Lời giải đề thi học sinh giỏi hóa học 9

DE SO 1 DE THI HOC S1I\IH GiOl HOA HOC LOfP 9, TP. HQ CHI MINH NAM HOC 1998 - 1999 Cdu I. Viet 3 phuang trlnli khdc nhau de dicu die muoi ZnCl2. Cdu II Viet phuang trinh phdn ling de bieu dlen chiiSi bien hoa sau: FeCls > Fe(0H)2 ^ FeSO, > Fe(N0s)2 FeCh > Fe(0H)3 > FesOs > Fe. Cdu III. Co 6 ong nghiem dugc ddnh so tii 1 den 6 chica cdc dung dicli: NaOH; (NHJ.SO,; NaoCOs; Ba(N03)2; PbiNO^)^; CaCl Hay cho bict 6'ng mang so ndo dUng chat ndo? Viet phdn icng minh hga. Biet rdng: a) Dung dich (2) cho ket tila trdng vai cdc dung dich (1), (3), (4). ''• b) Dung dich (5) cho ket tua trdng vai cdc dung dich (1), (3), (4). c) Dung dich (2) khong tgo ket tila vai dung dich (5). d) Dung dich (1) khong tgo ket tua vai cdc dung dich (3), (4). e) Dung dich (G) khong phdn i(ng vai dung dich (5). f) Dung dich (5) bi trung hoa bdi dung dich HCl. g) Dung dich (3) tgo ket tua trdng vai HCl, khi dun nong ket tua nay se tan. CduIV. a) Nong do dung dich bdo hoa KCl d 40°C la 28,57%. Tinh do tan cua dung dich KCl a ciing nhiet do. » ) Xdc dinh lugng AgNOs tdch ra khi Idm Ignh 2500 gam dung dich AgNOs bdo hoa a 60°C xuong 10°C. Cho biet do tan cua AgNOs o 60°C Id 525 gam, a 10°C Id 170 gam. Cau V. (A) la dung dich H2SO4; (B) la dung dich NaOH. • Trgn 0,3 lit (B) vai 0,2 lit (A) dugc 0,5 lit (C). Lay 20 nd (C), them mot it quy tim vdo thdy c6 mdu xanli. Sau do them tic tit dung dich HCl 0,05M tai khi quy tim ddi thdnh mdu tim thdy het 40 ??il axit. Trgn 0,2 lit (B) vai 0,3 lit (A) dugc 0,5 lit (D). Lay 20 ml dung dich (D), them mot it quy tim vdo thdy c6 mdu do. Sau do them tii tii: dung dich NaOH 0,1M tai khi quy ddi thdnh mdu tim thdy het 80 ml dung dich NaOH. Tim nong do mol/l dung dich (A) vd (B). I r(i niAi nc TUI unr CIMM nifii HHA Hfin Q Cdu VI. Xdc dinh cong thiJtc ciia hai oxit sdt A vd B, biet r&iig: • 23,2 gam (A) tan vica dil trong 0,8 lit HCl IM. • 32 gam (B) khi k/iii bdng Ho tqo thdnh sdt vd 10,8 gam IhO. L6\I Cdu I. * Zn + CI2 > ZnCla . ZnO + 2HC1 > ZnCl2 + H2O • ZnS04 + BaCl2 > ZnCh + BaS04i Cdu 11. Fe + 2IIC1 > FeCl2 + H2t FeCl2 + 2NaOH > Fe(0H)2^ + 2NaCl (trdng xanh) Fe(0H)2 + H2SO4 > FeS04 + 2H2O FeS04 + Ba(N03)2 > BaS04^ + Fe(N03)2 2Fe + 3CI2 > 2FeCl3 FeCla + 3NaOH > FeCOIDai + 3NaCl (ndu do) • 2Fe(OH)3 —> Fe203 + 3H2O Fe203 + 3C0 —> 2Fe + 3CO2T Cdu III. Theo cac duf kien de bai neu ra, cac lo diing cac hoa chat sau: f) => Lo so 5 diing: Na2C03 g) Lo so 3 diing: Pb(N03)2 d) Lo so 1 chufa: Ba(N03)2, lo 4 chuTa CaCl2. a) => Lo 2 chijfa: (NH4)2S04. e) =^ Lo 6 chijfa: NaOH. Phan irng: (NH4)2S04 + Ba(N03)2 > BaS04i + 2NH4NO3 (NH4)2S04 + Pb(N03)2 > PbS04i + 2NH4NO3 (NH4)2S04 + CaCl2 > CaS04i + 2NH4CI ' Cac phan ilng con lai hoc sinh tii vie't. Cdu rv. a) Goi S la do tan cua KCl d 40°C. Khoi lUcfng dung dich thu dUgfc: (S + 100) gam Nong do phan tram cua chat tan trong dung dich bao hoa: S + 100 S + 100 S + 100 - 3,5S » S = 40 (gam). I fi\l n£ Tui unr sluu Rini Hni Hnr. a '^)^ 60"C, trong 525 + 100 bang 625 gam dung dich c6 525 gam AgNOg va 100 gam H2O. Trong 2500 gam dung dich c6 x gam AgNOa va y gam nxidc: 2500 X 525 X = 625 = 2100 (gam) y = 2500 - 2100 = 400g (nxidc) Vay & 60°C trong 2500 gam dung dich c6 2100 gam AgNOs va 400 gam ni/dfc. CJ 10°C, CLf 100 gam ntfcJc h6a tan 170 gam AgNOa 400 gam niJcJc hoa tan z gam AgNOa 400x170 z = 100 = 680 (gam) AgNOa. Do do khoi liTgng AgNOa ket tinh khi lam lanh: 2100 - 680 = 1420 (gam) Cdu V. Phan ufng: H2SO4 + 2NaOH > Na2S04 + 2H2O (1) Lan thuf nhat quy tim c6 mau xanh chufng to diT NaOH. Them axit HCl de trung hoa NaOH dif: HCl + NaOH > NaCl + H2O Lan 2 lam quy tim c6 mau do chufng to H2SO4 dir. Them NaOH de trung h5a H2SO4 di/. 