Cho hang s6 Faraday F = 96500 a Tinh kh6i hrong dung dich giam trong qua trinh di~n pharr tren, b Tinh pH cua dung dich Y khi thoi gian di~n pban dung dich la 630 giay.. Coi th~ tich dun[r]
(1)so GIAO CUQC THI cAP TiNH GIA.:r._ ToAN TREN MAy TiNH CASIO, VINACAL NAM HQC 2012-2013 D1}C vA DAo T~O LAo CAl Mon thi: Hoa hoc (ChuO'ng trtnh rHPT) WONG nAN CHAM Thai gian thi: 150 phut (kh6ng kd thai gian giao dJ thi), Ngay thi: 17/0112013 nJ thi g6m 10 cdu, 08 trang, m6i cdu 05 aidm DE THI CHINH THUC co Can 1: Nguyen t6 A co loai d6ng vi vci cac d~e diem sau: Tong s6 khoi cua 41o~i d6ng vi 1ft 825 TAng s6 notron cua d6ng vi A3 va A4l6n han s6 notron cua AlIa 121 hat, Hieu s6 kh6i ella d6ng vi A2 va ~ nh6 hon hieu s6 kh6i cua hai d6ng vi Al va A3 la don vi T6ng s6 hat cua hai d6ng vi Al va A416n han t6ng s6 hat khong mang di~n cua d6ng vi Ai va A3 la 333 hat se kh6i cua A4 bfulg 33,5% tbng s6 kh6i cua d6ng vi Xac dinh s6 kh6i cua d6ng vi va so di~n tich hat nhan cua nguyen tir nguyen t6 A each giai Ket qua Di~m GQi N}, N2, N3, N4 Ian hrot la so notron cua cac dong vi Ab A2, A3,A4 GQi Z la s6 proton, electron cua cac d6ng vi d6 DiSu kien: Nt, N2, N3, N4, Z EN* , Thea d~ bai ta c6 cac phirong trinh: (1) 4Z+N1+N2+N3+N4=825 (2) N3+N4-Nl = 121 (3) Nz-N4 = N, - N3 - ~ (N, + N4) - (N2+ N3) = (4) 4Z+N1+N4-(N2+N3)=333 (5) N4 + Z = 0,335 (3Z + N, + N2 + N3) D~t NI + N4 = x; N2 + N3 = Y Thay vao (1), (3) va (4) ta co h~: + x + y = 825 4Z +x- y = 333 - {Z = 82 4Z { giai h~ ta duoc: X= N, +N4 =251 x- y =5 Y = N2 + N3 = 246 Thay Z= 82; N2 + N3 = 246 vao phuong trinh (5) ta e6: N4 + 82 = 0,335 (3.82+ N, + 246) ~ N4 -0,335Nl = 82,82 Ket hop voi ptr: N + N4 = 251 co h~: NI +N4 =251 { - O,335N, + N4 = 82,82 Thay Nj, N4 vao (2) - - ., h" giai ~: NJ N3 = 122 - = 126; N4 = N2 = 124 125 Al = 208; A2 = 206; A3 = 204; ~ = 207; Z=82 Can 2: Ldy 2,5 ml dung dich CH3COOH 4M r6i pha loang vao mroc lit dung dich A a) Hay tinh de) di~n Ii a cua CH3COOH va pH cua dung dich A Biet rfu1g tron§ 1ml dung dich A co 6,28.1018 ion va phan ill axit chua phan Ii Cho s6 Avogadro NA = 6,02.102 • b) Tinh hang s6 axit K, cua axit axetic (2) each giai Bi~m Giili: a) S6 mol axit = 0,0025.4 = 0,01 (mol) ; s6 molW = O,01.a S6 mol hat vi rno m1: 6;28.1018/(6,02.1023) S6 mol hat vi mo lit dung dich A: 6,28.1018.103/(6,02.1023) = 6,28.10-2/6,02 CH3COOH!:; CH3COO" + W bd(mol) 0,01 0 cb(mol) 0,01- 0,01.0, O,01.a 0,01.0, Theo dS bai: 0,01- O,Ol.a+ O,01.a + O,Ol.a = 6,28.10-2/6,02 => a = 0,043(189368) hay 4,3(189368)% b) N dng dQ cua cac ion va phan ill (i trang thai can bang la: [CH3COO-] = [H+] = 0,01 a (M) [CH3COOH] Va .y = O,Ol.