Cẩm nang hướng dẫn ôn luyện thi đại học 18 chuyên đề Hóa học: Phần 2

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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.

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Nội dung Text: Cẩm nang hướng dẫn ôn luyện thi đại học 18 chuyên đề Hóa học: Phần 2

Ca'm nang 6n luyCn thi dai hpc 18 chuy6n dS H6a hpc - Nguygn Van HSi Cty TNHH MTV DWH Khang Vijt C h u y e n de 8 + Phuang phdp dien hoa: No'i k i m loai can bao ve v o i 1 tarn k i m loai khac c6 t i n h k h u m a n h h o n . AN M(m K I M LOAI - IHltU €Hlt K I M LOAI V i d v , de bao v? v 6 tau bien bang thep, n g u d i ta gan vao v6 tau (phan c h i m 1. A N M O N K I M L O A I ^-^^ 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 a. K h a i n i ? m -' ' g r m nsffq oyt qsii 6v lorn S0;0 = t i h • dan, v 6 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 do 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. VI D U MAU b. Phan loai ^-''("OVi'v-r^ -.vrHf rtsrfr: 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 2 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 ra m g n h h o n . Ban chat d i ^ n hoa. •„-•!•> • ^ .i s-: • - f/.'^ - >5 ^ ff'^v*^ 4'^'^ cua h i ^ n t u p n g t r e n la ^^"t"' ' ^' + An mon hoa hoc: A. A n m o n dien hoa. ^' • ' ' B. A n m o n hoa hpc. 'i'i - - A n m o n hoa hpc la sy pha h u y k i m loai do k i m loai p h a n u n g hoa hoc v6i C. A n m o n k i m loai. D. A n m o n hpp k i m . axit hoac cac chat k h i (hoi) 6 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 Nhqn xet: T r o n g day d i e n hoa, cap > ——— —> T i n h o x i hoa Cu^* > H * 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. H2 Cu Vidu\ , ,.s;Vt«nf .^'-S'iffW • .£0.0 • FeS04 + H2 +€^q Zn + CuS04 > ZnS04 + C u nkqkQ ^ 3Fe + 2O2 '" > Fe304 - 'v .,gA^ ^ " ^ ^ ' ' " C u thoat ra b a m tren hat Z n nen thoa m a n 3 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 nhau A n m o n d i ^ n hoa la s u pha h u y k i m loai do 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 do xay ra 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. Vi 2: T r u o n g h p p nao sau day xay ra an m o n di§n hoa: "'^"^ ^- ' Co che an m o n d i e n hoa: •.^_.,f^y.y..:M.._y.^;_,^^ v \. Zn^* D. V 6 tau bien bSng thep n g a m t r o n g nuoc b i e n . ^ 8""*^ ' , 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 Laigidi: ^mi>nkdaoBoo.r a m chuyen t o i va b i o x i hoa tao thanh k h i H2: 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 2 H * + 2e > H2 khong k h i - » D e u bj an m o n hoa hpc. SOB^E EJJ9>1£ • • Ba d i e u k i ^ n de xay ra an m o n d i f n hoa: (1) Co hai k i m loai c6 ban chat L o a i B: K e m de t r o n g axit H2SO4 loang 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 6 tau bang thep la h p p k i m Fe - C (chua 2 d i | n cue c6 ban chat khac c. C a c h chong an mon k i m loai: nhau 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 + Phuomg phdp bdo ve be mat: (nuae bien) nen xay ra an m o n dien hoa. :,; 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: loai can bao ve. A. I , I I va I V . B.I, II vain. C.I,IIIvaIV. D. I I , I I I va IV. 1Q1
C^m nang 6n luy^n thi dgi hpc 18 chuySn 66 Hoa hpc - Nguygn Van HSi Cty TNHH MTV DVVH Khang Vi$t Lai gidi: p l E U CHE K I M LOAI Nhan xet: Trong cac qua trinh an mon dien hoa, k i m loai m ^ h han se bi an phtfcng phap nhif t Iuy?n ^ mon —> Fe bi an mon truoc khi no la chat k h u manh hon (dong vai tro la }Chdi niem: anot - cue am). IsThiet luy^n la phuang phap dieu che kim loai bang each dung cac chat khu Trong cac cap (I), (III), (IV) thi Fe deu manh han, rieng cap (II) thi Zn manh nhu C, CO, H2, A l de khvr oxit kim loai thanh kim loai a nhi?t dp cao. hon. pham vi dp dung • ' '•' •^DapanC. , . . , • , .:>;;.f:,,,, , ,,• Chat khu la CO, H2 se khvr dug-c cac oxit cua kim loai trung binh va yeu V i di^ 4: Gang bi an mon dien hoa trong khong khi am. Qua trinh kliir xay ra tren nhu: Oxit sat, (FeO, Fe203, Fe304), oxit dong, oxit chi (khong khu dugc cac •n- be mat aia gang la: ,fe^,^|,.,B,M>i.#«i: -)1»ft»l^s*&i^.»v:t 40H- ' B. 2H2O + 4e > O2 + 4 0 H - Chat khu la C: Khu dugc cac oxit cua kim loai tir Zn va tao thanh khi CO. C. Fe > Fe2* +2e D. Fe > Fe^* +3e Chat khu la Al: khu dugc cac oxit ciia kim loai tu Cr. Lai gidi: . Phtfcng phap thuy luy?n -i, Thep la hop k i m Fe-C (vdi mot so nguyen to'khac). K h i tie'p xiic voi khong Thuy luyen la phuang phap dieu che kim loai bang each dimg kim loai khi am (c6 chiia chat di^n li) se xuat hi^n qua trinh an mon di?n hoa. Fe la m^nh hon day kim loai yeu hon ra khoi dung dich muo'i. ' cue am (bi an mon), C la cue duong se xay ra su k h u hieu y: Do'i voi cac kim loai kiem va kiem tho (tru Be, Mg) khi tac dung vai dung dich muoi, chiing se tac dung vai nuoc -> Dung dich baza, bazo tao O2 + 2H2O + 4e > 40H- ^nfi^YGx6«tm-• thanh c6 the tac dung tie'p voi muo'i: —> Dap an A. Vi du cho N a kim loai vao dung dich CuS04 xay ra cac phuang trinh: Luu y: Cac em de chon riham B v i bo qua vai tro cua chat oxi hoa la oxi 2Na + 2H2O > 2 N a O H + H2 khong khi. 2NaOH + CuS04 > Na2S04 + Cu(OH)2^ V i du 5: Co 4 dung dich: a) H C l , b) CuCh, c) FeCb, d) H C l c6 Ian CuCh. Phttong phap di^n phan Nhiing vao moi dung dich mpt thanh Fe nguyen chat. So truong hop xuat Bien phan nong chay hien an mon dien hoa la: , Day la phuang phap dieu che cac kim loai manh. Cac kim loai kiem, kiem A.O. B. 1. C.2. D.3. tho thuang dieu che bang phuang phap dien phan nong chay muoi Laigiai, yu,
Ca'm nang 6n luyen thi dgl hoc 18 chuyen d8 H6a hpc - Nguyin Van Hi\ \jt.y MViiii ivi i V L ^ v v i i iMiuiiy viyt Lm giai: Alt D i n h luat Faraday ( t i n h lup-ng chat t h u d u g c 6 cac d i | n cue): m = ach 1: Giai theo p h u o n g t r i n h hoa hpc nP m: K h o i l u g n g chat t h u dug-c a d i ^ n cue (gam) FexOy + yCO — ^ xFe + y C 0 2 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. Mol: a ax ay n: So electron trao d o i m a ion d a n h a n de t^o t h a n h n g u y e n t u (phan tu) 2 npe ^'^ = 0 , 1 m o l ax = 0,1; ay = 0,15 ^ - = t u o n g u n g 6 d i e n cue. 56 y 3 O /;Sv;.„-v-V, I : C u o n g d p d o n g d i e n (ampe) t: t h a i gian d i # n p h a n (giay) O x i t sat la Fe203 V c o = 0,15.22,4^ = 3,36 l i t - ^ D a p an D . ?A i; u i F: H a n g so Faraday (F = 96.500) i ' . . ^ 1 . ...m,- . v gach2: I \.H:/sf i yjnij i>t;..>i 'i;''.* Lieu y: T u cong thue d i n h luat Faraday, so mol electron trao d o i t r o n g qua Vrno = Nhan xet: XXQ ^^xit) = " c o = "002 = 0,15 -> Vco2 = 3,36 3,36 ->• L o a i B v a C. t r i n h d i ^ n p h a n d u o c t i n h theo eong thue: n g = — .n = — M a t khae: n^^: TYQ = 0,5:0,15 = 2:3 -> Fe203 -> D a p an D . V i dv 4: N g a m m o t la k e m t r o n g 100 m l d u n g d i c h CuS04 a mol/1. Sau k h i Day la cong thue giiip giai nhanh chong nhieu bai tonffcfien phan. p h a n u n g ke't thue, lay la k e m ra k h o i d u n g dich, r u a sach, say k h o v a dem ViDVMAU _ • ,^.nn can, thay kho'i l u p n g la k e m g i a m 0,05 gam. Gia t r i cua a la V i dxf. 1: Cho l u o n g k h i H2 (du) qua h o n h g p cae o x i t C u O , Fe263, A I 2 O 3 , MgO A. 0,30. B. 0,25. C. 0,50. r.C/Ag D . 