hướng dẫn giải 99 bài tập học sinh giỏi hóa 9
HNG DN GII 500 BI TP HểA HC THCS DNH CHO HC SINH GII Dựng : - ễn luyn HSG mụn Húa hc cỏc cp - ễn luyn vo lp 10 Húa hc trng chuyờn GV: TRNG TH THO Hố 2016 Hng dn gii 500 bi Húa hc THCS hay v khú Bi Vit cỏc phng trỡnh phn ng xy cỏc trng hp sau: a) Ca tỏc dng vi dung dch Na2CO3 b) Na tỏc dng vi dung dch AlCl3 c) Fe tỏc dng vi dung dch AgNO3 d) Ba(HCO3)2 tỏc dng vi dung dch AlCl3 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 Bi t chỏy cacbon khụng khớ nhit cao thu c hn hp khớ A Cho A tỏc dng vi FeO nung núng thu c khớ B v hn hp rn C Cho B tỏc dng vi dung dch Ca(OH)2 thu c kt ta K v dung dch D un sụi D li c kt ta K Cho C tỏc dng vi dung dch HCl thu c khớ v dung dch E Cho E tỏc dng vi dung dch NaOH d, thu c kt ta F Nung F khụng khớ c mt oxit nht Xỏc nh A, B, C, D, K, E, F v vit cỏc phng trỡnh phn ng xy t t - t chỏy cacbon: C + O2 CO2, CO2 + C 2CO Hn hp khớ A gm CO, CO2 v N2 0 t - 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 CaCO3 + CO2 + H2O Ca(HCO3)2 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 khụng khớ: 4Fe(OH)2 + O2 2Fe2O3 + 4H2O Bi Hũa tan ht 20,88 gam mt oxit kim loi bng dung dch H2SO4 c núng, thu c dung dch X v 3,248 lớt khớ SO2 (sn phm kh nht, ktc) Xỏc nh oxit kim loi Gi cụng thc oxit kim loi l MxOy đặ c, t xM2(SO4)m + (mx-2y)SO2 + (2mx-2y)H2O (1) 2MxOy + (2mx2y)H2SO4 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 8/3 m 3 M 64 (Cu) 136 (loi) 56 (Fe) 8/3 Vy MxOy l Cu2O hoc FeO GV: Trng Th Tho Email: thaonguyenh81@gmail.com Page Hng dn gii 500 bi Húa hc THCS hay v khú Bi Ba dung dch A, B, C tha món: 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 CO2 thoỏt Tỡm A, B, C v vit cỏc phng trỡnh phn ng xy 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 A: H2SO4 hoc NaHSO4, B: BaCl2, C: Na2CO3 NaHSO4 + BaCl2 BaSO4 + NaCl + HCl BaCl2 + Na2CO3 BaCO3 + 2NaCl 2NaHSO4 + Na2CO3 2Na2SO4 + CO2 + H2O Bi Chn cht phự hp, ghi rừ loi cht v vit cỏc phng trỡnh phn ng theo s sau: - 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 - 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 Bi Hũa tan ht mt lng kim loi M dung dch H2SO4 20% (loóng, d 20% so vi lng cn phn ng), thu c dung dch cha mui trung hũa cú nng l 23,68% v axit d Tỡm M Gi húa tr ca M l n (1 n 3), chn mol M 2M + nH2SO4 M2(SO4)n + nH2 (1) mol mol n 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 1, 2n 100 588n gam Khi lng dung dch H2SO4 ó dựng l: 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 M 28 (loi) 56 (Fe) 84 (loi) Vy kim loi M l st (Fe) Bi Chia 26,32 gam hn hp A gm Fe, Mg, Al2O3 v oxit ca kim loi X cú húa tr thnh phn bng Phn tan ht dung dch HCl d, thu c 0,22 mol H2 Phn tỏc dng ht vi dung dch HNO3 loóng d, thu c khớ NO (sn phm kh nht), ú th tớch NO Fe sinh bng 1,25 ln Mg sinh Nu hũa tan ht lng oxit cú mi phn phi dựng va ht 50 ml dung dch NaOH 2M Bit ly m gam Mg v m gam X cho tỏc dng vi dung dch H 2SO4 GV: Trng Th Tho Email: thaonguyenh81@gmail.com Page Hng dn gii 500 bi Húa hc THCS hay v khú loóng d thỡ th tớch khớ H2 Mg sinh ln hn 2,5 ln X sinh Vit cỏc phng trỡnh phn ng xy ra, xỏc nh X v tớnh s mol mi cht mi phn Gi s mol Mg, Fe, Al2O3 v XO ln lt l a, b, c, d 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 ta cú: b=1,25ì a b=0,1 mol Ta cú: moxit = 13,16 (0,12.24 + 0,1.56) = 4,68 gam Nu ch cú Al2O3 tan dung dch NaOH thỡ: 0,1 n Al2O3 n NaOH 0, 05mol m Al2O3 = 5,1 gam > 4,68 gam 2 XO tan dung dch NaOH Theo (11, 12) v bi ta cú h: d 0, 05 c d 0, 05 0, 42 X 77, 102c (X 16)d 4, 68 d 86 X 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: Bi Bit axit lactic cú cụng thc l: Hóy vit cỏc phng trỡnh phn ng xy cho axit lactic ln lt tỏc dng vi: a) Na d b) CH3COOH c) Dung dch Ba(OH)2 d) Dung dch NaHCO3 va , cụ cn ly cht rn, cho cht rn tỏc dng vi vụi tụi xỳt nung núng a) CH3CH(OH)-COOH + 2Na CH3CH(ONa)-COONa + H2 H2SO4 đặ c,t0 CH3CH(OOC-CH3)-COOH +H2O b)CH3CH(OH)-COOH +CH3COOH c) 2CH3CH(OH)-COOH + Ba(OH)2 [CH3CH(OH)-COOH]2Ba + 2H2O d) CH3CH(OH)-COOH + NaHCO3 CH3CH(OH)-COONa +CO2 + H2O CaO,t CH3CH2OH + Na2CO3 CH3CH(OH)-COONa + NaOH GV: Trng Th Tho Email: thaonguyenh81@gmail.com Page Hng dn gii 500 bi Húa hc THCS hay v khú Bi Vit cỏc cụng thc cu to ng vi cụng thc phõn t C4H8 CH2 = CH CH2 CH3, CH3 CH = CH CH3,CH2 = C(CH3) CH3, Bi 10 Tớnh lng benzen ti thiu cn dựng iu ch 15,7 gam brombenzen Bit hiu sut phn ng theo benzen t 80% Fe,t Phng trỡnh phn ng: C6H6Br + Br2 C6H5Br + HBr (1) 0,1 0,1 (mol) 15,7 100 78 = 9,75 gam m(C6H6)= 157 80 Bi 11 Xỏc nh cỏc cht v vit cỏc phng trỡnh theo s chuyn húa sau: A (iu ch t ỏ vụi) B CH3CHO C Este Polime A iu ch t ỏ vụi A: CaC2, B: C2H2, C: CH3COOH/C2H5OH CaC2 + 2H2O C2H2 + H2O HgSO ,t0 CH3CHO C2H2 + H2O Mn ,t t0 2CH3CHO +O2 2CH3COOH hoc: CH3CHO+H2 C2H5OH H SO đặ c,t0 CH3COOCH2CH=CH2 + H2O CH3COOH+CH2=CH-CH2OH H SO đặ c,t0 C2H3COOC2H5 + H2O Hoc: C2H3COOH + C2H5OH xt,p,t [-CH2-CH(CH2-OOCH3)-]n nCH3COOCH2CH=CH2 xt,p,t [-CH2-CH(COOC2H5)-]n Hoc PVA Hoc: nC2H3COOC2H5 Bi 12 Thc hin phn ng este húa gia axit C xHyCOOH v ru CnH2n+1OH Sau phn ng tỏch ly hn hp X ch gm este, axit v ru t chỏy hon ton 13,2 gam hn hp X, thu c 12,768 lớt khớ CO2 (ktc) v 8,28 gam H2O Nu cng cho hn hp X nh trờn thỡ tỏc dng