(l-a) K = [CH3COO-][H+] [CH3COOH] a 1,949520047.10"5 = = a = 0,043(189368) hay 4,3(189368)% 2,5 Ka= (O,Ol.a)2 O,Ol(1-a) = 0,01 a?l(l-a) = 1,949520047.10"5 = 10-4,710072295 2,5 104,710072295 Cau 3: Cho 23,52 gam h6n hop kim loai Mg, Fe, Cu van 400 ml dung dich HN03 1,7M dung c6c X thay thoat m<)tkhi nMt khong mau, hoa nau khong Trang c6c X du m<)t kim loai chua tan h~t £)6 tiep tir tir dung dich H2S04 0,5M vao c6c X, eh~t tren lai bay cho d~n kim loai vira tan h~t thi m~t 440ml, thu duoc dung dich A Cho dung dich NaOH d~n du van 112 dung dich A, 19Ck~t tua sau d6 rna sach, nung k~t rna khong d~n kh6i hrong khong d6i thu diroc ch~t rim B nang 15,6g Tinh nAng de molll (CM) cua cac ion dung dich A? K~t qua each giai Khi cho h6n hop kim loai Mg, Fe, Cu tac dung voi RN03 1,7 M thu diroc dung dich va kim loai khong tan + Kim loai d6 la kim loai Cu + dung dich thu diroc chua mudi Mg2+, Fe2+, N03", Cu2+ eo th~ co (vi Cu du nen Fe3+ bi klnr het v~ Fe2 Them H2S04 van tiep tuc xay phan trng : 8W + 3Cu + 2N03 - + 3Cu2+ + 2NO + 4H20 Khi kim loai tan vira h~t nrc hi Fe2+ chua bi oxi hoa len Fe +' l Tong s6 mol ion W: 0,4 1,7 + 0,5.2.0,44 = 1,12 (mol) G9i x, y, z l§.n hrot la s6 mol cua Mg, Fe, Cu co 23,52 gam h6n hop (x, y, z >0) V~y toan h<)qua trinh ta co: Mg + Mg2+ + 2e x Fe y + Fe2+ 2x + 2e 2y 4W + N03-+3e + NO + 2H20 1,12 0,84 OA Brem (3) Cu -7 CUZ+ + 2e z 2z Theo dinh lu~t bao toan e: 2x + 2y + 2z = 0,84 (1) Theo bai ra: 24x + 56y + 64z = 23,52 (2) Thanh phan cua dung dich A: Fe2+, Cu2+,Mg2+, SO/-, N03Y2dung dich A tac dung voi dd NaOH du -7 k~t tua: Fe(0H)2 y/2 mol; Mg(0H)2 X/2 mol; Cu(OH)2 zl2 mol Nung k~t tua khong d€n kh5i IU<JIlgkhong d6i thu diroc kch rna g6m: MgO (X/2 mol); Fe203 (y/4 mol); CuO (Z/2 mol) -7 phirong trinh: 20x + 40y + 40z = 15,6 (3) Giai h~ phuong trinh g6m (1), (2), (3) ta diroc k€t qua: x ='0,06 mol; y = 0,12 mol; z = 0,24 mol V~y dung dich A co 0,06 mol Mg2+, 0,12 mol Fe2+, 0,24 mol Cu2+, 0,22 mol S04 2- ~ Theo dinh lu~t bao toan di~n tich ~ s6 mol ion N03- dung dich A= 2(0,06+0,12+0,24) - 0,22x2 = 0,40 mol N03T6ng th6 tich cua dung dich Ala: 0,44 + 0,4 = 0,84 (lit) V~y nong dQ mol cua cac ion dung dich l§n hrot lit: [M~2+] = 0,06/0,84 = 0,071428571 (M) , [Fe l = 0,142857142 (M) [Cu2+] = 0,285714285 (M) [Sot] = 0,22/0,84 = 0,261904761 (M) [N03-] = 0,4/0,84 = 0,476190476 (M) [Mg2+] = 0,06/0,84 = 0,071428571 (M) [Fe2+] = 0,142857142 (M) [Cu2+] = 0,285714285 (M) [SOlO] = 2,5 = 0,22/0,84 