dap an vao chuyen hoa 10 dai hoc su pham ha noi 2011

8/6/2019 Dap an Vao Chuyen Hoa 10 Dai Hoc Su Pham Ha Noi 2011

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B GIO DC V O TO CNG HA X HI CH NGHA VIT NAMTRNG I HC S PHM H NI

---------------------------------c lp Tdo Hnh phc

----------------------------

P Nv THANG IM Mn thi: Ha HcVO KHI TRUNG HC PH THNG CHUYN 2011

Ni dung imCu I. (2,75 im)1. (1,0 im)- Dung dch X c khnng lm mt mu nc brom v lm mt mu thuc tm,vy trong dung dch X c mui st (II).- Dung dch X c khnng ha tan bt ng, vy trong dung dch X c mui st(III).Kt lun: oxit st l Fe3O4.Cc phng trnh phn ng:Fe3O4 + 4H2SO4 (long) FeSO4 + Fe2(SO4)3 + 4H2O6FeSO4 + 3Br2 2Fe2(SO4)3 + 2FeBr3(hoc: 2FeSO4 + 3Br2 + H2SO4 Fe2(SO4)3 + 2HBr)10FeSO4 + 2KMnO4 +8H2SO4 5Fe2(SO4)3 + 2MnSO4 + K2SO4 + 8H2OFe2(SO4)3 + Cu 2FeSO4 + CuSO43Fe3O4 + 28HNO3 (long) 9Fe(NO3)3 + NO + 14H2O2Fe3O4 + 10H2SO4(c nng) 3Fe2(SO4)3 + SO2 + 10H2O

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2. (0,75 im)Cch 1:

CuO + 2HCl CuCl2 + H2O- Cho hn hp tc dng vi dung dch HCl d:

Fe2O3 + 6HCl 2FeCl3 + 3H2O- Ly dung dch thu c cho tc dng vi Al d:3CuCl2 + 2Al 3Cu + 2AlCl3FeCl3 + Al Fe + AlCl3

- Lc ly cht rn cho tc dng vi HCl d, Al, Fe tan, cn li Cu.Fe + 2HCl FeCl2 + H2Al + 3HCl AlCl3 + 3/2H2

Cch 2:

3CuO + 2Al- Nhit nhm hn hp oxit vi Al d:

ot3Cu + Al2O3

Fe2O3 + 2Al

ot

2Fe + Al2O3- Ly hn hp rn ha tan trong dd HCl d, lc c Cu.Al2O3 + 6HCl 2AlCl3 + 3H2OFe + 2HCl FeCl2 + H2Al + 3HCl AlCl3 + 3/2H2

Cch 3:

Al + 3HCl AlCl3 + 3/2H2- Cho Al tc dng vi HCl, ly kh H2:

- Ly kh H2 kh hn hp oxit, sau ha tan sn phm trong HCl d c Cu:CuO + H2

otCu + H2O

Fe2O3 + 3H2ot

2Fe + 3H2OFe + 2HCl FeCl2 + H2

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0,253. (1,0 im)


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Xc nh c dung dch B l Ba(OH)2 do c phn ng vi CO2Ba(OH)2 + CO2 BaCO3 + H2O- Cho BaO vo dung dch H2SO4 long:BaO + H2SO4 BaSO4 + H2OBaO + H2O Ba(OH)2

Kt ta A: BaSO4- Cho Al d vo dung dch B:Ba(OH)2 + 2Al + 2H2O Ba(AlO2)2 + 3H2Kh E: H2, dung dch D: Ba(AlO2)2- Ly dung dch D cho tc dng vi dung dch Na2CO3:Ba(AlO2)2 + Na2CO3 BaCO3 + 2NaAlO2Kt ta F: BaCO3.

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Cu II. (2,25 im)1. (0,75 im)

Xc nh cu to ca A.nA= 14,4/144 =0,1 mol, nNaOH = 0,1.2 = 0,2 molTa c nA: nNaOH = 1: 2 v A + NaOH 1 mui + 1 ru=> A este 2 chc ca mt ru v mt axit => A c dng:R1-COOR2OOC-R1 (1); R1-OOC-R2-COOR1 (2);O C R1

O

C

R2

O

O (3)TH1: Cng thc (1) hoc (2): 2R1 + R2 = 144-88 = 56, ng vi C4H8, s cacbontrong R1, R2 bng nhau, c mt gc cha lin kt i; vy khng ph hp (3 gc

c s C bng nhau m tng bng 4C)TH2:

O C CH2

O

C

CH2

O

O

CH2

CH2

Cng thc (3): R1+ R2= 56 tng ng -C4H8- ; s cacbon trong R1, R2bng nhau. Vy chn R1 = R2 = C2H4

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2. (1,5 im)

danken/ankan = =Mankan

Manken2,625 Manken = 2,625.Mankan

iu kin thng hai hirocacbon l cht kh nn s nguyn t cacbon 4

Manken MC4H8 2,625.Mankan 56 Mankan 21,33 ankan l CH4:M = 16

Manken = 2,625.16 = 42 anken l C3H6

b. Cc phng trnh iu ch

CH2=CH-CH3 + H2ot

CH3-CH2-CH3

CH3-CH2-CH3ot

CH2=CH2 + CH4

1,0

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Cu III. (2,5 im)

1. (1,25 im)a. Gi x, y l s mol ca Na v Al trong m gam hn hp.

