Dap an de thi thu thpt quoc gia hoa hoc ha noi

Ki SO GIA.O DVC VA DAO T�O HA N C6 thi nghiern thoa man la ( ) , (4) Dap an C Cau : Phuong phap: Tinh ch.it h6a hoc cua Crom va hop ch.it Hurrng d�n giai: Khi cho Cr03 vao mroc se xay dong thoi phan (mg sau: Cr03 + H20 -* H2Cr04 2Cr03 + H20 -* H2Cr207 Dap an B Cau : Phuong phap: Bao toan nguyen t6 oxi: no(X) + 2n02 = 2nc02 + nH20 => no(X) => nx Goi s6 lien k�t pi X la k => Khi d6t chay x thi thu duce san pham: ncoz - nH20 = ( k - l)nx => k Bien luan => s6 lien k�t pi c6 th� phan (mg duce voi Br2 => ns-z Hurrng d�n giai: Bao toan nguyen t6 oxi: no(X) + 2n02 = 2nc02 + nH20 => nocx) = 0,6 mol Vi X la triglixerit => nh6m COO => nguyen tu oxi => nx = 1/6.nocx) = 0, mol Goi s6 lien k�t pi X la k => Khi d6t chay X thi thu duce san pham: ncoz - nH20 = ( k - l)nx => k = C6 lien k�t pi nh6m COO => c6 lien k�t pi g6c hidrocacbon (C=C) => ns-z = nx = , mol Dap an B Cau : Hurmg d�n giai: Cac kim loai kiem la: Li, Na, K Dap an B Cau : Phurrng phap: Cac phan (mg: HPoi- + Ol-l -+ P043- + H20 => Tinh duoc nH20 => Bao toan khoi hrong => kh6i luong Na3P04 => s6 mol => 3AgN03 + Na3P04 -+ 3NaN03 + Ag3P04 J => ffik§ttua Hurmg d�n giai: Cac phan (mg: H2P04- + 20H- -+ P043- + 2H20 HPoi- + Ol-l -+ P043- + H20 Ta thay: nNaOH = nH20 = 0,25 = 0,25 mol Bao toan khoi luong: mx + ffiNaOH = ffiNa3P04 + mnzo => ffiNa3P04 = , + , - , = 44,28g => nNa3P04 = 0,27 mol 3AgN03 + Na3P04 -+ 3NaN03 + Ag3P04 J => nAg3P04 = nNa3P04 = 0,27 mol => ffiAg3P04 = l , g Dap an B Cau : Phurrng phap: - Tu ti 1� mol ns-z : nx => s6 lien k�t pi trung binh cac ch�t - K�t hop du kien X la h6n hop => CTCT cua chat X - Tinh duce s6 mol cac ch�t X => ti 1� mol I X => Trong , g X thi ti 1� mol kh6ng d6i => s6 mol cac ch�t => Kh6i luong k�t tua Hurmg d�n giai: x nx = , : 22,4 = 0,4 mol; nBr2 = : = , mol => ns-z : nx = , = sf> lien k�t pi trung binh Cua ch.lt X => ch.it X c6 pi va pi phan tu Vi hidrocacbon la ch.it => s6 C :S => ch.it phai l a : CH=C-CH=CH2 (x mol) va CH=C-C=CH (y mol ) => x + y = 0,4 mol va 3x + 4y = , mol => x = , ; y = 0,25 mol s6 => x : y = ,1 : 0,25 = : (ti l� mol ch.it h6n hop X) - Trong , g X c6 CH=C-CH=CH2 (3n mol) va CH=C-C=CH (Sn m o l ) => mx = n + S n S O = , => n = , mol Khi phan (mg voi AgN03/NH3 thu duce h6n hop k�t tua gom: , mol AgC=C-CH=CH2 va , mol AgC=C-C=CAg => ffitua = 44,925g Dap an A Cau : Phurrng phap: - Tu s6 mol HCl => Dua vao s6 s6 mol 02 => s6 mol Cb mol k�t tua cho Z + AgNQ3 du => phan k�t tua g6m nhirng g i ? (C6 Ag hay khong ? ) => s6 mol cac ch.it X - X + HN03: Coi T