0,261904761 (M) [N03 -] = 0,4/0,84 = 0,476190476 (M) 2,5 Cau 4: D6t chay 3,2 gam sunfua kim loai M2S (M cac hop chfrt co s6 oxi hoa + va +2) oxi duo San pham r~n thu duoc dem hoa tan het m9t hrong vita du dung dich H2S04 39,2% nhan diroc dung dich rnuoi co n6ng dQ 1~~o% Bern the, ~ khi1 d nong muoi t." 4800 tinh lai at l'a - 107 r ~ ~ 01 /0 lam 1~ dung dich mudi th§.y tach 2,5 gam T'lID cong '" t h ire tiinh thxe muoi ;., tac h r each giai M2S + 202 ~ MO + H2S04 ~ r K~t qua (1) 2MO + S02 MS04 + H2O (2) D~t s6 mol M2S la x => (2M+32)x = 3,2 (*) Theo (1) va (2): nH2so4 =nMSO = 2nM2s = 2x(mol) = 2x; • Kh6i IU<JIlgdung dich H2SO4: 2x.98.100 500x(gam) mddH,SO ::: 39,2 Kh6i hrong dung dich mu6i = Kh6i hrong dung dich H2SO4 + kh6i IU<JIlgMO = 500x + (2M+32)x = (2xM + 532x) gam mMSO=(M + 96).2xgam Diem (4) ~+96).2x = 1600 => M = 64 ~ M la Cu 2Mx+532x 33.100 I - ~ Cong thirc mu6i sunfua la CU2S Thay M = 64 van (*) diroc x = 0,02 mol Kh6i luong CUS04 = 2.0,02.160 = 6,4 gam Kh6i Iuong dung dich CUS04 = 2.0,02.64 + 532 0,02= 2,56 + 10,64 = 13,2 gam Khi lam lanh tach fa 2,5 gam tinh th@, kh6i hrong dung dich giam 13,2 - 2,5 = 10,7 gam Cong thirc muoi sunfua 1ft CU2S 2,5 Kh6i hrong CUS04 dung dich 4800 _ 10,7 -Yo - 4,8 gam 107 Kh6i IUQ'ngCUS04 tach fa = 6,4 - 4,8 = 1,6 gam hay 1,6/160 = 0,01 mol Kh6i hrong H20 tinh th~ tach fa la 2,5 -1,6 hay 0,9/18 = 0,05 mol ~ tinh th~: nCUS04 : nH20 = 0,9 gam = 0, 01: 0, OS = : => Cong thirc tinh th8 muoi la CuS04.5H20 Cong thirc tinh th~ muoi la CuS04·5H20 2,5 Cau 5: C6 dung dich X g6m Fe2(S04)3 0, 100M; FeS04 0,0 10M va NaC12M Di~n phan 100 ml dung dich X voi cuong d(>dong di~n m(>tchieu khong d6i co 1= 9,650A voi di~n cue tro (co mang ngan) va thoi gian 100 giay, thu duoc dung dich Y Cho hang s6 Faraday F = 96500 a) Tinh kh6i hrong dung dich giam qua trinh di~n pharr tren, b) Tinh pH cua dung dich Y thoi gian di~n pban dung dich la 630 giay (Coi th~ tich dung dich thay d6i khong dang k~ sau di~n pharr) each giai Ket qua cr, (-) Catot (FeL+, Na+, FeJ+, (+) Anot (SO/", H2O) Fe3+ + l e ~ Fe2+ (1) 2Cr ~ Cl2 + 2e Fe2+ + 2e ~ Fe (2) H20 ~ Y202 + 2e + 2H+ (2') (1 ') 2H20 + 2e ~ H2 + 20H" (3) a) s6 mol e trao d6i a di~n eire thai gian dien phan hi: 9,65.100 = Ol(mol) 96500 ' S6 mol Fe + 100 ml dung dich la: 0,02 mol s6 mol Fe2+ 100 ml dung dich la: 0,001 mol mol 100 ml dung dich la: 0,2 mol Theo cac ban pharr (rug (1) va (1') thai gian di~n phan la 100s thi moi c6 0,01 mol Fe3+ bi di~n pharr va 0,01 mol bi dien phan n = e se cr cr f)i~m H2O) I (5) V~y ca~ plnrong trinh (2), (3), (2') deu chua xay V~y khoi hrong dung dich giam hi: 0,005.71 = 0,355 (gam) b) S6 mol e trao dbi di~n C\lC thai gian dien phan hi: n = 9,65.630 e 96500 Kh6i IUQ11gdung dich giam 0,355 (g) 2,5 = 063(mol) ' Thea cac ban phan irng (1), (2), (3): Fe3+, Fe2+ bi di~n phan h~t v~ Fe ~ s6 mol e cln 1:1: 0,02.3 + 0,001.2 = 0,062 (mol e) ~ s6 mol e dung cho phan irng (3) hi: 0,063 - 0,062 = 0,001 (mol e) Tai anot: s6 mol cr bi di~n phan l~ : 0,063 mol ~ cr du ~ khong xay phan irng di~n phan H2O V~y s6 mol OH- dung dich thu duoc sau di~n phan hi: 0,001 mol (theo (3» ~ [OH-] = 0,00110,1 = 0,01 M ~ [W] = 10-12 (M) ~pH= 12 , pH= 12 2,5 Can 6: Amoni hydrosunfua hi mQt hop chat khong b~n, d8 dang phan huy NH3(k) va H2S(k): NI4HS(r) ~ NH3(k)+ H2S(k) Cho biet cac s6li~u nhiet QQnghoc sau day tai 25°C: ' HO(kJ.mor1) S°(J.K-1.mor1) NH4HS(r) ~156,9 113,4 NH3(k) ~45,9 192,6 H2S(k) ~20,4 205,6 a) Tinh AIfl, ASo, AGOtai 25°C cua phan irng tren (theo J/mol hoac J/mol.K) b) Hay tinh hang s6 can bang K; tai 25°C cua phan irng tren (khong cAn ghi don YD c) Hay tinh hang s6 can bfug K, tai 35°C cua phan irng tren gift thi6t rang ca AHOva ASOkhong phu thuoc nhiet dQ d) Hay tinh ap sudt toan pMn binh chua n6u phan img phan huy dat' can bang tai 25°C B6 qua thS tich cua ~HS(r) Biet T = 273 + tOC;R = 8,314 J/mol.K each giai KSt qua a) AHo = ~20,4 ~45,9 + 156,9:=,90,6 (kJ.mor ) = 90 600 (J.mor ) ASo = 205,6 + 192,6 - 113,4 = 284,8 (J.K-1.mor1) AGo298K= mo - TASo = 5729,6 (J.morl) b) AGo298K = ~RTlnKp Thay s6 vao ta thu diroc K, = 0,099004826 c) AG308K= MIo ~ TASo = 90600 - 308.284,8 = 2881,6 (J.morl) => x', = 0,324551184 d) P(toanph§n)= P(NH3) + p(H2S) P(NH3) = p(H2S) = 0,5 P(toanphAn) (do co s6 mol bang nhau) Kp = P(NH3).P(H2S) = [0,5P(toanphAni = 0,099004826 => P(toanohan)= 0,629300646 (bar) l Diem l 1,5 1,5 Can 7: ChAt hiiu cc X chi chua (C, H, 0) D6t chay hoan toan gam X r6i cho san pham chay hAp thu h6t 100ml dung dich NaOH 20% (d=I,2g/ml) thi sau phan img thAy n6ng dQ NaOH (6) dung dich chi 14,125% dang thoi kh6i hrong binh NaOH tang them 4,6 gam Xac dinh cong thirc phan ella X, bi€t Mx <290 tu Cach ghii Ket qua Dat cong tlnrc phan tir cua Y la: CxHyOz co so mol gam la a (mol) (x, y, z nguyen dirong) CxHyOz + (x+ y/4 - zl2) O2 ~ xC02 + y/2 H20 (mol) a ~ a.