Phuong phap giai_nhanh_bttn-hoa

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  • 1. TS. PHM NGC SNPhng php gii nhanh bi tp trc nghimho hc LUYN THI I HCNh xut bn gio dc

2. Li ni u Cun Phng php gii bi tp trc nghim ho hc - Luyn thi i hc cung cp cho cc em h thng cc phng php mi gii cc bi tp trc nghim ho hc trong chng trnh THPT mt cch ngn gn, khoa hc v chnh xc. Ni dung cun sch gm 10 phng php gii nhanh vi h thng cc v v minh ha cng hng dn gii chi tit. Ni dung cc bi tp ho hc phong ph, a dng, cp nht. Tc gi hi vng rng cun sch ny s gip cho bn c hiu c bn cht cc phng php gii ton, c th x l linh hot v chnh xc cc bi ton ho hc, nhm t kt qu cao trong cc k thi. Qu trnh bin son khng trnh khi thiu st, tc gi rt mong nhn c nhng gp xy dng ca bn c cun sch c hon thin hn trong ln xut bn sau. Xin trn trng cm n ! TC GI 3. 1.PHNG PHP BO TON KHI LNG1. Nguyn tc Xt phn ng A + B C + D ta c ; mA + mB = mC + mD Mt s dng thng gp : - Hn hp oxit tc dng vi axit to mui: MO + HCl, H2SO4 long moxit + maxit = mmui + mnc Trong s mol nc c tnh theo axit. - Hn hp oxit tc dng CO, C hoc H2 : moxit mCO,C,H2 mran mCO2 ,H2OTrong s mol CO2, H2O c tnh theo CO, C v H2 - Hn hp kim loi tc dng vi axit gii phng H2. mkim loi + maxit = mmui + mH2 s mol H2 = 2HCl = H2SO4 - Hn hp mui cacbonat tc dng vi axit mmui (1) + maxit = mmui + mH2O + mCO2 2. Cc v d minh ho V d 1: (2007 - Khi A) Ho tan hon ton 2,81 gam hn hp gm Fe2O3, MgO, ZnO trong 500 ml axit H2SO4 0,1M (va ). Sau phn ng, hn hp mui sunfat khan thu c khi c cn dung dch c khi lng l A. 6,81 gam. B. 4,81 gam. C. 3,81 gam. D. 5,81 gam. Hng dn Fe2O3 + 3H2SO4 Fe2(SO4)3 + 3H2O (1) MgO + H2SO4 MgSO4 + H2O (2) ZnO + H2SO4 ZnSO4 + H2O (3) Theo cc pt ho hc (1, 2, 3): n H2O = n H2SO4 = 0,5 0,1 = 0,05 (mol) p dng nh lut bo ton khi lng: m hh mui khan = 2,81 + 98 0,05 18 0,05 = 6,81 (g). V d 2: Hn hp X gm Fe, FeO v Fe2O3. Cho mt lung kh CO i qua ng s ng m gam hn hp X nung nng. Sau khi kt thc th nghim thu c 64 gam cht A trong ng s v 11,2 lt kh B (ktc) c t khi so vi H2 l 20,4. Gi tr ca m l A. 105,6. B. 35,2. C. 52,8. D. 70,4. Hng dn Cc phng trnh ho hc ca phn ng kh oxit st c th c: 0t 3Fe2O3 + CO 2Fe3O4 + CO2 (1) 4. 