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Phương pháp kỹ thuật mới giải nhanh bài tập Hóa học (Tập 2: Hóa học vô cơ): Phần 1

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Tài liệu Kỹ thuật mới giải nhanh bài tập Hóa học (Tập 2: Hóa học vô cơ) giới thiệu tới người đọc phương pháp giải hay, giải nhanh nhằm giúp các em rèn luyện kỹ năng giải nhanh các dạng bài tập trong đề thi tuyển sinh Đại học - Cao đẳng. Mời các bạn cùng tham khảo nội dung phần 1 tài liệu.

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Nội dung Text: Phương pháp kỹ thuật mới giải nhanh bài tập Hóa học (Tập 2: Hóa học vô cơ): Phần 1

  1. 546.076 CtJ THANH TOAN K600T NGUYEN NGOC OANH KYTHUATMdfl GIAINHANHBAITAP HOA HOC I I 114 4 TAP 2 B 4 wo'c lOnil.., f HOA H O C VO C O i i i 4 ' ^ ^ - Tuyen chon cac ky thuat giai nhanh ve 7 chuyen de hoa v6 ca - Danh cho hoc sinh \6p 10, 11, 12 va luyen thi D H - CD OCR] NHA XUAT BAN DAl HOC i^UOC GIA HA NOT
  2. " X U THANH TOAN NGUYEN NGOC OANH KYTHUATMOfl GIAINHANHBAITAP HdA HOC TAP 2 HOA H O C V 6 C O - Tuyen chpn cac ky thuat giai nhanh ve 7 chuyen de hoa v6 ca - Danh cho hoc sinh Idp 10,11,12 va luyen thi D H - CD NHA XUAT B A NDAI H O C Q U O C GIA H AN O I
  3. C. KI THUAl GSAl NHANH 7 CHUVeN HOA VO CO L6INOIDAU Cac ban ddng nghidp va cac em hoc sinh than m6'n! CHUYEN ©11: KT T H U A T ©SAI N H A N H Trdn CO so phan tfch kl luong cac n6i dung kien thiic v^ ki nang nam cAc D^m BAI TJSip y g H A L O G E N trong khung chuong trinh thi, ca'u true, ma tran d6 thi va cac dang bai tap thucmg gap trong da thi tuyen sinh dai hoc, cao dang ciia B6 GD - DT, chung A . L I THUYI^T toi da biSn soan tap sach: "Kithudt moi gidi nhanh bai tap Hod hoc" 1. DAC D I £ M CIJA CAC NGUVfiN T6 HALOGEN Cu6'n sach mof ra m6t co h6i cho giao vian va cac em hoc sinh nhin Flo Qo Brom lot nhan lai mot each sau sac va toan dien cac da thi tuyan sinh DH - CD ki tii khi ap dung hinh thiJc thi trac nghiem, tilr do giao vian c6 dinh hirdng dung (F) (CI) (Br) (I) dan cho cac em hoc sinh va miic do kien thiic va cac dang bai tap; con cac ca'u hmh electron 16p 2s^2p' 3s^3p' 4s'4p' 5s^-5p' ngoai Cling em hoc sinh se vihig vang, tu tin hom de di tran con duomg den c6ng truong Ban kinh 0,64A° 0,99A° 1,14A° 1,33A° Dai hoc ma minh mo udfc. nguyan tif Tac gia cung khdng quan gili gam vao eu6'n sach cac phuong phap giai D6 am dien 3,98. 3,16 2,96 2,66 hay, giai nhanh, nham giiip cac em ren luyan ki nang giai nhanh cac dang 3,58 eV. 3,81 eV 3,56 eV 3,29 eV Ai lire electron bai tap trong di thi tuyen sinh DH - CD. Cuon sach se la nguori ban d6ng hanh than thia't vdri cac em hoc sinh trong qua trinh chinh phuc nhiing dinh Nang lugmg ion hoa 17,42 12,97 11,34 10,45 thif nhatl, (eV) cao vinh quang ciia tri thiic; cuon sach se la tai lieu qui cho cac ban d6ng nghiep trong qua trinh giang day. Nang luong 159 242 192 150 lien ket (kJ/mol) Da' cuon sach hoan thian hon, ra't mong nhan duoc su dong gop y kia'n • Nhiat d6 nong chay -219,6 -101,0 -7.3 113,6 chan thanh ciia cac ban d6ng nghiep va ciia cac em hoc sinh. ("C)" Chuc cac em dat dirge nhiau thanh tich cao trong cac ky thi sap t6i. Nhiat d6 s6i ("C) -188,1 -34,1 59,2 185,5 Xin tran trong cam cm ! '* Tiang thai (dkt) khi khi long ran T A C G I A Mau sac luc nhat vang luc do nau den tim Nhd Sach Kkang Viet xin tran trong giai thieu toi Quy doc gid vd xin long Mui rat d6c Xoc hdi kh6ng miJi nghe moi y kien ddng gop de cuon sach ngdy cdng hay hon, bo ich hon. Thu xin 4.10-' gtH ve: Ham iKong (% s6' 0,02 0,02 3.10-* nguydn tir) Cty TNHH Mot thanh vien - Dich Vu Van Hoa Khang Viet. 71- Dinh Tien Hoang, Fhuimg Dakao, Quan 1, TP HCM. Tel: (08) 3 9 1 1 5 6 9 4 - 3 9 1 1 1 9 6 9 - 3 9 1 1 1 9 6 8 - 3 9 1 0 5 7 9 7 - Fax: (08) 3 9 1 1 0 8 8 0 Email: khangvletbookstore@yahoo.com.vn
  4. II. TINH CHAT HOA HOC CUA DON CHAT III. DI^U cut TRONG P H 6 N G THI NGHlfeM VA S A N XUAT TRONG F2 Br, I2 C6NG NGHIEF Vdikim - Tac dung vod - Tac dung - Tic dung vdi - Tac dung Nguyen tac chung: Oxi hoa ion halogenua thanh nguyen tir: loai tat ca cac kim v2HBr03 6 H , 0 ^ khijr + lOHQ 2HIO3 + lOHCl Nhan xet F2 > C I 2 > Bo > h J Tinh oxi hoa gi5m dan 4
  5. I V . CAC H I D R O H A L O G E N U A VA A X I T H A L O G E N H I D R I C (HX) V. M U O I H A L O G E N U A HF HCl HBr HI Mu6'i florua Mu6'i clorua Mu6'i bromua Mu6'i iotua (CD (Br) (D Nhidt d6 s6i - D6u tan - Da s6' tan - Da s6' tan - Da s6' tan 19,5 -85,1 4-66,8 -35,4 Ti'nh (°C) (trir AgCl, (li ij AgBr, tan (trif CaFj,...) (triif A g l , Hgl,, D6 tan trong PbCl,,...) PbBr2,...) Pbl,, Mnlj). 1 lit nude Mu6'i Tan, mau Kh6ng tan, Kh6ng tan, Kh6ng tan, Tan v6 han 500 lit 600 lit 425 lit (bOPC) AgX. trSng. mau trang. mau vang nhat. mau vang. Tinh axit Manh hom Manh hon Phan Kh6ng xay F2 + 2C1 („6ngchdy) CI2 + 2Br-,,,, CI, + 2r„„ -> cua dd HX Yiru Manh ling vori ra. -> 0 2 + 2F ^ Br2 + 20" I2 + 2a- HCl HBr halogen Br2 + 2r„d) - > Tac dung I , + 2Br vdi dd AgCU AgBri Agli Khong Phan CaF2 + NaCV,+ 2NaBr„y + SKI + 5H2S04,a, AgNOj trdng vang nhat vang iJng vod HiSOiij) H2S04 4I2 + HjS + Tdc dung H,S04 250^ C ^ 2HC1 (k) 6HX + 2H3PO, RH + CI2 - > RCl + HCl 7