- Khi cho m m gam A tc dng vi nc d c cht rn khng tan, vy Al d.Na + H2O NaOH + 1/2H2x x 0,5xAl + NaOH + H2O NaAlO2 + 3/2H2x x x 1,5x 0,5x + 1,5x = 1,344/22,4 = 0,06 x = 0,03.- Cho 2m gam A tc dng vi dung dch NaOH d to ra: 20,832 lt H2- Cho m gam A tc dng vi dung dch NaOH d to ra: 20,832/2=10,416 lt H2Na + H2O NaOH + 1/2H20,03 0,03 0,015Al + NaOH + H2O NaAlO2 + 3/2H2y y y 1,5y 0,015 + 1,5y = 10,416 /(22,4) = 0,465 y = 0,3.Vy: mNa = 0,03.23 = 0,69g.

mAl = 0,3.27 = 8,1g.b. Dung dch B c NaAlO2 = 0,03 mol. nAl(OH)3 = 0,78/78 = 0,01mol.TH1: NaAlO2d.HCl + NaAlO2 + H2O Al(OH)3 + NaCl0,01 0,01 0,01

M(HCl) 0,01C 0,2M.0,05

= =

TH2: NaAlO2 ht.HCl + NaAlO2 + H2O Al(OH)3 + NaCl0,01 0,01 0,014HCl + NaAlO2 AlCl3 + NaCl + 2H2O0,08 0,02

M(HCl)0,09

C 1,8M.0,05

= =

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2. (1,25 im)V CO ch khc nhng oxit kim loi ng sau Al trong dy hot ng hahc ca kim loi, nn c 2 trng hp xy ra.

a. Trng hp 1: Kim loi phi tm ng sau Al trong dy hot ng ha hcca kim loi, oxit ca n b CO kh.

CuO + CO Cu + CO2 (1)Mol x x

MO + CO M + CO2 (2)Mol 2x 2x

3Cu + 8HNO3 3Cu(NO3)2 + 2NO + 4H2O (3)

x 8x/33M + 8HNO3 3M(NO3)2 + 2NO + 4H2O (4) 0,25


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2x 16x/3Ta c h : 80x + (M + 16).2x = 1,2

3

8x+

16

3

x

= 0,05 gii h cho x = 0,00625 v M = 40 (Ca).

Trng hp ny khng tho mn v Canxi ng trc Al trong dy

hot ng ha hc ca kim loi v CaO khng b kh bi CO.b. Trng hp 2 : Kim loi phi tm ng trc Al trong dy hot ng ha hcca kim loi v oxit ca n khng b CO kh. Do :

CuO + CO Cu + CO2Mol x x

MOMol 2x

3Cu + 8HNO3 3Cu(NO3)2 + 2NO + 4H2Ox 8x/3 2x/3

MO + 2HNO3 M(NO3)2 + H2O

2x 4x80x + (M + 16).2x = 1,28x

3+ 4x = 0,05 x = 0,0075 M = 24 (Mg) (tho mn)

Vy: M l Mg, V = (2.0,015/3).22,4 = 0,224 lt.

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0,25Cu IV. (2,5 im)1.(1,25 im)728 ml hn hp A:

nA = 0,728/22,4 = 0,0325 mol.Khi cho A qua dung dchbrom d c kh bay ra, l ankan (CnH2n+2).

n 2 n 2C Hn 0, 448 / 22, 4 0, 02 mol

+

= =

hidrocacbon khng non 0,0325 0,02 0,0125= =

2Brn 2 / 160 0, 0125= =

Ta c: nhidrocacbon khng no:2Br

n =1:1 hidrocacbon khng no l anken (CmH2m).

t chy hon ton 1456 ml hn hp A:

CnH2n+2 +3n 1

2

+O2 nCO2 + (n+1)H2O

0,04 0,04nCmH2m +

3m

2O2 mCO2 + mH2O

0,025 0,025mCa(OH)2 + CO2 CaCO3 + H2O0,075 0,075 0,075Ca(OH)2 + 2CO2 Ca(HCO3)20,04 0,08 0,04

Ca(HCO3)2ot

CaCO3 + CO2 + H2O

0,04 0,04 0,04n + 0,025m = 0,075 + 0,08 = 0,155 4n + 2,5m = 15,5Vi n = 2 C2H6

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m = 3 C3H6.

.851,01000

74).04,0075,0(x =

+=

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0,252.(1,25 im)a. Gi th tch cc cht trong hn hp B l x,y v z Th tch ca hn hp kh

B lx+ y + z = 27,4CH4 + 2 O2 CO2 + 2 H2Ox x

C3H8 + 5 O2 3 CO2 + 4 H2Oy 3y

CO + 1/2 O2 CO2z z

Th tch CO2thu c lx+ y + z = 27,4

x+ 3y + z = 51,42y = 24 y = 12

Thnh phn % v th tch ca C3H8 trong hn hp B l% C3H8 = (12.100/27,4) = 43,80%

b. Khi lng mol phn t trung bnh ca hn hp B l

284,27

1226,28

4,2712

4,2712.444,15.16

4,271212.44)(16

4,272812.4416

=>+=

++

=+++

=++

=

Mnitoz

M

zzzxzxM

Vy cng iu kin, 1 lt hn hp B nng hn 1 lt kh nit.

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Ghi ch: Nu hc sinh gii cch khc m ng th vn cho im ti a.

Cu IV.2.b nu CHGHI 1 lt hn hp B nng hn 1 lt nit th khng cho im.

dap an vao chuyen hoa 10 dai hoc su pham ha noi 2011

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