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7/29/2019 DE thi MTCT mon Hoa hoc

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Thi HSG Gii Ton Trn My TinhTnh Thi Bnh Nm 2012

Cu 1. Mt hp cht A ( M2X) cu to t ion M+ v X2-. Trong phn t A c tng s ht (e,n,p) l 140 ht,trong s ht mang in nhiu hn ht khng mang in l 44 ht. S khi ca ion M + ln hn s khi ion X2-

l 23. Tng s ht e,n,p trong ion M+ nhiu hn trong ion X2- l 31 ht. Xc nh CTPT ca A?Hng dn chm

Cch gii imHDG: Gi Z v N ln lt l s hiu nguyn t v s notron ca nguyn t M

Z v N ln lt s hiu nguyn t v s notron ca nguyn t X

1

Theo bi ra ta c:4Z + 2N + 2Z + N = 140 (1)(4Z + 2Z ) (2N + N) = 44 ( 2)( Z + N ) ( Z + N)= 23 (3)( 2Z 1 + N) ( 2Z + 2+ N) = 31

=> 2Z + N 2Z N = 34 (4)

2

Gii h => Z = 19, Z = 8; CTPT A l K2O 2

Cu 2. Tnh bn knh nguyn t Mg. Bit rng khi lng ring ca Mg l 1,74g/cm3, th tch cc qu cuchim 74% th tch ca ton mng tinh th v Mg = 24,31

Hng dn chmCch gii im

HDG: Vtt ( 1 mol Mg) =M 74

.d 100

=24,31 74

.1,74 100

1,5

V ( 1 nguyn t Mg ) = 23Vtt

6,023.10= 23

24,31.74

1,74.6,023.10 .100

1,5

T VMg =4

3 R3 R = 1,6.10-8cm = 16Ao

2,0

Cu 3. ha tan ht mt mu Zn trong dd HCl 20oC cn 27 pht. Cng mu Zn tan ht trong dd axit ni

trn nhit 40oC trong 3 pht. Hi ha tan ht mu Zn trong dd axit ni trn 65oC th cn bao nhiugiy?Hng dn chm

Cch gii im

HDG: Ta c40

20

V

V=

27

3= 9

1,0

Mt khc: V40/V20 =t

10kV

= k2 = 32 => k = 3

( tc l khi tng nhit thm 10oC tc phn ng tng 3 ln)

1,5

=>65

40

v

v

=t

10k

= k2,5 = 32,51,5

Do thi gian ha tan hon ton Zn 65oC l t = 3.60/32,5 = 11,547 giy 1,0

Cu 4. Khi trn 1,0 mol CH3COOH vi 1,0 mol C2H5OH v phn ng xy, lc cn bng ngi a thy tothnh 2/3 mol etyl axetat. Nu trn 1,0 mol CH3COOH vi 3,0 mol C2H5OH th lc cn bng s to ra baonhiu mol etyl axetat?Hng dn chm

Cch Gii imHDG: t th tch ca h l V( lt)- Khi trn 1 mol CH3COOH vi 1 mol C2H5OH

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CH3COOH + C2H5OH CH3COOC2H5 + H2O

[]cb 1/3V 1/3V 2/3V 2/3V

Ta c Kc =

2 2.

3V 3V

1 1.

3V 3V

= 4

Khi trn 1 mol CH3COOH vi 3 mol C2H5OH, gi x l s mol este to thnh

CH3COOH + C2H5OH

CH3COOC2H5 + H2O[]cb (1-x)/V (3-x)/V x/V, x/V

Ta c Kc =

x x.

V ' V '(1 x) (3 x)

.V ' V '

= 4 => 3x2 -16x + 12 = 0

2,0

Gii phng trnh ta c x1 = 4,4305 v x2 = 0,9028Do lng este khng th ln hn lng axit ban u nn x = 0,9028mol

1,0

Cu 5. Axit sunfuric l mt axit mnh nc in ly th nht, nc in ly th 2 c hng s in ly bng 10 -2.Tnh pH ca dd H2SO4 10-3M?