  6. V I . H O P C H A T C6 O X I CUA H A L O G E N B. C A C D A N G B A I T A P T R O N G D E T H I TUYJ^N S I N H D H - C D Nudrc Javen C l o r u a vol M u o i k a l i clorat Dang 1: Xdc dinh chat (halogen, muoi halogenua, kim loqi) Thanh NaCl + NaClO CaOCU: KCIO3 phan + H2O -1 +1 1 . L i thuyet van d u n g va phuomg phap giai: CI - Ca - 0 - CI * D6' giai cac bai tap dang nay c6 th^ sir dung cac phuomg phap giai nhanh: Dieu * C l 2 + 2NaOH„„,„,, * CI, + Ca(OH)„^, . Phuong phap xac d i n h chat dua vao cong thiirc t i n h p h a n tur k h o i : Gia six c6 * 3C1, + 6 K O H — ^ Che NaQ + NaQG chat cho electron la A . Ta c6: 5KC1 + K Q O j + 3 H , 0 + H2G + H2O ke * Didn phan dung m. * C I , + Ca(OH), * Di6n phan dung dich -.k dichNaa(15-20%) (v6i sOa) K C l 2 5 % a n h i a d6 6 dieu kifenthucmg, 70°C - 75°C. khdng CO mang ngan. - j f ^ CaOCl, m * 6C1, + 6Ca(OH), Vay: + H2O MA ^ ' ne(cho) —-> Ca(C103), + 5CaCl, + 6 H , 0 Trong do: m ^ : Khd'i lucmg chat A (gam) Ca(C103)2 + 2KC1 M ^ : Khd'i luong mol chat A (gam) k: s6' electron do m6t nguyen tir A nhucmg 2KC103i +Caa2 n^ (cho)- s6' mol electron do toan bo luong chat A nhircJng. \ Tinh - Tinh oxi hoa - Tinh oxi hoa - Kem ben nhifit: + Dua vao cac dir ki^n bai ra cho, xac dinh diroc cac dai lirong m ^ , k, n^fcho)- chat manh: manh: 2KCIO3 — ^ + Tinh gia t r i M ^ . 4 N a a O + PbS -> CaOCl, + 2HC1 PbS04 + 4NaCl 2KC1 + 30,T + Tir phan tir khoi xac dinh duoc cha^ cin t i m . f ——> Phirong phap t r u n g b i n h : C6ng thiic tinh nguyfin tir kh6'i trung bmh ciia hai - La mu6'i ciia ax it - Chat oxi hoa manh: CaCl, + a, + H2O ra't y€u: 5 K C I O 3 + 6P -> — = mhh k i m loai lidn tife'p trong ciing mot nhom A : M - Trong kh6ng khf N a c i o + CO, + ao 5KC1 + 3P2O5 ^hh -^HClO + NaHcbj KCIO3 + 6HC1 -> + Tinh cac luong chat da cho trong bai ra. 2CaOCl, + CO, KCl + 3 0 , + 3H,0 + Viet phuong trinh hoa hoc (so d6 hoa hoc) six dung k i hi6u trung binh. + H , 0 ->CaC03 + CaCi, + 2HC10 + Tinh dai lucfng trung binh thich hop. + Rut ra kd't luan va tinh cac dai luong theo ydu c^u. tfng - Sat trung - Tiy trSng soi, vai, - Thudc n6, ditm. gia'y. Phuong phap bao toan electron: Trong phan umg oxi hoa khir, t6ng s6' mol dung - T^y trSng sod, vai, - Diiu che O , trong gia'y. - T^y u6' chu6ng phong thf nghidm. electron do chat khir cho phai diing bang t6ng s6' mol electron do chait oxi hoa nhan: trai, c6'ng ranh, h6' E e (cho) = S e (nhan) n , h „ . N , ,ho = n „ h j „ . N , „^,„ rac. Trong do: n^ho- "nhan l^n 'ifcrt la s6' mol nguyen tir chat nhircmg, nhan electron (s6' - XiJr ly chat d6c, mol nguyen tir chat khir, chat oxi hoa). tinh che' dSu mo. Nccho. N , „hj„ \in luot la s6' electron m6t nguyen tur chat khir nhuomg, m6t nguyen tijr chat oxi hoa nhan. + Viet cac qua trinh oxi hoa, qua trinh khir (kh6ng cin vie't phuong trinh phan ling oxi hoa - khir). + Dat in, dua vao djnh luat bao to^n electron lap duoc phuong trinh dai s6'. It. Giai he phuong trinh, xac dinh duoc M => chat cin t i m . . Q
  7. - Phuong phap bao loan k h o i lucmg: Trong mdt phan ung hoa hoc, tdng khO'i luong Hu&ng ddn gidi ciia cac cha't san phsim bang long khO'i liromg cua cac chat tham gia phan ling. Trudmg hcrp 1: Hai mudi NaX va NaY d^u tao ke't tiia vdfi A g N O , . Gia sir c6 phan ung: aA + bB cC + dD Dat cong thiic chung ciia 2 mudi la N a X : I h e o djnh luat bao toan kh6'i luong ta c6: niA + nig = nic + NaX + AgNOj -> A g X 4- + N a N O , A p dung: Trong mot phan ung, c6 n cha't {ki ca chat phan ling va san pha'm), n6'u biet kh6'i luong ciia (n - 1) cha't thi tinh duoc khd'i lucmg cua ch&t con lai. (23 + X ) g (108+X)g * M o t s6' n6i dung cin chu y: 6,03 g 8,61 g ' - Cac k i m loai c6 nhieu hoa tri (Fe, Cr,...) tac dung voi CI2, axit H C l tao ra mu6'i Ta c6: (23 + X ) . 8,61 = 6,03. (108 + X ) X = 178 (khdng c6 halogen nao c6 clorua k i m loai C O hoa t r i khac nhau: ' " " trong t u nhien ma c6 M > 178 => loai truong hofp nay). 2Fe + 3 a . - > 2FeCl3 Tru&ng hop 2: X la flo, Y la clo: i Fe + 2HC1 -> FeQ, + H , Goi X , y \in luot la sd mol ciia NaF, NaCl. - - - Cac dung dich mudi halogenua nhu mu6'i clorua, bromua, iotua d^u tac dung voi Theo bai ra, ta c6: 42x + 58,5y = 6,03 (1) ' J4 i . A g N O j , con muoi florua khong tac dung duoc voi dung djch A g N O , : NaF + A g N O j > khdng xay ra . NaF + A g N O j : Khong xay ra NaCl + A g N O j > A g C U + NaNOj NaCl + A g N O j Agai + NaNOj NaBr + A g N O j ^ A g B r i + NaNO, y y (mol) Ta c6: 143,5y = 8,61 y = 0,06 (2) Nal + AgNO, A g l l + NaNO, V i vay cac bai tap h6n hop mu6'i halogenua tac dung vdri dung dich A g N O j tao TCr ( 1 , 2) ta c6: x = 0,06; y = 0,06. ra ke't tiia cSn chia hai trucmg hop d^ giai. 2. Cac t h i d u m i n h hoa: 6,03 Dap an diing la D. Thidu 1: Hoa tan hoan toan 1,1 gam h6n hop gom m6t k i m loai kiim X va m6t Thi du 3: Hoa tan hoan toan 6,645 gam hdn hop mudi clorua ciia hai k i m loai k i m loai ki6m thd Y (Mx < My) trong dung djch H C l du, thu duoc 1,12 lit khi ki^m thudc hai chu ky ke' tiep nhau vao nude duoc dung dich X . Cho toan bo H , (dktc). K i m loai X la dung djch X tac dung hoan toan vdi dung dich A g N O j (du), thu duoc 18,655 A. K. B. Rb. C.Li. D.Na. gam ke't tiia. Hai k i m loai ki6m tren la (Trich de tuyen sink Cao dang khoi A) A. R b v a C s B. N a v a K C. K va Rb D. L i va Na Hudfng din gidi (Trich de tuyen sinh Cao dang - Khoi A) Theobaira: nn^ =0,05 (mol) => 0,05