2NaOH + H2SO4 > Na2S04 + 2H2O Goi X, y lan lugt la nong do cua H2SO4 va NaOH. Theo cac phan ufng (1), (2), (3) ta c6 phiiofng trinh: 500 (2) (3) 0,3y - 2 X 0,2x = 0,05 x 0,04 x 0,3x - 0,2y 0,1 X 0,08 20 500 = 0,05 = 0,1 2 2 20 Giai he phuang trinh (a), (b) => x = 0,7M va y = 1,1M. Cdu VI. Goi cong thufc oxit sat la FexOy. Fe^Oy + 2yHCl > x FeClayx + yH20 0,4 (a) (b) (1) (mol) 0,8 ^^IfilAlflg THI HOC SINH GIO GIOI HOA HOC 9 Ta c6: IIHCI = 0,8 X 1 = 0,8 (mol) Theo de: 23,2 = — (56x + 16y) y <=> 56x = 42y <=> X _ 3 fx = 3 y 4 [y = 4 Vay oxit sSt (A): Fe304 (sdt tii oxit) • FexOy + yH2 > xFe + yH20 0,6 (mol) 0,6 (2) Ta c6: 10,8 18 = 0,6 (mol) x = 2 TheodI: ^(56x + 16y] =32 « - = -=>• y ^ ' y 3 [y = 3 => Cong thufc oxit (B): FeaOs- DE SO 2 DE THI CHOIV flOl TUYEN KQC SINH GIOIHOA HOC 9, qUAAl TAN BINH TP. HO CHJ MINH NAM HOC 1998 - 1S99 PHAN A. Li thuyet Cdu I. Chi dugc diing thuoc thii de nhgn biet cdc muoi sau: NH4CI, FeClo, FeCls, MgCl2, NaCl, AICI3 (Gidi thich vd viet phuang trinh phdn ling neu c6). Cdu II. Viet phuang trinh phdn ling bieu dien cdc bien hoa sau: a) Fe > Fe2(S04)3 > Fe(0H)3 > FcsOa >Fe > FeClz > FeCls >FeCl2 > AgCl. b) MgCOs > MgS04 > MgCl2 > Mg(0H)2 > MgO > MgCOs > CO2 > Ca(HC03)2 > CaCOs Cdu III. Chi dugc dung guy tini vd dung dich AgN03 c6 sdn, neu each phdn biet cdc dung dich: NaOH, NaCl, HCl, H2S, H2SO4. Cdu IV. Sdt nguyen chdt trong khong khi thi khong bi han gl, nhUng sdl CO tap chdt de Idu ngdy trong khong khi Igi bi han gl. Hay gidi thich hien tugng nay. a Lfll GIAI Oi THI HOC SINH GIOI HOA HOC 9 PHAN B. Bai toan Bai 1- Co mot hon hgp gom Na2S04 vd K2SO4 dugc trgn Idn theo tl Ic I : 2 ve so mol (1 mol Na2S04 yd.2 mol K2SO4). Hoa tan hon hgp vdo 102 gam nUac thi thu dugc dung dich A. Cho 1664 gam dung dich BaCL 10% vdo dung dich A. Lgc ket tua, them H2SO4 du vdo tiudc vUa Igc tin thdy tgo ra 46,6 gam ket tua. Xdc dinh nong do phdn tram cua NosSOj vd K2SO4 trong dung dich ddu. • Bai 2. Tinh CM cua dung dich H2SO4 vd NaOH, biet rdng 10 nd dung dich H2SO4 tdc dung vUa du vdi 30 ml dung dich NaOH. Neu lay 20 ud dung dich H2SO4 cho tdc dung vdi 2,5 gam CaCOs thi axit con du vd lugng du ndy tdc dung vita du vdi 10 nd NaOH. LdlGIAI PHAN A. Li thuyet Cdu I. - Trich m6i lo mot it lam mt\ thuf. - Cho dung dich NaOH dii Ian liicft vao cac mau thif tren va dun nhe; Mau thuf nao c6 khi miii khai bay ra la NH4CI: NaOH + NH4CI —> NaCl + NHgt + H^O Mau tao ket tiia trSng xanh, hoa nau do trong khong khi la FeC]2: 2NaOH + FeCl2 > Fe(0H)2i + 2NaCl trang xanh , 2Fe(OH)2 + - O2 + H2O > 2Fe(OH)3 2 Mau CO ket tija nau do la FeCls: ^ 3NaOH + FeCls > Fe(0H)3i + 3NaCl nau do ' Mau tao ket tua keo trSng sau d6 tan trong NaOH dif la AICI3: 3NaOH + AICI3 > Al(0H)3t + 3NaCl keo trdng A1(0H)3 + NaOH > NaAlOa + 2H2O Mau tao ket tua trSng la MgCl2: 2NaOH + MgCl2 ^ Mg(0H)2 + 2NaCl Mau khong c6 hien tJong gi la NaCl. Lfll GIAI THI unr emu ni/\i un» unr a ^ Cdu II. a) 2Fe + 6H2SO4 damdac > 1^02(804)3 + 3SO2T + 6II2O Fe2(S04)3 + 6NaOH > 2Fe(OII)3i + 3Na2S04 2Fe(OH)3 —> FeaOs + 3H2O FeaOg + SCO —> 2Fe + SCOat Fe + 2HC1 > FeCla + Hat 2FeCl2 + CI2 > 2FeCl3 2FeCl3 + Fe > SFeC^ FeCla + 2AgN03 > Fe(N03)2 + 2AgCU b) MgCOg + H2SO4 loang > MgSOi + CO^^t + HgO MgS04 + BaCla > MgCl2 + BaS04i MgCl2 + 2NaOH > MgCOIDai + 2NaCl Mg(0H)2 —> MgO + H2O MgO + CO2 —^ MgCOg MgCOa > MgO + COat 2CO2 + Ca(0H)2 > Ca(HC03)2 Ca(HC03)2 —> CaC03i + H2O + C02t hoac Ca(HC03)2 + 2NaOII > CaCOai + NaaCOg + 2H2O hoac Ca(HC03)2 + Ca(0H)2 2CaC03i + 2H2O Cau HI. Trich m6i lo mot it lam mau thtf. - Cho quy tim tam ni/dc Ian \\iat vao cac mau thijf tren: • Mau thuf nao l^m quy tim hoa xanh la NaOH. • Mau khong c6 hien tiiong la NaCl. • Nhufng mau l^m quy tim hda do: IICl, H2S, H2SO4 - Sau do cho dung dich AgNOa Ian li/gt vao cac mau lam quy tim hoa do. • Mau nao tao ket tua trSng la HCl: AgNOs + HCl > AgCli + HNO3 irdng • MSU nao CO ke't tua den la H2S: 2AgN03 + H2S > kgiSi + 2HNO3 • Mdu khong c6 hien tuong gi la H2SO4. 