va vi 150 ml dung dch NaOH 1M, thu c 3,84 gam ru Húa hi hon ton lng ru ny thỡ thu c th tớch hi ỳng bng th tớch ca 3,36 gam N2 (o cựng iu kin nhit v ỏp sut) it cỏc phng trỡnh phn ng, xỏc nh cụng thc este v tớnh hiu sut phn ng este húa H2SO4 đặ c,t0 CxHyCOOCnH2n+1 + H2O (1) CxHyCOOH + CnH2n+1OH CxHyCOOH + NaOH CxHyCOONa + H2O (2) t0 CxHyCOONa + CnH2n+1OH (3) CxHyCOOCnH2n+1 + NaOH (n+x+1)CO2+(2n+y+1)/2H2O(4) CxHyCOOCnH2n+1+(4x+6n+y+1)O2 (x+1)CO2 + (y+1)/2H2O (5) CxHyCOOH + (4x+y+1)/4O2 nCO2 + (n+1)H2O (6) CnH2n+1OH +3n/2O2 Gi s mol este l a (mol) Cú (0,12 a) (mol) CnH2n+1OH d, (0,15 a) (mol) CxHyCOOH (RCOOH) d 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 ru l CH3OH GV: Trng Th Tho Email: thaonguyenh81@gmail.com Page Hng dn gii 500 bi Húa hc THCS hay v khú 13, 0,57.12 0, 46.2 a = 0,08 16 Vy 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 rư ợ u(ban đầu) 0,12 H%theo rư ợ u Do: 1 1 0, 08 H% = 100% =66,67% 0,12 BTNT oxi: 2(0,15-a) + 2a + (0,12-a) = Bi 13 Chn 10 cht rn khỏc m cho 10 cht ú ln lt tỏc dng vi dung dch HCl cú 10 cht khớ khỏc thoỏt Vit cỏc phng trỡnh phn ng minh ho -Cỏc cht rn cú th chn ln lt l: Zn; FeS; Na2SO3; CaCO3; MnO2; CaC2; Al4C3; Na2O2; Mg3N2; Zn3P2 -Cỏc khớ iu ch ln lt l: H2 ; H2S ; SO2 ; CO2 ; Cl2 ; C2H2 ; CH4 ; O2 ; NH3 ; PH3 -Cỏc ptp: 1/ Zn + 2HCl ZnCl2 + H2 2/ FeS + 2HCl FeCl2 + H2S 3/ Na2SO3 + 2HCl 2NaCl + SO2 + H2O 4/ CaCO3 + 2HCl CaCl2 + CO2 + H2O t 5/ MnO2 + 4HCl c MnCl2 + Cl2 + 2H2O 6/ CaC2 + 2HCl CaCl2 + C2H2 7/ Al4C3 + 12HCl 4AlCl3 + 3CH4 8/ 2Na2O2 + 4HCl 4NaCl + O2 + 2H2O 9/ Mg3N2 + 6HCl 3MgCl2 + 2NH3 10/ Zn3P2 + 6HCl 3ZnCl2 + 2PH3 Bi 14 Vit phng trỡnh húa hc ( dng cụng thc cu to thu gn) thc hin cỏc bin húa theo s sau: (1) (2) Axetilen Etilen Etan (7) (5) (8) (3) (4) (6) P.V.C Vinylclorua icloEtan Etylclorua Cỏc ptp: HC CH + H2 t , Pd H2C = CH2 (1) t , Ni H2C = CH2 + H2 t0 HC CH + HCl H3C CH3 (2) H2C = CHCl (3) - [H2C - CHCl]n - (4) n(H2C = CHCl) t , xt H2C = CH2 + Cl2 H2C = CHCl + HCl t , xt ClH2C CH2Cl (5) ClH2C CH2Cl (6) as GV: Trng Th Tho Email: thaonguyenh81@gmail.com Page Hng dn gii 500 bi Húa hc THCS hay v khú H3C CH3 + Cl2 CH3 CH2Cl + HCl (7) H2C = CH2 + HCl CH3 CH2Cl (8) Bi 15 Bit axit lactic cú cụng thc l: Hóy vit cỏc phng trỡnh phn ng xy cho axit lactic ln lt tỏc dng vi: a Na d b CH3COOH c Dung dch Ba(OH)2 d Dung dch NaHCO3 va , cụ cn ly cht rn, cho cht rn tỏc dng vi vụi tụi xỳt nung núng a) CH3CH(OH)-COOH + 2Na CH3CH(ONa)-COONa + H2 H SO đặ c,t0 CH3CH(OOC-CH3)-COOH +H2O b) CH3CH(OH)-COOH +CH3COOH c) 2CH3CH(OH)-COOH + Ba(OH)2 [CH3CH(OH)-COOH]2Ba + 2H2O d) CH3CH(OH)-COOH + NaHCO3 CH3CH(OH)-COONa +CO2 + H2O CaO,t CH3CH(OH)-COONa + NaOH CH3CH2OH + Na2CO3 Bi 16 Ba dung dch A, B, C tha món: 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 CO2 thoỏt Tỡm A, B, C v vit cỏc phng trỡnh phn ng xy 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 A: H2SO4 hoc NaHSO4, B: BaCl2, C: Na2CO3 NaHSO4 + BaCl2 BaSO4 + NaCl + HCl BaCl2 + Na2CO3 BaCO3 + 2NaCl 2NaHSO4 + Na2CO3 2Na2SO4 + CO2 + H2O Bi 17 Chia 80 (g) hn hp X gm st v mt oxit ca st thnh hai phn bng nhau: Ho tan ht phn I vo 400 (g) dung dch HCl 16,425% c dung dch A v 6,72 lớt khớ H (ktc) Thờm 60,6 (g) nc vo A c dung dch B, nng % ca HCl d B l 2,92% a Tớnh lng mi cht hn hp X v xỏc nh cụng thc ca oxit st b Cho phn II tỏc dng va ht vi H2SO4 c núng ri pha loóng dung dch sau phn ng bng nc, ta thu c dung dch E ch cha Fe2(SO4)3 Cho 10,8 (g) bt Mg vo 300 ml dung dch E khuy k, sau phn ng xy hon ton thu c 12,6 (g) cht rn C v dung dch D Cho dung dch D tỏc dng vi dung dch Ba(OH)2 d, lc kt ta v nung n lng khụng i c m (g) cht rn F (trong iu kin thớ nghim BaSO4 khụng b phõn hu) Tớnh CM ca dung dch E v giỏ tr m t cụng thc ca oxit st l FexOy Cỏc phng trỡnh hoỏ hc: Fe + 2HCl FeCl2 + H2 (1) FexOy + 2yHCl xFeCl 2y + yH2O (2) x nHCl ban u 400.16, 425 6,72 1,8 (mol); n H2 0,3 (mol) 100.36,5 22, mddB = 400 + 40 0,3.2 + 60,6 = 500 (g) GV: Trng Th Tho Email: thaonguyenh81@gmail.com Page Hng dn gii 500 bi Húa hc THCS hay v khú nHCl d 2,92.500 0, (mol) 100.36,5 nHCl ó phn ng (1) v (2) = 1,8 0,4 = 1,4 (mol) T (1): nHCl = 2n H2 = 2.0,3 = 0,6 (mol) T (1): nFe = n H2 = 0,3 (mol) mFe = 0,3.56 = 16,8 (g) m Fex Oy = 40 16,8 = 23,2 (g) nHCl (2) = 1,4 0,6 = 0,8 (mol) 0, T (2): n Fex Oy 0,8 2y y 0, x (56x 16y) 23, ta cú: y y Vy cụng thc ca FexOy l Fe3O4 Cỏc pthh: 2Fe + 6H2SO4 Fe2(SO4)3 + 3SO2 + 6H2O (1) 2Fe3O4 + 10H2SO4 3Fe2(SO4)3 + SO2 + 10H2O (2) Fe2(SO4)3 + 3Mg 2Fe + 3MgSO4 (3) Cú th: Fe + Fe2(SO4)3 3FeSO4 (4) Ba(H)2 + MgSO4 BaSO4 + Mg(OH)2 (5) Cú th: Ba(OH)2 + FeSO4 BaSO4 + Fe(OH)2 (6) Mg(OH)2 MgO + H2O (7) t Cú th: Fe(OH)2 t hoc: 4Fe(OH)2 + O2 n Mg FeO + H2O (8) 2Fe2O3 + 4H2O (9) 10,8 0, 45 (mol) 24 Xột trng hp 1: Mg cha phn ng ht, Fe2(SO4)3 ht (3) khụng cú (4,6,8,9) t: n Fe2 (SO4 )3 300ml ddE l x T (3): nMg ó phn ng = 3x nMg cũn li = 0,45 3x T (3): nFe = 2x mFe = 2x.56 Ta cú pt: (0,45 3x).24 + 2x.56 = 12,6 x = 0,045 (mol) CM ca Fe2(SO4)3 ddE 0,045 0,15(M) 0,3 Khi ú ddD ch cú: MgSO4 v kt ta gm BaSO4 v Mg(OH)2 T (3): n MgSO4 3n Fe2 (SO4 )3 3.0,045 0,135 (mol) T (5): n BaSO4 n MgSO4 0,135 (mol) T (7): n MgO n Mg(OH)2 0,135 (mol) Giỏ tr ca m trng hp ny = 0,135.233 + 0,135.40 = 36,855 (g) Xột trng