  8. Thi du 4: Di hoa tan hoan loan 6,4 gam h6n hop g6m k i m loai R (chi c6 hoa trj Fe + 2Fea3 > BFeClj I I ) va oxit cua no cdn viira dii 400 m l dung dich H C l I M . K i m loai R la Dap an diing la A . A. M g B. Be C. Ba D . Ca Thi du 7: X la k i m loai thu6c phan nhom chinh nhom I I (hay nhom I I A ) . Cho (Trich de tuyen sinh Cao dang khoi A) 1,7 gam h6n hop g6m k i m loai X va Zn tac dung vdi luong d u dung dich H C l , Hudng dan gidi sinh ra 0,672 l i t khi H j ( d dktc). Mat khac, khi cho 1,9 gam X tac dung v6i T h e o b a i r a : HHCI = 0 . 4 . 1 = 0 , 4 ( m o l ) luong d u dung djch H^SO^ loang, thi thi tich khi hidro sinh ra chua d€n 1,12 lit ( 0 dktc). K i m loai X la O x i t ciia k i m loai R la R O (R c6 hoa t r i I I ) . K i hifiu chung cho k i m loai R va A. Mg. B. Sr. C. Ba. D . Ca. oxit R O la X . • (Trich De thi tuyen sinh Cao dang khoi A) PTHH: X + 2HC1 RCI2 + . . . ^' ' Hu&ng ddn gidi 0,2 0,4 (mol) S6 mol k h i H j trong hai thi nghiem: n H 2 = 0,03 (mol); ni^^ < 0,05 (mol). ^ M R . R O = M x = 6 , 4 / 0 , 2 = 32 PTHH: E(X,Zn) + 2HC1 > ECl^ + H, ViMx=32nen j ^ ^ ^ ^ ^16 56,67 => Mx < 56,67 (1) Dap an diing la PTHH: X + H , S 0 4 (loang) > XSO4 + H^ Thi du 5: Cho 1,9 gam h6n hop mu6'i cacbonat va hidrocacbonat cua k i m loai Ux < 0,05 (mol) n^^^< 0,05 (mol) k i ^ m M tac dung he't v6i dung dich H C l (du), sinh ra 0,448 1ft k h i ( 0 dktc). Suyra: M x > 1,9/0,05 = 38 (2) K i m loai M la Tilr(l,2)tac6: 38 < Mx < 56,67. A.Na. B. K . CRb. D.Li. Trong cac k i m loai nhom I I A chi c6 k i m loai canxi (M = 40) thoa man. (Trich De thi tuyen sinh Dai hoc khoi B) vay X la Ca. Hu&ng ddn gidi Dap an diing la D . Theobaira: nco2 = 0,448/22,4 = 0,02 (mol) Thi du 8: H 6 n hop X g6m M3CO3, MHCO3 va M Q ( M la k i m loai k i ^ m ) . Cho T a c o M h h = 1,9/0,02 = 95 . 32,65 gam X tac dung vijra dii vofi dung dich HCl thu duoc dung djch Y va c6 Suy ra: 2 M + 60 > 95 > M + 61 => 17,5 < M < 34. 17,6 gam C O , thoat ra. Dung djch Y tac dung vdi dung dich A g N O j d u duoc 100,45 gam ket tiia. K i m loai M la Vay M = 23 (Na). A.Rb B. Na C.Li D. K Dap an dung la A . Hu&ng ddn gidi Thi du 6: Cho k i m loai M tac dung v6i C I , duoc mu6'i X ; cho k i m loai M tac dung vdfi dung dich H C l duoc mu6'i Y . Ndu cho k i m loai M tac dung vdi dung S6 m o l khi CO,: nco^ = I M = o,4 (mol) dich mu6'i X ta cung duoc mu6'i Y . K i m loai M c6 t h ^ la 2 44 A . Fe. B.Al. CMg. D. Zn. S6' mol ke't tua A g C l : n^gci = 100,45 /143,5 = 0 , 7 ( m o l ) (Trich De thi tuyen sinh Cao dang khoi A) Goi X , y , z Idn luot la s6' mol M^COj, MHCO3, M C I trong 32,65 gam X . Hu&ng ddn gidi Bao toan cacbon ta c6: x + y = 0,4 Trong cac k i m loai da cho, M c6 th^ la Fe. That vay: Bao toan clo ta c6: 2x + y + z = 0,7 2Fe + 3Cl2 y 2FeCl3 A p dung djnh luat bao toan khd'i luong ta c6: Fe + 2Ha > FeCl2 + H2 13
  9. n i x + niHci = mMCi(Y) + "^002 + " I H 2 0 Thi du 10: Cho 19 gam h6n hop b6t g6m k i m loai M (hoa tri khdng d6i) va Zn (ti IS m o l tirong umg 1,25 : 1) vac binh dirng 4,48 lit khi C I , (dktc), sau cac => 32,65 + 36,5(2x + y ) = ( M + 35,5).(2x + y + z) + 17,6 + 0,4.18 phan ufng hoan toaii thu duoc h6n hop chat rSn X. Cho X tan h6't trong dung rr> 36,5(x + y ) + 36,5x = ( M + 35,5)(2x + y + z) = 7,85 dich H Q (dir) tha'y c6 5,6 lit khi H , thoat ra (dktc). K i m loai M la A. Na B. Ca C. M g D. A l Thay x + y = 0,4;2x + y + z = 0,7 vao, tadiroc: , Huatng ddn gidi 0 , 7 ( M + 3 5 , 5 ) - 3 6 , 5 = 22,45 = > 0 , 7 M - 3 6 , 5 x - - 2 , 4 • '- S6'mol k h i CI, va H , Mn lirot la: n ^ j = 0 , 2 ( m o l ) ; n^^^ = 0 , 2 5 ( m o l ) 2,4 + 0 , 7 M " ,, => X = Dat X la s6' m o l Z n => s6' mol k i m loai M la 1,25x 36,5 Taco: 6 5 x + M . l , 2 5 x = 19 ( l ) V',' V i 0 < x < 0,4 => M < 17,4 => M = 7 ( < 17,4). Vay k i m loai k i ^ m la L i . Theo nguydn t i c bao toan electron ta c6: Dap an diing la C. l,25n.x + 2x = 0,4 + 0,5 = 0,9 (2) si Thi dti 9: H6n horp X chira dong ihbi hai mu6'i natri cua hai halogen lien tife'p 65x + M.l,25x 19 65+l,25M 19 trong bang tuSn hoan. La'y mot luorig X cho tac dung vira dii \6'\0 m l dung TO ( 1 , 2 ) ta c6: dich A g N O , I M thi thu duoc 15 gam ket tiia. C6ng thiJc phan tijf ciia hai muoi l,25n + 2 0,9 l,25n,x + 2x 0,9 trong X la (cho F = 19; Q = 35,5; Br = 80; I = 127; Na = 23) => 25,75n + 38 = 58,5 + 1,I25M A. NaF va NaCl B. NaBr va N a l C. NaCl va N a l D. NaCl va NaBr =>l,125M = 23,75n-20,5 Hu&ng dan gidi N 1 2 3 M le 24 le S6 mol A g N O j : n^gNOj = 0 - ' ("loO • Ket luan Loai Mg Loai ' Trirorng hop 1: Hai halogen d^u tao diroc kfi't tiia v6i A g N O j . Dat c6ng thufc chung ciia mu6'i la NaX. Gia t r i phii hop la n = 2 => M = 24 (Mg) PTPU': N a X + A g N 0 3 -> A g X i + N a N O j Dap an diing la C. •> 0,1 -> 0,1 3. Cac bai tap t u l u y e n : Cku i : K i m loai nao sau day tac dung voi dung djch HCl loang va tac dung vdi khf ^ M . ^ = - - - 1 5 0 0,1 CI3 cho cung loai mu6'i clorua k i m loai? CI ( M = 35,5) A.Fe B.Zn C. Cu D. A g 108 + X = 1 5 0 = > X = 42 Hu&ng ddn gidi Br ( M = 80) K i m loai do la Z n : Hai halogen la Q (35,5) va Br (80). vay hai mu6'i trong X la NaCl va NaBr. Zn + 2HC1 >ZnCl2 + H2 Dap an dung la D. Zn + Cl2 >ZnCl2 Trucmg hop 2: Hai mu6'i la NaF va NaCl. Dap an diing la B. '^ N a F 4 - A g N O j -> kh6ng xay ra Cha v.