10 ifil RI4I fiF TH] Hfir SlUH f;inr unA unr n Cciu IV. - sat nguyen chat khong bi ban gi vi n6 dugc bao ve bori Idp Fe203 ben trong khong khi a nhiet do thu6ng. - Khi trong sat bi Ian tap chat, de lau ngay bi ban gi do xay ra su an mon kim loai tufc la bie'n sat thanh hop chat cua sat. Gidi thidi: Tren be mat kim loai c6 Idp nUdrc am, da hoa tan mot luong oxi nen chuyen Fe -> Fe^\ Va oxi hoa tan trong niidc theo qua trinh: O2 + 2H2O >• 4011" Sau do Fe^' ket hop vdri OH" > Fe(0H)2 (trdng xanh) Mot phan Fe(0H)2 bi oxi hoa tao Fe(0H)3 (nau do) 4Fe(0H)2 + O2 + 2H2O -> 4Fe(OH)3 Do do sat bi han gi c6 mau nau do. PHAN B. Bdi toun rr. - C% X m,, 10 X 1664 n o / 1^ Bai 1. Ta co: n3,,,^ = = = 0,8 (mol) "Na^so, • "K^SO, =1:2 Goi X la so mol ciia Na2S04 => Hj^^g^^ = 2x (mol) Cac phan ufng xay ra: BaCl2 + Na2S04 > BaS04i + 2NaCl (1) (mol) X <- X -> X BaCl2 + K2SO4 > BaS04i + 2KC1 (2) (mol) 2x 2x ^ 2x Vi khi them dung dich H2SO4 vao nLfdc loc lai tao ket tua nen trong niidc loc con dtf BaCla. BaCl2 + H2SO4 > BaS04i + 2HC1 (3) (mol) 0,2 <- 0,2 «o = ^ = 0,2 (mol) B^so,/,3, 233 Theo de: ng^ci, = ^-^ ^n^^l) ng^^i./d) + "BaCi,/(2) + ^Baa.,im = ^'^ » X + 2x + 0,2 = 0,8 => X = 0,2 Khoi lagng dung dich A la: nidungdich = mNa.so^ + "^K,SO, + "^H^O « mdung dich = 0,2 X 142 + 0,4 x 174 + 102 = 200 (gam) C%Na so = ^^^^^ X 100 = 14,2% ^^2^°" 200 1 74 X 0 4 Ml GIAI DE THI HOC SINH GIOI HOA HOC 9 11 Bai 2. Ta c6: n CaCO, 2.5 100 = 0,025 (mol) 4. Tafco 3 6'ng nghi&m deu dung dung dich bari clorua, ngUdi ta cho them vao: diig l:~dung dich kali cacbonat Ong 2: dung dicli natri cacbonat Ong 3: dung dich bac nitrat Sau do cho them axit nitric vao ca ba ong nghiem. Em hay cho biet dng ndo con chat ket tua. Gidi thich vd viet phiXang trinh phdn icng. 5. Mot hon hap NaCl vd MgCl2 them nuac vao hon hap ta c6 dung dich A. Them dung dich AgNOs vao dung dich A, khi phdn iing ket thuc logi bo chat ket tua trdng, phdn dung dich con Igi la dung dich D. Chia B lam 2 phdn: a vd b. Phdn a: Sau khi c6 can tiep tuc dun nong thl diigc mot hon Jigp khi C. Cho hon hap khi nay qua binh dUng KOH. Phdn b: Cho vdo lugng dU dung dich HCl thi thu dugc ket tua trdng D. Viet cdc phuang trinh phdn dng. 6. Lam the ndo de phdn biet cdc Ig hoa chat diidi day md khong dugc dung theni hoa chat ndo khdc: MgCl2, H2SO4, NaCl, CuSO^, NaOH. 7. Mot hon hap gom: CuO, FeO, AI2O3. Lam cdch ndo de tdch chiing ra khoi nhau. Cdu II. Bai todn Cho 26,Ig MnOo tdc dung vdi dung dich HCl c6 20 gam HCl. Cho hct khi do qua mot lit dung dich NaOH lodng dU. a) Lugng HCl nay c6 du de phdn dng het vdi Mn02 khong? b) Tinh nong do mol 11 cua muoi thu dUgc trong phdn I'ing giOa do vd NaOH. c) Nung qugng pyrit sdt 'de tgo ra SO2. Cho khi SO2 sue vdo dung dich chda 2 muoi tren. Sau do them vdo mot lugng du Ba(N03)2. Tim khoi lugng ket tua va khoi lugng pyrit can dung. Biet rdng lugng SO2 tdc dung vUa du dung dich muoi. (Cho Mr, = 55; O = 16; H = 1; CI = 35,5; S = 32; Fe = 56; Ba = 137) ^^F:. , 13 Phiin ufng: H2SO4 + CaCOa - (mol) 0,025 <- 0,025 H2SO4 + 2NaOH (mol) a -> 2a Trong 20 ml dung dich H2SO4 chufa 0,025 mol H0SO4 10 ml dung dich H2SO4 chtfa x mol H2SO4 10 X 0,025 -> CaS04 + COat + H2O Na2S04 + 2H2O X = 20 = 0,0125 (mol) Trong 