gom cac ion Cu2+, Bao toan e voi s6 mol NO => s6 Fe +, Fe3+ mol cac ion T => C% F e(NQ3 )3 Hurrng dfin giai: Y gom oxit va muoi Clorua Hoa tan X dn nnci = , = 0,4 mol (2H+ + - -+ H20) => no(Y) = 1h nnci = 0,2 mol => n02 = 0, mol => ncu = , - , = , mol Goi s6 mol C u v a Fe l�n luot l a x va y => mx = 64x + 56y = , g (1) Sau phan (mg voi axit HCl thi ncirz; = 0,4 + , = , mol => ffiAgCI = , , = , < , 5 g => k�t tua c6 Ag (do Fe2+ + A g + -+ Fe + + Ag) => nAg = 0,07 mol Bao toan e ca qua trinh phan (mg: 2ncu + 3nFe = 4n02 + 2nc12 + nAg 2x + 3y = , + , + 0,07 = 0,64 mol (2) Tu ( , ) => x = , ; y = , mol - Khi X + HNQ3 san pham gia str c6: , mol C u + ; t mol Fe3+ va (0, - t) mol Fe2+ nNo = , 8 : 22,4 = 0, mol Bao toan e: 2ncu2+ + 2nFe2+ + 3nFe3+ = 3nNo => , + ( , - t ) + 3t = , => t = 0,03 mol Ta c6: nnnos = 4nNo = , = 0,68 mol => ffidd HN03 = g Bao toan khoi luong: mx + maa HN03 = ffidd T + ffiNO => ffi d d T = , + - 0, = , g => C%Fe(N03)3 = , : , = 5, 14% (Gin nh�t voi gia tri 5%) Dap an D Cau : P h u rr n g phap: - Dua vao thi nghiern Ba(OH)2 + Y => s6 mol BaC03 kSt tua => s6 mol HC03- Y (Vi Cho axit tu tu vao X nen thir W phan irng l a : O H - + H + -+ H20 col- + H+ -+ HC03- HCOf + H + -+ C02 + H20) Bao toan C : ncoz bd + nNa2C03 = nBaC03 + ncoz Bao toan dien tich: nNa = nc1 + 2nso4 + nHc03 san pham (Y) => nNa2C03 = y => nNa Bao toan Na: nNa+ = nNaOH + 2nNa2C03 => nNaOH = x =>x: y Hurmg d�n giai: Text ncoz = , : 22,4 = 0,275 mol nnci = , = 0,2 m o l ; nnzsoa = , , = 0,06 m o l ; nco2sanphftm = , 8 : 22,4 = , mol - Khi Y + Ba(OH)2 thi kSt tua gom 0,06 mol BaS04 va BaC03 => nBaC03 = 0,23 mol = nHC03 Y (Vi Cho axit tu tu vao X nen thir W phan irng l a : O H - + H + -+ H20 col- + H+ -+ HC03- HCOf + H + -+ C02 + H20) Bao toan C : ncoz bd + nNa2C03 = nBaC03 + ncoz san pham => nNa2C03 = 0,075 mol => y = 0,075 M - Khi X + axit thi sau phan irng dung dich chi con: NaCl ; Na2S04 va NaHC03 Bao toan dien tich: nNa = nc1 + 2nso4 + nHc03 => nNa = , 5 mol Bao toan Na: nNa+ = nNaOH + 2nNa2C03 => nNaOH = , 5 - , = 0,4 mol = x (Y) = > x : y = 0,4 : 0,075 = , Dap an B Cau : Phurrng phap: X gom ch�t no don chuc mach D6t chay X: CnH2n02 + 02 => Illbinh -+ => CTPT chung: CnH2n02 hay RCOOR1 nC02 + nH20 tang = mco2 + nH20 => ncoz - X + KOH: RCOOR1 + KOH -+ RCOOK + R1 OH => nxon = nx - Ta thay: Ilanken = Ilancol = neste => Ilaxit Goi s6 Cacbon cua este va axit la a va b => BiSu thirc lien => a M gifra s6 mol C02 va s6 mol cac ch�t x b => CTCT cua ch�t ' Hurmg d�n giai: X gom ch�t no don chuc mach D6t chay X: CnH2n02 + 02 => Illbinh -+ => CTPT chung: CnH2n02 hay RCOOR1 nC02 + nH20 t an g = mco2 + nH20 = 6,82g => ncoz = nH20 = 0, 1 mol - X + KOH: RCOOR1 + KOH -+ RCOOK + R1 OH => nxon = nx = 0,04 mol - Ta thay: Ilanken = Ilancol = neste = , mol => Ilaxir " 0,04 - , = 0,025 mol Goi s6 Cacbon cua este va axit la a va b => ncoz = , a + 0,025b = , 1 =>a= ; b = => axit la CH3CQOH va este la CH3CQOC2Hs Dap an C Cau : Phurrng phap: Ly thuyet tong hop h6a hoc hiru ca Hurmg d�n giai: C2H402 c6 cac c6ng thirc du t Z la HCOOCH3 => Y la HOCH2CHO A s a i X tao dUQ'C lien k�t hidro lien phan tu nen c6 nhiet soi cao hon Z B sai.Y la ch�t hiru CO' tap chirc C sai Z c6 phan irng trang bac D dung Dap an D Cau 7 : Phurrng phap: Vi Z, T, E c6 cung s6 C (Va X va Y d�u c6 6C phan t u ) = > m6i ch�t c6 2C => Cong thuc du tao cua x Tinh duce s6 mol cac ch�t phan => bien luan tim CTCT cua Y => Thanh phan cac ch�t c6 Q => %m Hurmg dfin giai: Vi Z, T, E c6 cung s6 C (Va X va Y d�u c6 6C phan t tr } > > m6i ch�t c6 2C => X la C2HsOOC-COONH3C2Hs (d� thoa man di�u kien tren) nNaOH = 0,6 mol ; nc2HSOH = 0,3 mol ; nc2HSOH = 0,2 mol Dung dich Q chira ch�t cung s6 C (trong phai c6 (COONa)2) nen cac ch�t d�u gom 2C Do nc2HSNH2 > nc2HSOH => C2HsNH2 sinh them tu Y => Y phai la CH3COOCH2COONH3C2Hs => nx = 0,2 mol va ny = , mol => Q gom 0,2 mol ( C O O N a ) ; , mol CH3COONa; , mol HOCH2COONa => %mcmcooNa = , % Dap an D Cau : Phurrng phap: Thu W dien phan: - Catot: Fe3+ + I e +» Fe2+ Cu2+ + Z e +-> Cu Fe2+ + e -+ Fe 2H20 + 2e -+ 20H- + H2 - Anot: e - -+ C b + 2e 2H20 -+ 4H+ + 02 + 4e Dua vao thir W di�n phan , oxit thu duce gom oxit => ion nao du sau dien phan, qua trinh di�n phan dung lai => S6 mol cac ion c6 X => mmu6i khan Hurmg dfin giai: Thu W dien phan: - Catot: Fe3+ + I e +» Fe2+ Cu2+ + Z e +-> Cu Fe2+ + e -+ Fe 2H20 + 2e -+ 20H- + H2 - Anot: e - -+ C b + 2e 2H20 -+ 4H+ + 02 + 4e a giai doan nao - Dung dich sau dien phan + NaOH tao kSt tua Y => nung kSt tua Y thu duoc oxit kim loai => oxit la CuO va Fe203 => Dung dich sau dien phan chi c6 Cu2+, Fe2+, so> => Chit rin thu duce sau dien phan chi gom C u = > ncu = , : 64 = , mol mctct giam = mcu + mc12 = , g => ncu = 0,25 mol Bao toan e: nFe3+ + 2ncu2+ = 2nc12 => nres- = 0,2 mol => nFe2+ (sau dp) = 0,2 mol => Oxit kim loai gom CuO va Fe203 (nFe203 = Yi nFe2+ = 0, mol) Ta c6: ffioxit = mcuo + ffiFe203 => mcuo = 28g => ncuo = , mol = ncu2+du => Dung dich ban d�u c6: ( , + 0, = 0,5 mol) C u + ; 0,2 mol F e + ; , mol Cl, Soi­ Bao toan di�n tich: 2ncu + 3nFe = nc, + 2nso4 => nso4 = , 5 mol =>mmu6ikhan = mo, + ffiFe + mo + ffiS04 = l , g Dap an C Cau : Phurrng phap: Bao toan khoi luong => n02 => s6 mol O T Tinh duce nNaOH thong qua Na2C03 => Bien luan so sanh s6 mol Oxi T (theo de bai) so voi tnrorig hop nSu T chi chira nh6m chira nh6m chirc khac hay kh6ng => CTPT tong quat cua cac chit Bao toan nguyen t6 C => s6 C cua cac muoi T => CTCT cac muoi => CTCT cua este ban d�u - Bao toan khoi luong => khoi hrong ancol Z Bien luan s6 C ancol => CTPT ancol => CTCT cua este =>%m Hurmg d�n giai: Bao toan nguyen t6 Na: nNaOHpu=2nNa2C03 = , mol Bao toan khoi luong: ffiT + m02 = ffiNa2C03 + mco2 + mnzo=> n02 = 0,26 mol Bao toan : no(T) + 2n02 = 2nc02 + nH20 + 3nNa2C03 => no(T) = 0,32mol Gia SU muoi T d�u chi chira nh6m coo => no(T) = 2nNa = , mol < , mol => Phai c6 muoi chira nh6m OH => X la RCOOR1 (x m o l ) ; Y la ACOO-A1-COOB (y mol) T gom: RCOONa (x m o l ) ; ACOONa (y mol) va HO-A1-COONa (y mol) => nNaOH = x + 2y = , mol no(T) = 2x + 2y + 3y = , mol => x = 0,06 ; y = 0,04 mol D�t n, m, p la s6 C tuong irng voi muoi tren => Bao toan C : nc = 0,06n + 0,04m + 0,04p = nNa2C03 + ncoz = 0,07 + , coo => T => n + 2m + 2p = => n = ; m = ; p = (thoa man) => T gom CH3CQONa (0,06 m o l ) ; HCOONa (0,04 m o l ) ; HOCH2H4CQONa (0,04 mol) Bao toan khoi luong: mE + mNaOH = mz + mT => mT = , + , - , = 4,04g Z gorn R10H (0,06 m o l ) va BOH (0,04 mol) => mz = , M + 0,04.M2 = 4,04 => 3M1 + 2M2 = 202 => M1 = 46 (C2HsOH) va M2 = 32 (CH3QH) la nghiem nh�t X la CH3CQOC2Hs (0,06 mol) Y la HCOOC2H4CQOCH3 (0,04 mol) =>%my= 0,04.132: , = 50% Dap an B Cau : Phurrng phap: Ly thuyet cac ch�t VO CO' Hurrng d�n giai: ( ) AgNQ3 + HCl � AgCl t + HNQ3 (3) Cu khong phan (mg voi HCl (4) Ba(OH)2 + 2NaHCQ3 � BaCQ3 t + Na2C03 + 2H20 (5) Fe3Q4 + 8HC1 � 2FeCb + FeCb + 4H20 � x 2x 2FeCb + C u � CuCb + 2FeCb 2x � x => du x mol Cu (6) B a + 2H20 � Ba(OH)2 + H2 Ba(OH)2 + Ca(HCQ3)2 � CaCQ3 + BaCQ3 + 2H20 (9) 4Ba(OH)2 + Ab(SQ4)3 � 3BaSQ4 + Ba(Al02)2 + 4H20 => c6 thi nghiem thoa man Dap an A ... ancol Dun n6ng toan bo ancol tren voi H2S04 d�c thi thu duce , m o l anken nhat N�u d6t chay hoan toan X r6i h�p thu toan bo san pharn vao dung dich Ca(OH)2 du thi khoi hrong binh tang 6,82g (hieu... => F l a F e Dap an D Cau : Hurrng d�n giai: E c6 cong thirc C4Hs02 thuy phan tao C2HsOH nen E la CH3CQOC2Hs Dap an A Cau : Phurrng phap: Ly thuyet cacbohidrat Hurrng d�n giai: D� phan biet ch�t... + 4H20 (e) khong phai la phan (mg oxi h6a - khir (g) khong phai la phan (mg oxi h6a - khir => C6 thi nghiem xay phan (mg oxi h6a - khir Dap an A Cau : Hurrng dfin giai: Cac phan (mg: Ca(OH)2 +