- Phan ling ciia cac k i m loai v6i khi CI, va axit H C l : NaCl + A g N O j AgCl i + NaNOj FeClj 0,1 mol CuCl, «_ ± £ l l _ Cu - t H C ^ Khdng xay ra ' => "^Agca =0.1.143,5=14,35(gam) ^1.5(gam) AgCl ^ -C!2_ Ag—^Khangxayra. loai tnrcmg hop nay. 15
  10. C&u 2: Cho m6t lirong don chat halogen tac dung h6't vdd magie thu dirge 19 gam C&u 4: Cho dung djch chura 6,79 gam h6n hop g6m 2 mu6i K X , K Y (X, Y m 2 magie halogenua. Cung lugng don chat halogen do tac dung h6t vdi nhom tao ra nguydn t a mol (2) K Y + AgNO, ^ AgY + K N O 3 3X2 + 2A1 > 2AIX3 S6 mol h6n hop KX, K Y = ^'^^ ~ ^'^^ = 0,0264(mol) a mol > 2a/3 mol 108-39 ^ ^ Theobaira: a.(24 + 2X) = 19 vay: M h h - - ^ ^ = 2 5 7 , 4 ( g / m o l ) = ^ M x , y - 2 1 8 , 4 ( g / m o l ) ' ; 0,0264 ^ . ( 2 7 . 3 X ) = ,7,8 =,ll?itl^ = J L =,x=35.5 (a) vay khdng CO X, Y phu hgp. 3 ^ ^ 2.(27+ 3 X ) 17,8 ^ Trucmg hop X 1^ F, Y la CI: AgF tan n6n chi c6 AgCl kd't tiia Vay halogen do la clo (CU). Tir (2) ^ s6' mol KCl = s6' mol AgQ = 8,61 : 143,5 = 0,06 (mol) a=- =0,2 vay: mKci = 0,06 . 74,5 = 4,47 (gam) => m K P = 2,32 (gam) (24 + 2.35,5) Vay mei2 =71.a = 14,2(gam). %m(KF) = — x l O O % =34,17%. ' " ^ ^ 6,79 Dap an dung la B. Dap an diing la A. C&u 3: Cho 1,03 gam mu6'i natri halogenua (A) tac dung vdi dung dich AgNO, du Cau 5: M6t h6n hcyp X g6m 3 muoi halogenua ciia kim loai natri nang 6,23g hoa thi thu dirge mot kd't tua, ke't tiia nay sau khi phan iJng hoan loan cho 1,08 gam tan hoan toan trong nuoc dugc dung dich A. Sue khi clo du vao dung dich A roi bac. C6ng thiic ciia mu6'i A la c6 can hoan toan dung dich sau phan ung dugc 3,0525g mu6'i khan B. La'y m6t A. NaF. B.NaCl. C. NaBr. D. Nal. nua lucmg mu6'i nay hoa tan vao nude r6i cho phan ihig voi dung djch AgNO, du Hu&ng ddn gidi thi thu dugc 3,22875g ket tiia. Tim c6ng thiic ciia cac mu^i va tinh % theo khd'i I no lugng m6i mu6'i trong X. Sd'molAg: n^ = — - 0 , 0 l ( m o l ) Hu&ng ddn gidi Gia sii lugng mu6'i khan B thu dugc sau khi cho clo du vao dung djch A chi c6 Dat mu6'i natri halogenua la NaX. PTHH: NaX + AgNOj > AgXi + NaNOj NaCl-> n^aci = - ^ ^ = 0,0522 mol ' 58,5 ** ' 0,01 < 0,01 (mol) NaCl + AgNO, A g a i + NaNO, (1) AgX Ag + -X2 * 3 22875 Theo (1) -> n^^ci = "Aga = -2 = 0,045 mol < 0,0522 mol t 0,01 23 + M x - 1 0 3 =^ -80(Br) niNaF = m B - mNaci = 3,0525 - 0,045. 58,5 = 0,42 (g) , vay mudi A 1& natri bromua (NaBr). 0 42 Dap an diing la C. %NaF = ——.100% = 6,74%"'^'*''-»-""''«*"'-»-'"-^*":'^^^ 6.23 THiy mji TiwH oiNii Tmm 17
  11. Goi c6ng thiJc chung ciia hai mutfi halogen c6n lai 1^: NaY S6' mol I2 = ^ : ^ = 0 , 0 2 4 ( m o l ) 2NaY + a 2 - > 2 N a C l + Y2 (2) 0,024 T h e o ( 2 ) - > n ^ ^ Y = n N a a =0.045 (mol) > .^ - Theo phirong trinh s6' mol R Q O , = •"NaY "'x - niNaF = 6,23 - 0,42 = 5,81 (g) l,086x R+35,5+16x = R=29,25x~35,5 0,024 Do 66: Y= = 129,11 = 23 + M y ^ M y = 106,11 ' ' 0,045 , . yj^ Thoa man vdi x = 2; R = 23, kim loai k i i m la Na; mu^i \h NaQO^. -> ph^i c6 m6t halogen c6 M > 106,11 - » d6 iot. • v ^tf 2. Phuong trinh phan umg nhifit phan mu6'i: Vay c6ng thiic cua mud'i thiJ 2 Ilk Nal. .0 Dod6c6haitrucmghop: " '« f-*' NaClOj — ^ NaQ + O, * Tru&ng hop 1: NaF, NaCl vk Nal. Theo phuong trinh s6' mol NaQ = s6 mol NaQOj = 0,012 Goi a, b Mn lugt la s6 mol ciia NaQ Nal # => Kh6i luong NaQ = 0,012. 58,5 = 0,702 g. ^ ^ [58,5a + 150b = 5,81 [ 3 = 0,01027 Cau 7: Cho 50 gam dung dich M X ( M la kim loai ki6m, X 1^'halogen) 35,6% tac laco: < dung vcri 10 gam dung dich AgNOj thu duoc ke't tua. Loc kfi't tiia, duoc dung a + b = 0,045 [b = 0,03472 djch nude loc. Bie't n6ng d6 M X trong dung djch sau thi nghidm giam 1,2 ISn so n i N ^ = 58,5. 0,01027 = 0,6008 (g); niN,, = 150. 0,03472 = 5,208 (g) vdi n6ng d6 ban ddu. Vay: %Naa = ^ ^ 1 ^ . 1 0 0 % = 9,64% C6ng thurc mu6'i M X la 6,23 A. NaF. B.KI. C. L i Q . D.NaBr. ' • %NaF = 6,77%; %NaI = 83,59% Hudng ddn gidi * Tru&ng hop 2: NaF, NaBr va Nal. r,, , ^• 35,6x50 o , X Tac6: j l 0 3 a ' ^ 1 5 0 b ' = 5,81_^ ra' = 0,02 Theo bai ra: mMx = = 17,8 (gam) [a' + b' = 0,045 [ b ' = 0,025 100 ,0 - = M X + AgNO, -> M N O 3 + AgX i Ttif,^ = 103. 0,02 = 2,06 (g); niN„ = 150. 0,025 = 3,75 (g) X X X X (mol) 2 Ofi vay %NaBr = — . 1 0 0 % = 33.07% => m^gx = (108 + X).x; HIMX phin = (M + X).x 6,23 => mMXc6„i,i = 17,8 -(M + X).x %NaI = 141-100% = 60,19%; %NaF = 6,74%. => C% M X trong dung dich sau phan ihig la: 6,23 _17,8-(M+X).x ^ 3 5 ^ ^ j2o.(M + X ) =35,6.(108 + X ) C&u 6: Ho^ tan 1,086 gam mu6'i chiJa oxi ciia clo vdi m6t kim loai ki^m v^o nude, 6 0 - ( 1 0 8 + X).x 1.2 axit hod dung djch bang axit H2SO4 loang rdi them tiT dung dich K I vao cho de'n khi kh6ng c6n I , thoat ra thi thu duoc 6,096 gam I2. CQng luong mu6'i trdn dem M L i (7) Na(23) K(39) nhiet phan hoan to^n sau phan ihig con lai a gam mu6'i. X a (35,5) 12,58 4634,44 1. Xdc dinh c6ng thiic mu6'i? ' (') MlaLi,Xiaa mu6i M X la L i a . 2. Tfnha? Dap dn dung la C. Hu&ng ddn gidi 1. C6ng thiic mu6'i c6 dang RQO, (x = 1 -> 4). Phuong trinh phan umg: Rao, + 2xKI + xH,S04 ^ RQ + xlj + xK,S04 + xH^O 18 19
