Bồi dưỡng HSG Hóa Học 8

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    Bi dng HSG Ha Hc 8

    Cc dng bi tp ha hc chng trnh lp 8-THCS

    Chuyn 1. Bi tp v nguyn t, nguyn t ha hc1/ Nguyn t (NT):- Ht v cng nh , trung ha v in, to nn cc cht.Cu to: + Ht nhn mang in tch (+)(Gm: Proton(p) mang in tch (+) v ntron

    khng mang in ). Khi lng ht nhn c coi l khi lng nguyn t.+ V nguyn t cha 1 hay nhiu electron (e) mang in tch (-). Electron

    chuyn ng rt nhanh quanh ht nhn v sp xp theo lp (th t sp xp (e) ti a trongtng lp t trong ra ngoi: STT ca lp : 1 2 3

    Se ti a : 2e 8e 18e Trong nguyn t:- Sp = se = sin tch ht nhn = s th t ca nguyn t trong bng h thng tunhon cc nguyn t ha hc- Quan hgia sp v sn : p n 1,5p ( ng vi 83 nguyn t)

    - Khi lng tng i ca 1 nguyn t( nguyn tkhi )NTK = sn + sp- Khi lng tuyt i ca mt nguyn t( tnh theo gam )

    + mT= m e + mp + mn+ mP mn 1VC 1.67.10- 24 g,+ me 9.11.10-28 g

    Nguyn t c th ln kt c vi nhau nh e lp ngoi cng.2/ Nguyn t ha hc (NTHH):l tp hp nhng nguyn t cng loi c cng s p tronght nhn.- S p l s c trng ca mt NTHH.

    - Mi NTHH c biu din bng mt hay hai ch ci. Ch ci u vit di dng in hoach ci th hai l ch thng. l KHHH- Nguyn t khi l khi lng ca nguyn t tnh bng VC. Mi nguyn t c mtNTK ring. Khi lng 1 nguyn t = khi lng 1vc.NTK

    NTK =1

    khoiluongmotnguyentu

    khoiluong dvc

    m a Nguyn t = a.m 1vc .NTK

    (1VC =1

    12KL ca NT(C) (MC = 1.9926.10

    - 23 g) =1

    121.9926.10- 23 g= 1.66.10- 24 g)

    * Bi tp vn dng:

    1. Bit nguyn t C c khi lng bng 1.9926.10- 23 g. Tnh khi lng bng gam canguyn t Natri. Bit NTK Na = 23. (p s: 38.2.10- 24 g)2.NTK ca nguyn t C bng 3/4 NTK ca nguyn t O, NTK ca nguyn t O bng 1/2NTK S. Tnh khi lng ca nguyn t O. (p s:O= 32,S=16)3. Bit rng 4 nguyn t Mage nng bng 3 nguyn t nguyn t X. Xc nh tn,KHHHca nguyn t X. (p s:O= 32)4.Nguyn t X nng gp hai ln nguyn t oxi .b)nguyn t Y nh hn nguyn t Magie 0,5 ln .c) nguyn t Z nng hn nguyn t Natri l 17 vc .

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    Bi dng HSG Ha Hc 8Hy tnh nguyn t khi ca X,Y, Z .tn nguyn t, k hiu ho hc ca nguyn t ?5.Nguyn t M c s n nhiu hn s p l 1 v s ht mang in nhiu hn s ht khngmang in l 10. Hy xc nh M l nguyn t no?6.Tng s ht p, e, n trong nguyn t l 28, trong s ht khng mang in chimxp x 35% .Tnh s ht mi loa .V s cu to nguyn t .7.Nguyn t st c 26p, 30n, 26e

    a.Tnh khi lng nguyn t stb.Tnh khi lng e trong 1Kg st8.Nguyn t X c tng cc ht l 52 trong s ht mang in nhiu hn s ht khngmang in l 16 ht.a)Hy xc nh s p, s n v s e trong nguyn t X.b) V s nguyn t X.c) Hy vit tn, k hiu ho hc v nguyn t khi ca nguyn t X.9.Mt nguyn tX c tng sht e, p, n l 34. Sht mang in nhiu hn sht khngmang in l 10. Tm tn nguyn tX. Vscu to ca nguyn tX v ion c to

    ra tnguyn tX10.Tm tn nguyn tY c tng sht trong nguyn tl 13. Tnh khi lng bng gamca nguyn t.

