2A1 +2H2S O4 (1) AI2(SO4)^ +3H

Một phần của tài liệu phân loại và phương pháp giải nhanh bài tập hóa học vô cơ 12 (Trang 169)

D. axit axetic va axit propanoic '

2) 2A1 +2H2S O4 (1) AI2(SO4)^ +3H

Cdu 7: Xet cac phat bieu (a) : Dung (1) (b) :Dun (2) (c) : Dung (3) (d) : Sai, vi: < ( C , H , o 0 5 ) ^ + n H 2 0 - Tinh bot C , 2 H 2 2 0 „ + H 2 0 - y ± Saccaroza (e) : Dung (4) fructoza — ^ Dap an dung la B. H + ^nC6H,206 glucozo ->C6H,206+C6H,206 Glucoza fructoza > glucozo-> C H 2 0 H ( C H 0 H ) ^ C H 0 ^ A g (g): Sai, VI saccaroza khong tac dung v6i Hj.

Dap an dung la A-

Cdu 8: Trong tinh ths' phan tir (thi du tinh th^ H 2 0, t i n h thé I2, tinh th^ C O 2 , I 1),

cac phan tir lien két vdi nhau bang cac lien ket yeu nen tinh thé phan tir de nong chay, de bay hoị

=> Phat biéu C sai Dap an dung la C.

Cdu 9: Cac chát trong X (C2H4,CH4,C3H4,C4H4) c6 dac di^m chung la c6 cung s6' nguydn tijf H trong phan tir => Dat c6ng thiic chung la CnH4 .

Theobaira: M x = 1 7 . 2 = 34=>12n + 4 = 34

=>12n = 30=>n = 2,5(C2,5H4) " >'

C2,3H4 +3,502 ^ 2 , 5 C 0 2 + 2 H 2 O

0,05 ->0,125(mol)+0,l(mol)

Khd'i luang blnh tang bang kh6'i luang C02,H20 sinh ra trong phan ling chaỵ Do do: m = 0 1 ^ 0 2 + I T I H 2 O

=>m =0,125.44+ 0,1.18 = 7,3(g)

Dap an dung la D.

duJO: Cac thi nghidm:

(a) : Ca Cu, Ag deu phan ling v6i O3:

Cu + O3 CuO + O2 x

2 A g + O3 ^ A g 2 0 + O2

(b) : Ca Cu va Ag d l u phan ling v6i HNO3 dac: ~ ^ . f t i

Cu + 4 H N O 3 ^ C u(NO3 )^ + 2 N O 2 + 2 H 2 O , :

Ag + 2 H N O 3 -> AgNOj + NO2 + H 2 O , 11^.) r,

(c) : Ca Cu va Ag deu kh6ng phan ung v . v v (d) : Chi CO Cu tac dung (bi oxi hoa) con Ag khong tac dung. -T

Cu + 2FeCl3 -> CUCI2 + 2 F e C l 2

Dap an dung la

Chu y: Trong day didn hoa c6 thir tir: Cu^^ /Cu;Fé'^ /Fe2+;Ag* / Ag nen theo

quy tac anpha: ^ > j

Cu + 2Fê^->Cu2^+2Fe2-'

Ag"^ + Fê"^ Fe"'^ + Ag (kh6ng xay ra phan ung ngirac lai)

Cdu 11: PTHH: CHjCOOC^H, + 2NaOH -> CH3COONa + C^HjONa + H 2 O

Phenyl axetat khong c6 ancol

1) C H 3 C O O C H 2 C H = C H 2 + NaOH ^ C H 3 C O O N a + C H 2 - CH - C H 2 O H

anlyl axetat , ancol anlylic

TJZi nr>c I ~ I tiunii I uu/i

2) CH3COOCH3 + NaOH CH3COONa + C H 3 O H metyl axetat ancol metylic 3) HCOC2H3 + N a O H H C O O N a + C2H, O H

etyl fomat ancol etylic

4) ( C, , H 3| C O O ) 3 C3H5 + 3NaOH ^ C 3 H ,( O H) ^ + 3C,5H3,COONa ' ' ' tripanmitin glixerol

' < ^ ' ^ I ' ' Dap an diing la IJ.