  12. Dang 2: Xdc dinh luffng chat (tham gia, sdn phdm, cdn Iqi sau phan itng) Hu&ng ddn giai vd thdnh phan cdc chat trong hdn harp Goi s6 mol Fe304 la x => s6 mol Cu la 3x 1. L i thuy^'t v$n dung va phuong phap glai: T a c 6 : 2 3 2 x + 64.3x = 4 2 , 4 = > x = 0 , l * Cac phucfng phap giai nhanh duoc sir dung: Fe304 + 8HC1 2FeCl3 + FeCl2 + 4 H 2 O - Phuong phap bao toan electron: Trong phan ung oxi hoa khu, t6ng s6' mol 0,1 -> 0,2 A.-,J/-' 'rf^ electron do chat khu ciio phai diing bang tdng s6' mol electron do chat oxi hoa nhan: Cu + 2FcCl, CUCI2 + 2FeCl2 ''"'"'"^ ' S e ( c h o ) = E e ( n h a n ) => iicho-Necho = n„hi:,. N,„h,„ + Viet cac qua trinh oxi hoa, qua trinh khijf (khong cin vie't phucmg trinh phan 0,1 0,2 • •••• ling oxi hoa - khir). = > n c u ( d . , - 0 . 1 - 3 - 0 , 1 - 0 , 2 (mol) + Dat ^n, dua vao djnh luat bao toan electron lap duac phuong trinh dai s6'. Vay m = mcu (du) = 0 . 2 . 64 = 12,8(gam) • + Giai he phircmg trinh, xac djnh cac dai luong theo yfiu ciu ciia bai toan. Dap an dung la D . ' '' " '' - Phuong ph^p bao toan khoi luong: Trong m6t phan ung hoa hoc, tdng kh6'i luong Chu v: +) Fe304 = Fe203.FeO ' ciia cac chat san ph^m bang t6ng kh6'i luong ciia cac chat tham gia phan ung. +) Tuy Cu kh6ng tan trong H C l , nhung bj hoa tan bod FeCl,. A p dung: Trong m6t phan ung, c6 n chat (k^ ca chat phan ung va san ph^m), — —~— t ' ne'u bie't kh6'i luong cua (n - 1) chat thl tinh duoc khd'i luong ciia chat con lai. Thi du 2: Hoa tan hoan toan 2,7 gam h6n hop X g6m Fe, Cr, A l bang dung djch - Phuong phap su dung cac cong thurc giai nhanh: H C l du, thu duoc 1,568 lit khi H j (dktc). Mat khac, cho 2,7 gam X phan ling + C6ng thiic tmh kh6'i luong mu6'i clorua thu ducfc khi hoa tan hit h6n hop kim hoan toan vdi k h i du, thu duoc 9,09 gam mud'i. Khd'i luong A l trong 2,7 loai bang dung dich H C l : m„„,„, = m^i^^,^ + ^ I H H J gam X la bao nhiau? + C6ng thiirc tinh kh6'i luong mu6'i clorua thu duoc khi hoa tan h6't h6n hcyp oxit A . 1,08 gam. B. 0,27 gam. C. 0,81 gam. D . 0,54 gam. k i m loai bang dung djch H C l : m,|„„, = mh4„ + 27,5nHci (Trich de tuyen sink Cao dang nam 2012 - Khoi A) * M6ts6'chuy: Hu&ng ddn gidi - Fe304 khi giai bai tap c6 the qui doi thanh: FeO.Fe^Oj. Theo bai ra: H j j ^ = 0 , 0 7 (mol) FejO^ + 8HC1 - > 2FeCl3 + FeCl, + 4 H , 0 Goi x , y , z 1 & luot la so mol F e , C r , A l trong X. Do do h6n horp Cu, Fe304 c6 tha' bi tan hat trong dung djch H C l (Cu khdng tac =>56x + 52y + 27z = 2,7 (l) 3!::: dung vdi dung djch HCl nhung bj hoa tan boi dung dich FeClj): Cu + 2Fea3 CuCU + 2 F e a 2 *) X + dd H Q du: - M u d i clorua bj oxi hoa boi KMn04/H2S04: Fe + 2HC1 - > F e C l j + H j t Cr + 2HC1 ^ C r Q j + H2 t IOAICI3 + 6KMn04 + 24H2SO4 5Al2(S04)3 + 3K,S04 + 6MnS04 + 24H2O X -> X y -> y - K i m loai Fe (hoac Cr) tac dung voi CU va axit HCl tao ra san ph£m mu6'i clorua A l + 3Ha ^ AICI3 + l.SHj t khac nhau: 2Cr + 30, - > 2CrCl3 Cr + 2HC1 CTCXJ + H , z l,5z _ . 2. C a c thi du minh hoa: .; =>x + y + l , 5 z - 0 , 0 7 (2) *) X + C l , ( d u ) : ' • ' Thidu 1: Cho 42,4 gam h5n hcrp g6m Cu va Fe304 (c6 t i le so mol tuong ung la 3 : 1) tac dung vori dung dich H C l du, sau khi cac phan ung xay ra hoan toan Fe + - C l 2 - > F e a 3 Cr + - C l 2 ^ C r Q j 2 2 3 2 ^ c6n lai m gam chat ran. Gia trj ciia m la x->l,5x y->1.5y A . 6,4. B.9,6. C. 19,2. D . 12,8. \ (Trick de tuyen sink Cao dang khoi A)
  13. A1 + -C12 • AlCl, fhi du 4: D6't 16,2 gam hOn hop X g6m Al va Fe trong khi CI2 thu duoc h6n hop 2 ^ chat ran Y . Cho Y vao nude du, thu duoc dung dich Z va 2,4 gam kim loai. z—•l.Sz Dung dich Z tac dung duoc vdfi t6i da 0,21 mol KMn04 trong dung dich H2SO4 (kh6ng tao ra SO2). Phdn tram khd'i luong ciia Fe trong h6n hop X la A. 72,91%. ' B. 64,00%. C. 66,67%. ' D. 37,33%. =>x + y + z = 0,06 (3) (Trich di thi tuyin sink DH khoi B) Tir (1, 2, 3) =>x = 0,02; y = 0,02; z = 0,02 Hu&ng ddn gidi • Vay iTiAi/x -0,02.27 = 2,54(g) , , j) ? in Goi X , y l^n luot la s>6 mol Al, Fe phan ling. Dap an diing la D. Tac6: 27x + 56y = 16,2-2,4 = 13,8 (l) ' .K, Theo djnh luat bao toan electron ta c6: ' Thi du 3: H6n hap X c6 khd'i luong 82,3 gam g6m K Q O j , €3(003)2, CaQj va 3(x + y ) = 0,21.5=>x + y = 0,35 (2) K Q . Nhiet phan ho^n to^n X thu duoc 13,44 lit O2 (dktc), chat ran Y g6m CaQz va KCl. Toan b6 Y tac dung vira du vdi 0,3 lit dung djch K2CO3 IM thu Tir(l,2) =>x=0,2;y = 0,15. duoc dung dich Z. Luang KCl trong Z nhi^u ga'p 5 15n luong K C l trong X. Khdi lirang Fe trong X nam trong khoang: ' •' PhSn tram khd'i luong KCl trong X la 0,15.56 2cr vay %m^„(^plii2l!^. 18,10% 0 2 + 4e 20-^ f 82,3 Ag*+ le ^ Ag ' Dap an diing la A. Theo nguyfin tac bao toan electron, ta c6: '^^ ' ' ' 0,08.2 + 0,08.3 = 2x + 4y + Iz =>2x + 4y+ z = 0,4 (l) 99 23