0,20. ' n u n g a nhiet d o eao. Sau p h a n u n g h o n hgp ran con lai la: Lai giai: -O2 + 4 H * +2e (Qua trinh oxi hoa) T a n g - g i a m k h o i l u p n g : m o (trong F e 2 0 3 ) = 12 - 9,6 = 2,4 g a m 2 2 4 —> D a p an B. n o = - ^ = 0,15 m o l n Ghi nha cau tha: A n h la A n o t n h u o n g e, Em la Catot n h a n e t h o i ma. " 16 V i dy 6: D i e n p h a n 800 m l d u n g d i c h A g N O a 0,1M v o i d i e n cue t r a t r o n g t gia, rnpejOa ~ 0/05.160 = 8 gam -> D a p an A . 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 V i d\ 3: K h u h o a n toan m o t oxit sat FcxOy a n h i ^ t d p cao can v i r a d u V l i t kl^' 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, C O (dktc), sau p h a n u n g t h u dupe 5,6 gam Fe v a 0,15 m o l k h i CO2. Cong 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 t h i i c cua o x i t sSt v a gia t r i cua V la gam va c 6 k h i N O (san p h a m k h u d u y nhat ciia N"^"^) bay ra. Gia t r i eiia t la A. Fe304 va 3,36. , .., . ,,, B. Fe304 v a 4,48. - . m »• EOMS A. 0,8. B. 1,2. C.1,0. D.0,3. C. FeO v a 2,24. -s^ D . Fe203 va 3,36. ; .- . " 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: 3.BAITAP6NLUYEN Phan ling dien phan dung dich AgNOs: ' " . |. pai 1: "Sat tay" la sat dupe trang mot lop thiec mong de bao v§ sat khong b} an 2AgN03 + H2O —^Edd_). 2Ag + ^ 0 2 + 2HN03 mon, ten gpi eua phuong phap eho'ng an mon nay la: , . A. Phuong phap hoa hoe. B. Phuong phap d i f n hoa. , , Mol: 2a "'^ ' ' ' ' 2a ' 2a ' '^^' C. Phuong phap bao ve be mat. D. Phuong phap tao hpp k i m khong gi. Nhqn xet: Vi khoi lug-ng Fe tang them nen thanh Fe dupe A g bam vao-^ y Bai 2: Co 4 dung dieh rieng biet: CuS04, ZnCh, FeCb, AgNOa. Nhiing vao moi chiia A g N 0 3 - > AgNOs chua bi di^n phan het* l«l^/;v.' - dung dich mpt thanh N i . So truong hpp xuat hif n an mon di^n hoa la Cac phan ung khi cho Fe vao Y: A. 2. .B.4. C.3. D. 1. Fe + 4HN03 > Fe(N03)3 + N O + 2 H 2 0 , o y a n i ^ M ' Bai 3: Trong pin di^n hoa Zn-Cu, qua trinh khir trong pin la Mol: 0,5a 2a -> 0,5a f' A. Zn^* + 2e > Zn. B. Cu^* + 2e > Cu. ^ Can luu y Fe tac dung voi Fe(N03)3: / -l^ j ' C. Cu > Cu2* + 2e. D. Zn > Zn^* + 2e. Fe + 2Fe(N03)3 > 3Fe(N03)2 rVrt', Bai 4: Tien hanh boh thi nghi^m sau: , Mol: 0,25a •(- 0,5a /^^v r,-- '^ " - ^ ' ' ' , i » V i i i W.KWMi * (1) Nhiing thanh Fe vao dung djch FeCb; ^ Fe + 2AgN03 > Fe(N03)2 + 2Ag ' ' ' .ofex* (2) Nhiing thanh Fe vao dung dich CuS04; ^' ' Mol: b 'jsufTO-f 2b (3) Nhiing thanh Cu vao dung djch FeCb; ^^"^ "'"'^ ' " \ Theo bai: 2b.l08 - 56(0,75a + b) = 2,36. w: a'.: (4) Cho thanh Fe tiep xiic voi thanh Cu roi nhiing vao dung dich HCl. Matkhac: nAgNOa = 2a + 2b = 0,08 a =0,02 mol; b =0,02 moi. £ jo; So truong hpp xuat hi^n an mon di^n hoa la ne= 2a = 0,04 ^ = 0/04 mol ^ t = 2880 giay = 0,8 gio ^ Dap an A. A.l. B.2. C.4. D.3. " Bai 5: Trpn 32 gam Fe203 voi 13,5 gam A l roi nung 6 nhi^t dp eao, thu dupe Vi 7: Di?n phan dung dich X gom 0,16 mol NaCl va 0,24 mol CuS04 (dien cue hon hpp chat ran X. Cho X tac dung voi dung djch N a O H du, thu dupe 6,72 tra, mang ngan xop) den khi tha'y thoat ra 6 anot 2,688 lit khi (dktc) thi ngung lit khi H2 (dktc). Hi^u suat phan ling nhi?t nhom la: di|n phan. Khoi luong dung dich sau dien phan giam d i bao nhieu gam? ... A. 60%. B.50%. • C. 75%. D. 80%. A. 6,96 gam. B. 12,08 gam. C. 17,20 gam. D. 5,12 gam. Bii 6: Trpn m gam bpt A l voi 16 gam bpt Fe203 thu dupe hon hpp X. Nung LMgidi: nong X trong dieu ki?n khong c6 khong khi toi phan ling hoan toan, thu 2,688 „\ , dupe 21,4 gam chat ran Y. Cho Y tac dung voi dung dich HNO3 loang, d u nt = = 0,12 mol. , „ , v^,,, thu dupe V lit N O (san pham khu duy nha't, 6 dktc). Gia tri eiia V la: 22,4 0: A. 2,24. B.4,48. C. 6,72. D. 8,96. Phan ling "quy d o i " gia dinh: i*-^ v ^ 7: Tien hanh phan ung nhi^t nhom voi h6n hpp bpt gom A l va CuO, thu CuS04 + 2NaCl > CuCh + Na2S04 ^^nwib D dupe eha't ran X. Chia X thanh hai phan bang nhau. ,j i"-. Mol: 0,08 Khoi lupng dung dich giam = 64.0,16 + 0,08.71 + 0,04.32 = 17,2 gam. 27,6 gam X can vua d i i V lit khi CO (dktc). Gia trj eua V la Dap an C. A. 4,48. B.6,72. C. 8,96. D. 5,60. ,.CI .J 't.8 196
^ Cty TNHH MTV DVVH Khang Vigt CJm nang 6n luy^n thi dgi hpc 18 chuygn dg H6a hqc - Nguygn Van HJi THI/O'NG DAN-LCJI G I A I pai 9: Nhung mot la kim loai R (hoa tri 2) c6 khoi lugng 10 gam vao 200ml dung dich CuS04 0,1M cho den khi phan ling hoan toan thu duoc chat rAn X va Rai ^' dung dich Y chua 3,22 gam muoi. K i m loai M la Sn trang len Fe se bao phii be mat tranh cho Fe bi an mon D a y la A. Mg. . ^'^ B. Be. C . Fe. D. Zn. phvrang phap bao ve be mat —> Dap an C . pai 10: Cho hon hgp gom M g va Fe vao dung djch C u C h deh khi cac phan virig luu y: day cac em de chgn nham dap an B. C a n nho rang phuang phap hoan toan, thu dug-c dung dich X gom hai muo'i va chat ran Y gom hai kim dien hoa la de bao v§ kim loai, can ghep no vol kim loai m^nh hon. loai. H a i muoi trong X la 2: .:,..,..,,.v.,i if-y •••:> • r:- A n mon dien hoa. pai 11: N h u n g mot thanh M g vao 100ml dung dich gom Cu(N03)2 0,3M va + Nhung N i vao dung dich Z n C h thi khong xay ra phan ling - > N i khong bi AgNOa 0,2M. Sau mpt thoi gian lay thanh kim loai ra, rua sach lam kho, an mon. thay khoi luong thanh M g tang 2,72 gam. Khoi lugng M g da phan ung la + Nhung N i vao dung dich FeCla thi N i bi an mon hoa hgc do phan ling: A. 1,20 gam. B. 0,72 gam. C . 0,48 gam. D . 0,96 gam. Ni + 2FeCl3 > NiCh + 2FeCl2 pai 12: C h o 1,3 gam hot Z n vao 100ml dung dich chua hon hgp gom AgNOa —>• D a p an A. 0,2M va Cu(N03)2 0,5M. Sau khi cac phan ung xay ra hoan toan, thu dugc Bai3: dung dich X v a m gam chat ran Y . Gia tri cua m la: ^ Van dung cm tha: A n h la Anot nhuang e, E m la Catot nhan e thoi ma. ' A. 2,80. B.5,36. C . 4,08. D . 2,16. Trong pin Z n - C u , ta c6: • pai 13: Cho hon hgp bpt gom 1,3 gam Z n va 3,2 gam C u vao 200ml dung dich Anot (cue am), Z n nhuong electron thanh Zn^* (qua trinh oxi hoa Zn): ^ AgNOa a mol/1 sau phan ung hoan toan thu dugc 12,08 gam chat ran X gom Zn > Zn2- + 2e. " ' hai kim lo^i. Gia tri cua a la T'lMj Catot (cue duang): Ion Cu^* den nhan electron (qua M n h khii ion Cu^*): A. 0,20. B.0,30. C.0,15. • D . 0,50. Cu2* + 2e > C u . ' i A ^^:tt.: y ^_..i_-_Q|i:>,,iA igrm riE/iq tj.u pai 14: Cho 4,1 gam hon hgp X gom Z n va Fe vao 200ml dung dich CuS04 0,2M. Dap an B. ••:>iMJ%;' ' • • ' , Sau khi cac phan ihig xay ra hoan toan, Igc thu dugc 4,24 gam chat ran Y va BM4: . dung dich Z chiia hai muoi. Khoi lugng cua Z n trong X la: Nhan xet: 6 thi nghi^m 2 (Fe day C u v a C u bam vao Fe) v a 4 (Fe tiep xuc vol Cu) -> D e u sinh ra cap di|n c\fc Fe - C u , tiep xiic tryctiep vai nhau va ciing A. 0,65 gam B. 1,3 gam C . 