hp 2: Mg ht, Fe2(SO4)3 sau phn ng (3) cũn d: (4,6,7) hoc (4,6,8) xy 3 T (3): n Fe2 (SO4 )3 n Mg 0, 45 0,15 (mol) GV: Trng Th Tho Email: thaonguyenh81@gmail.com Page Hng dn gii 500 bi Húa hc THCS hay v khú T (3): n Fe 2 n Mg 0, 45 0,3 (mol) 16,8 (g) 3 Theo bi lng cht rn ch cú 12,6 (g) nh hn 16,8 (g) chng t (4) cú xy v lng Fe b ho tan (4) = 16,8 12,6 = 4,2 (g) 0,075 (mol) t (4): n Fe2 (SO4 )3 = nFe b ho tan = 0,075 (mol) Tng n Fe2 (SO4 )3 300 ml ddE trng hp ny = 0,15 + 0,075 = 0,225 (mol) Vy C M ca dung dch E 0, 225 0,75(M) 0,3 Khi ú: Kt ta thu c cho dung dch D phn ng vi Ba(OH)2 gm: BaSO4, Mg(OH)2, Fe(OH)2 Vi : n MgSO4 (3) = nMg = 0,45 (mol) T (4): n FeSO4 = 3nFe= 3.0,075 = 0,225 (mol) T (5): n BaSO4 n Mg(OH)2 n MgSO4 0,45 (mol) T (6): n BaSO4 n Fe(OH)2 n FeSO4 0,225 (mol) S mol kt ta ln lt l: n BaSO4 = 0,45 + 0,225 = 0,675 (mol) n Fe(OH)2 = 0,225 (mol), n Mg(OH)2 = 0,45 (mol) Khi nung kt ta trờn ta li phi xột trng hp: a) Nu nung chõn khụng: T (7): n MgO n Mg(OH)2 0,45 (mol) T (8): n FeO n Fe(OH)2 0,225 (mol) Giỏ tr ca m trng hp ny = 0,675.233 + 0,45.40 + 0,225.72 = 191,475 (g) b) Nu nung khụng khớ: T (9): n Fe2O3 1 n Fe(OH)2 0, 225 0,1125 (mol) 2 Vy giỏ tr ca m trng hp ny l: 0,675.233 + 0,45.40 + 0,1125.160 = 193,275 (g) Bi 18 Chia 26,32 gam hn hp A gm Fe, Mg, Al2O3 v oxit ca kim loi X cú húa tr thnh phn bng Phn tan ht dung dch HCl d, thu c 0,22 mol H2 Phn tỏc dng ht vi dung dch HNO3 loóng d, thu c khớ NO (sn phm kh nht), ú th tớch NO Fe sinh bng 1,25 ln Mg sinh Nu hũa tan ht lng oxit cú mi phn phi dựng va ht 50 ml dung dch NaOH 2M Bit ly m gam Mg v m gam X cho tỏc dng vi dung dch H 2SO4 loóng d thỡ th tớch khớ H2 Mg sinh ln hn 2,5 ln X sinh Vit cỏc phng trỡnh phn ng xy ra, xỏc nh X v tớnh s mol mi cht mi phn Gi s mol Mg, Fe, Al2O3 v XO ln lt l a, b, c, d 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) GV: Trng Th Tho Email: thaonguyenh81@gmail.com Page Hng dn gii 500 bi Húa hc THCS hay v khú 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 ta cú: b=1,25ì a b=0,1 mol Ta cú: moxit = 13,16 (0,12.24 + 0,1.56) = 4,68 gam Nu ch cú Al2O3 tan dung dch NaOH thỡ: 0,1 n Al2O3 n NaOH 0, 05mol m Al2O3 = 5,1 gam > 4,68 gam 2 XO tan dung dch NaOH Theo (11, 12) v bi ta cú h: d 0, 05 c d 0, 05 0, 42 X 77, 102c (X 16)d 4, 68 d 86 X m m 2,5 X 60 Mt khỏc theo (9, 10) v bi ra: 24 X Vy X l Zn (km) c = 0,03 mol v d = 0,02 mol Bi 19: Dung dch X v Y cha HCl vi nng mol tng ng l C1, C2 (M), ú C1 > C2 Trn 150 ml dung dch X vi 500ml dung dch Y c dung dch Z trung hũa 1/10 dung dch Z cn 10ml dung dch hn hp NaOH 1M v Ba(OH)2 0,25M Mt khỏc ly V1 lớt dung dch X cha 0,05 mol HCl trn vi V2 lớt dung dch Y cha 0,15 mol axit c 1,1 lớt dung dch Hóy xỏc nh C1, C2, V1, V2 n NaOH 0, 01.1 0, 01 (mol); n Ba (OH)2 0, 01.0, 25 0, 0025 (mol) Phng trỡnh húa hc: HCl NaOH NaCl H 2O (1) Mol : 0,01 0,01 2HCl+Ba(OH) BaCl 2H 2O (2) Mol : 0,005 0,0025 0,15C1 0,5C2 10.(0, 01 0, 005) 0,15 C2 0,3 0,3C1 (*) Mt khỏc, ta cú: V1 + V2 = 1,1 (lớt) 0, 05 0,15 0, 05 0,15 V1 ; V2 1,1 C1 C1 C2 C2 Thay (*) vo (**) ta c: 0, 05 0,15 1,1 C1 0,3 0,3C1 0,33C12 0,195C1 0, 015 C1 0,5M hoc C1 = 1/11 M * Vi C1 = 0,5 M C2 = 0,3 0,3.0,5=0,15 (M) (tha vỡ C1 > C2) 0, 05 0,15 0,1 (lớt); V2 (lớt) V1 0,5 0,15 GV: Trng Th Tho Email: thaonguyenh81@gmail.com Page Hng dn gii 500 bi Húa hc THCS hay v khú S mol C = 0,01 ; s mol B = 0,02 ; s mol A = 0,03 Khi lng hn hp = 0,01.90 + 0,02.60 + 0,03.30 = g (phự hp u bi) Vy CTCT ca C l: CH3 CH COOH hay CH CH2COOH | | OH OH Bi 74 cht hu c X, Y, Z, T u cú cụng thc phõn t : C3H6O3 C cht u cú kh nng phn ng vi dung dch NaOH, cho sn phm l cỏc mui, cht T cũn cho thờm mt cht hu c R Khi phn ng vi Na d mol X hay Y hoc R gii phúng mol H 2, mol Z hay T gii phúng 0,5 mol H2 Xỏc nh cụng thc cu to ca X, Y, Z, T, R, bit rng khụng tn ti hp cht hu c m phõn t cú t nhúm OH cựng liờn kt vi nguyờn t cacbon Vit phng trỡnh hoỏ hc ca X (hoc Y) vi : Na, NaOH,C2H5OH, ghi rừ iu kin nu cú X : HOCH2CH2COOH Y : CH3 CH COOH | OH Z : CH3OCH2COOH ; T : HCOOCH2CH2OH ; R : HOCH2CH2OH Cỏc phng trỡnh hoỏ hc ca phn ng : NaOCH2CH2COONa + H2 HOCH2CH2COOH + 2Na HOCH2CH2COONa + H2O HOCH2CH2COOH + NaOH H 2SO4 HOCH2CH2COOH+C2H5OH HOCH2CH2COOC2H5 + H2O Bi 75 Dn lung khớ CO d qua hn hp cỏc oxit : CaO ; CuO ; Fe3O4 ; Al2O3 nung núng (cỏc oxit cú s mol bng nhau) Kt thỳc phn ng thu c cht rn (A) v khớ (B) Cho (A) vo H2O (ly d) c dung dch (C) v phn khụng tan (D) Cho (D) vo dung dch AgNO3 (s mol AgNO3 bng 7/4 s mol cỏc oxit hn hp u), thu c dung dch (E) v cht rn (F) Ly khớ (B) cho sc qua dung dch (C) c dung dch (G) v kt ta (H) Vit cỏc phng trỡnh hoỏ hc ca phn ng xy ra, xỏc nh thnh phn ca (A), (B), (C), (D), (E), (F), (G), (H) Gi s mol mi oxit l a => s mol AgNO3 = 7a + Phn ng cho CO d qua hn hp cỏc oxit nung núng : to CO + CuO a mol a mol + CO2 a mol to 4CO + Fe3O4 a mol Cu 3a mol 3Fe + 4CO2 4a mol Thnh phn ca (A) : Cu = a mol ; Fe = 3a mol ; CaO = a mol ; Al2O3= a mol Thnh phn khớ (B) : CO2 = 5a mol ; CO d + Phn ng cho (A) vo nc d : CaO + H2O a mol Al2O3 + Ca(OH)2 a mol a mol Ca(OH)2 a mol Ca(AlO2)2 + H2O a mol GV: Trng Th Tho Email: thaonguyenh81@gmail.com Page 41 Hng dn gii 500 bi Húa hc THCS hay v khú Thnh phn dung dch (C) : Ca(AlO2)2 = a (mol) ; H2O Thnh phn (D) : Cu = a(mol) ; Fe = 3a (mol) + Phn ng cho (D) vo dung dch AgNO3 : Fe + 3a mol 2AgNO3 6a mol Fe(NO3)2 + 2Ag 3a mol 6a mol Cu(NO3)2 Cu + 2AgNO3 0,5a mol a mol 0,5a mol Thnh phn dung dch (E) : + 2Ag a mol Fe(NO3)2 = 3a mol ; Cu(NO3)2 = 0,5a mol ; H2O Thnh phn (F) : Ag = 7a mol ; Cu = 0,5a mol + Phn ng cho khớ (B) sc qua dung dch (C): CO2 + 3H2O a mol CaCO3 + 2Al(OH)3 + Ca(AlO2)2 a mol a mol 2a mol Ca(HCO3)2 a mol CO2 + CaCO3 + H2O a mol a mol Thnh phn dung dch (G) : nCa(HCO3)2 = a mol ; H2O Thnh phn kt ta (H) : nAl(OH)3 = 2a (mol) Bi 75: Vit phng trỡnh xy gia mi cht cỏc cp sau õy: A Ba v dd NaHCO3 C K v ddAl2(SO4)3 D Mg v ddFeCl2 B Khớ SO2 v khớ H2S D dd Ba(HSO3)2 v dd KHSO4 E Khớ CO2 d v dd Ca(OH)2 a 2Ba + 2H2O -> Ba(OH)2 + H2 Ba(OH)2 + 2NaHCO3 -> Na2CO3 + BaCO3 + 2H2O b 2H2S + SO2 -> 3S + 2H2O c 2K + 2H2O -> 2KOH +H2 6KOH + Al2(SO4)3 -> 3K2SO4 + 2Al(OH)3 KOH + Al(OH)3 -> KAlO2 + 2H2O d Ba(HSO3)2 + 2KHSO4-> K2SO4 +BaSO4+ SO2 + 2H2O d Mg + FeCl2 -> MgCl2 + Fe e CO2 + Ca(OH)2 -> CaCO3 + H2O CaCO3 + H2O + CO2 -> Ca(HCO3)2 Bi 76: Ch dựng mt thuc th hóy phõn bit cỏc dung dch sau: NaOH, CuSO4, Fe(NO3)3, Fe(NO3)2, NH4Cl, AlCl3 - Dựng qu tớm -> NaOH (qu xanh) - T NaOH nhn bit c cỏc cht cũn li: + Xut hin xanh -> CuSO4 -> Vit PT + Xut hin mõu -> Fe(NO3)3 -> Vit PT GV: Trng Th Tho Email: thaonguyenh81@gmail.com Page 42 Hng dn gii 500 bi Húa hc THCS hay v khú + Xut hin trng xanh, hoỏ nõu K2 l Fe(NO3)2 -> Vit PT + Cú khớ mựi khai -> NH4Cl -> Vit PT + Xut hin keo, tan dn -> AlCl3 -> Vit PT Bi 77: Cho 44,2g mt hn hp ca mui sunfỏt ca mt kim loi hoỏ tr I v mt kim loi hoỏ tr II tỏc dng va vi dung dch BaCl2 thu c 69,9g mt cht kt ta Tớnh lng cỏc mui thu c sau phn ng? Dn H2 d i qua 25,6g hn hp X gm Fe3O4, ZnO, CuO nung núng cho n phn ng xy hon ton Sau phn ng thu c 20,8g cht rn Hi nu ho tan ht X bng dung dch H2SO4 thỡ cn bao nhiờu gam dung dch H2SO4 20% a Gi kim loi hoỏ tr I l A, hoỏ tr II l B => Cỏc mui sun fỏt: A2SO4, BSO4 A2SO4 + BaCl2 => 2ACl + BaSO4 (1) BSO4 + BaCl2 -> BCl2 + BaSO4 Theo PT ta thy nBaCl nBaSO4 (2) 69,9 0,3mol 232 => p dng L TBKL: Tớnh c m mui sau phn ng = 36,7g Gi x, y, z l s mol Fe3O4, ZnO, CuO (x,y,z>0) => 232x + 81y + 80z = 25,6 - Vit c phn ng => Lp PT: Mkim loi = 168x + 65y + 64z = 20,8 -> nO (oxớt) = 4x + y + z = 0,3 mol - Vit PT: oxớt + H2SO4 => nH2SO4 = nO = 0,3 mol -> mH2SO4 = 0,3 x98 = 29,4g => md2H2SO4 = 147g Bi 78: Cho 16,4g hn hp M gm Mg, MgO v CaCO3 vo dung dch HCl d thỡ thu c hn hp khớ cú t hi so vi H2 l 11,5 Cụ cn dung dch sau phn ng c 30,1g hn hp mui khan a Tớnh lng cỏc cht hn hp M? b Nu cho hn hp M trờn vo dung dch H2SO4 c núng d thu c 4,48l hn hp X gm khớ ktc cú lng 10,8g thỡ X gm nhng khớ gỡ? a - Vit ỳng cỏc PT - Lp c cỏc PT i s, gii chớnh xỏc - Tớnh c lng cỏc cht M b - Vit ỳng mi PT c 0,25 x = 0,75 - Khng nh X cú CO2 H2S hoc SO2 Tỡm Mkhớ cũn li = 64 -> Kt lun l SO2 Vy hn hp khớ X gm SO2 V CO2 Bi 79 Ho tan m gam kim loi M bng dung dch HCl d thu c V lớt khớ H2 (ktc) Cng hũa tan m gam kim loi trờn bng dung dch HNO3 loóng d thu c V lớt khớ NO (ktc) GV: Trng Th Tho Email: thaonguyenh81@gmail.com Page 43 Hng dn gii 500 bi Húa hc THCS hay v khú a Vit cỏc phn ng xy ra? b M l gỡ? Bit lng mui Nitrat gp 1,905 ln mui Clorua Kim loi M phn ng vi HCl cú hoỏ tr n (m, n N) m n Kim loi M phn ng vi HCl cú hoỏ tr m 2M +2nHCl = 2MCln + nH2 (mol) x x nx 3M + mHNO3 -> 3M(NO3)m + mNO + H2O (mol) x x mx Vỡ VNO VH nNO nH (1) (2) mx nx n (chn n = 2; m = 3) m Mt khỏc KL mui nitrat = 1,905 ln lng mui clorua Nờn: mM ( NO3 ) 1,905mMCl M 56( Fe) Bi 80 Hn hp A gm kim hoi l Mg v Zn B l dung dch H2SO4 cú nng l x mol/l TH1: Cho 24,3g (A) vo 2l dung dch (B) sinh 8,96l khớ H2 TH2: Cho 24,3g (A) vo 3l dung dch (B) sinh 11,2l khớ H2 (Cỏc th tớch khớ o ktc) a Hóy chng minh TH1 thỡ hn hp kim loi cha tan ht, TH2 axớt cũn d? b Tớnh nng x mol/l ca dung dch B v % lng mi kim loi A? a - Gii thớch c TH1 d kim loi, TH2 d axớt Vit ỳng PT -Xột TH2 lõp h phng trỡnh 65x+24y = 24,3 x =0,3 % Zn = 80,25% x+y = 0,5 y = 0,2 % Mg = 19,75% Xột TH1, nH2SO4 = nH2 = 0,4 mol =>CM H2SO4 = x M = 0,2M Bi 81 Trong s sau: Mi ch cỏi l tờn mt cht Xỏc nh cỏc cht ng vi mi ch cỏi A,B,C,D,F, G v vit phng trỡnh hoỏ hc +X Fe +Y +Z A B C +D C + HCl HHCl A +E F t0 G +L Fe A; FeCl2 B: FeS C: FeCl3 D: Cl2 F: Fe(OH)3 G: Fe2O3 Fe + 2HCl FeCl2 + H2 Fe + S t0 FeS 2Fe + 3Cl2 2FeCl3 2FeCl2 + Cl2 2FeCl3 GV: Trng Th Tho Email: thaonguyenh81@gmail.com Page 44 Hng dn gii 500 bi Húa hc THCS hay v khú FeS + 2HCl FeCl2 + H2S FeCl3 + 3NaOH Fe(OH)3 + NaCl 2Fe(OH)3 t0 Fe2O3 + 3H2O Fe2O3 + 3H2 t0 2Fe + 3H2O Bi 82 a) vi mu CaO khụng khớ mt thi gian sau ú cho vo dung dch HCl Vit cỏc phng trỡnh hoỏ hc cú th xy ra? b) Nung núng st khụng khớ mt thi gian thu c cht rn A Ho tan A bng H2SO4 c núng d c dung dch B v khớ C cú mựi hc Khớ C tỏc dng vi dung dch NaOH c dung dch D Dung dch D va tỏc dng vi dung dch KOH, va tỏc dng vi dung dch BaCl2 Cho dung dch B tỏc dng vi dung dch KOH Xỏc nh A,B,C,D v vit cỏc phng trỡnh hoỏ hc a) CaO + CO2 CaCO3 CaO + H2O Ca(OH)2 Ca(OH)2 + CO2 CaCO3 + H2O CaO + 2HCl CaCl2 + H2O Ca(OH)2 + 2HCl CaCl2 + 2H2O CaCO3 + 2HCl CaCl2 + H2O + CO2 b)A: Fe3O4 cú th cú Fe d Dung dch B: Fe2(SO4)3; H2SO4 d; C: SO2 