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gii chi tit thi th i hc - m 302Cu 1: Nguyn t X c lp electron ngoi cng l lp M

nguyn t X c 3 lp electron nguyn t X chu k 3- Hp cht kh vi hiro dng XH4 nguyn t X c 4 electron ha tr X nhm IVA(Khng th l nhm B v nguyn t X ch c 3 lp electron)- Oxit cao nht c dng XO2 nguyn t X c 4 electron ha tr

X nhm IVAVy X l nguyn t Si (1s22s22p63s23p2; SiH4; SiO2). S hiu nguyn t ca Si l 14.p n ng l A

Cu 2: Cc nguyn t cng chu k (theo th t t tri sang phi): 3Li; 8O; 9FCc nguyn t cng nhm (theo th t t trn xung di): 3Li; 11NaTrong cng mt chu k, bn knh nguyn t ca cc nguyn t gim dn. Trong cng mt nhm (nhm

A) bn knh nguyn t ca cc nguyn t tng dn. Do , dy nguyn t c sp xp theo chiu bn knhnguyn t tng dn l: F, O, Li, Na.

p n ng l CCu 3: Da vo trng thi lai ha ca nguyn t trung tm v s lng obitan lai ha tham gia lin kt,c th d on c cu trc hnh hc khng gian ca phn t

- Xt cc phn t:+ BF3: T cng thc cu to phn t c 3 lin kt (zich ma), khng c cp electron cha chia B lai ha sp2 BF3 c cu trc tam gic (3 nguyn t F nm 3 nh; B nm tm ca tam gic)+ H2O: T cng thc cu to phn t c 2 lin kt (zich ma), c 2 cp electron cha chia Oc lai ha sp3. V ch c 2 AO lai ha tham gia lin kt vi 2 nguyn t H phn t H2O c cu trcgp khc (gc, ch V)+ BeH2: T CTCT phn t c 2 lin kt (zich ma), khng c cp electron cha chia Be c laiha sp phn t BeH2 c cu trc thngp n ng l B

Cu 4: Cc PTHH (ion rt gn)(1)5Fe2+ + MnO4- + 8H+ 5Fe3+ + Mn2+ + 4H2O(2)2Fe3+ + 2I- 2Fe2+ + I2T (1) Fe3+ c tnh oxi ha yu hn MnO4-T (2) I2 c tnh oxi ha yu hn Fe

3+Suy ra, tnh oxi ha MnO4

- > Fe3+ > I2p n ng l C

*Ch : Chiu ca phn ng oxi ha kh:Cht oxi ha mnh + cht kh mnh cht oxi ha yu hn + cht kh yu hn

Oxh1 + kh1 oxh2 + kh2oxh1 > oxh2; kh1 > kh2

Cu 5: S phn ng: kC2H3Cl + Cl2 C2kH3k-1Clk+1 + HCl

Theo bi ra % Cl = 35,5(k + 1) 63,96

= k = 324k + 3k - 1 + 35,5(k + 1) 100

p n ng l BCu 6: Cc kh lm nht mu dn dung dch brom (mu vng) c th do cc kh phn ng vi nc brom,to ra cc cht khng mu (khng c to ra sn phm c mu m hn), c th do cc kh (nh hi nc,dung mi hu c) pha long nc brom hoc chuyn brom vo pha dung mi hu c (dung dch Br2/dungmi hu c c mu nht hn mu dung dch Br2/nc)Xt cc phng n:

A. Loi, v CO2 khng phn ng vi nc bromB. Tha mn

SO2 + Br2 + 2H2O H2SO4 + 2HBrkhng mu

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H2S + 4Br2 + 4H2O H2SO4 + 8HBr5Cl2 + Br2 + 6H2O 10HCl + 2HBrO3

khng muC. Loi,V HI phn ng vi Br2 to ra I2 c mu m hn:

2HI + Br2 I2 + 2HBrvng nu

D. Loi, v l do nh phng n C

p n ng l B*Ch : Kh khng mu lm m dn dung dch Br2 (mu vng) l kh hiro iotua (HI). Do nc iot (to ra)c mu m hn nc bromCu 7: Phn ng dng: aA + bB ....