  14. 2H* + - > H2O O2 + 4e -> 20^" Hu&ng ddn gidi 3 36 0,24->0,12 0,06 0,12 Theobaira: nci2 = r T 7 = 0,15(mol) 22,4 =>y = 0,06 (2) K2Cr207 + 14HC1 2KC1 + 2CrCl3 + 3CI2 + 7H2O Ag* + l e - > Ag J. Ag+ + Cr ^ AgCli 0,05 0,7 (2x +0,24) =>a = 0,05; b = 0,7 => 108.Z + 143,5.(2x + 0,24) - 56,69 Ddp &n diing la A. =>108z + 287x = 22,25 (3) Thi du 8: Cho m gam h6n hop X g6m FeO,Fe203,Fe304 vao m6t luong vira du Tiir(l,2, 3) =>x=0,07;y = 0,06;z = 0,02 dung djch H Q 2M, thu diroc dung dich Y c6 ti 16 s6' mol Fe^"" va Fe^^ la 1: 2. Chia Y thanh hai ph^n bang nhau. C6 can phSn m6t thu duoc m, gam mu6'i Vay % V c , , h , = % n c , , / h h = - ^ ^ ^ ^ ^ ^ ^ ^ = 53,85% khan. Sue khi clo (du) vao phfo hai, c6 can dung dich sau phan ling thu duoc •^ a2/hh ci2/hh 0,07 + 0,06 m2 gam mu6'i khan. Bid't mj - m, = 0,71. Th^ tich dung djch HCI da dung la Dap an diing la C. A. 320 ml B. 80 ml C. 240 ml D. 160 ml (Trich de tuyen sinh Cao dang khoi A) Thidu 6: Cin t6'i thi^u bao nhidu gam NaOH (m,) va (mj) de' phan ling hoan tokn vdi 0,01 mol CrQj. Gia tri cua m, va mj iSn luot la Hu&ng ddn gidi A. 3,2 va 1,065 B. 3,2 v^ 0,5325 Vi Fe304 = FeO.Fe203 => coi h6n hop X chi gom FeO (x mol) va FejOj C. 6,4 va 0,5325 D. 6,4 vk 1,065 (y mol) (Trich de thi du bi dai hoc) Ta c6: n^^c\2 = "FeO = x (mol); npecij = 2nFe203 = 2y (mol) Hu&ng ddn gidi V i ti le s6 mol Fe2+ va Fe^* la 1 : 2 PTHH: =>nFeCi2 :"FeCi3 = 1:2 => X : 2y = 1:2 =>x = y . 16NaOH + 2Cra3 + 3CI2 ^ 12NaCl + 2Na2Ci04 + 8H2O CI2 + 2FeCl2 2FeCl2 0,08
  15. Cdchl: Bao toan electron 3. Cac bai tskp tu luyen: +7 +2 -1 Mn + 5e CI2 + 2e C&u 1: Cho 2,13 gam h6n hop X g6m ba kim loai Mg, Cu \k A l 6 dang b6t tac Mn 2a dung hoan toan \cn oxi thu dugc h6n hop Y g6m cac oxit c6 kh6'i luong 3,33 0,02 ^ 0 , 0 1 (mol) 0,1 0,1 (mol) gam. The' ti'ch dung dich HCl 2M vifa dii d^ phan umg he't voi Y la S6 mol H Q bj oxi hoa la 0,10 mol A. 90 ml. B. 57 ml. C. 75 ml. D. 50 ml. Dap an diing la A. (Trich De thi tuyen sinh Dai hoc khoi A) Cdchl: Sir dung phuong trinh ion rut gpn: Hu&ng ddn gidi 2Mn04 + I6H+ + lOCF ^ IMn^^ + 5CI2 + 8H2O Ta C O so d6 phan ulig: ^' * •' * ' ' ' ' ''^'Pi i,"'- ^ •. 0,02 -> 0,1 (mol) Kimloai+oxi > oxit ..fi:,Z7s iFi i;. . _,2 5- = > HHCI (bi oxi hod) = 1^.,- (bj oxi ho4) = 0,10 (mol) ( 0 + 2e > O) ^,,a Dap an diing la A. Dodo: m^imioai + moxi = m,,,;, Chu y; Trong phan ung nay, HCl dong hai vai tro: . => m„,i = m„,,i, - mk,^,o,i = 3,33 - 2,13 = l,2g -1 12 + Bj oxi hoa (chat khijr): 2C1 -> Clj + 2e + M6i tnrcmg (tao mu6'i clorua) => Ho = — = 0,075 (mol) = n 2 v ' ' Do do kh6ng ti'nh diroc luong H Q bi oxi hoa theo phucfng trinh phan ling dang Qua trinh hoa tan oxit vao dung djch axit: phan i\X. -2 Thi du 10: Didn phan dung dich CuQj voi difin cue tro, sau m6t thoi gian thu O + 2H" > H.O , = 2 . n 2 = 2 . 0,075 = 0,15 (mol) duoc 0,32 gam Cu b catot va m6t lucfng khi X b anot. HSfp thu hoan toan lugtng khf X tr^n vao 200 ml dung dich NaOH (d nhiet d6 thucmg). Sau phan = > HHCI = n ^+ = 0,15 (mol) ,_ ung, ndng d6 NaOH con lai la 0,05M (gia thie't th^ tich dung dich kh6ng thay ddi). N6ng d6 ban ddu ciia dung dich NaOH la vay: HO = ^ = 0,075 lit = 75 (ml). A.0,15M. B.0,2M. C.0,1M. D. 0,05M. Dap an diing la C . (Trich De thi tuyen sink Dai hoc khoi A) C^u 2: Di hoa tan hoan toan 2,32 gam h6n hcpfp g6m FeO, Fe304 v^ FcjO, (trong Hu&ng ddn gidi do s6' mol FeO bang s6' mol Fe^O^), cSn diing vita dii V lit dung djch H Q I M . 0,32 Gia trj cua V la S6 mol kim loai Cu: n^^ - = 0,05(mol); 64 A. 0,08. B.0,16. C.0,18. D. 0,23. S6' mol NaOH du sau phan ihig: nN,oH(d„) = 0,05.0,2 = 0,01 (mol). (Trich De thi tuyen sinh DH khoi A) ^ CuCU > Cu + a, Hu&ng ddn gidi 0,005 0,005 (mol) V i s6' mol FeO bang s6' mol FcjOj ndn qui d6i h6n hap thanh Fe304 (FeO.FcjOj): a. 2NaOH NaQ + NaQO + H.O 0,005 0,01 (mol) S6 mol Fe304: np,304 = ^ - 0,01(mol) ^ => T6ng s6' mol NaOH ban ddu: PTPIT: Fe304 + 8HC1 > FeCl^ + 2FeCl3 + 4H2O 2^nN,OH (bandiu) = HNaOH (p/x) + "NaOH (du) = 0,01 + 0,01 = 0,02 (mol) 0,01 -> 0,08 (mol) 0,02 Vay n6ng do ban ddu ciia dung dich NaOH: CN,OH (ban d4u) = = 0,1M 0,2 Suy ra th^ ti'ch dung djch HCl cSn dung: V,