1,95 gam. D . 2,6 gam. nhung trong mgt dung dich di|n li Xuat hi|n an mon di|n hoa -> D a p an B. pai 15: Di^n phan 200ml dung dich gom AgNOs 0,5M va Cu(N03)2 I M bang Lieu y: 6 thi nghi?m 1 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 trong thai gian 4 gia. Kho'i lugng kim loai thoat ra 6 catot la: |,, Fe + 2FeCla > 3FeCl2 -fy?^ vt;;/''^ A. 20,4 gam. B. 23,6 gam. C . 10,2 gam. D . 11,8 gam/' 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 J?/ trong thai gian 2 gio v a i dong difn c6 cuong dg la 1,34A. Biet h i f u suat ciia nAl= ^ = 0,5 mol; n^^^o, = ^ = ^'^ "H2 ~ = 0'^ qua trinh di|n phan la 100%. Khoi lugng kim loai thoat ra tai catot la: A. 6,4 gam. B. 3,2 gam. C . 12,8 gam. D . 9,6 gam. 2A1 + Fe203 — ^ AI2O3 + 2Fe 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 m 2a - > nAidu = 0 , 5 - 2 a . tro, mang ngan xo'p) den khi khoi lugng dung dich giam di 17,5 gam thi K h i cho X + N a O H du: • ^ , , ngung di^n phan. The rich khi (dktc) thoat ra 6 anot la Al + NaOH + H2O > NaAICh + | H 2 U S , ^ • U 0 > A. 1,12 lit. B. 1,68 lit. C . 2,24 lit. D . 2,80 lit. 199
Ca'm nang On luy^n thi dgi hpc 18 chuy§n 6i Hoa hgc - Nguyjn VSn Hai Cty TNHH MfV DVVH Khang Vi$t AI2O3 + 2NaOH > 2NaA102 + 2H2O Bai 9: 3 _^ Nhan xet: Trong cac dap an, kim loai c6 khoi lugng mol Ion nha't la Z n Mian xe^: n H , = as ^ - (a5-2a) = 0,3-> a = 0,15. : - ; • H2 ' 2 ^ ' ' ' '^^^ •••1 /Till j^nrv^' H M ^ T ? > "CUSO4 -> C u S 0 4 p h a n u n g h e t . , . H = ^ ^ . 1 0 0 % = 75% ^ D a p a n C , * 0,2 ^ Phan ling hoa hpc: n; i" ^ < i > 1 , Bai6: » fr^. M + CuS04 > MSO4 + C u 1^.1 N/zfl« A:et: Truoc het can ap dung bao toan khoi lugfng: m x = m y Mol: 0,02 MMSOA = - ^ = 161 ^ M = 65(Zn) ^ D a p an D . ' 0,02 A , ; - i So do phan ling: Bai 10: Al^ FeaOg X ^ " ^ ° 3 > AI3^, pe^^. ; r ^ , Nhan xet: Y gom 2 kim l o ^ —> Y chiia Mg hoac Fe Mpt kim loai con d u - > C u C k het Loai B va C . Nhan thdy: t u tr^ng thai dau tien den trang thai cuoi ciing thi: So'oxi hoa ciia M|t khac: Do tinh khii M g > Fe —> Mg phan ling truac roi moi den Fe. Fe khong doi (luon la +3), con so oxi hoa ciia A l tang tit 0 len +3. ,, ,^ Gia sii M g con d u —> Fe con nguyen va C u bi " d a y " ra het - > Chat rSn X A p dyng bao toan electron: ng=3nAi =3nNO HC* gom 3 kim lo^ii —> Dieu gia sii la sai. -> n ^ o = r»Al= 0/2 mol -> V = 0,2.22,4 = 4,48 lit -> Dap an B. Mg phan ling het va Fe phan ling mpt phan —> Y gom Fe d u va C u . Bai7: -> Hai muoi trong X la M g C h va F e C h -> Dap an A . Nhan xet: Bai 11: K h i cho phan 1 + N a O H , tat ca A l va AI2O3 deu chuyen vao hp-p chtYt " A g N 0 3 = 0,02 mol; ncu(N03)2 = 0,03 mol. •? > NaA102 -> A p d y n g bao toan nguyen to'Na va A l ta c6: nNaOH="NaA102 = " A l nAi= 0,1.1 = 0,1 mol. Nhan xet: T i n h oxi hoa Ag* > Cu^* AgNOa phan ling trudc roi moi den Cu(N03)2. »i>UV!iJ C/»?T S S > , • , f TlfWC{" &fo «r--..v. So do phan ling: A l , C u O — ^ X ) A l ^ \. Phan ling hoa hpc: \ iri* JOP; Nhan thdy: T u trang thai dau tien den trang thai cuoi cung thi: So oxi hoa M g + 2AgN03 > Mg(N03)2 + 2Ag ciia C u khong doi (luon la +2), con so'oxi hoa cua A l tang tu 0 len +3. Mol: 0,01 < - 0,02 . 0,02 -> mtang = 2,16-0,24 = l,92g. A p dyng bao toan electron: = Sn^i = S n ^ o M g +CU(N03)2 > Mg(IJ03)2 + C u -> n ^ o = " A r 0,1 mol ^ V = 0,1.22,4 = 2,24 lit. Mol: a Dap an A . Theo bai: m Mg tang = 2,72 1,92 + 40a = 2,72. Bai8: -> a = 0,02 mol — > mMg (pu) = 24.(0,01 + a) = 0,72 gam. Nhan xet: ^,v,,Wj./«-i5 «ir/;, -> Dap an B. K h i cho X + H2SO4 thi oxit chuyen thanh muoi sunfat va mot ion O^" trong Bai 12: ' ' ^'^'^ ' = 0,02 mol; n^^j^o^ = 0,02 mol; ncu(N03)2 = 0,05 mol. oxit dupe thay the bang mpt ion S O 4" . ' Phuong trinh hoa hpc: ^^^^^^ ^^^^^ .^^ 1 mol 02- > 1 mol S O ^ ~ - > Khoi lu(?ng tang 96 - 1 6 = 80 gam. ' Zn + 2AgN03 > Zn(N03)2 + 2 A g j « ; ; a mol tang 59,6-27,6 = 32 gam. Mol: 0,01 0,02 32 -> a = — =0,4 mol -> n o (x) = 0,4 mol. Zn + Cu(N03)2 > Zn(N03)2 + C u 80 Mol: 0,01-> 0,01 -> 0,01 >«k;, ... K h i cho 27,6 gam X tac d y n g voi C O thi: no (x) = n c o = 0,4 mol. - > mY = 0,02.108 + 0,01.64 = 2,80 gam - > D a p an A . r j c i t i ( ' • • ; c': V c o = 0,4.22,4 = 8,96 lit - > Dap an C . ' ^4 ;sh , ..•iii,;,;£;*.!'?.;iK
dm nang On luyQn thi dgi hgc 18 chuySn dg H6a hpc - Nguygn Van HSi Cty TNHH MTV D W H Khang Vijt Bai 13: So m o i CuS04 phan u n g = 0,15 m o l . ::.,,f:-./:..J.!A.f./'-:'.,-' nzn = 0,02 m o l ; ncu = 0,05 m o l . if, fi m i ) j .m qf ^ • mcntot tang = m A g + m c u = 0,1.108 + 0,15.64 = 20,4 gam. Cachl: - ^ Or ^ D a p an A . N/iflM xet: X g o m hai k i m loai c6 chua Z n hoac C u —> A g N O s he't. Bail6: ' ' ^' -""'-^ • • umUMYl MLM ^, 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 X gom Ag IsJhanxet: . anlAnyH va C u ( d u ) . , 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 + 2AgN03 > Zn(N03)2 + 2 A g - -> :h-: , , ^ It 1,34.2.3600 , molelecti-on: n g = - = ^^^^^ =0,1 m o l . . Mol: 0,02^ 0,04 0,04 f . , „ . Cu + 2 A g N 0 3 > Cu(N03)2 + 2 A g X chua: n^^2+ = 0,15mol; n^^. = O ,lm ol; n^+ = O ,lm ol; ng^2- = 0,15mol. Mol: X V - 2x ^ 2x . Catot: C u - + 2e — ^ Cu ^ j^.,^^, " mx = mcudu+ mAg = (3,2 - 64x) + 108.(0,04 + 2x) = 12,08 Mol: 0,05 0,05 r.-..^ . - > X = 0,03 m o l - » n^gNOa " " ^'^ ^'^^ " ^'-^ - » m = 0,05.64 = 3,2 g a m - > D a p an B. - > a = 0,50 m o l -)• D a p an D . . , , Bai 17: , , ^ , Cach 2: G p i : ncu (pu) = x m o l . Bao toan electron: 2n2n + 2ncu(ph4n ijng) = "^g+ Phan u n g q u y d o i gia d m h : • «. , j , T-/-.-,-r r>i.,tjr 4- 2.0,02 + 2x = 0,2a - > 0,02 + x = 0,la. ,4 . m . . ;^a;j i » « . v i CuS04 + 2KC1 > CuCh + K2SO4 " M a t khac: mx = 64(0,05 - x) + 108.0,2a = 12,08 - > x = 0,03; a = 0,5. ? (BH • Mol: 0,1 0,1 0,1 - » D a p an D . - > 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 . " '' T h u t u d i e n p h a n n h u sau: , 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 < CuCl2 Cu + CI2 "-"•••"•^ ^-^^^^"'^ 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 Mol: 0,1 0,1 0,1 mgiam = 0,1.(64+ 71)= 13,5 g a m . ZnS04 v a FeS04. ^.. - _ 'tidrt gnv .m- -> K h o i l u g n g d u n g d i c h con g i a m tie'p 17,5 - 1 3 , 5 = 4 g a m . G o i so Z r ban d a u = a m o l ; = b. N h a n tha'y: r^yi^^:^^^ +. nFe(p.) gj./j cuso4 + H20 - i p ^ Cu + io2 + H2SO4 , * a mol Zn > a m o l C u -> K h o i l u g n g g i a m (65-44)a = a. t' Mol: a a 0,5a ''•TDfeMS ^ cO:dr b m o l Fe > h mol Cu -> tang (64-56)b = 8b 64a + 32.0,5a = 4 - > a = 0,05 m o l . ,.^14 4 r;(HO)!/ Theobai: mt5ng = 4 , 2 4 - 4 , 1 =0,14 g a m 8 b - a = 0,14. J- Vc,2 + V02 = (ai + 0,025).22,4 = 2,8 l i t . L a o c J i o ioinv Bao toan electron: 2 n ^ 2+ ~^^Zn'^ ^'^Fe(pu) -» Dap anD.wi malBlHqlonyrtq [dmx • , \nnU laix w -> a04 = a + b. ,. g ^ a = 0,02; b = 0,02 mzn = 0,02.65 = 1,3 gam. " ""^ 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