10 ml dung dich NaOH chijfa 2a mol NaOH 30 ml dung dich NaOH chufa y mol NaOH 2a X 30 y = = 6a (mol) Na2S04 + 2H2O 10 H2SO4 + 2NaOH (mol) 0,0125 -> 0,025 Vi phan ufng trung hoa nen 6a = ^' (mol) 6 "2SO4 " 0,01 0,025 = 1,25M ^6 2,5 0,03 M OE SO 3 DE THI HOC SINH GiOl HOA HOC 9 QUAN 3 TP. HQ CHJ MINH NAM HOC 1998 - 1999 Cdu 1. Li thuyet 1. Tit 7 Ig hoa chat sau, em c6 the dieu chc nhUng chat khi nuo? Axit sunfuric; natri hidroxit; amoni nitrat; canxi cacbonat; natri sunfit; sdt sunfua vd kim logi keni. 2. Ta H2SO4 CO may cdch de dieu che CaSOJ 3. Viet cong thUc vd ten ggi 3 muoi diing trong nong nghiep (phdn dam, phdn Idn vd plidn kali). Hay gidi thich tgi sao ngUdi ta khong trgn tro bep vdi phdn dam de ban rugng? 19 Ldl GIAI Cdu I. Li thuyet 1. Ta dieu che duac cac chat khi sau: CaCOa 1000°C hoac Zn + H2SO4 loang Zn + 2H2S04dac FeS + H2SO4 — -> CaO + COat > ZnS04 + H2t —> ZnS04 + S02t + 2H2O -> FeS04 + H2St hoac NaOH + NH4NO3 Na2S03 + H2SO4 NH4NO3 > NaOt + 2H2O NH4NO3 > NaNOs + NHat + H2O > Na2S04 + SOst + H2O >^°°°^ ) N2t-H^02t + 2H20 2 2. Dieu che CaS04 tif H2SO4: Cdch 1: Tac dung vdri canxi: H2SO4 + Ca Cdch 2: Tac dung vdri CaO: H2SO4 + CaO Cdch 3: Tac dung v6i bazcJ Ca(0H)2: H2SO4 + Ca(0H)2 > CaS04 + 2H2O Cdch 4: Tac dung vcri muo'i cija canxi: H2SO4 + CaC03 > CaS04 + COat + H2O Chu y: Con nliieu cdch klidc, xin nJiuang ban doc! 3. Cong thufc ba muoi dung trong nong nghiep: • CO(NH2)2 (46% nita): Ure (phan dam) • Ca(H2P04)2: Supe phot phat kep (phan Ian) [Ca(H2P04)2 > CaS04 + Hat CaS04 + II2O hoac supe pho't phat dcfn 2CaS0, • KRO'i. kali nitrat (phan kali) hoac KCl: kah clorua. Phan dam de bi phan hiiy trong moi trtfdng kiem, vi tro bep mang tinh kiem nen ngi/di ta khong tron phan dam vcfi tro bep. 4. Ong 1: BaCl2 + K2CO3 > BaCOgi + 2KC1 2HNO3 + BaCOg > Ba(N03)2 + C02t + H2O Ong 2: BaCIa + Na^CO^ > BaCOgi + 2NaCl 2HNO3 + BaCOa -> Ba(N03)2 + C02t + H2O dng 3: BaCla + 2AgN03 > Ba(N03)2 + 2AgCli AgCl + HNO3 Trong 3 ong nghiem thi ta thay ket tija AgCl khong tan trong axit mac dti la axit HNO3 vi the trong ong 3 con chat ket tua. 5. Phan ufng: NaCl + AgN03 > AgCU + NaNOg MgCl2 + 2AgN03 > 2AgCU + Mg(N03)2 Sau khi Ipc bo ket tua thi dung dich (B) gom: NaNOs, Mg(N03)2 va AgNOs dif. Phdn a: Khi dun nong dung dich B: NaNOs —> NaNOa + -02t 2 2Mg(N03)2 —> 2MgO + 4N02t + O-^t 2AgN03 —> 2Ag + 2N02t + 02t 2NO2 + 2K0H —> KNO3 + KNO2 + H2O Phdn b: Khi cho HCl du vao dung dich B thu diJOc ket tua. Dieu nay cho ta thay rang trong dung dich con AgNOs dif. HCl + AgNOs > AgCli + HNO3 (D) 6, Ta nhan thay: trong tat ca cac dung dich tren thi chi c6 mot dung dich CO mau xanh la: CUSO4, cac dung dich con lai la khong mau. Trich moi lo mot it lam mau thijf. - Cho dung dich CUSO4 Ian lifot vao cac mau thuf tren: • Mau thCf cho ke't tiia mau xanh la NaOH: CUSO4 + 2NaOH > Cu(0H)2i + Na2S04 • Cac mlu con lai khong c6 hien tu'ong. - Dung NaOH lam thuoc thiif, cho Ian liiot vao cac mau ihxi con lai: • Mau thLf CO ket tua mau trSng la MgClg: 2NaOH + MgCla > Mg(0H)2 + 2NaCl • Mau thii tao dung dich trong suot va toa nhiet manh la H2SO4. 2NaOH + H2SO4 > Na2S04 + 2H2O • Mau khong c6 hien tugng la NaCl. 7. Tach CuO, FeO, AI2O3 ra khoi nhau: CuO AI2O3 FeO NaOH da ^NaAlOg CuO_ FeO^ +C0, + HC1 du -> Al(OH), ->A1,03 CuCl^ FeCl,, [Cu(NH3)J(OH)2 +HC1 ->CuCL +NaOH du + NH.OH dil -^Cu(OH), Fe(OH), cliaii kliong -^FeO Phan ufng: • AI2O3 + 2NaOH > 2NaA102 + H2O NaAlOz + CO2 + 2H2O > Al(0H)3i + NaHCOg 2A1(0H)3 - CuO + 2HC1 FeO + 2HC1 -> AI2O3 + 3H2O —> CUCI2 + H2O -> FeCl2 + H2O > Cu(0H)2i + 2NH4CI. CuClz + 2NH4OH — Cu(0H)2 + 4NH3 > [Cu(NH3)4](OH)2 (xanh tim, tan) [Cu(NH3)4](OH)2 + 