0t Fe3O4 + CO 3FeO + CO2 (2)0t FeO + CO Fe + CO2 (3) Nhn xt: Cht rn A c th gm 3 cht Fe, FeO, Fe3O4 hoc t hn, iu quan trng l s mol CO phn ng bao gi cng bng s mol CO2 to thnh Gi x l s mol CO2 to thnhnB =11, 2 = 0,5 (mol) 22, 444x + 28(0,5 x) = 0,5 20,4 2 = 20,4 x = 0,4 (mol) Do n CO phn ng = 0,4 (mol) p dng nh lut bo ton khi lng: m = mA + mCO2 mCO = 64 + 44 0,4 28 0,4 = 70,4 (g). V d 3: Ho tan hon ton 5 gam hn hp 2 kim loi bng dung dch HCl thu c dung dch A v kh B. C cn dung dch A th c 5,71 gam mui khan. Tnh th tch kh B (o ktc). Hng dn Gi 2 kim loi cho l X v Y 2X + 2m HCl 2XClm + m H2 (1) 2Y + 2n HCl 2YCln + n H2 (2) Theo (1, 2): n HCl = 2 n H2 p dng nh lut bo ton khi lng: 5 + 36,5 2 n H2 = 5,71 + 2 n H2 n H2 = 0,01 (mol) Vy VH2 (ktc) = 0,01 22,4 = 0,224 (l). V d 4: (2009 - Khi A)Cho 3,68 gam hn hp gm Al v Zn tc dng vi mt lng va dung dch H2SO4 10%, thu c 2,24 lt kh H2 ( ktc). Khi lng dung dch thu c sau phn ng l A. 101,68 gam. B. 88,20 gam. C. 101,48 gam. D. 97,80 gam. Hng dn 2Al + 3H2SO4 Al2(SO4)3 + 3H2 (1) Zn + H2SO4 ZnSO4 + H2 (2) T (1, 2): n H2SO4 = n H2 = mdd H2SO4 =2, 24 = 0,1 (mol) 22, 498 0,1 100 = 98 (g) 10p dng nh lut bo ton khi lng: mdd sau phn ng = mhh + mdd H2SO4 mH = 3,68 + 98 2 0,1 = 101,48 (g). 2 5. V d 5. Cho t t mt lung kh CO i qua ng s ng m gam hn hp gm Fe, FeO, Fe3O4, Fe2O3 nung nng, kt thc phn ng thu c 64g st, kh i ra gm CO v CO2 cho sc qua dung dch Ca(OH)2 d c 40g kt ta. Vy m c gi tr l A. 70,4g B. 74g C. 47g D. 104g Li gii : Kh i ra sau phn ng gm CO2 v CO d cho i qua dung dch Ca(OH)2 d : CO2 + Ca(OH)2 CaCO3 + H2O 40 0,4 (mol) 1000,4 (mol) S phn ng: FeO Fe2O3 Fe3O4+CO Fe+CO228.0,4 + m = 64 + 44.0,4 m = 70,4g V d 6. Ngi ta cho t t lung kh H2 i qua mt ng s ng 5,44 gam hn hp gm FeO, Fe3O4, Fe2O3, CuO nung nng, kt thc phn ng thu c m gam hn hp cht rn A v 1,62 gam H2O. Vy m c gi tr l A. 4g B. 5g C. 4,5g D. 3,4g Li gii : nH2 nH2O 0, 09 (mol) S phn ng: FeO H2 + Fe2O3 A + H2O Fe3O4 CuO 0,09.2 + 5,44 = m + 1,62 m = 4g V d 7. Cho 35g hn hp Na2CO3, K2CO3 tc dng va vi dung dch BaCl2. Sau phn ng thu c 59,1g kt ta. Lc tch kt ta, c cn dd thu c m(g) mui clorua. Vy m c gi tr l A. 38,3g B. 22,6g C. 26,6g D. 6,26g Li gii : S phn ng: Na 2 CO3 K 2 CO3 + BaCl2 BaCO3 +NaCl KClnBaCl2 nBaCO3 0,3 (mol)p dng nh lut bo ton khi lng: mhh mBaCl2 m m dd m = 35 + 0,3.208 59,1 = 38,3 (g) 6. V d 8. Cho 4,48g hn hp Na2SO4, K2SO4, (NH4)2SO4 tc dng va vi 300 ml dung dch Ba(NO3)2 0,1M . Kt thc phn ng thu c kt ta A v dung dch B. Lc tch kt ta, c cn dung dch thu c m(g) mui nitrat. Vy m c gi tr l A. 5,32g B. 5,23g C. 5,26g D. 6,25g Li gii : S phn ng: Na2SO4 K2SO4 + Ba(NO3)2 BaSO4 + (NH4)2SO4NaNO3 KNO3 NH4NO3nBa(NO3 )2 nBaSO4 0, 03 (mol) ; 4, 48 7,83 6,99 mB mB 5,32 (g)V d 9. Ho tan 2,57g hp kim Cu, Mg, Al bng mt lng va dung dch H 2SO4 long thu c 1,456 lt kh X (ktc), 