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 9 2NaHC03 Na2C03 + CO2T + H2O - *• iff KIM LOAI KIIIM - KIltM THO - NHOM IVluoi clorua 1. K I M L O A I K I E M . >, phan leng dim phan: 2NaCl * 2Na + CI2T a. KimloaiMem ^rv.'/-., vv,;:,r • • ^ 2NaCl + 2H2O "P""''"" > 2NaOH + C h t + H2t + Tac dung vai nude ^ + TflcdwM^^ 2NaOH + H 2 t f. Muoi nitrat ^ ; Phan ung nhiet phan / ^ ^^^^^^^^^ , ^ j,^^,, , , + Tac dung v&i dung dich axit ;, 2Na + 2HC1 > 2NaCl + H 2 t 2KN03 2KNO2 + O2 . , J _^ • f it 'I.XI N e u N a d u : 2Na + 2H2O > 2NaOH + H 2 t + Tinh oxi hod (khi c6 mat axit) + Tac dung vai dung dich muoi Cho Cu vao dung dich hon h(?p KNO3 va H2SO4 loang, Cu tan dan tao Cho K k i m loai vao dung dich CuS04 xay ra cac phuong trinh: thanh dung dich mau xanh theo phuong trinh: ^ 2K + 2H2O > 2KOH + H2t 3Cu +2NO3 + 8H* ^ 3Cu2- + 2NOt+4H20 2KOH + CuS04 > K2SO4 + Cu(OH)2i VIDUMAU • >'•' ' b. Hidroxit,,. „„,„.„:„.„ rfe:)|'iWomr.O.':^ Vi dii 1: Hoa tan hoan toan 1,7 gam hon hop gom k i m loai kiem M va oxit ciia + Tac dung vdi oxit axit ; no vao nuoc, thu dugc 300ml dung dich chiia mQt chat tan c6 nong dp 0,2M CO2 + 2NaOH > NaiCQh + H2O va 0,224 lit khi H2 (dktc). K i m loai M la ^,^^1^ j-j ^,,, CO2 + NaOH > NaHCOa A.Na. B.Li. C.Cs.,jy^|j;^XcmP'lfe)e(*Oe)::. + Tac dung vdi dung dich muoi Lcrigidi: ^^_^>„c,,^.a MM.' Fe(N03)3 + 3NaOH > Fe(OH)3i +3NaN03 Cac phuang trinh hoa hpc: + Tac dung vdi hap chat luang tinh M + H2O > M O H + ^ H 2 t ' ^ i l ^ ' 0 . H '+ AI2O3 + 2NaOH > 2NaA102 + H2O " " do Mol: 0,02 0,02 0,01 ^^^^^^^^^^^^^^ Al(OH)3 + NaOH > NaAI02 + 2H2O c. Muoi cacbonat M2O + H2O ^ 2MOH f^iapi^^^ + Phan ieng thuy phan (quy tim ~^ xanh; phenolphtalein ~^ hong) >0 Mol: x 2x * " Ta c6: = 0,3.0,2 = 0,06 0,02 + 2x = 0,06 - > x = 0,02. col' + H2O nMOH HCO3 + OH- Mat khac: 0,02M + (2M + 16).0,02 = 1,7 M = 23 (Na) - > Dap an A. Tac dung v&i dung dich axit ^1 du 2: Hoa tan hoan toan 1,794 gam kim loai kiem M vao 200 m l dung dich Cho tu tu dung dich axit HCl vao dung djch Na2C03 H2SO4 0,1M. Co can dung dich sau phan ung thu dugc 4,36 chat ran khan Y. Na2C03 + H C l > NaHC03 + NaCl Kim loai kiem M la ifiui >• NaHC03 + H C l > NaCl + C 0 2 t + H2O A.Rb. , . B.K. C.Na |oe D.Li. d. Muoi hidrocacbonat Lcrigidi: , f^Q + Tinh chat ludng tinh Cac phan ung hoa hoc: NaHC03 + HCl > NaCl + C02t + H2O 2M + H2SO4 ^ M2SO4 +H2 va M +H2O ^ MOH + ^H2 NaHCOs + NaOH ^ NaiCO^ + H2O 205
Hoa TiQc - NQuyen van Hai Cty TNHH MTV D W H Khang Vijt Nhdn xet: Bao toan kho'i lugng -> m y = ITIM + "^coZ- + ^ Al(OH)3 + OH- > AIO2 + 2H2O O H -> m _ , . = 4,36 - 1,794 - 0,02.96 = 0,646 g a m - > n ^ „ . = 0,038 moi. u Mol: 0,04 0,04 r f-,,, r „•.*„ . , , O H O H _^ a = 0,08.78 + 0,1.233 = 29,54 gam Dap an A. 1 794 -> nM = 2nH,so. = 0,078 M= ^ = 23 (Na). yf du 5: C h o 100ml dung dich H3PO4 a mol/1 vao 100ml dung dich K O H 2 M M H2t>U4 OH 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. C . 0,5. D.0,8. V i d u 3: H o a tan hoan toan 4,6 gam N a vao SOOjnl H2SO4 0,1M, thu dugc khi Lai gidi: i,«Tr rr, r» ,• .- r r . . 7. 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. B.7,1. C.14,2. D . 18,2. K O H phan ung he't. Laigidi: . " O H - ( K O H ) = ^'2' "H-iH3P04) = ^'^^ nH20 = H K O H = 0,2. Hfga = 0-2 mol; nH2 = 0/1 mol; nH2S04 = 0,05 mol. So do phan ling: H3PO4 + K O H > Muoi + H2O. 3 ,• , 2Na + H2SO4 > Na2S04 + H 2 t Bao toan khoi lugng: 0,la.98 + 0,2.56 = 16,42 + 0,2.18 - > a = 0,9. y^,,, ,. Mol: 0,1 0,05 0,05 —> Dap an A. Luu y: N a con d u se tiep tuc phan ung voi nuoc: ^^^^ ^ Vi dv 6: C h o 1,7 gam hon hgp X gom hai kim loai (thupc hai chu ki ke tiep - Na + H2O ^ NaOH + -Hi trong nhom l A ) tac dung voi nuoc d u , thu dugc 0,672 lit khi hidro (dktc). 2 Hon hgp X gom: .ji^^ n.3i>l -tikAn \M i i i b VBf;f ^ Mol: 0,1 '•^•'•'••al A.LivaNa. B. N a v a K . C.KvaRb. D.RbvaCs. ^ m = mNa2S04 + rnNaOH= 142.0,1 + 40.0,1 = 11,1 gam —> D a p an A . 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 Ggi 2 kim loai lien tiep nhom l A la M , '. (du), thu dugc dung dich X v a 4,48 lit H2 (dktc). C h o 300ml dung dich _ • _ iOnmA * , iOaMyi'' Al2(S04)3 0,2M vao X, thu dugc m gam ket tua. Gia tri cua m la M + H2O > MOH + -H2t 2 t. iO'ABVi < A. 29,54. B. 6,24. C . 24,86. D . 9,34. Mol: 0,06 ^ ,,, Ba(OH)2 + H2 OsH OiM 0,2M. Trung hoa dung dich X boi dung dich Y , tong khoi lugng cac muoi Mol: a I': a a .>h!.•( • dugctaorala ,^^,,| ^ ^ A t " " = 0,2 mol -> 2a = 0,2 ^ a = 0,1 mol. - aO,0 = SAc.*- •*>:• A. 9,40 gam. B. 8,69 gam. C . 7,98 gam.^^ D. 9,90 gam. = HNaOH + 2nBa(OH)2 = 0,2 + 2.0,1 = 0,4 mol Laigidi: ' .{ ifiol f Cac phan ung hoa hoc: ; ' i f B fiu> r Ba2- + SO^- >BaS04^ ' .8 .d>I Na + H2O > Na* + OH" + 4H2 ^ '' Mol: 0,1 0,1 0,1 --^J^i^J 2 A13^ + 30H- > Al(OH)3 4 , g r m nB/!q Di' K + H2O > + OH- + - H 2 Mol: 0,12 0,36 0,12 , :'}H.%
Ca'm nang fin luy$n thi dgi hpc 18 chuy§n de 116a hqc - Nguygn Van H i ! Cty TNhH 1TV DWH Khang Vigt Nhdn xet: n ^ „ . = 2 n H , = 0,1 m o l . 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 : OH ^ + OH" > H2O M a t khac, n o n g d o H C l gap 3 Ian H2SO4 -> T r o n g c i i n g m p t the tich thi Ivlol: 0,02 0,02 i.^.,,*-i •, 1, i , u, , n H c i = 3nH2S04- G o i n H c i = 3a -> nH2S04 = a- ' " , \ Ba2- + SO4" > BaS04i -> T r o n g Y: n , . = 5 a n i o l ; n i_ = aniol;n =3aniol/ ' M o l : 0,03 0,03 0,03 TrunghoaXboiY: + OH" > H2O AP* + 30H" > Al(OH)34 - > 5 a = 0,l ^ a = 0,02mol. * 0,02 M o l : 0,02 - > 0,06 -> K h o i Ivtqmg m u o i t h u d u g c = 5,35 + m^,- + ^cr,2- ''' '' SO^ Al(OH)3 + O H ' - AIO2 + 2H2O '•'1 ,j =5,35 + 0,06.35,5 +0,02.96 = 9,40 gam. M o l : 0,01 < - 0,01 ->DapanA. "I'" --'i-. • D a p an C. So m u o i t r o n g day k h i n h i ^ t p h a n hoan toan t h i tao ra so m o l k h i I o n h o n so V i dv 10: H o a tan hoan toan m gam h o n h o p X g o m L i , N a , K v a o nude, t h u m o l m u o i p h a n l i n g la: 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 A . 2. B.3. C.4. ^ D.5. :d . chat ran k h a n . Gia t r i cua m la: .(••A:Ab) cp.tid IciA J;'Cd,i' Laigidi: ' i'^-^ -mddo A. 0,94. B. 1,28. t')'-«tf+ C. 1,62. :>fd Iv D . 1,26. 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 .•>:'-a.,.f:,/ Laigidi: , M.Mm)itx>iA.A • 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. Gpi cong thue c h u n g ciia cac k i m loai la M . 'X M Uv Cac p h a n ving nhiet phan: 1 M + H2O MOH + - H2 2 2KMn04 ^ K2Mn04 + Mn02 +02t M o l : 0,04 mx = 1,62 + 0,02.2 - 18.0,04 = 0,9i g a m DM • - f 3 H + rf^^ CU(N03)2 CuO + 2N02t + -Oi't - > D a p an A ••' ..;.(,{:••'• ^ 0~ ,.(y-)n'^ m.im.n 2 V i d\ 1 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 , CaC03 > CaO + C02t •:>rl .nofi, cn.fijjf ?L,c o d " } :\ . thoat ra 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 KCIO3 KCI + - 0 2 t ran k h a n . Gia t r i ciia m la 2 •it'.M. A. 19,80. B. 15,25. C. 22,45. D . 17,80. j AgN03 ^ A g + N02t_+-at_^^^^^,,^^ Laigidi: i ?;[ 3 92 —> D a p an B. n H , = - ^ ^ — = 0,175. • ni c n III 1., . 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 6 i lOOml 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 Bao A p dtoan u n g electron: =n2nH2 d i n h luat tru g hoa=dien: 0,35 -> n x == 2n= 0,35. 1.0,35 2- + l^ir^u- ^ • Giatricuaala ' ptj^si fMsnij! miiq-.:. A. 6,99. B.4,66. , C. 7,77. D.8,55., , = 0,35 - 0,2 = 0,15 m o l . OH Laigidi: Bao toan k h o i l u g n g : T r o n g X: n 2+ = 0,03 m o l ; n + = 0,03 m o l ; n = 0,09 m o l . n + =m in = m xx +1^502- + mgQ + n ^ o H - " ^'^^ ^'^"^^ 0,15.17 = 17,8 g a m . T r o n g Y: n ^ ,3+ = 0,02 m o l ; n ^ ^ 2 - = 0,04 m o l ; n „ + = 0,02mol. A' H jd-iS. + D a p an D . 208 209