2HC1 > CuCh + 4NH3T + H2O CUCI2 + 2NaOH > CuCOIDal + 2NaCl Cu(0H)2 > CuO + H2O FeCl2 + 2NaOH Fe(0H)2 Cdu II. Bai todn Ta c6: n chankhong 26,1 -> Fe(0H)2i + 2NaCl FeO + H2O MnO., 87 = 0,3 (mol): n^j^j = 20 0,55 (mol) MnOa + 4HC1 (mol) 0,55 CI2 + 2NaOH (mol) 0,1375 36,5 > MnCla + Cl2t + 2H2O 0,1375 -> NaCl + NaClO + H2O 0,1375 a) Lifgng HCl nay c6 du de phan lifng het vcfi Mn02 khong? TCr phan ufng (1) ta c6 ti le: n MiiO, = 0,3> n -»CuO (1) (2) HCl 0,1375 1 4 4 Vay lirang HCl nay khong dii de phan ufng het liicfng Mn02 da cho hay IICl phan ijfng het va Mn02 con dii. b) The tich dung dich thu dUcfc chinh 1^ the tich cua NaOH tufc la: Vduiigdich = 1 (lit) So' mol cua NaCl trong phan ufng (2): 0,1375 (mol) C^ = C„ = = 0,1375M c) Phan ufng: 4FeS2 + IIO2 > 2Fe203 + 8S02t (i) (mol) ^ 0,1375 2 SO2 + NaClO + 2NaOH > Na2S04 + NaCl + H2O (2) (mol) 0,1375 <- 0,1375 -> 0,1375 Ba(N03)2 + Na2S04 > Ba^O^i + 2NaN03 (3) (mol) 0,1375 ^ 0,1375 Txi (3) ^ ne^so, = ^i^.,so, = 0.1375 (mol) mg^so^ = 0,1375 x 233 = 32,04 (gam) TO (1), (2) :^ np^3 = ^ = (mol) ^so, _ 0,1375 2 " 2 0,1375 => mpes, = -^-r— X 120 = 8,25 (gam). DE SO 4 OE THI HOC SiNH GIOI HOA HOC 9, CAP TP. HO CHI MINH NAM HOC 1999 - ZOOO Cdu I. Khi cho kirn loai vdo dung dich mud'i c6 the xdy ra nhitng phan ling gi? Cho vi du minh hoa. Cdu II. Viet cdc phuang trinh phdn vCng theo chuSi bien hoa sau: FeCh > Fe(0H)3 > FezOs Fe > FesO^ <^ ^ FeCl2 > Fe(0H)2 > FeO Cdu III. Mot nhd hoa hoc dieu che dugc 3 mdu kirn loai gidng nhau ve dang ben ngoai (mdu sdc) va da tim dugc phuang phdp phdn biet nhanh chong. dng lay cdc mdu kim log.i cho tdc dung vdi^axitjm_d^g dich NaOH, ket qua do dugc ghi tron^^M^:^,^^"^ 6SNH THUAN Lfll RlAi n£ Tui unn SIUH Rini HflA HtlR 9 — • _ . JO 17 Thuoc thii Kim loai I Kim loai II Kim loai III Axit HCl - + + Axit HNO3 + - + Dung dich NaOH - + + Trong do ddu (+) de chi trudng hap kim loai hoa tan, ddu (-) chi triiang hap kim loai khong tdc dung vai dung dich kicm hay axit. Hay xdc dinh kim loai nghien ciiu, viet phuang trinh phdn ling vd gidi thich vi sao kim loai khong tdc dung vai cdc chat dd cho. Cdu TV. Chi diing kim loai, hay nhgn biet cdc dung dich sau day: HCl, HNO3 ddc, AgNOa, KCl, KOH. Vii't cdc phuang trinh phdn dug xdy ra trong qua trinh nhgn biet. Cdu V. Hoa tan oxit cua kim logi hoa tri II trong mot lilgng vita dil dung dich H2SO4 20% thi thu dicgc dung dich mudi c6 nSng do 22,69c. Xdc dinh kim logi do. Cdu VI. Co mot hon hap gom Na2S04 vd K2SO4 dilgc trgn Idn iheo ti le I : 2 ve so mol. Hoa tan hon hap vdo 102 gam nUac thi thu dilgc dung dich A. Cho 1664 gam dung dich BaClz 10% vao dung dich A, xudt hien ket tua. Loc bo ket tua, them H2SO4 dU vdo nilac Igc thi tlidy tgo ra 46,6 gam ket tua. Xdc dinh nong do phdn tram cua A^a^SOj vd K2SO4 trong dung dich A ban ddu. (Cho Na = 23; S = 32; K = 39; Ba = 137; CI = 35,5). Ld\I Cdu I. Khi cho kim loai vao dung dich muoi c6 the xay ra: - Neu kim loai cho vao khac kim loai trong muoi thi xay ra phan uTng the - oxi hoa khijf. Fe + CUSO4 > FeS04 + Cu - Neu kim loai trung v6i kim loai trong muoi thi xay ra phan tfng tiT oxi hoa. Fe + 2FeCl3 > SFeClg Cdu II. Phan ufng: 3Fe+ 202 ^° > Fe304 Fe304 + 8HC1 > FeCla + 2FeCl3 + 4H2O FeCl3 + 3NaOH > FeCOIDgi + 3NaCl 2Fe(OH)3 —> FezOs + SHsO FeCla + NaOH > Fe(0H)2i + 2NaCl Fe(0H)2 TT-^rt > FeO + H2O chan khong Cdu III Theo de bai thi: Kim loai I: Ag (Bac) Kim loai II: Al (Nhom) Kim loai III: Zn (Kem) • Ag khong phan uTng vdi HCl vi Ag dufng sau hidro trong day hoat dong kim loai. Ag khong phan ufng vdi NaOPI vi khong phai la kim loai iLfdng tinh. • Al khong tac dung vdi HNO3 vi Al bi thu dong