D: NaHSO3 v Na2SO3 3Fe + 2O2 t0 Fe3O4 2Fe + 6H2SO4 ,núng Fe2(SO4)3 + 3SO2 + 6H2O 2Fe3O4 + 10H2SO4 c, núng 3Fe2(SO4)3 + SO2 + 10H2O SO2 + NaOH NaHSO3 SO2 + 2NaOH Na2SO3 + H2O 2NaHSO3 + 2KOH Na2SO3 + K2SO3 + 2H2O Na2SO3 + BaCl2 BaSO3 + 2NaCl Fe2(SO4)3 + 6KOH 2Fe(OH)3 + 3K2SO4 H2SO4 + 2KOH K2SO4 + 2H2O Bi 83 a) Cho 100 gam CaCO3 tỏc dng vi dung dch HCl d Khớ thu c cho i qua dung dch cha 60 gam NaOH Tớnh lng mui to thnh b) Nhn bit cht rn sau : NaCl, Na2CO3, BaCO3, BaSO4 ch bng dung dch HCl a) CaCO3 +2HCl CaCl2 + H2O + CO2 (1) 1mol mol CO2 + 2NaOH Na2CO3 + H2O (2) x 2x CO2 + NaOH NaHCO3 (3) y y S mol CaCO3 : 100:100 = mol Theo PTHH (1) suy s mol CO2 l mol S mol NaOH l: 60 : 40 = 1,5 mol T l s mol CO2 : s mol NaOH l: 1,5 => To thnh hai mui Gi s mol CO2 tham gia phn ng l x S mol CO2 tham gia phn ng l y GV: Trng Th Tho Email: thaonguyenh81@gmail.com Page 45 Hng dn gii 500 bi Húa hc THCS hay v khú ta cú h: x+ y = 2x + y = 1,5 Gii h ta c x = 0,5 mol; y = 0,5 mol Suy s mol Na2CO3 l 0,5 mol S mol NaHCO3 l 0,5 mol Khi lng Na2CO3 l: 0,5 106 = 53 gam Khi lng mui NaHCO3 l: 0,5.84 = 42 gam b) Dựng dung dch HCl: - Nhn NaCl tan, BaSO4 khụng tan - Nhn Na2CO3 v BaCO3 tan v cú khớ thoỏt Na2CO3 + 2HCl 2NaCl + H2O + CO2 BaCO3 + 2HCl BaCl2 + H2O + CO2 - Cho hai cht rn Na2CO3 v BaCO3 vo dung dch HCl t t n d, n khụng cũn thy khớ thoỏt na, nu cht rn no tip tc tan thỡ ú l Na2CO3, nu cht rn no khụng tan thỡ ú l BaCO3 Bi 84: Cho hn hp Mg v Cu tỏc dng vi 200 ml dung dch cha hn hp hai mui AgNO3 0,3M v Cu(NO3)2 0,25M c dung dch A v cht rn B.Cho dung dch A tỏc dng vi dung dch NaOH d, lc kt ta nung khụng khớ n lng khụng i c gam hn hp hai oxit Ho tan B vo H2SO4 c núng thu c 0,896 lớt khớ ktc - Vit cỏc phng trỡnh hoỏ hc xy - Tớnh lng Mg, Cu hn hp ban u a) (1) Mg + 2AgNO3 Mg(NO3)2 + 2Ag (2) Cu + 2AgNO3 Cu(NO3)2 + 2Ag (3) Mg(NO3)2 + 2NaOH Mg(OH)2 + 2NaNO3 (4) Mg(OH)2 t0-> MgO + H2O (5) Cu(NO3)2 + 2NaOH Cu(OH)2 + 2NaNO3 (6) Cu(OH)2 t0 CuO + H2O (7) 2Ag + 2H2SO4 c t0-> Ag2SO4 + SO2 + 2H2O (8) Cu + 2H2SO4 c -t0-> CuSO4 + SO2 + 2H2O b) Vỡ dung dch A tỏc dng vi NaOH d, to kt ta, nung kt ta thu c hai oxit nờn A ch cha hai mui l Mg(NO3)2 v Cu(NO3)2 => AgNO3 ht Nu khụng xy phn ng (2) thỡ lng hai oxit ti a s l: 0,03.40 + 0,05 80 = 5,2 g < g - Vy xy phn ng (2) v Mg ht n AgNO3 = 0,2.0,3 = 0,06 mol n Cu ( NO3 ) = 0,2.0,25 = 0,05 mol - t s mol Mg phn ng l x (mol) Theo PTHH (1), (3), (4): n MgO = n Mg = x (mol) Khi lng MgO l 40.x Theo PTHH(1): n AgNO = 2n Mg = 2x ; => S mol AgNO3 phn ng (2) l 0,06 2x S mol Cu(NO3)2 phn ng (2) l : 0,06 x 0,03 x n Cu ( NO3 ) = n AgNO3 = 2 Theo PTHH (2), (5), (6) : n CuO = n Cu = 0,03 x S mol CuO Cu(NO3)2 d to l 0,05 mol Khi lng CuO l : (0,03 x + 0,05).80 = (0,08 x).80 Pt lng hai oxit l : 40.x + (0,08 x).80 = x = 0,01 Khi lng Mg l 0,01.24 = 0,24 g GV: Trng Th Tho Email: thaonguyenh81@gmail.com Page 46 Hng dn gii 500 bi Húa hc THCS hay v khú Theo PTHH (1), (3): n Ag = n AgNO3 = 0,06 mol n Ag = 0,03 mol t s mol Cu ban u l y Theo PTHH (3): n Cu phn ng = n AgNO3 = 0,02 mol S mol Cu d l: y 0,02 Theo PTHH (9): n SO2 = n Cu = (y 0,02) S mol SO2 theo bi l: 0,896 : 22,4 = 0,04 Ta cú phng trỡnh: 0,03 + (y 0,02) = 0,04 y = 0,03 mol Khi lng Cu l: 0,03.64 = 1,92g Theo PTHH (8): n SO2 = Bi 85 Hũa tan hon ton 18,2 gam hn hp hai mui cacbonnat trung hũa ca hai kim loi húa tr I bng dung dch HCl d, thu c 4,48 lit khớ CO2 (ktc) Hai kim loi ú l nhng kim loi no nhng kim loi di õy? Gi hai kim loi cn tỡm l X, Y ta cú cỏc PTHH: X2CO3 + 2HCl 2XCl + H2O + CO2 (1) Y2CO3 + 2HCl 2YCl + H2O + CO2 (2) S mol CO2 l 4,48 : 22,4 = 0,2 mol Theo PTHH (1) v (2) : S mol hai mui = s mol CO2 = 0,2 mol Khi lng mol trung bỡnh ca hai mui l:18,2: 0,2 = 91g Ta cú pt: (2X + 60 + 2Y+60) : = 91 => X+Y = 31 Hai kim loi cn tỡm l: Na v Li Bi 86 Cho m gam hn hp Na v Fe tỏc dng ht vi axit HCl Dung dch thu c cho tỏc dng vi Ba(OH)2 d ri lc ly kt ta tỏch ra, nung khụng khớ n lng khụng i thu c cht rn nng m gam Tớnh % lng mi kim loi ban u - PTHH xy cho m gam hn hp Na v Fe tỏc dng vi HCl: 2Na + 2HCl 2NaCl + H2 (1) Fe + 2HCl FeCl2 + H2 (2) - PTHH xy cho dung dch thu c tỏc dng vi Ba(OH)2 d: FeCl2 + Ba(OH)2 Fe(OH)2 + BaCl2 (3) - PTHH xy nung kt ta khng kh: t 2Fe2O3 + 4H2O 4Fe(OH)2 + O2 o (4) - Gi m = mFe + mNa = 100 gam m Fe O 100gam n Fe O 3 100 0,625mol 160 2.0,625 1,25mol - Theo PTHH (4): n Fe(OH ) 2.n Fe O - Theo PTHH (3): n FeCl n Fe(OH ) 1,25mol - Theo PTHH (2): n Fe n FeCl 1,25mol m Fe 1,25.56 70gam 2 2 GV: Trng Th Tho Email: thaonguyenh81@gmail.com Page 47 Hng dn gii 500 bi Húa hc THCS hay v khú - Vy %Fe = 70% ; % Na = 30% Bi 87: T cỏc cht KMnO4, BaCl2, Zn cú th iu ch c cỏc khớ no? Vit cỏc phng trỡnh húa hc xy (ghi rừ iu kin phn ng nu cú)? 