S GIO DC & O TOTNH B RA VNG TU

THI CHNH THC

K THI CHN HC SINH GII OLYMPIC 27/4

LP 11 THPT, NM HC 2014 2015

MN THI: HO HC

Thi gian lm bi: 180 pht

Ngy thi: 05/ 03/ 2015

( thi c 02 trang)

Cho : H= 1; C= 12; N = 14; O= 16; P= 31;S= 32; Cl= 35,5; Cu= 64; Zn= 65; Se= 79; Ag = 108

Fe=56; Al=27; Na=23; Ca=40; Ba=137; Mg=24; K=39.

Ghi ch: Th sinh khng c s dng bng tun hon cc nguyn t ho hc.

Bi I (4 im):Cu 1. Cho 1 lt dung dch axit yu HA c in li , hng s cn bng Ka v nng C0 mol/l.

a. Chng minh rng mi quan h gia Ka v C0 l

b. Cn pha long dung dch ban u bao nhiu ln in li ca axit tng gp i?

Cu 2. Trn 20,00 ml dung dch H3PO4 0,50 M vi 37,50 ml dung dch Na3PO4 0,40 M, ri pha long bng nc ct thnh 100,00 ml dung dch A. Tnh pH ca dung dch A.

Cu 3. Nu hin tng quan st c, gii thch v vit phng trnh phn ng xy ra khi:

a. Trn dung dch FeCl3 vi dung dch Na2CO3.

b. Trn dung dch Al2(SO4)3 vi dung dch K2S.

Bi II (4 im):

Cu 1. Nhit phn hon ton m gam mui nitrat ca kim loi R thu c gam oxit kim loi v hn hp kh. Xc nh cng thc ho hc ca mui nhit phn.

Cu 2. Chia 97,5 gam kim loi Zn thnh hai phn bng nhau. Ho tan phn I vo 1,5 l dung dch cha HCl 0,9M v NaNO3 0,2M thu c V1 lt (ktc) hn hp kh (X) gm hai kh khng mu, trong c mt kh khng mu ho nu ngoi khng kh (l sn phm kh duy nht ca N+5). Cho phn II tan ht vo dung dch cha NaNO3 v KOH, un nng thu c V2 lt (ktc) hn hp kh (Y) gm hai kh khng mu. Ton b lng kh (Y) c th kh c ti a 36 gam CuO. Tnh V1 v V2.Bi III (4,0 im)

Cu 1. Mt hirocacbon A khi tc dng vi lng d nc brom, to nn dn xut tetrabrom. cha 75,83% khi lng brom. Khi un si A vi dung dch kali pemanganat, c mt axit sunfuric ch to ra axit cacboxylic mt nc. A c th tham gia phn ng hirat ho trong iu kin thch hp. Xc nh cng thc cu to ca A v vit cc phng trnh ho hc minh ho cc phn ng nu trn.

Cu 2.

a. 3-metylbut-1-en tc dng vi axit clohiric to ra cc sn phm, trong c A l 2-clo-3-metylbutan v B l 2-clo-2-metylbutan.Bng c ch phn ng, hy gii thch s to thnh hai sn phm A v B.

b. 2-metylbut-2-en phn ng vi axit clohiric. Trnh by c ch phn ng, cho bit sn phm chnh v gii thch?Bi IV (4,0 im):

Cu 1. t chy hon ton hirocacbon X (mch h). Hp th ton b sn phm chy vo dung dch cha 0,15 mol Ca(OH)2 th thu c 12 gam kt ta trng, ng thi khi lng phn dung dch gim 1,38 gam.

a. Xc nh cng thc n gin nht ca X.

b. Xc nh cng thc phn t ca X, bit hiro ho hon ton 2,46 gam X cn dng 1,344 lt kh H2 (ktc) c xc tc Ni.

c. Xc nh cng thc cu to ca X, bit:

X CO2 + CH3COOH + HOOC- CO-CH3.