Cdu 12: Chat beo nay g6m c6:

- Triglixẹ-it: ( R C O O ) ^ C 3 H 5 , , I M J W- , )f f -t^Har^,,: '

- A x i t b e o t u d o : RCOOH V^n

So mol KOH de trung hoa axit beo tir do: 200 7

" • ^ ^ ^ " 1000 56 " = "NaOH

RCOOH + NaOH-> RCOONa + H2O

SlWfnk 0,025 <-0,025 ^ 0 , 0 2 5 ( m o l )

; f, H ( R C 0 0) ^ C 3 H s+ 3 N a 0 H ^ 3 R C 0 0 N a + C3H5(0H)3

X -> 3x(mol) i

Khdi lirong mu6'i tang so vai khdi luong chat beo: ni,g = 0,025.(23 - 1 ) + x.(3.23 - 41) = 0,55 + 28x - Theobaira: m,g =207,55-200 = 7,55(g)

=>0,55+ 28x = 7,55 ^ x = 0,25

Vay khoi liromg NaOH da tham gia phan ling: '

i"NaOH = (0,025 + 3x).40 = (0,025 + 3.0,25).40 = 3 Igam

. iCAuJ.h:i^ .Kttfi,,<V..:;ạ'.^-'• Dap an diing la C.

Cdu 13: PTHH:

CgHjONa + HCl ^ C ^ H ^ O H i + N a C l (XJ ( Y ) ( Z ) natri phenolat axit clohidric phenol

Dap an dung la C.

Cdu 14: Theobki ra: nfjaOH = ' 2 / 4 0 = 0,3(mol)

Tathay n x : n N „ o H- 0 , 1 5 : 0 , 3 = 1:2

Mat khac X la este dom chilc nan X la este cua phenol: RCOOC^H,

RCOOCgH, + 2NaOH RCOONa + CgH^ONa + H2O

0,15 -> 0,15 -> 0,15(mol)

Taco: 0,15.(R + 67 + 116) = 29,7 ..-...^^ip i i ; / , , fv, • , : , ( ; • • , ,

=>R = 15(CH3) ' i)fyM-i\tni li>] on,'! ị. :, : , .

Ca'u tao cua X dang este cua phenol: '

1) CH3COOCf,H3 , . 2) 0-HCOOC6H4CH3

3) m - HCOOC6H4CH3 • 4) p - HCOOC6H4CH3

Dap an diing la

Chu y; Este dcfn chuc X ma n^^ : nf^^QH = 1 : 2 thi do la este cua phenol dang RCOOC(,H^:

RCOOCgHs + 2NaOH ^ RCOONa + C^H^ONa + H2O o, ..>,m

Cdu 15: Theo bai ra: n^|Q^ =0,4x(mol)

" A M S O . ) , = 0 , 4 y ( m o l ) ; nNaOH =0,612(mol)

"AI(OH)3 = 8,424/78 = 0,108(mol);nBaso4 =0,144(mol) * T i m y :

AI2 (SO4 )^ + 3BaCl2 ->3BaS04 i +2AICI3

0,4y 0,144(mol) ^ 0,4ỵ3 = 0 , 1 4 4 ^ y = 0,12(n^,2(sô)^ =0,4.0,12 = 0,048mol

* Tim x: *V

V i n^aOH > 3.nA,(oH)3 Al(OH)3 bi hoa tan bot.

AICI3+ 3 N a 0 H ^ Al(OH)3 i + 3 N a C l 0,4x -> l,2x -> 0,4x

AI2 (SO4 )3 + 6NaOH ^ 2Al (OH)^ i +3Na2S04 0,048 -> 0,288-> 0,096 A l ( O H) 3 + N a O H - > N a [ A l ( O H ) ^ ] . _ a a (mol) ; Taco: • x + 0,096-a = 0,108 l,2x + 0,288 + a=0,612 ii>l,6x =0,336 =>x = 0,21 . vay x: y = 0,21:0,012 = 7:4 0 , 4 x - a = 0 , 0 1 2 l,2x + a = 0,324 Dap an dung la D.