Cty TNHH MTV DVVH Khang Vigt CSim nang 6n luy$n thi dgi hpc 18 chuyen dg H6a hgc - NguySn v a n HSi 2. K I M L O A I K I E M T H O V i dii 12: Hoa tan hoan toan 6,65 gam hon hop muo'i clorua cua hai kim lo^ij , Kim loai kiem tho , . , kiem thuoc hai chu ki ke tiep nhau vao nuoc duoc dung dich X. C h o X tg^ 4. Tac dung vai phi kim • dung voi dung dich AgNOs (du), thu dugc 14,35 gam ket tua. Hai kim \Q^^ kiem tren la 2Mg + O2 ^ 2MgO ^ , 3Mg + N2 ^ MgaNa A.RbvaCs. B. N a v a K . C.LivaNa. D. K v a R b . + Tac dung vai nuoc -.r- . i r. Laigiai: Cf nhiet do thuong. Be khong phan ung voi nuoc, Mg phan ling cham. Cac "AgCl ^ mo\. G o i cong thijfc chung ciia hai muo'i la M C I . , kim loai con lai k h u manh nuoc tao thanh hidro: Phan ung hoa h o c : . + •!('>)!/ Ca + 2H2O ^ Ca(OH)2 + H2 • r C D i o / . MCI + AgNCh > MNO3 + A g C l i 4i..=K;;n Ba + 2H2O > Ba(OH)2 + H2 ' ^ Moi: 0,1 0,1 8-;:.,!Gj,) • + Tac dung vai dung dich axit ' . * MCI = ^ = 66,5 M = 31 Mg + H2SO4 (/oflMg) ^ MgS04 + H 2 t mi 0,1 r^i -> H a i k i m loai la N a va K —>^DapanB. 5'MV V i du 13: C h o 4,86 gam hon hop X gom muoi cacbonat va hidrocacbonat ciia b. Hidroxit Tjjnoi kim loai kiem M tac dung het voi dung dich H C l (du), sinh ra 1,12 l i t khi + Tac dung vai exit axit v inV ,0 (dktc). K i m loai M la CO2 + Ca(OH)2 > C a C 0 3 4 + H2O .(rCOfiD) idv K C * A.Na. B.K. CRb. D.Li. 2CO2 + Ca(OH)2 * Ca(HC03)2 Laigiai: ^^^^ + Tac dung vai hap chat luang tinh n c o 2 = - ^ = 0 , 0 5 mol. Cac phan ling hoa hQc: > >J,QQ • 2Al(OH)3 + Ba(OH)2 > Ba(AlC)2)2 + 4H2O c. Canxi cacbonat • ''''''HP''TfiO-if---T^f^riVvoriAt^ M2CO3 + 2HC1 > 2MC1 + CO2 + H2O ''^^ + Tac dung vai dung dich axit K MHCO3 + H C l > M C I + CO2 + H2O -A,:..,., ^^y.i-m Mian x e f : n x = n c o 2 = 0,05 mol. kmqk CaCh + C02t + 2H2O — 486 ' ' ' ' CaCOs + CO2+ H2O ^ Ca(HC03)2 {Nuac chay da man) ^ Mx = 7 7 ^ = 97,2 ^ M H C O 3 < 97,2 < M2CO3. + Phan ung nhiet phan O . ' - v ^ ; :''-: 't!,,? - > M + 61< 97,2 < 2 M + 60 ^ 18,6 < M < 36,2 ^ M la N a ^ Dap an A. „ 1200"C , • \ V i dy 14: C h o 11,1 g a m hon hgp X gom Na, K va Ba vao 200ml H C l I M , thoat CaCOs ^ C a O + CO2 (Phan ung nung voi) ra 2,8 lit khi H2 (dktc) va dung djch Y . Co can Y thu dugc m gam chat ran d. Canxi hidrocacbonat — ...Qki^ < khan. G i a tri c u a m la c Tinh chat luang tinh -M + — - iDHS ( A . 27,75. B. 19,05. C . 17,20. D . 18,05. ^ Ca(HC03)2 + 2 N a O H > CaCOs + Na2C03 + 2H2O '^S'-^^'' Phan ung phan huy S^"'.yy&"^' Laigidi: nH2 = 0,125. B a o toan electron: ne=2nH2 = 0,25 -> nx = ne=0,25. Ca(HC03)2 '"""'"'^ > C a C 0 3 i + CO2 + H2O - ' i • {Tao thqch nhU a hang dong, can da voi a noi hai, am dun nude) ( ; A p dung dinh l u a t trung hoa di#n: 0,25 = l n ^ , _ + I n U OH Nuoccung H . ^ n ^ ^ . = 0,25 - 0,2 = 0,05 mol. Khai niem: Nuoc cung la nuoc c6 chiia nhieu ion kim loai hoa trj 2 n h u Mg^*, Ca^^.... Bao toan khoi l u g n g : m = mx + m + m ^ „ . = 11,1 + 0,2.35,5 + 0,05.17 = 19,05 gam - > Dap an B- N u o c mem la nuoc khong chua ho^c c6 chua it cac ion tren. ^1 Ori 211 210
CtyTNHH MTV DWH Khang Vigt dm nang On luy^n thi dgi hpc 18 chuy6n dg H6a hqc - NguySn V5n Hi\ Bao toan nguyen to CI: Phan loai nude cieng 2.7,18 2.4,27 88,55.C Nuoc cxing tarn thoi: Chua ion goc axit H C O 3 , khi dun soi se mat tinh 2nBeCl2 + 2nMgCl2 - " H C l +" C = 10 80 95 36,5(100-C) Cling do cac muo'i hidrocacbonat bi phan hiiy: , Dap an B. dun nong , . ''' ' • > i Yi du 3: Hoa tan hoan toan 3,84 gam hon hgp gom Fe va kim loai M (hoa tri 2) Ca(HC03)2 CaCO.U + CO2 + H2O trong dung dich H2SO4 loang du, thu dugc 2,24 lit khi H2 (dktc). Mat khae, Nuoc Cling vTnh cuu: chua ion goc axit CI", SO 4" . De lam mem nuoc ciing 1,8 gam M tan hoan toan trong 160ml dung dich HCl IM. Kim loai M la vTnh cuu, thuong cho ket tiia cac ion kim loai: j^,, j ^.^^ A.Mg. B. Ca. C. Zn. D. Be. CaCh + NaaCOs ^ CaCOa^ + 2NaCl f Lai giai: Nuoc Cling toan phan: Co ca tinh ciing tarn thoi va tinh ciing vinh ciiu (c6 Cac phuong trinh phan ling: chiia ion goc axit HCO 3 , Ch, SO 4 " ) . 1' / Fe + H2SO4 > FeS04 + H2 ' r : M + H2SO4 — > MSO4 + H2 • j,()gj;^a' + VI D V M A U Nhan thay: npe + nivi = n^j = 0,1 mol V i d u 1: Hop chat nao ciia canxi dugc dung de due tugng, bo bgt khi gay xuong? ^ M=^'^=38,4 MM < 38,4 < MFe Loai B va C. / 'i. 0,1 A. Thach cao nung (CaS04.H20). C. Voi song (CaO). ' • • " ''onn ' 10.(1 1,8 B. Da voi (CaCOs). 5 D. Thach cao song (CaS04.2H20). Mat khae: Mx > = 22,5 Loai D Dap an A. fit' 0,08 Laigidi: V i dii 4 : Hai chat dugc dung de lam mem nuoc ciing vTnh cuu la , || Nhan xet: Nhac den thach cao, cac em c6 the loai ngay B va C. A. Na2C03 va HCl. B. Na2C03 va Na3P04. De CO the diing due tugng, bo bot khi gay xuong, nguoi ta can nung thach C. Na2C03va Ca(OH)2. D. NaCl va Ca(OH)2. cao song de tao thanh thach cao nung. Thach cao nung moi c6 kha nang hiit LM giai: nuoc va dong Cling—> Dap an C. Nhan xet: De lam mem nuoc ciing vTnh eiiu bang phuong phap hoa hgc, V i d u 2: Hoa tan hoan toan hon hgp X gom Be va Mg bang mot lugng vua dii nguoi ta them vao nuoe ciing cac muo'i tan chtia goc axit c6 the ket tiia voi dung dich HCl C%, thu dugc dung dich Y. Nong do phan tram ciia BeCh va ion kim loai hoa tri 2 (Mg^*, Ca^*,...). MgCh trong Y Ian lugt la 7,18% va 4,27%. Gia trj cua C la Cac goc axit do thuong la goc cabonat, goc photphat. - > Dap an B. A. 5. B. 10. C.20. D.15. ' V i d^ 5: Cho 2,74 gam Ba vao 100 gam dung dich CuS041,6 %, thu dugc khi X Lai giai: va ket tua Y. Nung Y 6 nhiet do cao den khoi lugng khong doi thu dugc m Phan ling hoa hoc: > gam chat ran. Gia tri eiia m la Be + 2HC1 > BeCh + H2 A. 2,33. B.3,31. C. 5,64. ' ''* D. 3,13. Mg + 2HC1 > MgCh + H2 Lai giai: Xet voi 100 gam dung dich Y: , r ^ ; s)u/l nB.-, = 0,02mol; ncuS04 =0,01 mol. "
