 2 N a H C 0 3 Vay c h o n A C a u 41: G p i n l a so' C m o i a x i t va a, b I a n liicft l a so m o l 2 a x i t t r e n , t a c6 he: - + b = 0,2 9 na + n b = 0,6 De y r a n g ~ + < a + b < a + 2b n e n 0,2 < a + b < 0,4 0,6 0,6 0,6 3 > n > 1,5 o 0,2 ^ - > a^+—b > 0,4 V a y n = 2 tufc X l a CH3COOH v a Y l a H O O C - C O O H T h a y n = 2 vao he t r e n g i a i difpc a = 0 , 2 m o l ; b = O , l m o l . Do do % H O O C - C O O H = = 42,86(%) 6 0 . 0 , 2 + 90.0,1 Vay chon D C a u 42: G o i n , m I a n l i i g t l a so' C cua ancol da chufc ancol k h o n g no, c6 mot l i e n k e t d o i , m a c h h o da cho. T h e o de t a c6 0,07n + 0,03m = 0,23. C h u y r a n g t a p h a i c6 n > 2 va m > 3 n e n c h i c6 n = 2 va m = 3 l a p h u hop. V a y h 6 n h o p g o m 0,07 m o l C2H6O2 va 0,03 m o l C3H6O, do do m„jec = 18(0,21 + 0,09) - 5,4 g a m . V a y chon A C a u 4 3 : D a y cac c h a t deu c6 k h a n a n g t h a m gia p h a n uTng t h u y p h a n t r o n g d u n g d i c h H2SO4 d u n n o n g l a : saccarozcr, t i n h bot v a xenlulozcf. V a y chon B C a u 44: Cac p h a n ijfng x a y r a l a : Fe + CUSO4 > FeSOi + Cu S n + CUSO4 > S n S O , + Cu V ^ y chon B C a u 4 5 : NhOfng p h a t b i e u d u n g l a : - T r o n g b a n g t u a n hoan cac nguyen to hoa hoc, crom thupc chu k i 4, n h o m VIB. - T r o n g cac h p p c h a t , so' o x i hoa cao n h a t cua c r o m l a +6 - K h i p h a n ufng vdri k h i CI2 dii, c r o m tao r a h o p c h a t c r o m d H ) . V a y chon B C a u 46: Do NO2 l a o x i t a x i t n e n ngiicfi t a d u n g b o n g c6 t a m nifofc v o i t r o n g la bazo de giOT b d t l a i NO2 k h o n g cho t h o a t r a ngoai. Vay c h o n B 144
Cau 47: Ta c6 nxMx = nyMy nen ny = " ^ ^ ^ _ 1.0,925.2 ^ Q,925mol MY 10.2 Do do so mol Hg da phan ufng = 1 - 0,925 = 0,075mol. Vay chon C Cau 48: Cac dung dich lam xanh quy tim la CH3-CH2-NH2, H2N-CH2-CH(NH2)-COOH. Vay chon C Cau 49: Ta CO phan ufng Al + 4 H N O 3 > A1(N03)3 + NO + 2H2O nen a : b = 1 : 4 . Vay chon D Cau 50: Goi a, b la so mol CuO va AI2O3 thi a, b cung la so mol C U S O 4 va Al2(S04)3. 80a + 102b = 25,5 Ja = 0,255 Vay ta c6 he: 160a + 342b = 57,9 l b = 0,05 Do do %Al203 ^>02.0,05.100 ^ 20(%). Vay chon C 25,5 Cau 51: Ta c6 phan ufng: 6FeS04 + KzCrzOy + 7H2SO4 > 3Fe2(S04)3 + K2SO4 + Cr2(S04)3 + 7H2O Vay a : b = 6 : 1. Vay chon D Cau 52: Cac phat bieu diing la: - De xijf ly thuy ngan roi vai, ngifdi ta c6 the diing bot liJU huynh . - Khi thoat vao khi quyen , freon pha huy tan ozon - Trong khi quyen, nong do CO2 viiOt qua tieu chuan cho phep gay ra hieu ling nha kinh. - Trong khi quyen , nong do NO2 va SO2 vi/cft qua tieu chuan cho phep gay ra hien tifcfng mifa axit Vay chon C Cau 53: Cac phat bieu dung la: - Glucozcf CO kha nang tham gia phan ufng trang bac - Sif chuyen hoa tinh bot trong ca the ngufcJi c6 sinh ra mantozcf - Mantorazo c6 kha nang tham gia phan ufng trang bac Vay chon A Cau 54: Vi 0,6 mol Ag* phan ufng nhtfng chi tao 0,4 mol Ag chufng to day 1^ andehit dcfn chufc c6 noi ba dau mach. Theo de 13,6 gam tufc 0,2 mol C H s C - C H ^ - CHOda phan ufng viia du vdi dung dich chufa 0,6 mol AgNOa trong NH3 theo phufomg trinh: CH = C - C H , - CHO + SAgNOa + 4NH3 + H2O > CAg = C - C H , - COONH4 + 3NH4NO3 + 2Ag Vay chon C 145