  16. Cau 3: De oxi hod hoan to^n 0,01 mol C r a , th^nh K 2 C r 0 4 bang CI, khi c6 mSt Cau 5: Cho 9,12 gam h6n hop g6m FeO, FcjOj va Fe304 tac dung vdi dung djch KOH, lirong t6'i thi^u va KOH tucmg ling Ik HCl (du). Sau khi cac phan urig xay ra hoan to^n, duoc dung djch Y, c6 can Y A. 0,015 mol va 0,04 mol. B. 0,015 mol va 0,08 mol. thu diroc 7,62 gam FeClz va m gam FeQ,. Gia trj ciia m la C. 0,03 mol va 0,04 mol. C. 0,03 mol va 0,08 mol. A. 9,75. B. 8,75. C. 7,80. D. 6,50. (Trich De thi tuyen sink DH khoi A) (Trich De thi tuyen sinh Dai hoc khd'i B) Hudng ddn gidi Hu&ng ddn gidi 7 62 ' ' ' f - PTPU": ICrGj + 3 0 , + 16KOH ^ 2K2Cr04 + 12KC1 + 8H,0 SdmolFeQ,: n p ^ a 2 = - [ ^ = *^'06('"°') ^ .- 0,01 mol Vi Fe304 = FeO.FcjOj, nfen c6 th^ coi h6n hop g6m FeO (x mol) va FcjOj TheoPTPU": n^^ = - .ncK:i3 = -.0,01 = 0,015 (mol) (y mol). HKOH = 8- "ocb = 8- 0,01 = 0,08 (mol) v Theo bai ra, ta c6: 72x + 160y = 9,12 (1) Vay cin t6'i thieu 0,015 mol CI2 va 0,08 mol KOH. PTPU*: FeO + 2HC1 > FeCU + HjO X X Dap an dung la B. Cau 4: Nung h6n hop b6t g6m 15,2 gam CTJOJ va m gam A l & nhifit d6 cao. Sau Fe^Oj + 6 H a > 2FeCl3 + 3H2O ^ khi phan ihig hoan toan, thu duoc 23,3 gam h6n hop ran X. Cho toan b6 h6n hop y 2y X phan ling vdi axit HCl (du) thoat ra V lit khi H2 (6 dktc). Gia tri ciia V la n Feci2 = X = 0-06 (mol) (2) A. 3,36. B. 4,48. C. 7,84. D. 10,08. TO (1, 2) ta c6: X = 0,06; y = 0,03 (mol) (Trich Delhi tuyen sink DH - khd'i B) vay m = m p^ci, = 2.y. 162,5 = 2.0,03.162,5 = 9,75 (gam) Hudng ddn gidi 15,2 Dap an diing la A. S6' mol Cr^O,: nc^o3 = T T T = 0, '(mol) Cau 6: Cho m gam h6n hop X gdm A l , Cu vao dung dich HCl (du), sau khi V.€x 152 So 6.6 phan ling: A l + CfjOj -> H6n hop sau phan ihig thiic phan ling sinh ra 3,36 lit khi (o dktc). Neu cho m gam h6n hop X trSn vao mot luong du axit nitric (dac, nguoi), sau khi ka't thuc phan umg sinh ra 6,72 lit •"0203 + '"AI = nihhsp khi NO, (san phdm khir duy nha't, do 6 dktc). Gia tri ciia m 1^ => niA, = mhh,p - m C ^ Q J = 23,3 - 15,2 = 8,1 (g) A. 11,5. B. 10,5. C. 12,3. D. 15,6. (Trich De thi tuyen sinh Dai hoc - khoi B) => UA, = — = 0,3 (mol) Hu&ng ddn gidi PTPLT: Cr,03 + 2A1 > 2Cr + AI2O3 Theobaira: n^^ = ^ =0,15(mol); n^Q= ^ = 0,3(mol). ^ 22,4 ^ 22,4 Saupir: 0,1 0,1 (mol) 0,2 0,1 Sau phan ling nhiet nh6m c6: 0,1 mol A l (du); 0,2 mol Cr; 0,1 mol AI2O3. * X + HCl: 2A1 + 6HC1 > 2AICI3 + 3H2t X l,5x ; f ; 2 A 1 + 6Ha > 2k\C\^ + 3H2T ft, 0,1 - 0,15 (mol) Cu + HCl > khdng phan ung , . Cr + 2 H a > C r a , + HjT = 1,5. x = 0,15 =>x = 0,1 (mol). 0,2 0,2 (mol) * X + HNO3 Zn H2 = 0,15 + 0,2 = 0,35 (mol) Al + HNOj^ij, „g„oi) > bj thu d6ng hoa => =0,35. 22,4 = 7,84 (lit) . Cu + 4HN03, Cu(N03)2 + 2NO2 + 2H2O Dap an dung la C. y
  17. n NO2 = 2y = 0,3 => y = 0,15 (mol) vay CM(„a, = x/0,1 = 0.05/0,1 = 0.5M. Dap &n dung la B. Vay m = niA, + mc„ = 0,1.27 + 0,15.64 = 12,3 (g) c a u 9: Hoa tan he't 7,74 gam h6n hop b6t Mg, A l b ^ g 500 ml dung djch h6n hop Dap an dung la C. H Q I M va H2SO4 0,28M thu dugc dung djch X va 8,736 lit khi H2 (Jb dktc). C6 Chu v: * A l , Fe, Cr bi thu d6ng hoa trong HNO3 dSc, ngu6i. can dung djch X thu duoc luong mu6'i khan la Cau 7: Dung dich X chiia h6n hop g6m NazCOj 1,5M K H C O 3 I M . ^fh6 tir tv? A. 77,86 gam. B. 25.95 gam. C. 103,85 gam. D. 38,93 gam. timg giot cho de'n hfi't 200 ml dung djch H Q I M v^o 100 ml dung djch X, sinh (Trich Di thi tuyen sinh Cao ddng khd'i A) ra V 1ft khi (6 dktc). Gia trj cua V la • ,vi' Hudng ddn gidi A. 4,48. B. 1,12. C.2,24. D. 3.36. Theo b^i ra: n^a = 0.5. 1 = 0.5 (mol); ^ ^'l ' * ^ (Trich Be thi tuyen sinh Dai hoc khd'i A) nH2S04 = 0.5. 0.28 = 0,14 (mol); nH^ = 8.736/ 22.4 = 0,39 (mol). Hu&ngddngim Theo b^i ra: n^a = 0,2.1 = 0,2 (mol); n NajCOa = 0,1.1,5 = 0,15 (mol). Ta thafy = "82804 + ^na^ => Axit phan ling v&a he't voi kim loai. ' ' n K H c o 3 = 0.11 =0,1 (mol) ' Sord6: Kim loai+ Axit > mud'i + Hj ''^ ' ,. Khi nho tir tCr dung djch H Q (H") vao X thi: - % Na^COj + HCl > NaCl + NaHCO, =>m = 7,74 + 0,5.36.5 + 0.14.98 - 0,39.2 = 38.93 (gam). 0,15 > 0,15 > 0.15 (mol) Dap An diing 1^ D. Sau phan ling nay, luong HCl con lai la: n^am = 0,2 -0,15 = 0.05 (mol) Cau 10: Cho m gam b6t A l vao c6c chura V lit dung djch NaOH 2M, sau phan ling hoan tohn cho tiep dung djch HCl vao c6c 66 de'n khi cha't ran tan he't thafy cdn Do d6. xay ra qua trinh: diing 800 ml dung djch H Q I M va c6 3.36 lit khi thoat ra (cj dktc). Gia tri cua m HCO3 + H* > CO:t + H2O va V lin luot la A. 6.075 va 0.0625 B. 6.075 va 0,2500 0.05 < 0.05 — > 0,05 (mol) C. 7,425 va 0.0625 D. 3,375 v^ 0,2500 v a y V = Vco2 = 0,05.22,4 =1.12 (lit). Hu&ng ddn gidi Dap an dung la B. TTieo bM ra: HHCI - 0.8.1 = 0.8 (mol) c a u 8: Khi cho 100 ml dung djch KOH I M vao 100 ml dung djch HCl thu duot Khi thoat ra la H^: n^^^= 3.36 / 22.4 - 0.15 (mol) dung djch c6 chiia 6.525 gam chat tan. Ndng d6 mol (hoac mol/1) ciia H Q trong dung djch da dung 1^ 2Al + 2 N a O H + 6H2O ^ 2NaAI (OH)^ + 3H2 t A.0.75M. B.0.5M. C. I M . D. 0.25M. X -> x -> x (mol) (Trich Dim tuyen sinh Cao dang khoi A) NaAl(OH)^ + 4HC1 NaQ + AICI3 + 4H2O Huong ddngidi Theo bai ra: UKOH = 0,1.1 = 0.1 (mol). X - > 4x Phuong trinh phan ung: KOH + H Q > K Q + HjO 2A1 + 6HC1-).2A1C13+3H2 t Gia thid't KOH phan img he't => n^a = I I K O H = 0.1 (mol) y ->3y -). l,5y => mKc, = 0,1. 74,5 = 7,45 (g) > 6.525 (g). Tu ..v. , r4x + 3y = 0.8 Suy ra, KOH phan irng chira he't (HQ he't. chat tan c6 K Q v^ KOH). Theobairataco:-^ ^ ix> x = 0.125; y = 0.1 [1.5y=0.15 Goi X , y Idn luot 1^ s6' mol KOH phan ling voi HCl va c6n du. Vay: m = (x + y).27 = (0,125 + 0,l).27 = 6.075(g) J Theo b i i ra. ta c6: x + y = 0.1 (1) - i 74.5. X + 56. y = 6,525 (2) V = x/2 = 0,125/2 = 0,0625(1) Tir (1. 2) ta giai ra duoc x = 0.05. J>5pdndunglaA.