Cty TMHH M i V D V V I ! Khang Vigt dm nang 6n luy$n thi dgi hpc 18 chuy6n d6 H6a hpc - Nguyjn Van HSi V i du 6: Tron 200ml dung dich X gom NaOH 0,1M va Ba(OH)2 0,1M voi lOOrril Lai gidi: dung dich Y gom H2SO4 0,1M va MgS04 0,2M, thu dugc m gam ket tua. Gia j^'COs + 2HC1 > M'Ch + CO2 + H2O i tri cua m la 1 M2CO3 +.2HC1 > 2MC1 + CO2 + H2O " '••fc.J'j A. 5,24. B. 5,82. C. 4,66. D. 6,99. 1 //t«nxef: nco2= nH20 = 2"HCl =0,02. ' , TrongX: n„ ,+= 0^02 m o l ; n^,+ = 0,02 m o l ; n ^ ^ _ = 0,06 mol. B^o toan khoi lugng: 2,42 + 0,04.36,5 = m + 0,02.44 + 0,02.18 TrongY: n 2+ = 0,02mol; n o_ = 0,03 m o l ; n^+= 0,02mol. , t m = 2,64 gam Dap an B. Vi du 9: Cho 1,92 gam M g tac dung voi dung dich HNO3 (du). Sau khi phan Cac phuong trinh phan ling khi pha trpn: . ;, , , ^ . ling xay ra hoan toan thu dugc 0,224 lit khi N2O (dktc) va dung dich Y chua + OH" > H2O ^ • mgammuoi. Giatricuamla „ „ ^ ,.!(,, , 'i\ < ^oe^H ^ A. 12,16. B. 11,36. C.5,68. ^ ^ ' D . 12,96. Mol: 0,02 0,02 Lai gidi: s 1 Ba^* + SO 4 > BaS044' nMg= O'O^ '^N20= 0,01 mol. Mol: 0,02 0,02 0,02 iy':.*Uy. Chatkhu: M g - 2e > Mg^2 ; --i'-i ^^ Mg2* + 2 0 H " > Mg(OH)2l •">MM 4~K8e.»Y|^--M Mol: 0,02 ^0,04 -> 0,02 v S^VAK (GiC': Chat oxi hoa: 2N*^ + 8e > N2O; rMiu\^\xn'/UK\ Vay: a = 0,02.58 + 0,02.233 = 5,82 gam - > Dap an B. M :5Bi:M ;• 2N^-'^ + 8e > NH4Na O d A S ' ^ " — sOE + L . V i d v 7: Hap thu hoan toan 2,688 lit khi CO2 (dktc) vao 2,5 l i t dung djch Ap dyng bao toan electron: 2 n ^ g = 8 n N 2 0 + 8nNH4N03 '"^^'^ Ba(OH)2 a mol/1, thu dugc 15,76 gam ket tua. Gia tri cua a la _ 2.0,08-8.0,01 _ DHd+ i A. 0,032. B. 0,04. C. 0,048. D. 0,06. "NH4NO3 g =0,01. Lai giai: -> m = mMg(N03)2"^NH4N03 = 0,08.142 + 0,01.80 = 12,16 ->• Dap an A. nco2 =0,12mol; nBaC03 = ^ ' ^ ^ m o l . " ; , ^.„,r,-.. .v, , ^ Vi dy 10: De hoa tan hoan toan 3,2 gam hon hgp X gom k i m loai R (thugc Cach 1: Giai theo phuong trinh nhom IIA) va oxit cua no can vua du 200ml dung dich H C l I M . K i m loai R Ba(OH)2 + CO2 ^ BaCOs + H2O Mol: 0,08 0,08 ^ 0,08 A. Ba. B. Be. C. M g . D . Ca. ^• Ba(OH)2 + 2 C O 2 ^ Ba(HC03)2 , ^.^. Lbigidi: 5^.^, , - Mol: 0,02 0,04 '! . Cac phuong trinh phan ung: , , nBa(OH)2 " mol a = 0,04 M Dap an B. ^ R + 2HC1 > RCI2 + H2 5^^^!,, ^^^g^ ^i^^^^^ : Cach 2: •^ih^A M^-Vi: - ^vftH m l • ^ RO + 2HC1 > RCI2 + H2O ri'K Bao toan nguyen to C: nco2 = ^BaCO^ + 2nBa(HC03)2 nBa(HC03)2 = Bao toan nguyen to Ba: nBa(0H)2 = "BaCOg + nBa(HC03)2 = 04 + ^han xet: nx = - nHci = 0,1 mol -» M x = — = 32 Kit-ai ^ 2 0,1 -> a = 0,04M Dap anB. ;())sJj + cO- -^R R = 24(Mg) -> Dap an C. loai kiem M va k im loai kiem tho M ' bang dung dich HCl d u . Sau phan 11: Sue 0,448 lit khi CO2 (dktc) vao 100 m l dung djch X g o m Ba(OH)2 ling, thu dugc dung djch X va 0,448 lit khi (dktc). Co can X thu dugc m gam O'lM va N a O H 0,1 M . Sau khi cac phan ung hoan toan thu dugc m gam ket muoi khan. Gia t r i cua m la tua. Gia tri cua m la A. 2,20. B.2,64. C. 2,30. D. 2,34. ^•1,970. B. 2,364. ' C. 0,985. D . 1,379. " 215
Ca'm nang On luy^n thi dgi hge 18 chuy6n d § H6a hgc - Mytiyeii V i n HJi Cty i M i l l i M I V DVVH KhangVi^t Lot gidi: J, N h o m oxit " O H - " '^NaOH+ 2nBa(OH)2 = O'^l + 2.0,01 = 0,03 m o l + Tinh chat luang tinh: Trong X: " B a 2 + = 0,01 mol; n^^^ = 0,01 mol. ^ , ^, ^^ • ^. M2O3 + 2 N a O H ^ 2NaA102 + H2O J ' t - ^'"'^^ '• " ' ' n. Nhan thay: = = 1,5 > 1 Phan ling tao ra 2 muoi: Cacbonat VP "CO2 0,02 4A1 + 3O2 — ^ 2AI2O3 " 2Al(OH)3 — ^ AI2O3 + 3 H 2 O N h o m hidroxit hidrocacbonat. , . . • + Tinh chat luang tinh: Ap dung cong thuc tinh nhanh so'mol C O 3" : Al(OH)3 + N a O H > NaAXOi + 2H2O "co2- = "OH--"C02 = 0,03-0,02 = a O l mol. „„,^ .^.^^^ + Phan ung nhiet phan £(|i05:)r;A i :,;;.•,>•',.;- ^ N h a n thay: n^^2- = n^^z^ ^ n^^co^ = " ^ ^ 2 - = a O l mol. 'O'iW « 2Al(OH)3 — ^ — A . I 2 O 3 + 3H2O + ^ m = 0,01.197 = 1,97 g a m D a p an A. + Dieu che . •, • • TumuoiAl(III) '-'"-^^ ' ' ' " \ ^ & 8 ; , 4 ; ^ ' n . . - ' 3. N H O M . .,,...,„../ , „ • ; ; AICI3 + 3NH3 + 3 H 2 O > A l ( O H ) 3 ® + 3NH4CI a. N h o m ' '-'I'M*--A - §M ;WfU: + Tdc dung voi phi kirn iKOfUxo'; AlCl3 + 3 N a O H Al(OH)3@ + 3 N a C l '* Luu y: Ket tua A l ( O H ) 3 « se tan dan neu cho N a O H dvr: x :% q ^ , g « . . . 4A1 + 3O2 — ^ 2AI2O3 2A1 + 3Cl2 — ^ 2AICI3 Al(OH)3 + N a O H ^ NaA102+2H20 , / >• + i?(JiO)!:saf + Tdc dung v&i dung dich ax it . nuwjw*. • . ^. 1 ^ . T u muoi aluminat KC 2A1 + 6HC1 > 2AICI3 + 3 H 2 t UiM'~BO. ,•.„.„.— f_ NaA102 + C O 2 + 2 H 2 O ^ Al(OH)3@ + NaHC03.^£ + i ( H O ) f i S A l + 4HN03(loang) > A1(N03)3+ N O t + 2 H 2 O NaA102 + H C l + H 2 O > Al(OH)3© + N a C l ^ f.^"' 2A1 + 6 H 2 S 0 4 ( d a c ) > Al2(S04)3 + 3SO21 + 6H2O Ltfw 1/: Ke't tua A l ( O H ) 3 ^ se tan dan neu cho H C l d u : Luu y: N h o m bi thu dpng hoa (khong tan) trong dung dich HNO3 dac, A l ( O H ) 3 + 3HC1 > AICI3 + 3 H 2 O i .= f; nguQi v a H2SO4 dac, ngupi. /vne d. Phen chua + Tdc dungvoi dung dich kiem .srf! 8 - sf Phen nhom la mot loai muoi kep c6 cong thuc K2S04.Al2(S04)3.24H20 hay Al + N a O H + H2O > NaA102 + - H 2 t Viet gon la KA1(S04)2.12H20. 4 r-f*.?. • ,X cr':v 2 Phen chua c6 ung dung lam trong nuoc, cam mau vai sgi,... j.; j ,, + Phan ung nhiet nhom «,",;UH:VJ '-^ : O nhift dp cao, nhom k h u dugc nhieu oxit kim logi n h u Fe203, Cr203,. VIDVMAU thanh kim loai: ^ ^ , ^ ,, * V i dy 1: D e dieu che dupe 84 gam Fe tu Fe203 (du) bang phuong phap nhi?t 2A1 + Fe203 — ^ AI2O3 + 2Fe , '-tfj^ftx . nhom v a i hi^u suat cua phan ung la 90% thi khoi lug-ng bpt nhom can dung + Sdnxuainhom , ^.nif/jVi H,' . ' . ...^ • \. r toithieu la: ' " * Di$n phan nong chay AI2O3 voi criolit (3NaF.AlF3 hay Na3AlF6) trong binh A. 81,0 gam. B. 54,0 gam. C . 40,5 gam. D . 45,9 gam. di^n phan voi hai di?n eye bang than chi, thu dugc nhom: • Lcrigidi: ^..^ , ;,„. 2AI2O3 '^""^ > 4A1 + 3O2 Phuong trinh phan ling: 5,, >;|lftX'''S,0 =• , i r ' thuong n h u C , C O , H2. 216 217
Cam riaiig On l u y § n thi dai hgc 18 chuygn dg Hoa hgc - Nguyin van H5i Cty TMHH MTV DVVH Khang Vigt „, . , , 84 ^ _ , 100 5 , p u n g dich X gom: Na* = 0,2 mol; Ba^* = 0,1 mol; O H - = 0,4 mol. •Mi t Theo phan ung: np^ = — = 1,5 mol n^l = np^ = — mol fOii tron X voi dung dich gom: AP+ = 0,12 mol; S O 4" = 0,18 mol: ^ . -> mAi = 45 gam Dap an D . Ba2- + S O 4 ' > BaS04i s 0,1 . .R;; . ii: Al(OH)3>L • A.aSO. B. 0,10. C.0,20. * D.0,15/"" Mol: 0,12 - > 0,36 0,12 » r l fioH ^>mi nc. Lcngtat: Al(OH)3 + OH- > AIO2 + 2H2O 4 4 - o-n ' . ' Cac phan u n g hoa hoc: + ;(}{';/), Mol: 0,04 H2O . ICS n Lot gidi: * Mol: 0,02 0,02 ! Cac phan ung hoa hoc: --t>5*'.^«>b'W^ AlO 2 + H^ + H 2 O > Al(OH)3l . • 1 A . u'Uti iii'i' / n o f i N a + H2O > NaOH + -H2T ., . . . . , Mol: 0,02 0,02 0,02 Al(OH)3 + 3H* > AV* + 3H2O ' oi'i / M ,«f /• Mol: 2x -> 2x x Mol: 0,01 < - 0,03 Ba + 2H2O — ^ Ba(0H)2 + H2t Ba2^ + S O ^ > BaS04l •.'.'1 .t/< ..; . Mol: 0,02 0,02 0,02 U "hi) , u ' r > f ^' Mol: X - > X X >' Theo bai: nH2 = 0,2 ^ 2x = 0,2 ^ x = 0,1. ' C ; ij m = mAi(OH)3 + mBaS04 = 0,01.78 + 0,02.233 = 5,44 gam ^ Dap an D . 218 219