Biu thc tnh tc dng: v = k.(CA)x.(CB)

yTrong : k: hng s tc (hng s khi nhit khng thay i)CA; CB ln lt l nng ban u ca cht A, Bx, y: l cc s thu c t thc nghimTheo bi ra ta c: 2,0 = k.(0,5)x.(1,0)y (1)

36 = k.(1,0)x.(3,0)y (2)72 = k.(2,0)x.(3,0)y (3)

Ly (3) : (2) c: 2 = 2x x = 1

Ly (3) : (1) c: 36 = (4)1.3y y = 2Thay x = 1, y = 2 vo (1) c: 2,0 = k.(0,5)1.(1,0)2 k = 4Vy v2 = k.(0,5)

x.(2,0)y = 4.(0,5).(2,0)2 = 8 (mol. l-1.s-1)p n ng l C

Cu 8: Xt cc phng n:A. loi, v c phng trnh ion rt gn:

H+ + Cu(OH)2 Cu2+ + H2O

B. loi, v c phng trnh ion rt gn:2H+ + SO4

2 + Ba2+ + 2OH- BaSO4 + 2H2OC. loi, v c phng trnh ion rt gn:

HF + OH- F- + H2O

(HF l axit yu)D. ng: 2H+ + 2OH- 2H2Op n ng l D

Cu 9: Gi x, y ln lt l s mol Al v Fe2O3 trong XTa c: 27x + 160y = 10

Theo bi ra: NO2,24

n = = 0,1 (mol)22,4

Ta c: e (cho) = 3x e (nhn) = 0,1.3 = 0,3 (mol)

3x = 0,3 x = 0,1

mAl = 0,1.27 = 2,7 (gam) 2 3Fe O Al

= 10 - m = 10 - 2,7 = 7,3 (gam)

Vy %2 3Fe O

= 7,3

.100 = 73%10

p n ng l B

Cu 10: Theo bi ra: nNaOH = 0,075.1 = 0,075 (mol)

S phn ng: MCO3ot MO + CO2

Theo nh lut bo ton khi lng:

3 2MCO MO COm = m + m

2CO

m =3MCO MO- m = 13,4 - 6,8 = 6,6 (gam)

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2CO

n =6,6

= 0,15 (mol)44

V2

NaOH

CO

n

n=

0,075 11

0,15 2 to ra mui NaHCO3 (CO2 d)

PTHH: CO2 + NaOH NaHCO3mol 0,075 0,075Vy mmui =

3NaHCOm = 0,075.84 = 63 (gam)

p n ng l CCu 11: Nhit si ca cht hu c ph thuc vo:

+ Lin kt hiro (lin kt hiro ca axit bn hn ca ancol)+ Khi lng phn tXt cc cht:- imetyl ete (CH3OCH3, T): khng c lin kt hiro, M = 46- Ancol etylic (C2H5OH, Z): c lin kt hiro (yu hn axit), M = 46- Axit axetic (CH3COOH, Y): c lin kt hiro (mnh hn ancol), M = 60- Axit propionic (C2H5COOH, X): c lin kt hiro, M = 74Vy nhit si tng dn theo dy: T < Z < Y < X

p n ng l A*Ch : - Cht c lin kt hiro lin phn t c nhit si cao hn cht khng c lin kt hiro.

- Cc cht khng c lin kt hiro (hoc cng c lin kt hiro), cht c phn t khi ln hn th cnhit si cao hnCu 12: t hirocacbon X l CxHy (y 2x + 2). Cc PTHH xy ra:

CH42o+O

tCO2 + 2H2O

mol 0,1 0,1 0,2

CxHy2o+O

txCO2 + 2

yO

2

mol 0,2 0,2.x 0,1.yCO2 + Ca(OH)2(d) CaCO3 + H2O

mol (0,1 + 0,2.x) (0,1 + 0,2.x)Ta c:

3 2 2CaCO CO H O dd gimm - (m + m ) = m

(0,1 + 0,2.x).100 [(0,1 + 0,2.x).44 + (0,2 + 0,1.y).18] = 13,6 11,2.x 1,8y = 11,6Ta c bng:

x 1 2 3y < 0 6 12 > 2.3 + 2KL Loi C2H6 Loi

Vy X l C2H6. p n ng l A.Ch : Khi lng phn dung dch bnh ng nc vi trong gim

2 23

dd gim CO H OCaCO= m - (m + m )