Hng dn chmCch gii im

HDG: Nc 1 in ly hon tonH2SO4 H+ + HSO4-

[]cb 10-3 10-3 10-3

1,0

S in ly ca nc s 2: HSO4- H+ + SO42-

[]cb 10-3 x (10-3 + x) x

Ta c Kc =3

3

(10 x).x

(10 x)

+

= 10-2 => x2 + 0,011x 10-5 = 0

2,0

Gii phng trnh bc 2, ta c x1 = 8,44.10-4 v x2 = -0,012( loi)

=> [H+] = 10-3 + 8,44.10-4 = 1,844.10-3=> pH = 2,7342

2,0

Cu 6. X l hh hai axit gm 2 axit hu c n chc mch h lin tip nhu trong dy ng ng v ancoletylic. Chia X thnh hai phn bng nhau. Phn 1 tc dng vi kim loi Na d thu c 3,92 lt H2 ( kct). Phn2 t chy hon ton cn va 25,2 lt O2 ( ktc), sn phm chy cho qua bnh ng dd Ba(OH)2 d thy khilng bnh tng 56,7g v trong bnh c 177,3g kt ta. Xc nh CTcc axit trong hh X ?Hng dn chm

Cch gii imHDG: Gi a v b l mol ca 2 axit v C2H5OH trong 1.2 hh XCTPT TB cu 2 axit l CxHyO2 hoc RCOOH

PTP: 4CxHyO2 + ( 4x + y 4) O2 4xCO2 + 2yH2OC2H5OH + 3O2 2CO2 + 3H2OCO2 + Ba(OH)2 BaCO3 + H2O

Phn ng th: RCOOH + Na RCOONa + 1/2H2C2H5OH + Na C2H5ONa + 1/2H2

1,0

S mol oxi = 1,125 mol => ( x + y/4 1).a + 3b = 1,125 (1)S mol CO2 = s mol BaCO3 = 0,9 => ax + 2b = 0,9 (2)S mol H2O = ( 56,7 -0,9.44) : 18 = 0,95=> ya + 6b = 1,9 (3)S mol H2 = 0,175 => a + b = 0,35

2,0


3/4

Gii h 1, 2, 3, 4 thu c a = 0,15; b = 0,2mol. x = 3,33333, y = 4,6667Vy CTPT ca hai axit l C3H4O2 v C4H6O2

2,0

Bi ny c th p dng bo ton nguyn t cho oxi 2a + b + 2.nO 2 = 2.nCO2 + nH2Oa + b = 0,35.

Tm c a, b s c KLPT TB ca hai axit => kt qu

Cu 7. x phng ha hon ton 19,4g hh hai este n chc, mch h A v B cn 200ml dd NaOH 1,5M.Sau khi phn ng xy ra hon ton c cn dd thu c hh hai ancol ng ng k tip nhau v mt mui khan

duy nht. Tm CTCT v tnh % khi lng mi este trong hh ban u?Hng dn chm

Cch gii imHDG: Gi CTPT TB ca 2 este l RCOOR RCOOR + NaOH RCOONa + ROH

1,0

Ta c nNaOH = nRCOOR = 0,2.1,5 = 0,3=> R + 44 + R = 64,67

R + R = 20,67. Vy hai anol c CTCT l CH3OH v C2H5OH axit HCOOHVy CTCT 2 este l HCOOCH3 v HCOOC2H5

2,0

Gi x v y l s mol ca HCOOCH3 v HCOOC2H5 trong hh

x + y = 0,3 v 60x + 74y = 19,6.Gii h =>. x = 0,2, y = 0,1%mHCOOCH3 = 61,85%. v %mHCOOC2H5 = 38,15%

2,0

Cu 8. Cho 15,28g hh X gm Cu v Fe vo 1,1 lt dd Fe2(SO4)3 0,2M. Phn ng kt thc thu c dd Y v1,92g cht rn Z. Cho Z tc dng vi dd H2SO4 long khng c kh thot ra. Tnh khi lng cc cht 15,28ghh X ?Hng dn chm

Cch gii imHDG : Gi x v y l s mol ca Cu v Fe c trong 15,28g hh X

S mol Fe2(SO4)3 = 0,22 molCc ptp Fe2(SO4)3 + Fe 3FeSO4Fe2(SO4)3 + Cu 2FeSO4 + CuSO4

Cho Z vo dd H2SO4 long khng c kh bay ra, vy Z l Cu cn d.