trong moi trUcfng HNO3 dac nguoi. Cdu IV. Trich moi lo mot It lam mau thijf. - Cho bot kim loai Cu dti Ian luat vao cac mau thtf tren: +) Mau thii nao dung dich id khong mau chuyen sang xanh la AgNOs. Cu + 2AgN03 > Cu(N03)2 + 2Ag +) Mau thuf nao vifa tao dung dich mau xanh va c6 khi nau do bay ru la HNO3. Cu + 4HNO3 dac > Cu(N03)2 + 2N02t + 2II2O - Sau do cho dung dich viia thu di/oc a tren Ian liicft vao cac mau con lai +) Mau CO ket tua mau xanh la KOH. Cu(N03)2 + 2K0H > 2KNO3 + Cu(0H)2i • +) Cac mau con lai khong c6 hien tugng. - Cho dung dich KOH vao 2 mau con lai, mau nao c6 phan ufng toa nhiet la HCl KOH + HCl > KCl + H2O Mau con lai la KCl. Cdu V. Goi kim loai hoa tri II la A va c6 a mol => oxit la: AO. AO + H2SO4 > ASO4 + H2O (mol) a ^ a a Theode: «,,„,^.=^ "^dd Hi,S04 a(A + 96)100 C%ddASO, =— ™dd AS04 Ma: a(A + 96)xl00 a x 98 x 100 <=> • = a(A + 16) + 22,6 20 => A = 24: Magie (Mg). Cdu VI. Ta c6: 03 „, = 10 x 1664 ^ ^ ^^^'2 100 X 208 '^NajSO,, • '^K2S04 =• 1 • 2 Goi X la so' mol cua Na2S04 => n^^g^^ = 2x (mol) Cac phan ijfng: BaCl2 + Na2S04 > BaS04i + 2NaCl (1) (mol) X <- X -> X BaCls + K2SO4 > BaS04l + 2KC1 (2) (mol) 2x <- 2x 2x Klii them dung dich H2SO4 vao lo niidfc loc thi tao ket tua nCfa nen trong nude loc con dii BaCl2. BaCl2 + H2SO4 > BaS04i + 2HC1 (3) (mol) 0,2 <- 0,2 nBaso,„3, = H = 0,2 (mol) Theo de: ng^Q^ ban ddu = 0,8 (mol) TCr (1), (2), (3) => x + 2x + 0,2 = 0,8 =:> x = 0,2 Khoi lugng dung dich (A) la: md.ngdich = m^^^^^^ + m^^^^^^ + m^^^ => mdungdich = 0,2 X 142 + 0,4 x 174 + 102 = 200 gam 142 X 0,2 200 174x0,4 C%Na so = —^—— 100 = 14,2% NaaSO^ 200 C%„ „„ = ^ X 100 = 34,8%. '^2^^i 200 DE SO 5 OE THI HOC SINH GlDl HOA HOC, QUAN 9 (VONG 2) TP. HCM NAM HOC 19S9 - 2000 Cdu I. Cho chuoi phuang trinh plidn ling sau: +A, I, +B, CaCOs CaCOs Tim cdc chat A, B, C, X, Y, Z (Id cdc chat khdc nhau vd khdc CaCOs)- Viet cdc phuang trinh phdn ting. Cdu II. Dung mot kim loai de nhdn biet cdc lo dung dich sau: FeCh; FeCh; HCl; BaCh; (NH^2S04; AlCh; NH4CI Cdu III. Cho Na vdo 2 dung dich mud'i Al2(S04)3 vd CuSOj thi thu dagc khi A, dung dich B vd ket tua C. Nung ket tila dugc chat rdn D. Cho H2 di qua D nung nong dUgc chat rdn E. Hda tan E vdo dung dich HCl thi thdy E tan mot phdn. Gidi thich. Viet phuang trinh phdn iCng. Cdu IV. a) Cho 0,25 mol CuO tan het trong dung dich II9SO4 20% dcm nung nong lugng vila dii, sau do Idm ngugi dung dich den 10°C. Tinh khoi lugng tinh the CUSO4.5H2O tdch ra khoi dung dich. Biet do tan CUSO4 a 10°C Id 17,4g. b) Clio biet do tan CaSO^ Id 0,2g (a 20°C) vd khoi lugng rieng cua dung dich bdo hda coi bdng Iglml. • Tinh do tan cua CaSO^ theo nong do mol. • Khi trgn 50 ml dung dich CaCl2 0,012M vai 150 ml dung dich NaoSO^ 0,004M a 20°C thi c6 ket tila xudt hien khong? Cdu V. a) Hda tan hodn todn 1 hidroxit cua kim logi M bdng mot lugng vita dii dung dich HCl 10%. Sau phdn ling thu dugc dung dich A. Them vdo dung dich A mot lugng vUa dii dung dich AgNOs 20% thu dugc dung dich muoi c6 nong do 8,965%. Xdc dinli cong thvCc hidroxit tren. b) Khi phdn tich 2 oxit vd 2 hidroxit tUong iCng cila ciuig mot nguyen to hda hoc dugc so lieu sau: ti so thdnh phdn % ve khoi lugng ciia 0x1 20 trong 2 oxit do bdng —. , 2T Tl so thdnh phdn % ve khoi lugng ciia nhdm Iiidroxit trorig hai 107 hidroxit do bdng . Hay xdc dinh nguyen to' do. 135 Cdu VI. Cho X, Y Id hai dung dich HCl c6 nong do khdc nhau. Lay V ml dung dich X tdc dung vai-AgNOs du tao thdnh 35,876g ket tila. De trung hda V ml dung dich Y can 500 ml dung dich NaOH 0,3M. a) Khi trgn V (lit) dung dich X vol V (lit) dung dich Y thu dUgc 2 (lit^ dung dich Z. Tinh CM dung