Cau 55: Dat. cong thufc Z la CxHyO^Nt Ta CO nc - 0,06 m o l ; n n - 0,14 m o l ; n N = 0,02 m o l v a no - (0,06.2 + 0,07) - 2.0,075 - 0,04 m o l Do do X : y : z : t = 0,06 : 0,14 : 0,04 : 0,02 = 3 : 7 : 2 : 1 V a y Z CO cong thufc p h a n tuf C3H7NO2 vd?i so' m o l = 0,02 m o l P h a n ijrng x a y r a : X + 2H2O > 2Y + Z 0,02mol 0,04mol 0,04mol 0,02mol 4.06 + 0.04.18-89.0,02 Rut r a M Y = — = 75 . V a y X l a g l y x i n . V a y chon A 0,04 Cau 56: C h o n D Cau 57: T h e p cacbon de t r o n g k h o n g k h i a m b i a n m o n d i e n hoa do c6 sif tao p i n d i e n hoa F e - C , t r o n g do Fe l a ciJc a m v a C l a cifc diiong. V a y c h o n B C a u 58: V i d u n g d i c h c6 chuTa 0,05 m o l AgNOs chuTng to da c6 0,05 m o l A g bi oxi hoa. 108.0,05.100 Vay %Ag = - 4 5 ( 7 r ) . Vay chon D 12 Cau 59: T a c6 sa dd C r "'"'•'"^ > C r C l , ^M'-I'N"""-^-^ ) Na[Cr(OH) J V a y chon' D Cau 60: G o i a, b, c I a n lifat l a so' m o l 3 ancol t r e n t r o n g m g a m X . T a c6 he: a + 2 b + 3c = 0,7 a = 0,05 2a + 3 b 4 4c = l b = 0,10 3 2 k a + 4 6 k b + 9 2 k c = 80
JDE S O 2 DE THI TUYEN SINK DAI HOC NAM 2013 M 6 n t h i : H O A , k h o i B - M a d e : 537 Cho b i e t n g u y e n tuf k h o i cua cac n g u y e n to': H = 1; L i = 7; Be = 9; C = 12; N = 14; O = 16; N a - 2 3 ; M g = 2 4 ; A l = 2 7 ; S = 32; CI = 35,5; K = 39; Ca ^ 40; Fe - 56; Cu = 64; Z n = 65; B r ^ 80; Rb = 85; Sr = 88; A g = 1 0 8 ; Cs = 133; B a = 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 cau, tic cau 1 den cau 4G) Cau 1: H 6 n hop X gom ancol metyhc, e t y l e n gUcol. Cho m g a m X - p h a n l i i i g hoan loan vdi N a d i i , t h u dtfcfc 2,24 lit k h i H2 (dktc). D o t chay hoan t o a n m g a m X , thu di/oc a g a m CO2. G i ^ t r i cua a la A. 8,8 B . 6,6 C. 2,2 D . 4,4. Cau 2: Cho 0,76 g a m h 6 n h o p X g o m h a i a m i n dcfn chijfc, c6 so' m o l b ^ n g n h a u , phan Lfng h o a n t o a n vdfi d u n g d i c h H C l dif, t h u difgc 1,49 g a m m u o i . I v h d i lifcrng cua a m i n c6 p h a n tijf k h o i nho h o n t r o n g 0,76 g a m X l a A. 0,45 g a m . B . 0,38 g a m . C. 0,58 g a m . D . 0,31 g a m . Cau 3: H o n h o p X g o m FeO, FezOg v a Fe304. Cho k h i C O qua m g a m X n u n g nong, sau m o t thori g i a n t h u dtfoc h o n h o p c h a t r a n Y v a h o n hcfp k h i Z. Cho toan bo Z vao d u n g d i c h C a ( 0 H ) 2 dil, d e n p h a n urng h o a n t o a n , t h u dtfcfc 4 gam k e t t i i a . M a t k h a c , hoa t a n h o a n t o a n Y t r o n g d u n g d i c h H2SO4 dac, nong (dif), t h u dufcfc 1,008 l i t k h i SO2 ( d k t c , san p h a m khijf duy n h a t ) v a d u n g dich chufa 18 gam m u o i . Gia t r i ciia m l a A. 7,12. B . 6,80. C. 5,68. D . 13,52. Cau 4: H o a t a n h o a n t o a n 1,28 g a m Cu vao 12,6 g a m d u n g d i c h HNO3 6 0 % t h u diroc dung d i c h X ( k h o n g c6 i o n N H ^ ) . Cho X tac d u n g h o a n t o a n vdfi 105 ml dung d i c h K O H I M , sau do loc bo ke't t u a diicfc d u n g d i c h Y . Co can Y di/gc chat r S n Z. N u n g Z d e n k h o i lirgng k h o n g d o i , t h u duac 8,78 g a m chat ran. N o n g do p h a n t r a m cua Cu(N03)2 t r o n g X l a A. 28,66%. B . 30,08%. C. 27,09%. D . 29,89%. Cau 5: Cho g i a t r i do a m d i e n ciia cac nguy.en t o : F (3,98); O (3,44); C (2,55); H (2,20); N a (0,93). H o p chat nao sau day l a h o p chat ion? A. N a F . B. CH4. C. H2O. D . CO2. Cau 6: Cho m g a m m o t o x i t sSt p h a n uTng viia du vdfi 0,75 m o l H2SO4, t h u difoc dung d i c h chi chufa m o t m u o i duy n h a t va 1,68 l i t k h i SO2 ( d k t c , s a n p h a m khijf duy n h a t cua S"^). G i a t r i cua m la A. 24,0. B . 34,8. C. 10,8. D . 46,4. 147