Đề thi học sinh giỏi môn hóa 9 tỉnh quảng trị năm học 2015 2016(có đáp án)

S GIO DC V O TO QUNG TR THI CHNH THC Cõu í 1 Cõu 1 2 3 HNG DN CHM K THI CHN HC SINH GII VN HểA LP 9 THCS Khúa thi ngy 15 thỏng 3 nm 2016 Mụn thi: HểA HC Ni dung a) Ca + 2H2O Ca(OH)2 + H2 Ca(OH)2 + Na2CO3 CaCO3 + 2NaOH b) 2Na + 2H2O 2NaOH + H2 3NaOH + AlCl3 3NaCl + 2Al(OH)3 NaOH + Al(OH)3 NaAlO2 + 2H2O c) Fe + 2AgNO3 Fe(NO3)2 + 2Ag Fe(NO3)2 + AgNO3 Fe(NO3)3 + Ag d) 3Ba(HCO3)2 + 2AlCl3 2Al(OH)3 + 6CO2 + 3BaCl2 0 im 2,0 0 t t - t chỏy cacbon: C + O2 CO2, CO2 + C 2CO Hn hp khớ A gm CO, CO2 v N2 t0 - Cho A tỏc dng vi FeO: FeO + CO Fe + CO2 Khớ B gm CO2, N2; hn hp rn C gm FeO, Fe - Do un sụi D li c kt ta K nờn CO2 tỏc dng to hai mui: CO2 + Ca(OH)2 CaCO3 + H2O, 2CO2 + Ca(OH)2 Ca(HCO3)2 t0 Ca(HCO3)2 CaCO3 + CO2 + H2O Kt ta K: CaCO3, dung dch D: Ca(HCO3)2 - Cho C tỏc dng vi dung dch HCl: Fe + 2HCl FeCl2 + H2, FeO + 2HCl FeCl2 + H2O Khớ l H2, dung dch E gm FeCl2, HCl d (cú th) - Cho E tỏc dng vi dung dch NaOH d: HCl + NaOH NaCl + H2O, FeCl2 + 2NaOH Fe(OH)2 + 2NaCl Kt ta F l Fe(OH)2 - Nung F trong khụng khớ: 4Fe(OH)2 + O2 2Fe2O3 + 4H2O Gi cụng thc oxit kim loi l MxOy đặc, t 0 2MxOy + (2mx2y)H2SO4 xM2(SO4)m + (mx-2y)SO2 + (2mx-2y)H2O (1) 2(Mx+16y) gam (mx-2y) 20,88 gam 0,145 mol 2(Mx + 16y) (mx 2y) 2y = M = 72m 80 T l: 20,88 0,145 x 2y/x 1 2 8/3 m 2 3 3 3 M 64 (Cu) 136 (loi) 56 (Fe) 8/3 Vy MxOy l Cu2O hoc FeO 2,5 1,0 Cõu 2 1 2 Do A tỏc dng vi B thỡ cú kt ta BaSO4, B tỏc dng vi C thỡ cú kt ta xut hin, A tỏc dng vi C thỡ cú khớ thoỏt ra A: H2SO4 hoc NaHSO4, B: BaCl2, C: Na2CO3 NaHSO4 + BaCl2 BaSO4 + NaCl + HCl BaCl2 + Na2CO3 BaCO3 + 2NaCl 2NaHSO4 + Na2CO3 2Na2SO4 + CO2 + H2O - Oxit axit Axit: SO3 + H2O H2SO4 - Oxit axit Mui: SO3 + 2NaOH Na2SO4 + H2O - Axit Mui: H2SO4 + 2NaOH Na2SO4 + 2H2O - Mui Axit: AgNO3 + HCl AgCl + HNO3 1,0 1,75 2 Cõu í 3 Ni dung - Oxit baz Mui: Na2O + SO3 Na2SO4 - Oxit baz Baz: Na2O + H2O 2NaOH t0 - Baz Oxit baz: Cu(OH)2 CuO + H2O - Baz Mui: Cu(OH)2 + H2SO4 CuSO4 + 2H2O - Mui Baz: CuSO4 + 2NaOH Cu(OH)2 + Na2SO4 Gi húa tr ca M l n (1 n 3), chn 2 mol M 2M + nH2SO4 M2(SO4)n + nH2 (1) 2 mol n mol 1 mol n mol Theo (1): n(H2) = n(H2SO4) (phn ng) = n mol n.120 = 1, 2n mol Vỡ d 20% n(H2SO4) (ban u) = 100 98 ì 1, 2n Khi lng dung dch H2SO4 ó dựng l: ì 100 = 588n gam 20 Theo LBTKL: m(dung dch sau) = 2M + 588n 2n = 2M + 586n gam 2M + 96n 23,68 = M = 28n T l: 2M + 586n 100 n 1 2 3 M 28 (loi) 56 (Fe) 84 (loi) Vy kim loi M l st (Fe) im 1,5 Cõu 3 Gi s mol Mg, Fe, Al2O3 v XO ln lt l a, b, c, d trong mi phn Mg + 2HCl MgCl2 + H2 (1) Fe + 2HCl FeCl2 + H2 (2) Al2O3 + 6HCl 2AlCl3 + 3H2O (3) XO + 2HCl XCl2 + H2O (4) 3Mg + 8HNO3 3Mg(NO3)2 + 2NO + 4H2O (5) Fe + 4HNO3 