2.Phi kim R hp vi oxi to oxit cao nht cú cụng thc l R2O5 Trong hp cht ca R vi hiro thỡ R chim 82,35% lng Xỏc nh tờn nguyờn t R v vit cụng thc ca R vi hiro v oxi iu ch khớ oxi: 2KMnO4 K2MnO4 + MnO2 + O2 PNC - iu ch Cl2: BaCl2 - iu ch H2: Zn + H2SO4 (loóng) Ba + Cl2 ZnSO4 + H2 to - iu ch SO2: Zn + 2H2SO4 c ZnSO4 + SO2 + 2H2O Gi húa tr ca R hp cht vi hiro l n => CTHH l RHn R 100 82,35 R 4,67n - Ta cú %R Rn - Vỡ n l húa tr nờn ch nhn cỏc giỏ tr 1,2,3 n R 4,67 9,33 14 19 23 28 33 - Vi n =3, R=14=>R l nit,kớ hiu l N (0,5) - CT ca R vi hirụ l NH3, vi oxi l N2O5 Bi 88: Trỡnh by phng phỏp húa hc tỏch riờng cỏc mui t hn hp cht rn gm BaCl2, FeCl3 v AlCl3 Cú ba l ng ba cht rn KCl, NH4NO3, Ca(H2PO4)2 Hóy nhn bit mi l bng phng phỏp húa hc Cho hn hp mui vo cc ng dd NH3 d FeCl3 + 3NH3 + 3H2O Fe(OH)3 +3NH4Cl AlCl 3NH 3H O Al(OH) 3NH Cl - Lc tỏch Fe(OH)3, Al(OH)3 cụ cn dung dch ri nung núng nhit cao tỏch c BaCl2 t0 NH4Cl NH3 + HCl - Cho hn hp Fe(OH)3, Al(OH)3 vo dung dch NaOH d Al(OH)3 + NaOH NaAlO2 + 2H2O Fe(OH)3 khụng phn ng lc tỏch cho tỏc dng vi dd HCl d, cụ cn c FeCl3 Fe(OH)3+ 3HCl FeCl3 + 3H2O - Sc CO2 d vo dd NaAlO2 lc tỏch kt ta Al(OH)3 ri cho tỏc dng vi dd HCl d v cụ cn c AlCl3 NaAlO CO 2H O Al(OH) NaHCO Al(OH) 3HCl AlCl 3H O Trớch mi l ớt lm mu th ri hũa tan vo nc - Dựng Na2CO3 nhn Ca(H2PO4)2 vỡ to kt ta trng CaCO3 Na CO Ca (H PO ) CaCO3 NaH PO - Dựng AgNO3 nhn KCl vỡ to kt ta trng AgCl AgNO KCl AgCl KNO3 - Dựng NaOH nhn NH4 NO3 vỡ to khớ cú mựi khai NH3 GV: Trng Th Tho Email: thaonguyenh81@gmail.com Page 48 Hng dn gii 500 bi Húa hc THCS hay v khú NaOH NH NO3 NaNO3 NH H O Bi 89 Thớ nghim: Lm bay hi 60 gam nc t dung dch NaOH cú nng 15% c dung dch mi cú nng 18% Hóy xỏc nh lng dung dch NaOH ban u Cho m gam natri vo dung dch thu c thớ nghim trờn c dung dch cú nng 20,37% Tớnh m Gi lng dung dch NaOH ban u l m gam => mNaOH = 0,15m(g) - Khi lng dd NaOH sau lm bay hi nc l : m - 60(g) 0,15m 100 18 m 360(g ) m 60 mNaOH = 0,15.360 = 54g - Khi lng dd NaOH sau lm bay hi nc l: 360 60 = 300(g) - PTHH: 2Na +2H2O 2NaOH + H2 (1) x x 0,5x (mol) - Gi s mol Na m gam Na l x mol => nNaOH = nNa = x(mol) => mNaOH (1) = 40x(g) - Ta cú PT v nng dd sau phn ng: 54 40x C% NaOH 100 20,37% x 0,2 300 23x 2.0,5x m Na 0,2.23 4,6(g ) Bi 90 Nhỳng st nng 100 gam vo 500ml dung dch hn hp CuSO4 0,08M v Ag2SO4 0,004M Sau thi gian ly st cõn li v thy lng l 100,48 gam Tớnh lng kim loi bỏm vo st v nng mol cỏc cht dung dch sau phn ng Gi s th tớch dung dch thay i khụng ỏng k n CuSO 0,04mol ; n Ag2SO 0,002mol - PTHH: Fe Ag 2SO FeSO4 2Ag (1) Fe CuSO FeSO4 Cu (2) *TH1: Ch xy (1) t s nFe(p) = x(mol) = n Ag2SO4 ( pu) => mtng = 100,48 100 = 108.2x 56x x = 0,003 > n Ag2SO (loi) * TH2 : Xy c (1) v (2) - Theo (1) nFe = n Ag2SO = 0,002 mol nAg = 2n Ag2SO = 0,004 mol - Gi nFe(p2) = a(mol) = nCuSO = nCu => mtng = 100,48 100 = 108.0,004 + 64 a 56.0,002 56 a a = 0,02 - Vy lng kim loi bỏm vo st l : 108.0,004 + 64.0,02 = 1,712 gam - Sau p dd cú 0,04 0,02 = 0,02 mol CuSO4 d v 0,002 + 0,02 = 0,022 mol FeSO4 0,02 0,022 0,04M ; CM FeSO = 0,044M CM CuSO = 0,5 0,5 Bi 91 Cho 7,22 gam hn hp A gm Fe v kim loi M cú giỏ tr khụng i Chia hn hp lm hai phn bng Hũa tan ht phn I dung dch axit HCl thu c 2,128 lit H2 Hũa tan ht phn II dung dch HNO3 to 1,792 lớt NO nht Th tớch cỏc khớ ú ktc Xỏc nh kim loi M Tớnh % mi kim loi A 1 phn = 7,22 = 3,61(g); n H = 0,095 mol ; nNO = 0,08 mol GV: Trng Th Tho Email: thaonguyenh81@gmail.com Page 49 Hng dn gii 500 bi Húa hc THCS hay v khú Gi kim loi M cú húa tr l n - PTHH: Fe 2HCl FeCl2 H 2M 2nHCl 2MCln nH Fe 4HNO Fe( NO3 ) NO 2H O (1) (2) (3) 3M + 4nHNO3 3M(NO3)n + nNO +2nH2) (4) * TH1 : M khụng tỏc dng vi HCl (tc khụng xy (2)) - Theo (1) nFe = nH = 0,08 mol mFe = 0,08.56 = 4,48 > 3,61 (loi) * TH2 : M tỏc dng vi HCl (tc xy (2)) - Gi s mol Fe cú phn l x mol => mFe = 56.x (g) - Theo (1) : nH (1) = nFe = x(mol) 2 - Theo (2) : nM = n H2 ( 2) (0,095 x )mol n n => mM = 3,61 56.x = (0,095 x ).M (*) n - Theo (3) : nNO = nFe = x(mol) 3 - Theo (4) : nM = n NO (0,08 x )mol n n => mM = 3,61 56.x= (0,08 x ).M (**) n - T (*) v (**) => M(0,09M 0,81n) =0 => * M=0 (loi) * 0,09 M 0,81n = => M = 9n - Vi n=3 ; M = 27 => M l kim loi nhụm (Al) 3,61.3 0,19.27 0,05 Ta cú x = 56.3 2.27 0,05.56 => %mFe = 100 77,56% 3,61 => % mAl = 100 77,56 = 22,44 % Bi 92: Nờu hin tng xy v vit cỏc phng trỡnh phn ng cho mi thớ nghim sau: a) Cho kim loi Na vo dung dch AgNO3 b) Sc khớ SO2 t t cho ti d vo dung dch Ca(OH)2 c) Cho t t mi cht: khớ CO2, dung dch AlCl3 vo mi ng nghim cha sn dung dch NaAlO2 cho ti d d) Cho dung dch Na2CO3 vo dung dch FeCl3 a) Lỳc u bt khớ thoỏt ra,sau thy cú kt ta trng xut hin nhng khụng bn lp tc sinh cht kt ta mu en ( Ag2O) PTPU: 2Na + 2H2O -> 2NaOH + H2 AgNO3 + NaOH -> AgOH + NaNO3 2AgOH - > Ag2O + H2O b) Lỳc u cú kt ta trng xut hin, sau ú kt ta tan i, dung dch tr li GV: Trng Th Tho Email: thaonguyenh81@gmail.com Page 50 Hng dn gii 500 bi Húa hc THCS hay v khú SO2 + Ca(OH)2 -> CaSO3 + H2O SO2 + CaSO3 + H2O -> Ca (HSO3)2 c) To kt ta keo trng CO2 + NaAlO2 + H2O -> Al (OH)3 + NaHCO3 AlCl3+ 3NaAlO2 + 6H2O -> 4Al(OH)3 + 3NaCl d) To khớ khụng mu v kt ta mu nõu 3Na2CO3 + 2FeCl3 + 3H2O-> 6NaCl + 2Fe(OH)3 + 3CO2 Bi 93: Cú 15 gam hn hp Al v Mg c chia lm phn bng Phn th nht cho vo 600ml HCl nng xM thu c khớ A v dung dch B Cụ cn dung dch B thu c 27,9 gam mui khan Phn th cho vo 800ml dung dch HCl nng xM v lm tng t thu c 32,35g mui khan Xỏc nh % lng mi kim loi hn hp v x Tớnh th tớch hidro (dktc) thu c sau thc hin xong cỏc thớ nghim 15 + lng hn hp = 7,5 gam 2 + Nu thớ nghim m HCl d thỡ thớ nghim tng lng Axit -> Khi lng mui to phi khụng i (iu ny trỏi vi gi thit) Vy thớ nghim 1: Kim loi cũn d, Axit thiu +Nu ton b lng axit HCl thớ nghim to