2,0

=> n Cud = 1,92 : 64 = 0,03mol => nCu p = ( x 0,03)Theo cc ptp ta c x + y 0,03 = 0,22 => x + y = 0,25 (1)Theo bi ra ta c 64x + 56y = 15,28 (2)Gii h thu c x = 0,16mol v y = 0,09mol

2,0

=> mCu = 10,24g. mFe = 5,04g 1,0

Cu 9. Hh X gm Zn, Fe v Cu. Cho 46,25g hh X tc dng vi dd H2SO4 long d thu c 11,2 lt kh

H2(ktc). Mt khc bit 1,5mol hh X phn ng va vi 39,2 lt kh Cl2(ktc). Tnh % khi lng mi kimloi trong hh X?Hng dn chm

Cch gii imHDG: Gi x, y, z l s mol ca Zn, Fe, Cu trong 46,25g hh X

Gi s s mol cc cht c trong 1,5mol hh X gp k ln s mol cc cht trong 46,25g hh XS mol H2 = 0,5molS mol Cl2 = 1,75mol

Cc ptp xy ra:Zn + H2SO4 ZnSO4 + H2 (1)Fe + H2SO4 FeSO4 + H2 (2)

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Cu + H2SO4 khng phn ngZn + Cl2 ZnCl2 (3)2Fe + 3Cl2 2FeCl3 (4)Cu + Cl2 CuCl2 (5)Theo pt v bi ta c65x + 56y + 64z = 46,25 (6)x + y = 0,5 (7)k(x + y + z ) = 1,5 (8)

k(x + 1,5y + z ) = 1,75 (9)

2,0

Gii h pt ta c x = y = z = 0,25=> %mZn = 35,13% ; %mFe = 30,27% ; %mCu = 34,6%

1,5

Cu 10. Tin hnh in phn ( vi in cc tr, mng ngn xp)500ml dd X ( HCl 0,02M v NaCl 0,2M). Saukhi anot thot ra 0,448 lt kh ( ktc) th ngng in phn.

1) Tnh nng mol cc cht trong dd sau in phn2) Nu thi gian in phn l 24 pht, hiu sut in phn khng i bng 80% th cng dng in

cn dng bng bao nhiu?Hng dn chm

Cch gii im

1) Khi in phn dd X, HCl b in phn trc2HCl H2 + Cl2 (1)Khi HCl b in phn ht, NaCl b in phn

2NaCl + H2O H2 + Cl2 + 2NaOH (2)V nu NaCl ht th nc b in phn

1,0

Ta thy s mol HCl = 0,01molV s mol Cl2 ti a thot ra (1) = 0,01:2 = 0,005mol < s mol Cl2 ( anot) = 0,02 mol=> Chng t (2) xy ra.S mol NaCl = 0,5.0,1 = 0,1molGi s, NaCl in phn ht, theo(2) => s mol Cl2 = 0,05 => s mol kh to ra sau (1) = 0,02-0,005 = 0,015

=> Chng t NaCl cha in phn ht

2,0

Theo (2) S mol NaClp =s mol NaOH = 2.0,015 = 0,03mol=> Nng NaCl trong dd sau in phn l: (0,1 0,03) : 0,5 =0,14M=> Nng NaOH trong dd sau in phn l: 0,03 : 0,5 = 0,14M

1,0

2) Tnh cng dng inTheo nh lut Faray ta c

mCl2 =AIt

Fn.h% => I =

71.0,02.96500.2.100

71.24.60.80= 3,3507A

1,0

C th s dng nh lut bo ton electron v in tch

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