dich Z. b) Neu lay 100 nd dung dich X vd lay 100 nd dung dich Y cho tdc dung het vdi kim logi Fe thi lugng hidro thodt ra trong hai trUdng hgp lech iihau 0,448 (lit) (dktc). Tinh CM dung dich X, Y. LCII Riai nc Tu, „„o unA unr a 21 Cdu VII. Trgn Idn 10 ml dung dich HCl vai 20 ml dung dich HNO3 vd 20 nd dung dich H2SO4 thu dugc dung dich A, pha them II2O vdo dung dich A de diCgc dung dich B c6 the tich gap doi. Trung hoa 25 nd dung dich B can 8 ml dung dich NaOH 8% (D = l,25g/nd). Bern c6 can dung dich tqo thdnh dugc l,365g muoi khan. Neu cho 40 tnl dung dich B tdc dung vai mot liCgng dU dung dich BaCl-z thi thu dugc 0,932 ket tiia. a) Tinli CM dung dich axit ban ddu. b) Dung dich C chi'ia hon hap NaOH 0,8M vd Ba(0H)2 0,2M. Can bao nhieu ml dung dich C de trung hoa het 50 ml dung dich B? Cho: H= 1,N= 14, O = 16, Na = 23, S ^ 32; CI = 35,5, Co = 40, Fe = 56, Cu = 64, Ag = 108, Ba = 137 LCfl GIAI Cdu I. ThiTc hien chuoi phan ufng: CaCOg CaO lAl CO,, + NH011 ->Ca{OH), (Bl -^Na^CO, t-HCl ->CaCl, CaCO;, +CO.,*U.,0 ->NaHCO., Phan ijfng: CaC03 CaO + II2O iooo"c CaO + CO2T (A) (X) Ca(0H)2 (B) > CaCl2 + 2H2O Ca(0H)2 + 2HC1 — CaCl2 + NaaCOa > CaCO^i + 2NaCl CO2 + 2NaOH > Na2C03 + H2O (Y) NaaCOs + CO2 + II2O > 2NaHC03 (Z) -> CaCOyi + 2NaN03 + CO.T + II2O 2NaIiC03 + Ca(N03)2 — Cdu II. Trlch moi dung dich mot it lam mau thuf. Cho kirn loai Ba vao cac mau ihii tren, dau tiea c6 phan ufng: Ba + 2H2O Ba(0H)2 + H2T Mau nao cho ket tiia trfing xanh la FeCl2: Ba(01i)2 + FeCL — > FeCOIDai + 2BaCl2 22 I rii piAi nc Tui unr Qtwu nni unA unr Q M&u nao cho ket tua nau do la FeCla: 3Ba(OH)2 + 2FeCl3 > 2Fe(OH)3i + 3BaCl2 Mau CO ket tiia va khi miii khai bay ra la (NH4)2S04: Ba(0H)2 + (NH4)2S04 > BaS04^ + 2NH3T + 2H2O MSU cho ket tiia keo trang la AICI3: 3Ba(OH)2 + 2AICI3 > 2Al(OH)3i + SBaCla Mau chi CO khi mtii khai bay ra la NH4CI: Ba(0H)2 + 2NH4CI > BaCl2 + 2NH3t + 2H2O Mau cho dung dich trong suo't va toa nhiet la HCl: Ba(0H)2 + 2HC1 > BaCl2 + 2H2O Con lai la BaCla. 'Cdu III. Phan ufng: Na + 1120 -> NaOH + ^Hat 6NaOH + Al2(S04)3 2NaOH + CUSO4 — > 2Al(OH)3i + 3Na2S04 -> Cu(0H)2i + Na2S04 2A1(0H)3 - Cu(0H)2 — AI2O3 + H2 CuO + H2 - -> AI2O3 + 3H2O -> CuO + H2O ^—> Cu + H2O Vi E tan 1 phan nen E chufa: CuO, Cu, AI2O3 AI2O3 + 6HC1 CuO + 2HC1 - -> 2AICI3 + 3H2O -> CuCla + H2O CuO + H2SO4 — (mol) 0,25 -> 0,25 Khoi iLfoag dung dich H2SO4 20%: 0, 25 X 98 X 100% > CUSO4 + H2O 0,25 20 = 122,5 (gam) Khoi lugng CUSO4: 0,25 x 160 = 40 (gam) Khoi lu'gng dung dich sau phan ufng: '^duiig dich sau phan iMig — niCuO + '^^ddn^SO^ = 0,25 X 80 + 122,5 = 142,5 (gam) [...]... 30,4% = 69, 6% _ 30,4 2Y _ 69, 6 bY 60,8 69, 6 74,1 Dat 10 gam 5 50 = 100 gam B + 2D + 4E 2a 4a D i n h luat bao to^n khoi liicfng cho: M A X 2a = 22,86 x 7a a _ 2 b 5 => M A = 80 OE SO 9 BE THI HOC SINH GIOI HQA HOC 9, TP HAI PHONG (BANG A) NAM HOC Z001 - Z002 Cong thufc cua B la: X2Y5 Cdu J.- + Neu lay 2a mol A nhiet phan se tao t h a n h 7a mol k h i 74,1a 74,1a 50 a 2a (mol) 25,9b 25,9b 69, 6 X Y _... + NaaCOa Cong thile muo'i caebonat: MCO3 Phan ufng: MCO3 + H2SO4 > CaCOs^ + H2O CO2 + H2O + CaCOa IV - a ) Goi k i m loai hoa t r i 2 la: M 24,3 127 + 25,4- 18X 100 = 60,63 (gam) 160,63 gam dung dich 2 4 , 6 — ^ 1 8 x 127 + 1 8 X 127 3086,1 437,4x - 60,63 2 4 , 6 127 + 18x 127 + 1 8 X 127 + 1 491 , 498 18X - 265 19, 562x 127 + 1 8 X 1331 59, 754 + 18873,036x = -265 19, 562x + 308610 45 392 , 598 x = 175450,246... 