1,28g cht rn Y v dung dch Z. C cn Z thu c m gam mui khan, m c gi tr l A. 7,53g B. 3,25g C. 5,79g D. 5,58g Li gii: S phn ng : Cu MgSO4 + Cu + H2 Mg + H2SO4 Al2 (SO4 )3 Al m m (AlMg) m SO2 (2,57 1,28) 0, 065.96 7,53 (g) 4V d 9. Ho tan hon ton 3,72g hn hp 2 kim loi A, B trong dung dch HCl d thy to ra 1,344 lt kh H2 (ktc). C cn dung dch sau phn ng thu c mui khan c khi lng l A. 7,12g B. 7,98g C. 3,42g D. 6,12g Li gii : Theo phng trnh in li : nHnCl 2.1,344 0,12(mol) 22, 4mmui = mKL + mCl = 3,72 + 0,12.35,5 = 7,98 (g) V d 10. Nung m gam hn hp A gm 2 mui MgCO3 v CaCO3 cho n khi khng cn kh thot ra thu c 3,52g cht rn B v kh C. Cho ton b kh C hp th ht bi 2 lt dung dch Ba(OH)2 thu c 7,88g kt ta. un nng dung dch li thy to thnh thm 3,94g kt ta na. Nu cc phn ng xy ra hon ton th m c gi tr l A. 7,44g B. 7,40g C. 7,04g D. 4,74g Li gii: m = mB + mCO2 CO2 + Ba(OH)2 BaCO3 + H2O 2CO2 + Ba(OH)2 Ba(HCO3)2m = 3,52 + (7,88 3,94 2. ).44 7, 04 (g) 197 197 7. 2.PHNG PHP BO TON MOL NGUYN T1. Nguyn tc Tng s mol nguyn t ca mt nguyn t trc v sau phn ng lun bng nhau. Tnh s mol nguyn t ca mt nguyn t : nnguyn t A = x.nX = (s nguyn t A trong X).s mol X v d : nO = 4.nH2SO4 2. Cc v d minh ho V d 1: (2007 - Khi A) Ho tan hon ton hn hp gm 0,12 mol FeS2 v a mol Cu2S vo axit HNO3 (va ), thu c dung dch X (ch cha hai mui sunfat) v kh duy nht NO. Gi tr ca a l A. 0,04. B. 0,075. C. 0,12. D. 0,06. Hng dn Fe2 (SO4 )3 FeS2 HNO3 NO H 2O Cu 2S CuSO4S phn ng: 2FeS2 Fe2(SO4)3 0,12 0,06 Cu2S 2CuSO4 a 2a p dng nh lut bo ton nguyn t i vi S, ta c: 2 0,12 + a = 3 0,06 + 2a a = 0,06 (mol). V d 2: Dn t t V lt kh CO ( ktc) i qua mt ng s ng lng d hn hp rn gm CuO, Fe2O3 ( nhit cao). Sau khi cc phn ng xy ra hon ton, thu c kh X. Dn ton b kh X trn vo lng d dung dch Ca(OH)2 th to thnh 4 gam kt ta. Gi tr ca V l A. 1,120. B. 0,896. C. 0,448. D. 0,224. (Trch thi TSC nm 2008 - Khi A, B) Hng dn S phn ng:CuO t0 + CO CO2 (X) + hn hp rn Fe 2 O3 CO2 + Ca(OH)2 d CaCO3 + H2O p dng s bo ton s mol i vi nguyn t C: n C trong CO = n C trong CO2 = n C trong CaCO3 n C trong CO = n CaCO= 34 = 0,04 (mol) 100Vy V = 0,04 22,4 = 0,896 (l). V d 3: Ho tan hn hp X gm 0,2 mol Fe v 0,1 mol Fe2O3 vo dung dch HCl d c dung dch Y. Cho dung dch Y tc dng vi dung dch NaOH d thu c kt 8. ta. Lc kt ta, ra sch em nung trong khng kh n khi lng khng i thu c m gam cht rn Z. Gi tr ca m l A. 16,0. B. 24,0. C. 28,8. D. 32,0. Hng dn S cc phn ng: FeCl2 NaOH Fe HCl X Y Fe2O3 FeCl3Fe(OH) 2 t0 Z (Fe2O3) Fe(OH)3p dng s bo ton s mol i vi nguyn t Fe: n Fe trong Z = n Fe trong X = n Fe 2n Fe2O3 = 0,2 + 2.0,1 = 0,4 (mol) n Fe2O3 trong Z =1 n Fe trong Z = 0,2 (mol) Vy m = 0,2 160 = 32,0 (g). 