mui thỡ lng mui phi l 27,9.800 37,2 gam Theo u bi lng mui thu c l 32,35gam (37,2 > 32,35) thớ nghim 600 : axit HCl cũn d, kim loi ht Phng trỡnh phn ng: 2Al + 6HCl 2AlCl3 + 3H2 (1) Mg + 2HCl MgCl2 + H2 (2) Khi lng hn hp KL =7,5 gam ; Khi lng mui khan = 32,35 gam tng lng ( l lng Cl ca HCl ) = 32,35 - 7,5 = 24,85 gam 24,85 nHCl tham gia phn ng : 0,7mol nH = 0,35 mol 35,5 + VH = 0,35.22,4 =7,84 lit - S mol HCl tham gia phn ng thớ nghim 1: Nng mol dung dch axit (x) = n H2 = 27,9.0,7 0,6mol 32,35 0,6 1M 0,6 0,6 = 0,3 + VH = 0,3 22,4 = 6,72 lit Sau thớ nghim th tớch H2 thu c l :7,84 + 6,72 =14,56 lit 27a 24b 7,5 gi a,b l s mol ca kim loi Al v Mg hn hp t (1) (2) cú : 3a 2b 0,7 GV: Trng Th Tho Email: thaonguyenh81@gmail.com Page 51 Hng dn gii 500 bi Húa hc THCS hay v khú a = 0,1 mAl = 2,7 gam % Al = 36% b = 0,2 mMg = 4,8gam % Mg = 64% Bi 94: Hũa tan 6,58 gam cht A vo 100 gam nc thu c dung dch B cha cht nht Cho lng mui khan BaCl2 vo B thy to 4,66g kt ta trng lc b kt ta ta thu c dung dch C Cho lng Zn va vo dung dch C thy thoỏt 1,792 lit khớ H2(ktc) v dung dch D Xỏc nh cụng thc phõn t cht A 2.Tớnh nng phn trm cỏc cht dung dch D 1/ Dung dch B kt ta vi BaCl2,B cú th cú cỏc mui cú gc axit to kt ta vi Ba; hoc H2SO4 Dung dch C cú phn ng vi Zn cho khớ H2, vy C cú axit =>B phi l H2SO4 hoc mui M(HSO4)n Vy cht ban u cú th l : H2SO4 hoc SO3, hoc H2SO4.nSO3 hoc mui M(HSO4)n - Cỏc phng trỡnh phn ng : BaCl2 + H2SO4 BaSO4+2HCl (1) Zn + 2HCl ZnCl2 + H2 (2) Zn + H2SO4 ZnSO4 + H2 (3) 1,792 Theo phng trỡnh phn ng ta cú nH2SO4 = nH2 = 0,08mol 22,4 * Trng hp 1: A l H2SO4n H2SO4 = * Trng hp 2: A l SO3nSO3 = 6,58 0,067 0,08 ( Lai) 98 6,58 0,08225 0,08 80 (Loi) * Trng hp 3: A l H2SO4.nSO3 H2SO4.nSO3 +nH2O (n+1) H2SO4 6,58 0,08 Ta cú n =7 98 80n n Cụng thc phõn t A l H2SO4.7H2O * Trng hp 4: A l mui M(HSO4)n 2M(HSO4)n +nBaCl2 2MCln + 2nBaSO4 + 2nHCl Theo BTNT ta cú: 2M(HSO4)n 2nHCl nH2 => 0,16/n mol 0,08 mol MM(HSO4)n = 6,58: (0,16/n)= 41,125.n => loi 2/ Khi lng dung dch D l: 4,66.208 4,16 gam m BaCl2 = 233 mdd = 6,58 +100 + 4,16 + 0,08.65 - 0,08.2 - 4,66 = 111,12 gam nZnCl2 = nBaSO4 = 0,2 mol nZnSO4 = 0,08 - 0,02 = 0,06 mol C% ZnCl2 = 0,02.136 100 2,45% 111,12 GV: Trng Th Tho Email: thaonguyenh81@gmail.com Page 52 Hng dn gii 500 bi Húa hc THCS hay v khú C%ZnSO4 = 0,06.161 100 8,69% 111,12 Bi 95: Ch dựng mt hoỏ cht nht, hóy tỏch: a Tỏch FeO hn hp FeO, Cu, Fe b Ag2O hn hp Ag2O, SiO2, Al2O3 a Tỏch FeO hn hp FeO, Cu, Fe FeO Cu Fe Pt : Cu, Fe phn ng + FeCl b Tỏch Ag2O Ag2O SiO2 Al2O3 FeO khụng tan thu c FeO Cu + FeCl3 CuCl2 + 2FeCl2 Fe + 2FeCl3 3FeCl2 Ag2O khụng tan thu c Ag2O + NaOH to Pt: SiO2 + 2NaOH Al2O3 + 2NaOH SiO2 phn ng Al2O3 Na2SiO3 + H2O 2NaAlO2 + H2O Bi 96: Mt hn hp X gm cỏc cht: K2O, KHCO3, NH4Cl, BaCl2 cú s mol mi cht bng Ho tan hn hp X vo nc, ri un nh thu c khớ Y, dung dch Z v kt ta M Xỏc nh cỏc cht Y, Z, M v vit phng trỡnh phn ng minh Xỏc nh Y, Z, M: - t s mol mi cht = a(mol) K2O + H2O 2KOH ; a 2a KHCO3 + KOH K2CO3 + H2O a a a NH4Cl + KOH KCl + NH3 + H2O a a BaCl2 + K2CO3 BaCO3 + 2KCl a a Vy : Y l NH3 ; dung dch Z : KCl ; M : BaCO3 (mol) (mol) (mol) (mol) Bi 97: a, Cho 2,08 gam MxOy tan hon ton vo 100 gam dung dch H2SO4 4,9% Sau phn ng kt thỳc thu c dung dch Y cú nng axit l 1,056% Xỏc nh cụng thc húa hc ca oxit ú b, Dung dch CuSO4 100C cú tan l 17,4 (g); 800C cú tan l 55 (g) Lm lnh 1,5 kg dung dch CuSO4 bóo hũa 800C xung 100C Tớnh s gam CuSO4.5 H2O tỏch GV: Trng Th Tho Email: thaonguyenh81@gmail.com Page 53 Hng dn gii 500 bi Húa hc THCS hay v khú PTHH 0,039 mol T PTHH ta cú: CTHH ca oxit l Fe2O3 i: 1.5 kg = 1500 (g) - Xột t0 = 800C; t: mCuSO4 ( 800C) = x (g) x 100 x 532,258 1500 x 967,742( g ) Ta c: 55 = -> mH O t nCuSO4 H O tách = a(mol) -> x> (g) mCuSO4 tỏch = 160 a (g) mH O tỏch = 5.a.18 = 90 a(g) Xột t0 = 100C 532,258 160a Khi ú 17,4 100 a 2,52(mol ) 967,724 90a => mCuSO4 H O tách = 2,52.250=630 (g) Bi 98: a/ Vit phng trỡnh phn ng ca Ba(HCO3)2 vi mi cht sau : Ca(OH)2, HNO3, K2SO4, KHSO4, H2SO4, dung dch ZnCl2 b/ Vit phng trỡnh phn ng th hin cỏc phng phỏp khỏc iu ch mui ZnCl2 a/ Ba(HCO3)2 + Ca (OH)2 -> BaCO3 + CaCO3 + H2O Ba(HCO3)2 + 2HNO3 -> Ba (NO3)2 + H2O + CO2 Ba(HCO3)2 + K2 SO4 -> BaSO4 +2 KHCO3 Ba(HCO3)2 +2 KH SO4 -> BaSO4 + K2SO4 + H2O + 2CO2 Ba(HCO3)2 + H2SO4 -> BaSO4+ H2O + 2CO2 Ba(HCO3)2 + ZnCl2-> Zn (OH)2 + BaCl2 + CO2 b/ KL + Ax Zn+ HCl -> ZnCl2 + H2 KL + PK Zn + Cl 2-> ZnCl KL + M Zn + CuCl2 -> ZnCl2 + Cu Ax + M ZnCO3 + HCl -> ZnCl + H2O + CO2 M+ M Zn SO4 + Ba Cl2 -> Ba SO4 + Zn Cl2 Oxit + Ax ZnO + 2HCl -> ZnCl + H2O Bazo+ Ax Zn ( OH)2 + HCl -> Zn Cl2 + H2O Bi 99: GV: Trng Th Tho Email: thaonguyenh81@gmail.com Page 54 Hng dn gii 500 bi Húa hc THCS hay v khú a/ Cú mu kim loi :Ba, Mg, Fe, Ag, Al ch dựng dung dch H2SO4 loóng (khụng c dựng cht khỏc ) Hóy nhn bit kim loi trờn b/ Tỏch cỏc cht hn hp gm: SiO2, ZnO, Fe2O3 a/ Ly mi mu kim loi lng nh cho vo cc ng dung dch H2SO4 loóng - Cc no khụng cú khớ bay lờn l Ag ( khụng tan) - Cc no cú khớ bay lờn v cú l Ba Ba + H2SO4 -> Ba SO4 + H2 (1) - Cỏc cc cú khớ : Al, Mg, Fe 2Al + 3H2SO4 -> Al2 (SO4)3 +3 H2 (2) Mg + H2SO4 -> MgSO4 + H2 (3) Fe + H2SO4 -> FeSO4 + H2 (4) Thờm tip Ba vo cc cú phn ng (1) thỡ xy phn ng cú sau Ba + H2O -> Ba (OH)2 + H2 (5) Lc kt ta c dung dch Ba(OH)2 - Ly lng nh mi kim loi cũn li cho tỏc dng vi dung dch Ba(OH)2 nhn c Al vỡ cú phn ng to khớ Al + H2O + Ba (OH)2 -> Ba(AlO2)2 + H2 (6) ng thi cho Ba (OH)2 vo dung dch mui ca kim loi cũn li ( phn ng v 4) Ta nhn c st vỡ kt ta i mu khụng khớ Fe SO4 + Ba (OH)2 -> Ba SO4 + Fe (OH)2 (7) Fe (OH)2 + O2 + H2O -> Fe ( OH)3 (8) Trng xanh Nõu Cũn li kt ta khụng i mu l Mg(OH)2-> nhn c Mg b/ Hũa tan hn hp HCl d tỏch c SiO2 ZnO + 2HCl -> Zn Cl2 + H2O Fe2 O3 + HCl -> FeCl3 + 3H2O + Dung dch mui lc + NaOH d: HCl + NaOH -> NaCl + H2O Zn Cl + NaOH -> Zn (OH)2 + NaCl Zn ( OH) + NaOH -> Na2ZnO2 + H2O Fe Cl + NaOH -> Fe (OH)3 + NaCl + Lc tỏch kt ta nung nhit cao Fe( OH) -t> Fe2O3 + H2O tỏch c Fe2O3 Sc CO2 vo dung dch cũn li cú phn ng Na2ZnO2 + CO2 + H2O -> Zn ( OH) + NaHCO3 Nung kt ta tỏch ZnO Zn (OH) -t>ZnO + H2O õy l mt phn bn nhỏp ca b ti liu: HNG DN GII 500 BI TP HểA HC THCS DNH CHO HC SINH GII ca thy Trng Th Tho biờn son Mong quý thy cụ giỏo, cỏc em hc sinh, quý bc ph huynh gúp ý b ti liu c hon thnh tt hn Mi chi tit gúp ý xin gi v a ch Email: thaonguyenh81@gmail.com Xin chõn thnh cm n! GV: Trng Th Tho Email: thaonguyenh81@gmail.com Page 55 [...]... ra ta c: a = 0,15 (mol) ; b = 0,1(mol) nBa = 81,515 : 137 = 0, 595 (mol) PTHH: Ba + 2H2O Ba(OH)2 + H2 nBa(OH)2 = nBa = 0, 595 (mol) 3Ba(OH)2 + Al2(SO4)3 3BaSO4 + 2Al(OH)3 PTHH(mol): 3 1 3 2 B (mol): 0, 595 0,1 => d Al2(SO4)3 =>nBaSO4 = nBa(OH)2 = 0, 595 (mol); nAl(OH)3 = 2/3nBa(OH)2 = 0,5 19. 2/3 = 1, 19/ 3 => m = 0, 595 233 + 1, 19/ 3.78 = 1 69, 575(g) Bi 54 Cú 3 mui A, B, C u kộm bn bi nhit Bit rng : - Mui... c: y= a = 56 0,005a + 72(0,01175a 3z) + 160z a= 4000 z 9 % Fe = 5600x Thay a = 200x vo ta c : a % Fe = 5600 x = 28% 200 x % Fe2O3 = 16000z 4000 z Thay a = vo ta c : a 9 GV: Trng Th Tho Email: thaonguyenh81@gmail.com Page 19 Hng dn gii 500 bi tp Húa hc THCS hay v khú % Fe2O3 = 16000 z = 36% 4000 z 9 % FeO = 100 28 36 = 36% Bi 41: Cho 39, 6g hn hp gm KHSO3 v K2CO3 vo 400g dd HCl 7,3%, khi xong phn... R2Om (1nm3; n, m N*) mn to 2R(OH) n O 2 R 2O m nH 2O (1) 2 Khi lng cht rn gim i 9 ln a 9 a 8a mgim i = m R 2Om a m R (OH)n m R 2Om 9 8 9 9 mR (OH)n 2(R 17n) 9 => R 136n 72m m R 2 Om 2R 16m 8 n m R Kt lun 1 1 64 1 2 -8 1 3 -80 2 2 128 2 3 56 3 3 192 Loi Loi Loi Loi Tha món Loi Kim loi R l st, cụng thc hiroxit: Fe(OH)2 to 4Fe(OH) 2 O 2 2Fe 2O3 4H 2O (2) Gi x l s mol ca H2SO4 phn ng... R2Om (1nm3; n, m N*) mn to 2R(OH) n O 2 R 2O m nH 2O (1) 2 Khi lng cht rn gim i 9 ln a 9 a 8a mgim i = m R 2Om a m R (OH)n m R 2Om 9 8 9 9 n m R Kt lun mR (OH)n m R 2 Om 1 1 64 1 2 -8 1 3 -80 2 2 128 2 3 56 3 3 192 Loi Loi Loi Loi Tha món Loi 2(R 17n) 9 R 136n 72m 2R 16m 8 Kim loi R l st, cụng thc hiroxit: Fe(OH)2 t 4Fe(OH) 2 O 2 2Fe 2O3 4H 2O o (2) Gi x l s mol ca H2SO4 phn ng vi... THCS hay v khú 2C2 H5OH 2Na 2C2 H5ONa H 2 Mol : x 0,5x 2C3H5 (OH)3 6Na 2C3H 5 (ONa)3 3H 2 Mol : y mH2 2(0,5x 1,5y) x 3y 1,5y mT 46x 92 y 2,5 2,5 14y m H2 mT x 3y (46x 92 y) x 100 100 3 14y 46 46x 3 %mC2H5OH 100% 70% 46x 92 y 46 14y 92 y 3 %mC3H5 (OH)3 100% 70% 30% Bi 22: Thc hin phn ng este húa gia axit C xHyCOOH v ru CnH2n+1OH Sau phn ng tỏch ly hn hp X ch gm este, axit... ancol etylic 92 o Cho 10 ml X tỏc dng ht vi Na thỡ thu c bao nhiờu lớt khớ (ktc)? Bit khi lng riờng ca ancol etylic l 0,8 g/ml v ca H2O l 1,0 g/ml Cỏc phng trỡnh phn ng : 2C2H5OH + 2Na 2C2H5ONa + H2 (1) 2H2O + Na 2NaOH + H2 (2) Tớnh s mol ca ancol v nc trong 10 ml X: 10 .92 .0,8 n C2H5OH = = 0,16 mol 100.46 10.8.1 2 n H 2O = = mol 100.18 45 1 2 Theo (1;2) VH 2 = (0,16+ ).22,42, 29 lớt 45 2 Bi 29 Bit A l... THCS hay v khú 2C2 H5OH 2Na 2C2 H5ONa H 2 Mol : x 0,5x 2C3H5 (OH)3 6Na 2C3H 5 (ONa)3 3H 2 Mol : y mH2 2(0,5x 1,5y) x 3y 1,5y mT 46x 92 y 2,5 2,5 14y m H2 mT x 3y (46x 92 y) x 100 100 3 14y 46 46x 3 %mC2H5OH 100% 70% 46x 92 y 46 14y 92 y 3 %mC3H5 (OH)3 100% 70% 30% Bi 51 Nờu phng phỏp v v hỡnh mụ t quỏ trỡnh iu ch khớ clo trong phũng thớ nghim? Vit phng trỡnh húa hc minh... loi Gi x l s mol H2SO4 Cụng thc húa hc ca mui cacbonat l RCO3 RCO3 + H2SO4 x x mRSO4 = (R + 96 )x mdd RSO4 = mRCO3 + mdd H2SO4 mCO2 = (R + 60)x + RSO4 x + CO2 x + H2O 98 x.100 44x 14, 7 (3R 2048) x 3 Theo bi ra ta cú : = C% dd RSO4 = ( R 96 ) x.100 = 17 (3R 2048) x 3 Gii ra ta c : R = 24 Vy R l Mg Bi 39 Cho 16,8 gam hn hp gm 2 oxit kim loi (thuc phõn nhúm chớnh nhúm II) tan ht trong nc to thnh... mol C2H6, 0,08 mol C2H4, 0,12 mol C2H2, 0,62 mol H2 v 0,1 mol CH4 Phn trm theo th tớch mi khớ chớnh l phn trm theo s mol 0,1 => % C2H6 = 100% = 9, 8% 1,02 Tng t: % C2H4= 7,8% ; % C2H2 = 11,8% ; % H2 = 60,8% ; %CH4 = 9, 8% (mol) Bi 59 Hn hp X gm 0,15 mol CH4, 0, 09 mol C2H2 v 0,2 mol H2 Nung núng hn hp X vi xỳc tỏc l niken sau mt thi gian phn ng thu c hn hp khớ Y gm CH 4, C2H4, C2H6, C2H2 d v GV: Trng Th... i 9 ln, ng thi thu c mt oxit kim loi Hũa tan hon ton lng oxit trờn bng 330ml dung dch H2SO4 1M, thu c dung dch X Cho X tỏc dng vi dung dch Ba(OH)2 d, sau khi phn ng hon ton thu c m gam kt ta Tớnh a, m, bit lng axit ó ly d 10% so vi lng cn thit phn ng vi oxit t cụng thc ca hiroxit l R(OH)n, cụng thc oxit l R2Om (1nm3; n, m N*) mn to 2R(OH) n O 2 R 2O m nH 2O (1) 2 Khi lng cht rn gim i 9 ln a 9