Cu 13: V nhh 2, m > 3):

m 3 (C3H4) 4 (C4H6)n 4 (C4H8) 2 (C2H4)KL Khng c phng n p n A

p n ng l ACu 14: p n ng l ACu 15:

- Cht c nhit si thp nht: HCOOCH3 (este, khng c lin kt hiro)- Cht c nhit si cao nht: CH3COOH (axit cacboxylic, lin kt hiro mnh nht)- Xt cc phng n (A, B, C, D) ta thy ch c phng n D tha mn c iu nyp n ng l D

*Ch :- ost (CH3CH2CH2OH) >

o

st ((CH3)2CHOH)

- S dng phng php loi tr giiCu 16: t chy hon ton a mol axit hu c Y thu c 2a mol CO 2 trong phn t Y c 2 nguyn t C

trung ha a mol Y cn va 2a mol NaOH trong phn t Y c 2 nhm -COOH. Vy cuto ca Y l HOOC-COOH. p n ng l BCu 17: p n ng l CCu 18: gii nhanh bi ny, ta c th lp lun nh sau:

- Cc cht A, B, C tha mnX + NaOH H2N-CH2-COONa +

Ta thy trong X c 1 nguyn t N nX = 2.2N

n = 2.0,56

= 0,05 (mol)22,4

- Suy ra: S nguyn t C trong phn t X: nC = 2COX

n 3,36/22,4= = 3

n 0,05

Vy X l H2N-CH2-COO-CH3. p n ng l CCu 19: S phn ng: nC5H8+2S 5n 8n-2 2-2H C H S

% mS =2.32 2

=12.5n + 1.(8n - 2) + 32.2 100

6400 = 2.(68n + 62) 3200 = 68n + 62 68n = 3138 n = 46Vy trung bnh 46 mt xch isopren c mt cu ni isunfua (-S-S-)p n ng l B

Cu 20: S phn ng: M 2+O M2On ; M +HCl (d) MCln

Cc qu trnh xy ra:

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M Mn+ + ne16,2

(mol)M

16,2.n

(mol)M

16,2.n

e (cho) = (mol)M

O2 + 4e 2O-2

mol 0,15 0,15.4

2H+ + 2e H22.13,44

22,4

13,44

22,4

2.13,44

e (nhn) = 0,15.4 + = 1,8 (mol)22,4

Do , ta c: 16,2.n

= 1,8 16,2.n = 1,8M 9n = MM

Ta c bng:

n 1 2 3M 9 (Be) 18 27 (Al)KL Loi Loi Tha mn

(n = 1, M = 9 (Be) loi v Be ha tr n = 2)(n = 2, M = 18 khng c kim loi no tha mn)Vy M l Al. p n ng l B

Cu 21: phn bit 3 dung dch KOH, HCl, H2SO4 (l) c th dng mt thuc th l BaCO3:BaCO3 + KOH khng phn ngBaCO3 + 2HCl BaCl2 + CO2 + H2OBaCO3 + H2SO4 BaSO4 + CO2 + H2O

p n ng l CCu 22: Cc oxit ca cc kim loi c tnh kh trung bnh v yu b H

2kh thnh kim loi:

CuO + H2ot Cu + H2O

Fe2O3 + 3H2ot 2Fe + 3H2O

ZnO + H2ot Zn + H2O

Cc oxit ca cc kim loi c tnh kh mnh (kim loi kim, kim th, nhm) H2 khng kh c thnhkim loi:

MgO + H2ot khng xy ra

Sau phn ng hn hp rn cn li c Cu, Fe, Zn, MgOp n ng l B

Cu 23: C th dng thuc th: qu tm- Cho qu tm vo cc mu th:

+ Mu lm qu tm ha : NH4Cl, (NH4)2SO4 (nhm I)+ Mu lm qu tm ha xanh: NaOH, Na2CO3 (nhm II)+ Mu khng lm i mu qu tm: BaCl2

- Cho BaCl2 vo mu th nhm I:+ Mu to ra kt ta trng l (NH4)2SO4

BaCl2 + (NH4)2SO4 BaSO4 + 2NH4ClBaCl2 + NH4Cl

+ Mu cn li cha NH4Cl- Cho BaCl2 vo 2 mu th nhm II:

+ Mu to ra kt ta trng l Na2CO3:

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BaCl2 + Na2CO3 BaCO3 + 2NaClBaCl2 + NaOH

+ Mu cn li l NaOHp n ng l B

Cu 24: Ch c s dng nhiu trong cng nghip ch to cquy (lm in cc trong cquy ch). p nng l A*Ch : cquy ch: Loi cquy ph bin nht

-

Anot lm bng cc li hp kim ch ng bt ch xp, mn- Catot lm bng cc li hp kim ch ng ch (IV) oxit PbO2- Cht in li l dung dch H2SO4 38%Cc phn ng in cc xy ra khi cquy phng in: anot: Pb(r) + SO4

2 PbSO4 + 2e catot: PbO2(r) + SO4

2 + 4H+ + 2e PbSO4 + 2H2OPhn ng tng: Pb + PbO2 + 2H2SO4 2PbSO4 + 2H2OSau mt thi gian hot ng, sut in ng gim, lc dng dng in mt chiu ngoi npin, ngha l thc hin phn ng oxi ha kh ngc li:

2PbSO4(r) + 2H2O dng in Pb + PbO2 + 2H2SO4

Khi np in qu nhanh, c th xut hin H2 v O2 in cc

anot: 2H2O O2 + 4H+

+ 4e catot: 2H+ + 2e H2 V c th gy n do: 2H2 + O2 2H2O

Cu 25: Theo bi ra, ta c: P + E + N = 43V P = E 2P + N = 43 N = 43 2P (1)

Mt khc: 1 N

P 1,5

N P

N 1,5P (2)

T (1), (2) ta c: 43 2P P 43 3P P 14,3T (1), (3) ta c: 43 2P 1,5P 43 3,5P P 12,3Do : 12,3 P 14,3 P = 13 hoc P = 14 (P N*)

+ Nu P = 13: 1s22s22p63s23p1 (kim loi) c 3 electron ha tr

+ Nu P = 14: 1s22s22p63s23p2 (Si, phi kim, c 4 electron ha tr loi v phi kim)p n ng l C

Cu 26: Quy tc Hun: Trong cng mt phn lp, cc electron c in vo cc obitan sao cho tng selectron c thn l cc i

Cu hnh electron vi phm quy tc Hun: 1s12s2p2x2p1

yPhi l 1s12s22p1x2p

1y2pz

1. p n ng l ACh : Gi thit xem cc electron ln lt c in vo cc obitan npx, npy v npz

Cu 27: Trong tinh th kim cng, nguyn t C c lai ha sp3 do :+ Mi nguyn t C c s nguyn t ln cn gn nht l 4 (v 4 AO lai ha sp 3 hon ton nh nhau s

lin kt vi 4 nguyn t C khc, 4 nguyn t C ny nm 4 nh ca mt t din u)

+ Gc lin kt CCC = 109o28 (bng gc gia cc AO lai ha sp3)

p n ng l DCu 28: Xt M + 2E+ M2+ + 2E(xy ra trong dung dch nc; M, E l cc kim loi)Da vo nguyn tc (chiu) ca phn ng oxi ha kh:

Oxh1 + kh1 Oxh2 + kh2Oxh1 > Oxh2kh1 > kh2

Suy ra: + M c tnh kh mnh hn E+ E+ c tnh oxi ha mnh hn M2+

Lu : + M, E l cc kim loi Ch c tnh kh (khng c tnh oxi ha)

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+ E+, M2+ l cc ion kim loi (c mt ha tr duy nht) ch c tnh oxi haNi dung ng l: Kim loi M c tnh kh mnh hn kim loi E. (Kim loi M y c kim

loi E ra khi dung dch mui). p n ng l B

Cu 29: Theo bi ra: nCu = NaOH (d)0,32

= 0,05 (mol); n = 0,05.2 = 0,01 (mol)64

CuCl2pdd Cu + Cl2

mol 0,005 0,005

Cl2 + 2NaOH NaCl + NaClO + H2Omol 0,005 0,01

NaOH (banu) NaOH (p/) NaOH (d)n = n + n = 0,01 + 0,01 = 0,02 (mol)