Fe(NO3)2 + NO + 2H2O (6) Al2O3 + 6HNO3 2Al(NO3)2 + 3H2O (7) XO + 2HNO3 X(NO3)2 + H2O (8) Mg + H2SO4 MgSO4 + H2 (9) X + H2SO4 XSO4 + H2 (10) Al2O3 + 2NaOH 2NaAlO2 + H2O (11) XO + 2NaOH Na2XO2 + H2O (12) a + b = 0,22 a=0,12 mol Theo (1,2,5,6) v bi ra ta cú: 2 b=1,25ì 3 a b=0,1 mol Ta cú: moxit = 13,16 (0,12.24 + 0,1.56) = 4,68 gam Nu ch cú Al2O3 tan trong dung dch NaOH thỡ: 1 0,1 n Al2O3 = n NaOH = = 0, 05mol m Al2O3 = 5,1 gam > 4,68 gam 2 2 XO tan trong dung dch NaOH Theo (11, 12) v bi ra ta cú h: d < 0, 05 c + d = 0, 05 0, 42 X < 77, 6 102c + (X + 16)d = 4, 68 d = 86 X 2,5 1,5 m m > 2,5. X > 60 24 X Vy X l Zn (km) c = 0,03 mol v d = 0,02 mol Mt khỏc theo (9, 10) v bi ra: Cõu 4 1 a) CH3CH(OH)-COOH + 2Na CH3CH(ONa)-COONa + H2 1,25 3 Cõu í Ni dung im H SO đặc,t 0 2 4 b) CH3CH(OH)-COOH +CH3COOH ơ CH3CH(OOC-CH3)-COOH + 2 3 4 H2O c) 2CH3CH(OH)-COOH + Ba(OH)2 [CH3CH(OH)-COOH]2Ba + 2H2O d) CH3CH(OH)-COOH + NaHCO3 CH3CH(OH)-COONa +CO2 + H2O CaO,t 0 CH3CH(OH)-COONa + NaOH CH3CH2OH + Na2CO3 CH2 = CH CH2 CH3, CH3 CH = CH CH3,CH2 = C(CH3) CH3, Fe,t Phng trỡnh phn ng: C6H6Br + Br2 C6H5Br + HBr (1) ơ 0,1 0,1 (mol) 15,7 100 m(C6H6)= . .78 = 9,75 gam. 157 80 A iu ch t ỏ vụi A: CaC2, B: C2H2, C: CH3COOH/C2H5OH CaC2 + 2H2O C2H2 + H2O HgSO4 ,t0 C2H2 + H2O CH3CHO 0,5 0 0,5 Mn ,t t 2CH3CHO +O2 2CH3COOH hoc: CH3CHO+H2 C2H5OH 2+ 0 0 H SO đặc,t 0 2 4 CH3COOH+CH2=CH-CH2OH CH3COOCH2CH=CH2 + H2O ơ 1,25 H SO đặc,t 0 2 4 Hoc: C2H3COOH + C2H5OH C2H3COOC2H5 + H2O ơ xt,p,t 0 nCH3COOCH2CH=CH2 [-CH2-CH(CH2-OOCH3)-]n xt,p,t Hoc: nC2H3COOC2H5 [-CH2-CH(COOC2H5)-]n Hoc PVA 0 5 H SO đặc,t 0 2 4 CxHyCOOH + CnH2n+1OH CxHyCOOCnH2n+1 + H2O (1) ơ CxHyCOOH + NaOH CxHyCOONa + H2O (2) t0 CxHyCOOCnH2n+1 + NaOH CxHyCOONa + CnH2n+1OH (3) CxHyCOOCnH2n+1+(4x+6n+y+1)O2 (n+x+1)CO2+(2n+y+1)/2H2O(4) CxHyCOOH + (4x+y+1)/4O2 (x+1)CO2 + (y+1)/2H2O (5) CnH2n+1OH +3n/2O2 nCO2 + (n+1)H2O (6) Gi s mol este l a (mol) Cú (0,12 a) (mol) CnH2n+1OH d, (0,15 a) (mol) CxHyCOOH (RCOOH) d trong 13,2 gam X Ta cú: nru ban u = n N2 = 3,36 / 28 = 0,12mol , n CO2 = 0,57mol , n H2O = 0,46mol Theo (1, 2, 3) ta cú: naxit ban u = nmui = nNaOH = 0,15 (mol) Ta cú: 3,84 / 0,12 = 32 = 14n + 18 n = 1 ru l CH3OH 2,5 13, 2 0,57.12 0, 46.2 a = 0,08 16 Vy trong 13,2g X: 0,04(mol) CH3OH, 0,07(mol) RCOOH, 0,08(mol) RCOOCH3 Ta cú: 0,04.32 + 0,07(R + 45) + 0,08(R + 59) = 13,2 R = 27 l C2H3Vy CTPT ca este l C2H3COOCH3 axit(ban đầu) 0,15 r ợu(ban đầu) 0,12 = > = H%theo r ợu Do: 1 1 1 1 0, 08 .100% =66,67% H% = 0,12 - Thớ sinh cú th lm cỏch khỏc, nu ỳng vn t im ti a trong mi cõu. Nu thiu iu kin hoc thiu cõn bng hoc thiu c hai thỡ tr mt na s im ca PTHH ú. - Lm trũn n 0,25 im. HT. BTNT oxi: 2(0,15-a) + 2a + (0,12-a) = 4