    11.Mt nguyn tX c tng sht l 46, sht khng mang in bng 815

    sht mang

    in. Xc nh nguyn tX thuc nguyn tno ? vscu to nguyn tX ?12.Nguyn tZ c tng sht bng 58 v c nguyn tkhi < 40 . Hi Z thuc nguyntho hc no. Vscu to nguyn tca nguyn tZ ? Cho bit Z l g ( kim loihay phi kim ? ) (p s :Z thuc nguyn tKali ( K ))

    Hng dngii : bi 2p + n = 58 n = 58 2p ( 1 )Mt khc : p n 1,5p ( 2 ) p 58 2p 1,5p gii ra c 16,5 p 19,3 ( p :

    nguyn )Vy p c thnhn cc gi tr: 17,18,19

    P 17 18 19N 24 22 20NTK = n + p 41 40 39

    Vy nguyn tZ thuc nguyn tKali ( K )

    13.Tm 2 nguyn tA, B trong cc trng hp sau y :a) Bit A, B ng ktip trong mt chu kca bng tun hon v c tng sin tch htnhn l 25.b) A, B thuc 2 chu kktip v cng mt phn nhm chnh trong bng tun hon. Tngsin tch ht nhn l 32.14141414: Trong 1 tap hp cac phan t ong sunfat (CuSO4) co khoi lng 160000 vC. Cho

    biet tap hp o co bao nhieu nguyen t moi loai.

    3.Sto thnh ion (dnh cho HSG lp 9)

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    Bi dng HSG Ha Hc 8t cu trc bo ha ( 8e lp ngoi cng hoc 2e i vi H ) th cc nguyn tcthnhng hoc nhn thm electron to ra nhng phn mang in - gi l ion

    * Kim loi v Hiro : nhng e to ion dng ( cation)M ne M n + (Ca 2e Ca 2 + )

    * Cc phi kim nhn e to ion m (anion)X + ne X n- ( Cl + 1e Cl1- )

    * Bi tp vn dng:1.Hp cht X c to thnh tcation M+ v anion Y2-. Mi ion u do 5 nguyn tca2 nguyn tto nn. Tng sproton trong M+ l 11 cn tng selectron trong Y2- l 50.Xc nh CTPT ca hp cht X v gi tn ? ng dng ca cht ny trong nng nghip .Bit rng 2 nguyn ttrong Y2- thuc cng phn nhm trong 2 chu klin tip ca bngtun hon cc ng.t.

    Hng dn gii :t CTTQ ca hp cht X l M2YGis ion M+gm 2 nguyn tA, B :

    ion M+

    dng : AxBy+

    c : x + y = 5 ( 1 )x.pA + y.pB = 11 ( 2)Gis ion Y 2-gm 2 nguyn tR, Q : ion Y2- dng : RxQy

    2- c : x + y = 5 (3)xpR+ y.pQ= 48 (4 ) do se > sp l 2

    T( 1 ) v (2) ta c sproton trung bnh ca A v B : 11 2,25

    p= =

    1 trong AxBy+ c 1 nguyn tc p < 2,2 ( H hoc He ) v 1 nguyn tc p > 2,2

    V He khng to hp cht ( do tr) nn nguyn tc p < 2,2 l H ( gisl B )

    T( 1 ) v ( 2) ta c : x.pA + (5 x ).1 = 11

    pA=

    6

    1Ap x= +

    ( 1

    x < 5 )X 1 2 3 4pA 7(N) 4(B) 3(Li) 2,5 (loi)

    ion M+ NH4+

    khng xc nh ion

    Tng t: sproton trung bnh ca R v Q l : 48 9,65

    p= = c 1 nguyn tc sp z = 1

    Vy cng thc ca A l C3H80

    * Bi tp vn dng:

    +Trng hp cha bit PTK

    Tm c CTHH n gin1:t chy hon ton 13,6g hp cht A,th thu c 25,6g SO2v 7,2g H2O. Xc nhcng thc ca A

    2222: ot chay hoan toan m gam chat A can dung het 5,824 dm3O2(ktc). San pham coCO2va H2O c chia oi. Phan 1 cho i qua P2O5thay lng P2O5tang 1,8 gam. Phan2 cho i qua CaO thay lng CaO tang 5,32 gam. Tm m va cong thc n gian A. Tmcong thc phan t A va biet A the kh (k thng) co so C 4.3: t chy hon ton 13,6g hp cht A, th thu c 25,6 g S02v 7,2g H20. Xc nhcng thc A

    +Trng hp bit PTK Tm c CTHH ng1: t chy hon ton 4,5g hp cht hu c A .Bit A cha C, H, O v thu c 9,9g khCO2v 5,4g H2O. lp cng thc phn t ca A. Bit phn t khi A l 60.2: t chy hon ton 7,5g hyroccbon A ta thu c 22g CO2v 13,5g H2O. Bit tkhi hI so vi hyr bng 15. Lp cng thc phn t ca A.3: : t chy hon ton 0,3g hp cht hu c A . Bit A cha C, H, O v thu c224cm3kh CO2

    (ktc) v 0,18g H2O. lp cng thc phn t ca A.Bit t khi ca A ivi hiro bng 30.