2V d 4: Cho 2,13 gam hn hp X gm ba kim loi Mg, Cu v Al dng bt tc dng hon ton vi oxi thu c hn hp Y gm cc oxit c khi lng 3,33 gam. Th tch dung dch HCl 2M va phn ng ht vi Y l A. 57 ml. B. 50 ml. C. 75 ml. D. 90 ml. (Trch thi TSH nm 2008 - Khi A) Hng dn MgO Mg O2 X Cu Y CuO Al O Al 2 3 MgO + 2HCl MgCl2 + H2O (1) CuO + 2HCl CuCl2 + H2O (2) Al2O3 + 6HCl 2AlCl3 + 3H2O (3)Ta c mO / hhY = 3,33 2,13 = 1,2 (g) hay n O / hhY =1, 2 = 0,075 (mol) 16Theo (1, 2, 3): n HCl = 2 n O / hhY = 2 0,075 = 0,15 (mol) Vy Vdd HCl =0,15 = 0,075 2(l) = 75 (ml). V d 5: Cho mt mu Na lu trong khng kh, b chuyn ho thnh hn hp rn X gm Na, Na2O, NaOH, Na2CO3. Ho tan hon ton hn hp X bng H2SO4 long, sau phn ng thu c dung dch Y. Lm bay hi nc t t thu c 8,05 gam tinh th Na2SO4.10H2O. Khi lng mu Na l A. 0,575 gam. B. 1,15 gam. C. 2,3 gam. D. 1,725 gam. Hng dn Na 2 O NaOH H2SO4 kk dd Y (dd Na2SO4) Na2SO4.10H2O Na X Na 2 CO3 Na 8,05 Ta c n Na 2SO4 .10H2O = = 0,025 (mol) n Na ca mu Na = n Na trong 3222 n Na 2SO4 .10H2O = 0,05 (mol). Vy khi lng mu Na l: 0,05 23 = 1,15 (g).tinh th= 9. V d 6: Cho hn hp A gm ba kim loi X, Y, Z c ho tr ln lt l 3, 2, 1 v t l s mol ln lt l 1 : 2 : 3, trong s mol ca X bng x mol. Ho tan hon ton A bng dung dch c cha y mol HNO3. Sau phn ng thu c dung dch B khng cha NH4NO3 v V lt hn hp kh E ( ktc) gm NO2 v NO. Biu thc tnh y theo x v V l A. 8x +V . 22, 4B. 6x +V . 22, 4C. 5x +V . 22, 4D. 10x +V . 22, 4Hng dn Ta c n hh E =V (mol) 22, 4Theo bi n X : n Y : n Z = 1 : 2 : 3 M n X = x (mol) nn n Y = 2x (mol), n Z = 3x (mol) S phn ng: X , Y , Z + HNO3 X(NO3)3, Y(NO3)2, ZNO3 + NO2 , NO + H2O x 2x 3x x 2x 3x S mol mi mui bng s mol mi kim loi tng ng Da vo s bo ton s mol i vi nguyn t N: n HNO3 = n N = 3 n X(NO3 )3 + 2 n Y(NO3 )2 + n ZNO3 + n NO2 + n NO = 3. x + 2. 2x + 3x +V V = 10x + (mol). 22, 4 22, 4V d 7. Cho hn hp A gm 0,1 mol Cu, 0,2 mol Ag tan va ht trong V lt dung dch HNO3 1M thu c dung dch X v hn hp Y gm 2 kh NO, NO2 ( n NO n NO2 0,1 mol ). V c gi tr l A. 1 ltB. 0,6 ltC. 1,5 ltD. 2 ltLi gii S phn ng: Cu Ag+HNO3 Cu(NO3 )2 AgNO3+NO NO2+ H2 Op dng s bo ton nguyn t Cu, Ag ta c : nCu = nCu(NO3 )2 = 0,1 mol v nAg = n AgNO3 = 0,2 mol p dng cho nguyn t N : n N (HNO3 ) = n N(Cu(NO3 )2 AgNO3 NO NO2 ) n(HNO3 ) = 2nCu(NO3 )2 n AgNO3 n NO n NO2 n(HNO3 ) = 2.0,1 + 0,2 + 0,1 + 0,1 = 0,6 mol VHNO3 0,6 0,6(lit) 1V d