Ca'm nang 6n luygn thi dai hpc 18 chuySn 6i H6a hpc - Nguygn Van Hai Cty TNHH MTV DVVH Khang Vi§t V i d\ 5: H o a tan hoan toan m gam hon hop gom N a v a A l vao nude, thq A l tan va sui bgt khi: A l + N a O H + H2O -> NaA102 + - H 2 t dugc k h i H2 v a dung dich X trong suo't. Them tu tu dung dich H C l IJvj vao X, khi het 100ml thi bat dau xuat hien ket tua; khi het 150ml hoap ]y[g khong phan ung. 350ml thi deu thu dugc a gam ket tua. G i a tri cua m va a Ian lugt la Vi d|i 7: Nho tu tit den d u dung dich N a O H vao dung dich AICI3. H i e n tugng A . 7,3 v a 3,9. B. 7,3va7,8. C.5,0va3,9. D . 5,0 va 7,8. xay ra la 1 Laigidi: . ' i^i.- A. Co ket tua keo trang va c6 khi bay len. ..m Cac phan ung hoa hoc: J • ij; < B. Co ket tiia keo trang, sau do ket tua tan. Na + H2O > NaOH + -H2 ' ' ,f.jw>l- C. C h i CO ket tua keo trang. •!" i H 1 ,4 * .' M 1 > ' D. Khong CO ke't tiia, c6 khi bay len. Al + N a O H + H2O > NaA102 + - H 2 .^.(i/Ji Lai gidi: AICI3 + 3 N a O H > A l ( O H ) 3 i {keo trang) + 3 N a C l ' NMn xet: D u n g dich X trong suo't —> A l tan het. j.'! , ' y Al{OH)3 + NaOH(dM') > NaA102 + 2H2O • i tCtet^: + . fA NaOH + HCl > NaCl + H2O , , / - > D a p an B. \'-''f.H Mol: 0,1 < - 0,1 Vi dv 8: C h o hon hgp gom N a va A l c6 ti 1 | so mol tuang ung la 1 : 2 vao NaA102 + H C l + H2O >Al(OH)3i + NaCl . nuoc (du). Sau khi cac phan ung xay ra hoan toan, thu dugc 8,96 lit khi H2 Mol: 0,05 0,05 (dktc) v a m gam chat ran khong tan. G i a tri cua m la - > a = 0,05.78 = 3,9 gam - > Loai phuang an B va D . > A. 10,8. ^ B.5,4. C.7,8. D . 43,2. , , Nhqn thdy: K h i cho 350ml dung dich H C l vao X, toan bp NaA102 se chuyen Led gidi: he't thanh ke't tua A l ( O H ) 3 , va sau do bi hoa tan mot phan trong H C l : NaA102 + H C l + H2O > Al(OH)3i + NaCI Cac phan ung hoa hgc: * \ ' Mol: X X X . . • /^'-"iA Na +H2O — > NaOH + 1H2 ,' „ . ^ui. AICI3 + 3H2O Mol: a -> a b,5a "J Mol: NaA102 + | H 2 (2) :,s^c ' 3 ''"[' Mol: a m = 0,2.23 + 0,1.27 = 7,3 gam -> Dap an A. Chat ran la A l d u = 2a - a = a mol - > m = 27.0,2 = 5,4 gam D a p an B. V i du 6: C h i dung dung dich N a O H de phan bi^t dugc cac chat rieng bif t trong Vi d v 9: Dot chay hoan toan 8,7 gam hon hgp Mg, A l v a Z n trong khi O2 (du) nhom nao sau day? ' 1, i. thu dugc 15,1 gam hon hgp oxit. The tich khi O2 (dktc) da tham gia phan A. L i , Na,Be. B. AI2O3, A l , Mg. C. Zn, A l , Na. D . Fe, C u , Ag. UTig la Laigidi: A , f A. 4,48 lit. B. 8,96 lit. C . 17,92 lit. D . 11,20 lit. Loai A : L i , Na, Be deu tac dung v6i H2O trong dung dich N a O H v a sui bpt Lai gidi: f.: khi. O bai nay, cac e m ap diang bao toan khoi lugng se cho ket qua rat nhanh: Loai C : Z n , A l , N a deu tan trong dung dich N a O H va sui bgt khi. nikimio?! + moxi =moxit -> m o x i = 15,1 - 8,7 = 6,4 gam. Loai D : F 3 , C u , A g deu khong tac dung voi dung djch N a O H . ^ n o 2 = — = 0,2 mol = 4,48 lit J/ - > D a p an B. H i e n tuong n h u sau: AhOa tan: AI2O3 + 2 N a O H > 2NaA102 + H2O ~> Dap an A. , ,.. , , • .-v, 220 221
ca'm nang 6n luyQn Ihi dai hoc 18 chuySn dS H6a hpc - Nguyin Van H5i Cty TNHH MTV DWH Khang Vigt V i d\ 10: H 6 n h o p X g o m a m o l N a va b m o l A l . C h o X vao m o t l u o n g V i dv 12: C h o 1,62 g a m A l tac d u n g v o i d u n g d i c h HNO3 (loang, d u ) . Sau k h i n u a c t h i thoat ra V l i t k h i . M a t khac, n e u cho X vao d u n g d i c h N a O H (d^.^ p h a n l i n g xay ra hoan toan t h u d u g c 0,224 l i t k h i N2 (dktc) v a d u n g d i c h Y t h i d u o c 1,75V l i t k h i (cac the tich d o 6 c i m g d i e u k i e n nhiet do, ap sua't^ chua m g a m m u o i . Gia t r i ciia m la M o i quan he g i u a a va b la A . 19,18. B. 12,78. C. 13,58. D . 16,78. A . a = 2b. B. 3a = b. C. a = b. D . 2a = b . Lddgidi: lb nAi=~=0'06mol;n^,= ^ = 0 , 0 1 mol. ,^ - , + K h i cho X tac d u n g v o i nude ( d u ) : ^,, , ,^• .^.^ ^^^^^ , N a + H2O > NaOH + -H2 . g t i k f o a i fiu) B>l N2; 2N*5 + Se > NH4NO3 Mol: a —> a 0,5a Bao toan electron: 3 n ^ i = 10 n + 8 nNH^NOi D e n day cac e m l u u y, A l se tan m p t p h a n t r o n g N a O H v u a sinh ra: 3.0,06-10.0,01 , ^ , , . • • Al + NaOH + H20 > NaA102 + - H 2 - > nNH4N03 ^ =0,01 m o l . Ei,n'f h ' u b / w r *i4 v.:' = 2 Mol: a C h i c6 C u p h a n i m g i^'^ Bai 6: M o t coc n u a c c6 chiia cac i o n : N a * (0,02 m o l ) , Mg^* (0,02 m o l ) , Ca^* (0,04 A l b i t h u d p n g , k h o n g tan): m o l ) , C I - (0,02 m o l ) , S04~' (0,01 m o l ) va H C O J (x m o l ) . D u n soi coc n u a c tren cho d e n k h i cac p h a n u n g xay ra hoan toan t h i n u o c con l a i t r o n g coc Cu + 4HN03 > Cu(Na)2 + 2NO2 + 2H2O A. La nuoc mem. B. Co t i n h c u n g vTnh c u u . M o l : 0,15 m = ai.27 + ai5.64 = 12,3 gam - > Dap an C. " '^^ '^ C. Co t i n h c u n g toan phan. D . Co t i n h c u n g t ^ t h a i . 223
Cty TNHH MTV DWH Khang Vl^t cam nang on luy^ii thi dai lipc 18 chuyen de Hoa hgc - Nguyjn Van Hki C O 2 (du) vao Y thu duac a gam ket tua. /. Bai 7: Mot coc dung dich chua mot lugng nhieu cac ion: Ca^*, Mg^*, HCO3 Gia tri cua m va a Ian lupt la , • dnhi i"r . •••i C\-, SO l" . Nude trong coc tren thuoc loai gi? A. 8,3 va 7,2. B. 11,3 va 7,8. C. 13,3 va 3,9. D. 8,2 va 7,8. A. Nuoc mem. B. Nuoc cung vTnh cuu. pai 1^* ^''^ ^^"^ 400ml dung dich H C l 2M, thu C. Nuoc Cling tam thoi. D. Nuoc cxing toan phan. (Jupc dung djch X. Cho tung gipt den he't V lit dung djch N a O H 2 M vao X, Bai 8J Tien hanh phan ung nhiet nhom hon hop gom A l va 16 gam Fe203 de'n thu dupe 7,8 gam ket tiia. Gia trj Ian nhat cua V la ' khi phan ung hoan toan, thu dupe hon hop ran X. Cho X tac dung vua dii A. 0,45. B.a35. 0 ^ C. 0,25. J ; = D. 0,05. ' voi V l i t dung djch NaOH I M , sinh ra 3,36 lit H2 (dktc). Gia tri cua V la Bai 17: Hoa tan hoan toan 3,78 gam A l bang dung dich H N O 3 loang (du), thu A. 0,30. B.0,10. C.0,20. D. 0,15. dupe dung dich X va 0,448 lit (dktc) hon hpp khi Y gom N 2 O va N2. Ti khoi Bai 9: Cho khi C O (du) di vao ong s u nung nong dung hon hop X gom ctia Y so vai hidro la 18. Co can dung dich X, thu dupe m gam chat tin khan. AI2O3, Fe304, CuO thu dupe chat ran Y. Cho Y vao dung dich NaOH (du), Gia tri cua m la khuay k i , tha'y eon lai phan khong tan Z. Phan khong tan Z gom A . 29,82. B. 32,22. C. 31,42. D. 33,02. A. AI2O3, Fe, Cu. B. Fe304,Cu. C . Fe, Cu. D . A l , Fe, Cu. 5. H I / C ! K N G D A N - L 6 I GIAI ^ ij/t ' - • Bai 10: Day gom cae oxit deu bi A l khu 6 nhi^t dp cao la: Biil: A. FeO, CuO, Cr203. B. PbO, K2O, SnO. J'IATJ NaHCOs + N a O H ^ Na2C03 + H 2 O C. FeO, MgO, CuO. D . Fe304, SnO, BaO. Al(OH)3 + N a O H > NaAlCh + 2H2O ' Bai 11: Nhi^t nhom hon hpp gom a gam bpt A l va b gam Fe203, thu dupe hon hop X. Hoa tan X trong H N O 3 du, thu dupe 2,24 lit khi N O (dktc). Gia tri Ch + 2 N a O H > NaCl + NaClO + H2O ,., .....r fr cua a la: i'-nOi'' ^ •''a ^ / ' r r f ^ o i ff«;,n,-ii rv!+nf;! j NH4CI + NaOH > NaCl + N H 3 + H 2 O , f, - A.2,70. wniVffi- B. 5,40. C . 1,35. D.8,10.iG: HF + N a O H > NaF + H2O ' H0^| , Bai 12: Tien hanh phan ung nhif t nhom voi m gam hon hpp X gom A l va ~> Dap anB. .{, >fn l O . O - . j ' Fe304 den phan ung hoan toan, thu dupe eha't ran Y. Cho Y tac dung v a i BAi2: • . • , . i dung dich NaOH (du) thu dupe dung dich Z, phan khong tan va.3,36 lit k h i ^'•'f) Tie-" " O H - " " N a O H + n K O H = 0'2 + 0,1 = 0,3 mol • -Kt; .... 