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    Bi dng HSG Ha Hc 84:t chy 2,25g hp cht hu c A cha C, H, O phi cn 3,08 lt oxy (ktc) v thuc VH2O=5\4 VCO2 .Bit t khi hi ca A i vi H2 l 45. Xc nh cng thc ca A5: Hyro A l cht lng , c t khi hi so vi khng kh bng 27. t chy A thu cCO2v H2O theo t l khi lng 4,9 :1 . tm cng thc ca A

    S:S:S:S: A la C4H103: Bit thnh phn phn trm v khi lng cc nguyn t, cho bit NTK, phn t

    khi.Cch gii:

    - Tnh khi lng tng nguyn ttrong 1 mol hp cht.- Tnh smol nguyn ttng nguyn ttrong 1 mol hp cht.- Vit thnh CTHH.Hoc: - t cng thc tng qut: AxBy

    - Ta c t l khi lng cc nguyn t: yMBxMA.. = BA%%

    - Rt ra tlx: y = MAA% : MBB% (ti gin)

    -

    Vit thnh CTHH n gin: (AaBb )n = MAxBy

    n =

    MAxBy

    MAaBb nhn n vo h s a,b ca cng thc AaBbta c CTHH cn lp.

    Vi d.Mot hp chat kh Y co phan t khoi la 58 vC, cau tao t 2 nguyen to C va Htrong o nguyen to C chiem 82,76% khoi lng cua hp chat. Tm cong thc phan tcua hp chat.Gii : - t cng thc tng qut: CxHy

    - Ta c t l khi lng cc nguyn t: ..MC xMH y =

    %%

    CH

    - Rt ra tlx: y = %CMC : %HMH =82,76

    12 :17,24

    1 = 1:2- Thay x= 1,y = 2 voCxHyta c CTHH n gin: CH2

    - Theo bi ra ta c : (CH2 )n = 58 n =5814

    = 5

    Ta c CTHH cn lp : C5H8

    * Bi tp vn dng:1: Hp cht X c phn t khi bng 62 vC. Trong phn t ca hp cht nguyn t oxichim 25,8% theo khi lng, cn li l nguyn t Na. S nguyn t ca nguyn t O vNa trong phn t hp cht l bao nhiu ?2: Mot hp chat X co thanh phan % ve khoi lng la :40%Ca, 12%C va 48% O . Xac

    nh CTHH cua X. Biet khoi lng mol cua X la 100g.3:Tm cng thc ho hc ca cc hp cht sau.a) Mt cht lng d bay hi, thnh phn t c 23,8% C, 5,9%H, 70,3%Cl v c PTKbng 50,5.b ) Mt hp cht rn mu trng, thnh phn t c 4o% C, 6,7%H, 53,3% O v c PTKbng 180.4:Mui n gm 2 nguyn t ho hc l Na v Cl Trong Na chim 39,3% theo khilng . Hy tm cng thc ho hc ca mui n, bit phn t khi ca n gp 29,25 lnPTK H2.5555: Xac nh cong thc cua cac hp chat sau:

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    Bi dng HSG Ha Hc 8

    a) Hp chat tao thanh bi magie va oxi co phan t khoi la 40, trong o phan tramve khoi lng cua chung lan lt la 60% va 40%.

    b) Hp chat tao thanh bi lu huynh va oxi co phan t khoi la64, trong o phantram ve khoi lng cua oxi la 50%.

    c) Hp chat cua ong, lu huynh va oxi co phan t khoi la 160, co phan tram cua

    ong va lu huynh lan lt la 40% va 20%.d) Hp chat tao thanh bi sat va oxi co khoi lng phan t la 160, trong o phantram ve khoi lng cua oxi la 70%.

    e) Hp chat cua ong va oxi co phan t khoi la 114, phan tram ve khoi lng cuaong la 88,89%.

    f) Hp chat cua canxi va cacbon co phan t khoi la 64, phan tram ve khoi lngcua cacbon la 37,5%.

    g) A co khoi lng mol phan t la 58,5g; thanh phan % ve khoi lng nguyen to:60,68% Cl con lai la Na.

    h) B co khoi lng mol phan t la 106g; thanh phan % ve khoi lng cua cacnguyen to: 43,4% Na; 11,3% C con lai la cua O.

    i) C co khoi lng mol phan t la 101g; thanh phan phan tram ve khoi l ng cacnguyen to: 38,61% K; 13,86% N con lai la O.

    j) D co khoi lng mol phan t la 126g; thanh phan % ve khoi lng cua cacnguyen to: 36,508% Na; 25,4% S con lai la O.

    k) E co 24,68% K; 34,81% Mn; 40,51%O. E nang hn NaNO3 1,86 lan.l) F cha 5,88% ve khoi lng la H con lai la cua S. F nang hn kh hiro 17 lan.

    m)G co 3,7% H; 44,44% C; 51,86% O. G co khoi lng mol phan t bang Al.n) H co 28,57% Mg; 14,285% C; 57,145% O. Khoi lng mol phan t cua H la 84g.

    6666. Phan t canxi cacbonat co phan t khoi la 100 vC , trong o nguyen tcanxi chiem40% khoi lng, nguyen to cacbon chiem 12% khoi lng. Khoi lng con lai la oxi.Xac nh cong thc phan t cua hp chat canxi cacbonat?7777. Mot hp chat co phan t khoi bang 62 vC. trong phan t cua hp chat nguyen