Vy CNaOH(ban u) =0,02

= 0,1 M0,2

. p n ng l C

Cu 30: Sc kh HI (khng mu) vo dung dch Br2 (mu vng) sau mt thi gian thy dung dch m muhn:

2HI + Br2 2HBr + I2khng mu Vng khng mu nu m

p n ng l B

Cu 31: Theo nh lut tc dng khi lng: V = k.(CA).(CB)

x

Khi tng nng A, B ln gp i th:v = k.(2.CA).(2.CB)

x = 2.2x.k.(CA).(CB)x = 2x+1.v

Theo bi ra: v/v = 16 2x+1 = 16 = 24 x = 3p n ng l C

Cu 32: p n ng l BCu 33: PTHH:

oxt, t

6 7 2 3 3 () 6 7 2 2 3 2n nC H O (OH) + 3nHNO C H O (ONO ) + 3nH O

gam 162.n 3n.63gam 8,1 a

a = 3

HNO

8,1.3n.63 9,45= 9,45 (gam) n = x = = 0,15 (mol)

162.n 63

p n ng l BCu 34: Cc PTHH xy ra:

Ba + 2H2O Ba(OH)2 + H22Al + Ba(OH)2 + 6H2O Ba[Al(OH)4]2 + 3H2 Dung dch X cha Ba[Al(OH)4]2; Ba(OH)2 dCO2 + Ba(OH)2 BaCO3 + H2OCO2(d) + BaCO3 + H2O Ba(HCO3)2(tan)2CO2 + Ba[Al(OH)4]2 2Al(OH)3 + Ba(HCO3)2Vy sau phn ng ch thu c Al(OH)3 kt ta. p n ng l B

Cu 35: Theo bi ra:2 2CO H O

n = 17,6/44 = 0,4 (mol); n = 108/18 = 0,6 (mol)

PT t chy (tng qut): CxHy 2+O xCO2 + 2

yO

2

Theo nh lut bo ton nguyn t:

X Y 2 2C H C H C (CO ) H (H O)m = m + m = m + m

= nC. 12 + nH.1 =2CO

n .12 + 2.2H O

n .1 = 0,4.12 + 2.0,6.1 = 6,0 (gam)

p n ng l B

Cu 36: PTP: CxHy2o+O

txCO2 + 2

yO

2

V nguyn t C, H c bo ton nn:

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x y x y x y 2 2C (C H ) H (C H ) C H C(CO ) H(H O)m + m = m = m + m

m =11 9

.12 + .2.1 = 4,0 (gam)44 18

. p n ng l A

Cu 37: Theo bi ra: nhh = 10,752/22,4 = 0,48 (mol)Ta c:

2C CO hhn = n /n = 1,92/0,48 = 4

Suy ra, trong phn t anken v ankaien u c 4 nguyn t cacbon

CTPT ca C4H8 v C4H6. p n ng l BCu 38: p n ng l CCu 39: Hp cht a chc l nhng hp cht c t hai hay nhiu nhm chc ging nhau

- C2H6O2 ancol a chc: HOCH2-CH2OH- C2H2O2 (n = 2) anehit a chc: OHC-CHO: anehit oxalic- C2H2O3: khng c hp cht a chc, bn, mch h- C3H6O2 (n = 1): khng c hp cht a chc, bn, mch hVy c tt c 2 cht. p n ng l B

*Ch : Cc cht nh CH3CH(OH)2, HO-CH2-CH=CH-OH, khng bn, -OHC-COOH lhp cht hu c tp phc

Cu 40: Theo bi ra: etyl axetat NaOH8,8

n = = 0,1 (mol); n = 0,2.0,2 = 0,04 (mol)8,8

PTP: CH3COOC2H5(l) + NaOH (r) CH3COONa (r) + C2H5OH (l)Ban u: 0,1 0,04Phn ng: 0,04 0,04 0,04 0,04Sau phn ng: 0,04Cht rn thu c sau phn ng l: CH3COONa (C2H5OH, CH3COOC2H5 khi c cn b bay hi)Vy m = 0,04.82 = 3,28 (gam). p n ng l C

Cu 41: p n ng l DCu 42: Cc dung dch c pH < 7 (mi trng axit):

1. C6H5NH3Cl2. ClH3N-CH2-COOH3. HOOC-CH2-CH2-CH(NH2)-COOHC 3 dung dch c pH < 7p n ng l C

Cu 43: p n ng l BCu 44:

Nhn ra Ag v ch c Ag khng tc dng vi H2SO4 long- Nhn ra Ba v c kh bay ln v c kt ta trng:

Ba + H2SO4 BaSO4 + H2 - Cc kim loi cn li (Mg, Al, Fe) tc dng vi H2SO4 long, u c kh bay ln:

Mg + H2SO4 MgSO4 + H2 2Al+ 3H2SO4 Al2(SO4)3 + 3H2 Fe + H2SO4 FeSO4 + H2

-

Cho Ba d tc dng vi H2SO4 long:Ba + H2SO4 BaSO4 + H2 Ba + 2H2O Ba(OH)2 + H2

Lc tch kt ta BaSO4, c dung dch Ba(OH)2 (ly lm thuc th nhn bit cc dung dch cn li)Cho t t dung dch Ba(OH)2 ti d vo dung dch MgSO4, Al2(SO4)3, FeSO4:

MgSO4 + Ba(OH)2 Mg(OH)2 + BaSO4 Trng Trng

Al2(SO4)3 + 3Ba(OH)2 2Al(OH)3 + 3BaSO4

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Al(OH)3 + Ba(OH)2 Ba[Al(OH)4]2Tan

(to kt ta sau kt ta b tan bt)FeSO4 + Ba(OH)2 Fe(OH)2 + BaSO4

Trng xanh TrngNh vy c th nhn bit c tt c cc cht

p n ng l D

Cu 45: Gi s mol mi cht l xCc PTHH xy ra:Na2O + H2O 2NaOH

mol x 2xNaHCO3 + NaOH Na2CO3 + H2O

mol x x xNa2CO3 + BaCl2 BaCO3 + 2NaCl

mol x x 2xNaOH + NH4Cl NH3 + NaCl + H2O

mol (2x x) x xVy dung dch thu c ch cha NaCl. p n ng l BCu 46: Chiu gim tnh oxi ha ca cc ion l

Ag+ > Fe3+ > Cu2+ > Fe2+p n ng l B

Cu 47: PTHH: Fe + HNO3 (c, ngui) khng phn ngFe2O3 + 6HNO3 2Fe(NO3)3 + 3H2OFe3O4 + 10HNO3(c) 3Fe(NO3)3 + NO2 + 5H2O

NuFeO + 4HNO3(c) Fe(NO3)3 + NO2 + 2H2O

Suy ra, c th nhn ra c Fe v Fe2O3. p n ng l CCu 48: Trong pin kh Lclngs, v km ng vai tr cc m (anot):

Zn Zn2+ + 2e (phn ng in cc anot)

p n ng l CCu 49: Da vo thay i s oxi ha ca cc nguyn t ca cc cht trong s phn ng phn ng oxiha - kh hay khng

- Cc cht c s oxi ha thp phn ng vi HNO 3 c, nng; H2SO4 c, nng u l phn ng oxi ha - kh- Phn ng c s tham gia ca cc n cht u l phn ng oxi ha - kh- Phn ng to ra n cht t hp cht phn ng oxi ha khp n ng l D

*Ch : Hon thnh cc PTHH trn:a. FeO + 4HNO3 (c, nng) Fe(NO3)3 + NO2 + 2H2O

b. 2FeS + 10H2SO4 (c, nng) Fe2(SO4)3 + 9SO2 + 10H2Oc. Al2O3 + 6HNO3 (c, nng) 2Al(NO3)3 + 3H2Od. Cu + 2FeCl3 (dd) CuCl2 + 2FeCl2

e. CH3CHO + H2oNi, t CH3CH2OH

f. CH2OH(CHOH)4CHO + Ag2O 3Mt CH2OH(CHOH)4COOH + 2Ag

g. CH2=CH2 + Br2 CH2Br-CH2Brh. C3H7O2OH + HO-Cu-OH + HOO2H7C3 (C3H7O2O)2Cu + 2H2OPhn ng to phc u khngphi l phn ng oxi ha - kh

Cu 50: PTHH:Cl2 + 2KOH KCl + KClO + H2O

mol a a

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3Cl2 + 6KOH 5KCl + KClO3 + 3H2O

mol b5b

2

Theo bi ra: a =5b

3 3a = 5b

b 3= = 0,6

a 5

p n ng l BCu 51: Xt cc phng n:

A. ngB. Sai, v: Br2 + 5Cl2 +6H2O 10HCl + 2HBrO3

(Cl2 c tnh oxi ha mnh hn Br2)

C. Sai, V: 3Cl2 + 6NaOHo100 C 5NaCl + NaClO3 + 3H2O

D. Sai, v I2 + H2 2HI (thun nghch, v HI km bn)p n ng l A

Cu 52: Cht bt dng thu gom thy ngn (c hi) b ri vi l bt lu hunh (mu vng nht, cn gi ldim sinh)

PTHH: Hg (l) + S (r) HgS (rn)p n ng l D

Cu 53: t k hiu chung cho cc kim loi l M, c ha tr n.M + 2nHNO3 M(NO3)n + nNO2 + nH2O2M + 2nH2SO4 M2(SO4)n + nSO2 + 2nH2O

Nhn xt:-

23

2-24

NONO

SOSO

n (mui nitrat) = n = 0,1 (mol)

n = n = 0,02 (mol)

Vy m = mkim loi + - 2-3 4NO SO(mui) + m (mui)

= 6 + 0,1.62 + 0,02.96 = 14,12 (gam)p n ng l A

Cu 54: Trong phng php phn tch chun , mi php chun tin hnh ba ln ly kt qu trungbnh ca ba ln

p n ng l ACu 55: BTX gm: Benzen, Toluen v Xilen etilen khng thuc nhm BTX. p n ng l C

Cu 56: Than hot tnh c cu to rt xp, nn c kh nng hp th mnh cc cht kh v cht than trongdung dch

Than hot tnh c dng nhiu trong mt n phng c...p n ng l A

Cu 57:

2Ba(OH)n = 0,1.0,125 = 0,0125 (mol)

nHCl = 0,4.0,05 = 0,02 (mol)PTHH: Ba(OH)2 + 2HCl BaCl2 + 2H2OBan u: 0,0125 (mol) 0,02 (mol)Phn ng: 0,01 (mol) 0,02 (mol)Sau phn ng 0,0025 (mol) 0

Theo PTHH:2Ba(OH) HCl

1n (p/) = n = 0,01 (mol)

2

2Ba(OH)

n (d) = 0,0125 - 0,01 = 0,0025 (mol)

Ba(OH)2 Ba2+ + 2OH-

0,0025 0,005 (mol)

[OH] (d) = -20,005

= 0,01 = 10 (M)0,1 + 0,4

7/30/2019 Gii chi tit 302

11/11

11

pOH = -lg (10-2) = 2 pH = 14 - 2 = 12p n ng l D

Cu 58: Theo bi ra : nNO =1,344

= 0,06 (mol)22,4

Ta c:2.(11,36 - 56.x)

3x = + 0,06.316

24x = 11,36 - 56x + 1,44 80x = 12,8 x = 0,16 (mol)

Vy 3 3Fe(NO ) Fen = n = x = 0,16 (mol) 3 3Fe(NO ) = 0,16.242 = 38,72 (gam) p n ng l A

Cu 59: V X, Y, Z l 3 hirocacbon k tip nhau trong dy ng ng, nn:MY = MX + 14MZ = MY + 14 = MX + 28Theo bi ra: MZ = 2.MX MX + 28 = 2.MX MX = 28 (C2H4) Y l C3H6

C3H6 + 2 2 29

O 3CO + 3H O2

0,1 (mol) 0,3 (mol)CO2 + Ca(OH)2 (d) CaCO3 + H2O0,3 (mol) 0,3 (mol)

3CaCO= 0,3.100 = 30 (gam)

p n ng l ACu 60: iu kin c ng phn hnh hc ca ankaien (c 2 ni i trong phn t hirocacbon) cng gingnh ca anken: nguyn t C ni i phi lin kt vi cc nguyn t hay nhm nguyn t khc nhau

Xt cc cht:+ CH2=CH-CH2-CH2-CH=CH2 loi+ CH2=CH-CH=CH-CH2-CH3 tha mn+ CH3-C(CH3)=CH-CH3 loi

+ CH2=CH-CH2-CH=CH2 loiVy ch c 1 cht c ng phn hnh hcp n ng l B