01 H2 (dktc). Sue khi C O 2 (du) vao Z, thu dupe 39 gam ket tua. Gia tri ciia m l a Trong X: n^, + = 0,2 mol; n^+ = 0,1 mol. A. 45,6. B.48,3. C . 36,7. D . 57,0. Bai 13: Tien hanh phan ung nhif t nhom vai 24,1 gam hon hop gom A l va el: ^!^2H1 = M = 3 > 2 ~^ O H - du, phan ung chi ra tao muoi Nhqn xel: = ' Fe203 den phan ung hoan toan, thu dupe chat ran X. Chia X thanh hai phan nroo 0,1 "CO2 bangnhau. + Phan mpt phan ung vira du vai 150ml dung dich N a O H I M . cacbonat. o -r -! ro n - « * Pfvn «Jf • PfiO » 4 + B : + De hoa tan het phan hai can vua du dung dich chua a mol HCl. Phuang trinh ion: C O 2 + 2 0 H - ^ CO3" + H2O gfj'y r. Gia tri cua a la Mol: 0,1 0,2 0,1 A. 0,40. > B. 0,65. C . 0,35. D . 0,55. V gom: Na* = 0,2 mol; K* = 0,1 mol; O H " = 0,1 mol va CO 3" = 0,1 mol. Bai 14: Hoa tan hoan toan 0,3 mol hon hpp gom A l va AI4C3 vao dung dich Bao toan khoi lupng: a = m^^+ + + m ^ ^ . + m^^2- ^n-/' ' . K O H (du), thu dupe a mol hon hpp khi va dung dich X. Sue k h i C O 2 (dif) S-^t^ = 0,2.23+ ai.39 + a i . l 7 +0,1.60 = 16,2 gam. / ! 1 vao X, lupng ket tua thu dupe la 46,8 gam. Gia tri eua a la A. 0,55. B.0,60. C.0,40. D . 0,45. ->DapanC. * Bai 15: Hoa tan hoan toan m gam hon hpp X gom Na20 va AI2O3 vao H 2 O thi' ^ai3: nc„= — = 0,05 mol. . .nartko ^ 'Mr^'l ^ dupe 200ml dung dich Y chi chua chat tan duy nhat c6 nong dp 0,5M. Tho' 64 .ry:,..^ 225
C^m nang 6n luygn thi dji hqc 18 chuy6n ii H6a hqc - Nguygn VSn H&\ Cty TNHH MTV DVVH Khang Vigt Nhdn xet: D u n g d i c h X c h i i a m u o i K N O 3 va H2SO4 l o a n g - > Can giai theo So' m o l di^n tich d u o n g = So' m o l dien tich am phuong trinh ion. ei t'w; o , 1.0,02 + 2.0,02 + 2.0,04 = 1.0,02 + 2.0,01 + 1.x ^ x = 0,1. i;> i / \ n + = 2nH2S04 = 2.0,06 = 0,12; n = nKNOj = 0,04. ,i,Crw Den day, cac em luU y: Khi dun nong, muoi hidrocacbonat cua kim loai hoa tri 2 bi phan huy: P h u o n g t r i n h i o n r u t gon: Mg2^ + 2 H C O ; > M g C O s ^ + CO2 + H2O '' 3Cu + 8H* + 2NO3' ^ 3Cu2* + 2N0''' + 4H2O dun nong . , " f V>ri. ? A i. ' M o l : 0,045 0,12"^ 0,03 \::\> • 0,03 Ca^- + 2 H C O 3 C a C O s i + CO2 + H2O ' ' ' ^ V = 0,03.22,4 = 0,672 l i t ^ D a p an D . h> i 1 N h a n thay: n^^2+ + \^2+ = 0,06 M g ^ Ca^* con du, H C O 3 he't. ' '' Bai4: .j, ,.u v, A ...ib ynu,") Cac p h a n u n g hoa hpc: > i- ' N u o c t r o n g co'c sau k h i d u n c6 chua cac ion: Mg^^ Ca^*, C h v a SO 4 " -> ThuQC lo^i n u o c c u n g v i n h c u u - > D a p an D. M + HCl ^ MCI + -H2 va M +H2O ^ MOH + -H2 2 2 Bai 7: ' ' ^"''"^'''''' ' ^ Nhdn xet: Bao toan k h o i l u g n g my = + + A U «Jj * Nhqn xet: D61 chieu v o i d i n h nghia, m a u n u o c t r o n g co'c c6 chua ca t i n h Cling tarn t h a i (chua H C O 3 ) va c6 ca fa'nh c u n g vTng c u u (chua CI", SO^"). ^ m ^ ^ . = 3,316 - 1,794 - 0,04.35,5 = 0,102 g a m ^ n ^ ^ . = 0,006 m o l . -> ThuQC loqii n u o c c u n g toan phan - » D a p an D. , 1 794 nM= "HCl + n = 0,046 M = = 39 (K) ^ D a p an B. Bai8: ' 0,046 Sir nFe203 = ^ = 0,lmol ^ npe = 0 , 2 m o l . Bai 5: Nhqn xet: Bai nay neu cac e m viet p h u o n g t r i n h p h a n t u se gap n h i e u kho Nhdn xet: K h i cho X + d u n g d i c h N a O H —> Sinh ra H2 nen t r o n g p h a n u n g khan. D o vay, nen giai d u a theo cac p h u o n g t r i n h i o n . nhiet n h o m t h i A l con d u va FezOa het. ' " ' " "' ' ' " O H - " "NaOH + 2nca(OH)2 = 0,03 + 2.0,01 = 0,05 m o l t . T r o n g X: 2A1 + Fe203 > 2Fe + AI2O3 ^ n 2+ = 0,01 m o l ; n + = 0,03 m o l . M o l : 0,2 ai 1 - 0,2 0,1 ' 3 _ Q H _ = 9J91 > 1 Phan l i n e tao ra 2 m u o i : Cacbonat va hidrocacbonat. Al + N a O H + H2O > NaA102 + - H 2 dAm mmi^finml HCO3 CO2 + 2 0 H - ^ C 0 3 " + H 2 0 ^N.OH= | n H 2 + 2nAi203 = ^ + 2.ai =0,3mol V = 0,31it. ^ . , Mol: a a a Mol: b 2b b ->Dap^nA. „ .^^p i^f^ Ta c6: a + b = 0,03; a + 2b = 0,05 - > a = 0,01 m o l ; b = 0,02 m o l . Bai 9: Phan umg tao ket tua: Ca^* + C O 3 " ^ CaCOa §nw. Luu y: K h i C O k h o n g k h u d u g c AI2O3 nen k h i cho k h i C O d u tac d u n g v o i ^ m = aOl.lOO = 1,0 g a m D a p an B. X, xay ra 2 p h a n u n g : Fe304 + 4 C O — ^ 3Fe + 4C02 " ^ •r f . g l ^ ( v , , , ^ : T i n h nhanh: n^^2- = n ^ j ^ - - nco2 = 0,05 - 0,03 = 0,02 m o l . CuO + CO — ^ C u + CO2 ' iV/ D o n^^2+
Ca'm nang On luy?n thi djii hpc 18 chuyfin dg H6a hpc - Nguygn Van Hii Cty 1NHH MTV DVVH Kliang Vi$t Bai 10: Fe203 + 2A1 — ^ AI2O3 + 2Fe Luu y: Al chi khu dug-c cac oxit kim loai dung sau Al trong day di§n hoa. Mol: 0,1 -> 0,2 0,1 0,2 , -> Loai B, C, D vi Al khong khu dugc K2O, MgO va BaO -> Phan 2 gom: n^i = 0,05 mol; nFe= 0,1 mol; n^ijOa = 0/05 mol. —> Dap an A. ., . - > HHCI = 3 n A i + Znpe + 6nAi203 = 0,65 mol. • ' i ' Cac phan ung hoa hoc: ^.^^^j^ ^ . ; e ' L ^ . . . . . , ^ ^ 1 , Dap an B. >r r r 1- w - 2A1 + 3?eO — ^ AI2O3 + 3Fe,^^.,^^.^;, s--*,. ^ Bai 14: 2A1 + 3CuO Ah03 + 3Cu „ ^^g,,^ So do phan ung: ^ 2A1 + Cr203 ^ AI2O3 + 2Cr ' Al, A I 4 C 3 KAIO2 ^ ^ ° ? ^ " 2 0 > Al(OH)3 ' Bai 11: ^ Bao toan nguyen to Al: n^, + 4nAi4C3 = nAi(OH)3 ^ "Al + 4nAi4C3 = OA Cac qua trinh hoa hoc xay ra theo so do sauf'^J''' f ' Theo bai: nAi +nAi4C3 = 0,3 -> nAi =0,2; nAi4C3 = 0,l. ^ 3 '- J Al, FejOj — ^ X ) Fe(N03 )3 + AKNOg )3 + NO V$y: a = - nAi + 3nAI4C3 "0,6 ,. N/zfln xet: So oxi hoa ciia Fe khong thay doi a trang thai dau va cuoi (deu la -> Dap an B. Fe*^), con so'oxi hoa ciia Al tang tir 0 len +3. Bai 15: rO,0,8^-M..- Bao toan electron: 3 n A i = 3 n N o —>• nAi = nwo = 0,1 mol. a = 0,1.27 = 2,70 gam -> Dap an A. ' Nhan xet: Khi cho X vao nuoc, Na20 se tac dung ngay vai nu6c: Bai 12: Na20 + H2O > 2NaOH af^faqfi^i Nhqn xet: Bai nay ra't nhieu thi nghiem nen neu cac em giai theo phuong AI2O3 CO tinh luong tinh nen tan trong dung dich NaOH vvra tao ra trinh phan ung se mat nhieu thai gian. j^^, . Al + NaOH + H2O > NaA102 + - H 2 t So do duong di ciia Al trong cac qua trinh hoa hpc nhu sau: ^Fe304,t° ^ ^j^o^ ^ J^^^JQ^ ^ ^^^^^^^ 2 39 Mol: 0,1 0,1 0,1 T h 6 i k h i C 0 2 : NaA102 + CO2 + 2H2O > Al(OH)3 +NaHC03 Bao toan nguyen to Al: n^i (x) = nAi(OH)3 mol. t ri*^ij4:'>, Mol: 0,1 0,1 2 -> a = mAi(OH)3 = 0,1.78 = 7,8 gam; ., Mat khac n^i = —^H2 ^'^ ~^ tham gia phan ung nhiet 3 va m = m N a 2 0 + mAI2O3 = 0,05.62 + 0,05.102 = 8,2 gam. nhom ^ m =la m^i 0,4 mol: + mFe304 = 0,5.27 + 0,15.232 =£48,3 gam Dap an 3.^^ M -> Dap an D. Bai 16: > Bai 13: 8A1 + 3Fe304 — ^ 4AI2O3 + 9Fe 5 , qeU ^ nAi = 0,2 mol; nnci = 0,8 mol; nAi(OH)3 = 0,10 mol. Mol: "NaOH0,4 -> 0,15 = 0'15.1 = 0,15 mol., Dung dich X gom: AlCh = 0,2 mol; HCl = a2 mol. ^ ; 1, ,5 . N/zfln xet: Khi cho phan 1 + dung dich NaOH, tat ca Al va AI2O3 deu tac £ach 1: Giai theo phuang trinh hoa hpc: dung va chuyen thanh NaA102. NaOH + HCl > NaCl + H2O , ;v,: , ; . Do vay, bao toan nguyen to Al, ta c6: nAi= nj^jaOH = O'^S mol. AICI3 + 3 N a O H > Al(OH)3i + 3NaCl . ^ ^ Ban dau: n ^ r 2.0,15 = 0,3 mol -> nFe203= " ^ ^ ^ ^ ^ ^ ^ ^ Al(OH)3 + NaOH > NaA102 + H2O 99S -> VnNaOH= -> n,^^. = 0,45 OH = 0,20+ litnQ^.0,60+ = Dap an A. 0,9moI 0,10 = 0,90 mol.