(x+ y/4 - zl2) a.x a.y/2 nNaOHban dfiu = 100.1,2.20/ (100.40) = 0,6 (mol) nNaOHcon lai sau hiip thu = 124,6 : 14,125/ (100.40) = 0,44 mol ¢ nNaOHdophan img = 0,6 - 0,44 = 0,16 mol Vi NaOH du nen muoi thu duoc la mu6i trung hoa, CO2 + 2NaOH ~ Na2C03 + H20 0,08 +0,16 ax = n CO2 = 0,08 mol ay = (4,6 - 44.0,08) 19 = 0,12 mol a (l2x+y+ 16z) = => az = 0,12 mol ax: ay: az = x: y : z = 0,08 : 0,12 : 0,12 = : : CTDGN cua Y la: CZH303 Vi CTDGN la C2H303 ~ CT phan tir cua Y co dang: (CZH303)n (n nguyen dirong) Do Mx < 290 ~ 75n < 290 ~ n < 3,86 ~ n = hoac CTDGN cua Y lei: CZH303 f)i~m 2,5 hoac Nhung vi s6 nguyen ill H phai ch~n nen n = V~y CTPT cua Y lei: C4H606 CTPT cua Y lei: C4H606 2,5 Cau 8: Mot han hop X gam m(>t ankan, anken va mot ankin co th~ tich 1,792 (lit) dktc, dtroc chia lam phdn bang Phin 1: cho l(>icham qua dung dich AgN031NH3 du thi th~y th~ tich h6n hQ'P giam di 12,5% va co 0,735 gam k6t rna Phin 2: dem d6t chay hoan toan n)i cho san pham l(>iqua 9,2 lit dung dich Ca(OH)z 0,0125M thi thay 11 gam k6t rna L9C be k6t tua, dun nong dung dich thu diroc lai thiiy xu~t hien k6t tua, Xac dinh cong thirc phan nr cua cac hidrocacbon bi6t ankan va anken co cung s6 nguyen ill cacbon Ket_qmi Cach giai se mol cua h6n hop X: 1,792/22,4 = 0,08 mol ~ Tong s6 mol phan = phAn = 0,04 mol Vi cho 12 h6n hop qua dd AgN031NH3 th~y xuat hien k6t rna ~ ankin thuoc loai ank-l-in ~ nankin= 0,04xI2,5% = 0,005 (mol) ~ nketttia= 0,005 (mol) ~ Mketttia= 0,735/0,005 = 147 g/mol ~ Phan tir k6t rna chi chua nguyen tft Ag ~ CT k6t rna co dang R-C=CAg ~ MR + 24+108 = 147 ankin leiCH3G=CH Diem (7) ~MR= 15~R1aCH3 V~y ankin la CH30=CH S5 mol han hop anken + ankan = 0,04 -0,005 = 0,035 (mol) Phin 2: d5t chay 410 fa CO2 CO2 + Ca(OH)2 ~ CaC03 + HzO 0,11 0,11 0,11 S5 mol Ca(OH)z = 9,2.0,0125=0,115 mol Vi dung dich thu diroc sau loc bo k~t rna dun nong lai co k~t rna xudt hien ~ Phai c6 Ca(HC03)2 diroc tao 2C02 + Ca(OH)2 ~ Ca(HC03)z 0,01 0,005 V~y thea bai fa va phuong trinh ta tinh diroc t6ng s5 mol COzla: 0,01 + 0,11 = 0,12 mol S6 mol CO2 C3Iit chay sinh la: 0,005 = 0,015 mol ~ S6 mol CO2 d6t chay ankan+ anken 410 la: 0,12 - 0,015 = 0,105 mol ~ S6 nguyen illcacbon ankan = anken = 0,105 0,035 = V~y cong tlnrc phan ill cua ankan la : C3Hs cua anken: C3H6 ankan la : C3HS; cua anken: C3l4 Can 9: H6n hop A g6m chdt htru co mach he chua cung mQt loai nhom chirc hoa hoc Khi dun nong 47,2g han hop A voi hrong du dung dich NaOH thi thu duoc mQt nrou don chirc va 38,2 g han hop mu6i cua hai axit htru co don clnrc, k6 ti6p day d6ng dfutg M~t khac n6u d5t chay h6t 9,44g A cdn vita du 12,096 O2, thu diroc 10,304 COz Cac th6 tich dktc Xac dinh CTPT, vi~t CTCT cua cac ch~t co A each giai Di~m S5 mol O2 cAn dung d6 d6t chay 9,44 gam Ala: 12,096/22,4 = 0,54 mol S5 mol CO2 sinh la: 