  18. Cty TNHHMTV mvu KhangVift C&u 11: H6n h
  19. Huang ddn gidi prPlT: CrjOj +6HC1 ^ 2CrCl3 + 3 H 2 O V i npeo = npejOa ^ Co' h6n hop la Fe304 (FeO.FePj). X 6x 2x CuO + 2HCl->-CuCl2 + H 2 O J, , , "Fe304 - •^'"^^ =0,01(mol); HHCI =0,08(mol) y ^ 2y y PTPLT: Fe304+8HCl-^2FeCl3 + F e C l 2 + 4 H 2 0 Fe304 +8HCl->2FeCl3 +FeCl2 + 4 H 2 O 0,01 0,08 -> 0,02 ->0,01 z ^ 8z 2z z ^ =>nFea3 =0.02(mol); npecij =O.OKmol) Dung djch Y g6m: 2x mol CrClj; y mol CuQj 2z mol FeQj; z mol FeClj. Y + Ni: FeCl3+3AgN03 >3AgCl I+Fe(N03)3 Ni + 2FeCl3-).2FeCl2+NiCl2 , , 0,02 > 0,06 (mol) z/2 3x + y +,4z = 0,55 (1) ^ m = mAgci +mAg =(0,06 + 0,02).143,5 + 0,01.108-12,56(gam) z/2+ y/2 = 0,05 => z + y = 0,1 (2) Dap an dung la D. CAn tinh: m=mcK:i3 +mcua2 +'"FeCi3 +'"FeCi2 " C^u 16: Cho 17,5 gam h6n hop A l , Zn, Fe tac dung hoan toan vdi dung dich HCl du thu diroc dung djch X va V lit H , (dktc). C6 can dung djch X thu duoc 53 =>m = 158,5+ 135.^+162,5.7+127.- 2 2 gam mu6'i khan. VSy gia tri cua V la =>m = 158,5x +67,5y + 226z. v A. 8,96 B.5,60 C. 11,2 D. 6,72 Hit&ng ddn gidi =>m=67,5(y + z)+158,5(x + z) (3) Sodd: MCL -)-M"++nCr TO (1, 2) vao (3) duoc: m = 67,5.0,1+ 158,5.0,15=30,525(gam) Dap an diing 1^ A. Cau 18: Kh6i luong t6'i thi^u NaOH Q , d^ oxi hoa hoan t o ^ vdd 0,» ' ;nol > =>53 = 17,5 + m„o,„, NaCrOz thanh Na2Cr041^ luot la => nioiorua = 53 - 17,5 = 35,5 (g) ^ n^^_ = l(mol) A. 1.6gva l,065g. B. 3,2 g va 2,13 g. 2H++2e ^HjT C. 1,2 gva 1,065 g. D. 1,6 g v^ 2,13 g. HCl ->H^+cr Hu&ng ddn gidi 1 0,5(mol) PTHH: ' vay V = 0,5 .22,4= 11,2 (1ft) 2NaCr02 + ^Clj + 8 N a O H - > 2Na2Cr04 + 6NaCl + 4 H 2 O Dap an dung la C . 0,01 -> 0,015-> 0,04 (mol) Cau 17: D^ hoa tan hat h6n hop X g6m Cr^Oj, CuO, Fe304 cin vira du 550 ml dung djch HCl 2M, sau phan ling thu duoc dung dich Y. M6t nijfa dung dich Y tha'y mNaOH = 0,04.40 = l,6g; m p j = 0,015.71 = l,065g ^ ^ hoa tan he't t6'i da 2,9 gam b6t Ni. C6 can nifa dung dich Y con lai thi thu duofc E)ap an diing la A. bao nhiau gam mu6'i khan? Cau 19: Ho^ tan hoan to^n m gam h6n hop g6m Na, Ba, K v^o nu6c thu duoc x A. 30,525 gam B. 30,8 gam C. 61,6 gam D. 61,05 gam gam khi H,. Ne'u cho m gam h6n hop trfen tic dung vdd dung djch H Q du, r6i c6 Hu&ng ddn gidi can dung djch sau phan ling thi khd'i luong mud'i khan thu duoc 1^ Theo bki ra: HHCI =1,1 (mol); =0,05 (mol) A. (m + 17x)gam B. (m + 35,5x)gam Goi X , y, z \in luot la s6' mol ciia CrjOj, CuO, Fe304 c6 trong h6n hop X. C. (m + 142x)gam D. (m + 17,75x)gam ,, 35
  20. Ttnwaat mm gtat nhann BTH0irn^srTap7 -Timtanli Toan f)iiu ch6 cac hop chat c6 oxi ciia clo: Hu&ng ddn gidi K i hifiu Chung ba kim loai 1^ M (hoa trj chung 1^ n). CI2 + 2NaOH(,„j„g) > NaQ + N a Q O + H2O 0 2 + Ca(OH)2(b«„ > CaOClj + H2O M + n H j O ^ M (OH)^ + f " 2 '^ 302 + 6KOH — ^ 5KC1 + K O O j + 3H2O M + nHa^Ma„+|H2 t • 6 0 2 + 6Ca(OH)2 — ^ € 3 ( 0 0 3 ) 2 + 5 C a a 2 + 6H2O Lufong H2 sinh ra trong 2 phan ihig n^y bling nhau. 2. C a c thi d u m i n h hoa: 2 X 'rhidu I Chat diing d ^ 1 ^ kh6 khi 0 2 im \h Vay kh6i lucmg mudl thu duoc: m^i^ ,„,i + m.^™. = m + 35,5x (g) A . NajSOj khan. B. dung dich H2SO4 dam dac. Dap an dung la B. C CaO. D- dung djch N a O H dac. ^ (Trick di tuyen sinh Cao dang - Khoi A) Dang 3: Dieu che, tinh che Hu&ng ddn gidi vd phdn Met cdc halogen vd hap chdi cua halogen Cha't CO th^ dung di lam kh6 khi 0 2 im phai la: v + Chaft hao nude 1. L i t h u y e t vakn dung va phuorng phap giai: + Chat khdng tac dung vdd O j - D i l u ch^ cac halogen trong phong thi nghidm va trong c6ng nghidp: Thoa man d i ^ u ki6n tr6n la dung dich H2SO4 dam dac. Dap an dung la B . 4HX(,) + MnO. — ^ MnX^ + + 2H2O Thi du 2: C6 4 6'ng nghidm duoc danh s6 theo tu: t u 1, 2, 3 , 4 . M 6 i 6ng nghifim 16HX,,) + 2KMn04 > 2 K X + 2 M n X 2 + X2 + 8 H 2 O chiia m6t trong cac dung dich A g N O j , Z n 0 2 , H I , Na2C03. Bife't rang: 6HX(„ + K Q O j > KCl + 3X2 + 3H2O - Dung dich trong 6'ng nghifim 2 va 3 t^c dung duoc vdd nhau sinh ra ch&t k h i ; 2NaX + MnOj + 2H2SO4,,, — ^ Xj + MnS04 + Na.SO^ + 2 H 2 O - Dung dich trong 6'ng nghifim 2 va 4 kh6ng phan ling duoc vol nhau. Dung dich trong cac 6'ng nghifim 1, 2, 3 , 4 1 ^ lupt 1^: (X la a , Br, I) A. ZnCl2,Na2C03,HI,AgN03 B. A g N 0 3 , H I , N a 2 C 0 3 , Z n C l 2 2 N a a + 2H2O > 0 2 ! + H2t + 2 N a O H m^ng ngan C. A g N 0 3 , N a 2 C 0 3 , H I , Z n C l 2 D. ZnCl2,HI,Na2C03,AgN03 xtt" 4 H C I + O2 < > 2 a 2 + 2H20 (Trick de tuyen sinh Cao dang khdi A) Hu&ng ddn gidi 2NaBr + a2(ve«du) — ^ Br2 + 2 N a a - 6ng nghidm 2 va 3 la H I , Na2C03: 2NaI + Cl2(w«du) — ^ I2 + 2Naa Di^u chd' cac hidro halogenua H X : 2HI + Na2C03 2NaI + C O j t + H 2 O - ^ n g nghiem 2 va 4 la H I , Z n C l j . CaF2„) + H2S04(d) - i i ^ 2 H F t + CaS04 i H I + Z n C l 2 k h 6 n g xay ra 2Naa„ + H2S04,
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