Cu 2. Cho propan tc dng vi Cl2 (t l mol 1:1), trong iu kin c nh sng 250C), thu c hai dn xut monoclo l X v Y c t l tng ng bng 53% v 47% v s mol.

a. Vit phng trnh phn ng xy ra v xc nh X, Y.

b. Trnh by c ch phn ng to thnh X .

c. T kt qu trn, tnh t l kh nng phn ng th nguyn t hiro cacbon bc hai so vi hiro cacbon bc nht ca propan trong phn ng th vi clo ni trn.Bi V (4 im):

Hn hp A gm hai ancol n chc, cng bc, mch thng k tip nhau trong dy ng ng. Ly 6,97 gam hn hp A dn qua CuO nung nng, thu c cht rn B v dung dch X. Cho B tan ht trong trong dung dch HNO3 long thu c 2,24 lt (ktc) kh NO (sn phm kh duy nht). Chia dung dch C lm hai phn:

Phn I cho tc dng va vi 2,76 gam K2CO3 (phn ng khng sinh ra kh).

Phn II thm mt lng d CuSO4 khan lc k, lc tch phn rn thu c phn lng D.

Cho D tc dng vi Na (d) thu c 0,784 lt kh (ktc).

Tnh hiu sut chung ca qu trnh oxi ho cc ancol, bit khi lng CuSO4 tng 1,08 gam so vi trc khi cho vo phn II. Xc nh cng thc cu to hai ancol trong hn hp A.

--------HT---------

H v tn th sinh: Ch k gim th s 1:S bo danh:.HNG DN CCH GII THI

Bi I: (4,0 im)

Cu 1.

a. Cc bn vit phng trnh in ly t suy ra phng trnh tnh Ka, ri rt v mt v. Lu rng, v axit HA l axit yu nn 1 >> , nn 1 - coi nh bng 1.

b. Da vo biu thc va tnh, tng gp i, Ka khng i => C0 gim 4 ln

C0 = ; n0 khng i => V tng 4 ln => Cn pha long dung dch i 4 ln.

Cu 2.

Vit phng trnh H3PO4 + Na3PO4 => Na2HPO4 v NaH2PO4 ri tnh s mol tng mui

Nng ca chng trong dung dch A.

nh gi: NaH2PO4 l axit, Na2HPO4 l baz , rt ra biu thc tnh => pH (7,812)( nh gi chun xc hn v pH kh gn 7 ta nn nh gi c s phn li ca nc)

Cu 3.

a. CO2 + Fe(OH)3 (si bt kh + kt ta nu )

b. H2S + Al(OH)3 (si bt kh + kt ta trng hoc trng keo)

Bi II: (4,0 im)

Cu 1.

Gi ho tr ban u l a, ho tr lc sau l b

Phng trnh 2R(NO3)a R2Ob.

mR2Ob = mR(NO3)a v nR(NO3)a = 2nR2Ob => MR2Ob = MR(NO3)a. 2MR + 16b =

Bin lun ho tr a, b rt ra MR = 56 => R l Fe.Cu 2. nZn mi phn = 0,75 mol

Phn 1: Kh thot ra ho nu v l sn phm kh duy nht ca N+5 l NO => kh cn li ch c th l H2.

(HNO3 l axit mnh nn Zn kh ht vi NO3- trc, mi tip tc kh H+ d)

Vit phng trnh => V1 = 8,4 ltPhn 2:

4Zn + 7OH- + NO3- = 4ZnO22- + NH3 + 2H2O

Zn + 2OH- = ZnO22- + H2.

2NH3 + 3CuO = N2 + 3Cu + 3H2O

H2 + CuO = Cu + H2O

Lp h phng trnh theo s mol kh v s mol Zn => nNH3 v nH2 => V2 = 8,736 ltBi III: (4 im)

Cu 1: Xc nh cng thc cu to ca (A):

Cu 2: a. C ch u vit bnh thng, c ch sau c chuyn v Hb. Vit c ch nh bnh thng, sn phm chnh nh vo v tr 3

Bi IV: (4,0 im)

Cu 1a. C3H5.

b. C6H10.

c. CTCT:

CH2 = C(CH3) CH = CH CH3 v CH2 = CH C(CH3) = CH CH3.

Cu 2a. X: CH3 CHCl CH3; Y: CH3 CH2 CH2Cl

b. Vit c ch gm ba bc:

Khi mo:

Pht trin mch:

Tt mch:

c. Ta c %X = (vi a l kh nng phn ng th nguyn t H ca cacbon bc nht)

Rt ra a => t l l

Bi V (4 im)

Lu : hai phn c chia ra khng bng nhau nn lu t t l k.Trng hp 1: H = 60%, CH3OH v CH3CH2OH.Trng hp 2: H = %, CH3CH2OH v CH3CH2CH2OH. (hiu sut t tnh nh, t nhin qun mt hiu sut ri, i loi c gii l ra !!!)4

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