10,304/22,4 = 0,46 mol Theo dinh luat baa toan kh6i hrong: mH20 = 9,44 + 0,54 32 - 0,46.44 = 6,48 gam ~ nH20= 6,48/18 = 0,36 mol Theo dinh lu~t baa toan nguyen t5 oxi: S5 mol oxi co este la: 0,46.2 + 0,36 - 0,54.2 = 0,2 mol ~ s.3 mol este = 0,2/2 = 0,1 mol (vi nrou va muoi sinh fa la don chtrc ~ este don chirc) Ma s6 mol este = Deoz - nH20~ este la khong no don chirc, mach h6 va co lien k6t doi g6c hidrocacbon Theo bai fa axit la d6ng ding k~ tiep, va chi co nrou sinh fa nen este cling la d6ng ding k~ tiep Cong thirc trung binh cua este la C_H _ 02 " 2n-2 Kh6i hrong mol trung binh ella este: 9,44/0,1 = 94,4 glmol (8) - - n = 0,46/0,1 = 4,6 - CTPT cua este la : CJl602 va CSHg02 S6 mol este 47,2 gam la: 47,2/94,4 = 0,5 mols6 mol muoi = 0,5 mol - Kh6i lUQ11gmol cua muoi: 38,2/0,5 = 76,4 g/mol - mudi phai co khoi hrong mol < 76,4 g/mol - co nhAt HCOONa thea man (M= 68 gam/mol) - este Ia: HCOO-CH2 - CH=CH2 va CH3-COO-CHrCH=CH2 CTPT cua este la : C4H602 va CSHg02 CTCT cua este la: HCOO-CH2 CH=CH2 va CH3COOCH2CH=CH2 Can 10: m~ng co sa la don vi d.u true nho nh~t duoc l~p lai c~u true tinh th~ Phuong phap nhiSu xa tia X cho biSt rang mang co sa cua tinh the vang co c~u true l~p phirong Him dien (nrc la tam cua m6i nguyen nr n~m cac dinh cua hinh l~p plnrong va tam diem ella m6i mat) Canh ella () mang bang 0,408 nm , Tinh xem m6i () mang co sa chua bao nhieu nguyen Kh6i IUQ11grieng cua vang la 1,93.104 kg.m" Hay tinh the tich va kh3i hrong cua () mang l~p phuong co sa (theo gam) Tinh kh6i hrong cua mot n~uyen tU vang (theo gam) va s6 Avogadro Cho nguyen til kh3i cua vang la 196,97; lu= 1,6605.10- kg a ru /T~!/ "e; -a-i-:;; i vw each ghii KStqua Vi Au co cau true l~p phuong tam dien nen m6i mang chua nguyen tU kim loai Au : goc 1/8 nguyen tir = nguyen nr m~t 1/2 nguyen ill = nguyen ill - The tich cua mang l~p phuong eo sa = (0,408 1O-9 30 = 67,917312 10- m - Kh6i hrong ella mang l~p phirong co sa 30 26 = 1,93.10 67,917312.10= 131,080412 10- kg 23 = 131,080412 10- gam Tinh kh6i hrong ella mot nguyen yang va s6 Avogadro Cho nguyen tfr khoi cua vang la 196,97 - kh6i luong cua m9t nguyen nr yang la : 196,97 1,6605.10-24 = 327,068685 10-24 (gam) Co nguyen tU kim loai Au mang eo sa Di~m 1 i ru - I " 'A kh/ · 01 uang neng eua u = 1, 93 10 = 4.196,97.10- 30 NA·67,92.1O- - Kh3i hrong cua (> mang co so la: 131,080412.10-23 gam kh6i hrong ella mQt nguyen tir yang Ia 327,068685.10-24 (gam) INA = 6,010423723.1023 INA = 6,010423723.1023 Luu f: HS SIT dung each lam khac k~t qua dung v§n eho di€m t6i da -H~t8 (9)