0,75 mol x 18 g/mol = 13,5 gam iSlSi^li^ THI HOC SINH G|6'HOA HQC 9 d ) nc = Tii bang t r e n t a thay: = 1,66.10^^ (mol) D u n g dich n a o cho v a o t a o r a 4 I a n ket t u a l a d u n g dich Na2C03 va So n g u y e n tuf C = 6,02.10^^ n g u y e n ti:f/mol.l,66.10^ m o l • AgNOa (cap d u n g dich 1) D u n g dich n a o cho v a o t a o r a 3 I a n k e t t u a = 9, 993 .10^° n g u y e n tuf g) n la dung... ( l 6 0 ) + 0,03 98 x 1 0 0 = 11,6 X 29, 4 , (gam) b ) K h i m = 1,44 gam; V = 400 m l ; b = 440,83 gam Khoi lUctng dung dich luc sau: mdung dkh sau = 11,6 - 2 ,9 = 8,7 (gam) Trong (400 + 18x) gam Fe2(S04)3.xH20 t h i chufa 400 gam Fe2(S04)3 => Vay 2 ,9 gam Fe2(S04)3.xH20 t h i chura a gam Fe2(S04)3 Vay khoi lifong Fe2(S04)3 b i t^ch ra: 2 ,9 400 X 72 x +y 1160 = 2,001 400 + 18X c:> 35 ,98 2x = 360,4 => =>... B Idn liigt Id 30,4% vd 25 ,97 / : i 2SO3 > H2SO4 2H2SO4 + C u c) Cho hdn hap khi gom CO2, SOo Bdng pliuang phdp hoa hoc, hay tdch phdn ^ 450*^ C > BaSOsi + 2 NaCl Neu cong thi' ic phdn til cua A la XY2, thi cong thiic phdn tii cua B Id gi? Cdu 5: De gia tang nong do cila 50 gam dung dich CuSO^ 57c len gap luu Idn, CO boil hoc sinh da thUc Jiien bdng bdn cdch khdc nJiau: Hoc sinh D: them 50 gam dung dich... Na2S04 + 2H2O H2SO4 + 2 N a O H A p dung: ^ C% = m,, ddHjSO^ (gam) x 100 = m„ = 3 m^jXlOO C% 16,66 =— x100 98 , , = 17 (gam) b ) T i n h kho'i l u o n g s S t t h a m g i a p h a n uTng: DE SO 16 Tii p h a n tTng ( 9 ) : Fe + CuClz (mol) 1 (mol) DE THI TUYEN HOC SINH GIOI HQA HOC 9 (VONG 1), QUAN 9, TP HCM > FeCla + Cu 1 1 X NAM HOC 2 0 0 2 - 2003 1 X Theo de t h a n h sSt t a n g 0,8 gam n e n t a c6... khong tham gia phan I'Cng Cdu V KhvC hodn todn 4,64 gam mot oxit kim logi thi can 1, 792 lit khi CO (dktc) Neu lay todn bg lUgng kim logi thu diigc d tren cho vdo dung dich HCl du thi thu dugc 1,344 lit khi H2 (dktc) Xdc dinh cong thi' Cc hoa hoc cua oxit ndi tren - " a> Tinh nong do phan trdm cua dung dich H2SO4 6 ,95 M (D = 1, 39 g/ml) phuang trinh sau: 0,16 0,06C2 - 0,08Ci = 0,016 G i a i (a) v a (b)... g o m : N a C l tao t h a n h v a Na2C03 du Ttf chuoi > Fe(0H)3 0,1 < - 0,1 ^ R a f ! thien Fe < = ± Fe2(S04)3 208 BaCl, ^Ma OE THI HOC SINH GIOI HOA HOC LifP 9 (VONG 2 ) , QUAN THU flUfC, TP HCM 20 8 = — — = 0 , 1 (mol) m„„, (mol) ^^^^^^ = - ^ = 0 ' 1 6 ( m o l ) 10,4 va ^ m2 gam O2 thu dugc sdn phdm dich H2SO4 98 % khi tan h^t dugc bg khi A3 dugc tgo ra 2,3 gam muoi USO4.5H2O hdn hgp hap thu bdi 200... (gam) • 544 CaS04 n CO,, ( Q ) tac d u n g v d i CI2 chieu s a n g n e n ( Q ) l a : C2H6 136 = 544 " 250 -> Cu + C O 2 T Do do: (mol) 10^ 1,6 + 2H2O -> 2CO9T T a c6: 100 = 0,3539fc X1000 111 10' CaSO, %m Tir(l) 0, 392 .10'^ Cu(OH), 2 (mol) X 98 = 0, 392 .10^ 250 n CaSO,, m 200,1 CO + CuO > C u ( 0 H ) 2 i + CaS04 _ 10^ axetat) III > Ca(0H)2 •^4 TCrd) Cdu (mol) K h o i iLTcfng d u n g d i c h Boocdo t h... Iqnh xud'ng vd dung muoi Id 23% Xdc dinh cong thUc cua tinh the DE THI HOC SINH GIQI HOA HOC 9, CAP TP HO CHJ MINH NAM HOC 2002 - 2003 Cdu muoi trung hda MeS (kim loqi rV Cho m gam hon hgp CaCOs Me oxi du Hda tan chdt viia du dung dich H2SO4 29, 4% dp 34,5% the hidrat khi nay tdc dung viCa binh ciia hon hgp khi ban ddu ilng bdng mpt lugng 2 ,9 gam tinh hdn hgp khi gom H2 vd H2S sdn la nuac vd mpt chdt khi . FeaOs- DE SO 2 DE THI CHOIV flOl TUYEN KQC SINH GIOIHOA HOC 9, qUAAl TAN BINH TP. HO CHJ MINH NAM HOC 199 8 - 1S 99 PHAN A. Li thuyet Cdu I. Chi dugc diing thuoc thii de nhgn . " 0,01 0,025 = 1,25M ^6 2,5 0,03 M OE SO 3 DE THI HOC SINH GiOl HOA HOC 9 QUAN 3 TP. HQ CHJ MINH NAM HOC 199 8 - 199 9 Cdu 1. Li thuyet 1. Tit 7 Ig hoa chat sau, em c6 the. = 100% - 30,4% = 69, 6% - Trong B: TCrd), (2) X _ 30,4 _ X 60,8 2Y 69, 6 Y 69, 6 %X = 25 ,9% %Y = 74,1% aX _ 25 ,9 _ X 25,9b bY 74,1 Y 74,1a 60,8 25,9b a _ 2 69, 6 74,1a b 5 Cdu