T

Cdu 16: Sa 66 qua trinh siin xuát

Manhetit>Fe304 —> 3Fe> Gang

800 95

Kh6'i lirong Fe CO trong gang: nip^ = '•— = 760 (tán) ' r V ' / ( > ? . ' , ( ) ; . ( . , ; 100

, Vi lucmg sat bj hao hut 1% nen kh6'i lucmg sSt thuc te cSn la: ')'• '

760.100 . • . ftsbo;,' u: - F e 3 0 4 —^ 3 F e q ( ^ ^ , H : 232 tán 168 tán m, < - 767,68 tán ^68•: 168 > m . - m p, 3 0, = ' ' ' f ' / ^ ' - 1 0 6 0 , 1 3 (tán)

Khoi lucfng quang manhetit: x = m^^nhetii = '^^^'^"^"'^^ = 1325,16 (tán) 80

Dap an diing la C.

Chu y: Ta c6 the tinh nhanh nhu sau:

Manhetit (80%) ^ Fe304 (232) ^ 3Fe(l68) Gang(95%)

_ niFe3O4lQQ10Q _ mp,.232 D E J

^ " 80.(100-1) ^ " ^ ^ ^ 3 0 4 -

800.95 800.95.232.100.100

m p = =>x = ; - = 1325,16 (tan) 100 100.168.80.(100-1)

Cdu 17: Amin va ancol cung bac la:

C 6 H 5 N H C H 3 ( b a c l l ) v a C 6 H 5 C H ( 0 H ) C H 3 (bac I I )

Dap an dung la B.

Cdu 18: - Cac kim loai kiem din c6 cung kieu mang tinh thd' lap phuong tam khoi: Na,K

- Cac kim loai kiem th6 c6 ki^u mang tinh the khac nhau, trong do chi c6 kim loai Ba c6 kieu mang tinh the lap phucmg tam khoị

Vay cac kim loai do la Na, K, Ba

Dap an dung la C .

Cdu 19: Tac6: N C = - ^ ^ = — = 3

=> Phan i\x andehit va ankin deu c6 3 nguydn tiJf cacbon => ankin la C3H4

Cty TNHH MTV DWH Khang Viet

Mat khac 2.nH o 2.1,8x - , .

: HH = — = = 3,6<4 Phan tir andehit s6' nguydn tir H be thua 3,6 Phan tir andehit s6' nguydn tir H be thua 3,6

andehit la C3H2 (CH = C - CHO) C 3 H 4 ) 3 C Q 2 + 2 H 2 O C 3 H 2 -> 2a (mol) - i^ 2 _ ^ 3 C 0 2 + H 2 O -> b(mol) Ta c6: a + b = X 2a + b = l,8x •a = 0,8x;b = 0,2x „ 6.100% 0,2x.l00% vay %n = — = = 20% a + b X

Cdu 20: * Phdn I : n c o 2 =0,25mol; n^^o =0,35(mol)

V i riH^o > " c o 2 =>ancol no, ha, don chu:c: ROH

"ROH = " H 2 0 - n c o 2 =0,35-0,25 = 0,1 (mol)

n c o 2 _ 0,25

Dap an dung la

nc = - 2 , 5 J -f J

0,1 0,1

Hai ancol la: C 2 H 5 0 H ( X ) va C 3 H 7 0 H ( Y )

Goi x, y \in lirot la s6' mol X , Y trong m6i phSn. 'x + y = 0,1 :4 ^! ' .; ;:0! Taco: 2x + 3y ^ x =0,05; y = 0,05 — z, J x + y Phdnll: n „ , e= n N 2 =0,42/28 = 0,015(mol) Goi a, b \in luot la s6' mol X , Y tao estẹ

CjH^OH + C3H7OH -> C2H5 - O - C3H7 + H 2 O

2 C 2 H 5 O H- ^ C j H j - O - C 2 H 5 + H j O

2C3H7OH ->C3H7 - O - C3H7 + H 2 O , , ,

VI n3^„, = 2 . n „ „ =^a + b = 2.0,015 = 0,03 (0 Ma m^ste = '"C2H^_ + '"C3H^_ + " 1 - 0 - Ma m^ste = '"C2H^_ + '"C3H^_ + " 1 - 0 -

=>l,25 = ạ29 + b.43 + 0,015.16 => 29a + 43b = 1,01 (2) T i r ( l , 2 ) = > a = 0 , 0 2 ; b = 0,01

Hifeu suaft tao este cua X : H , ^ . = " ' " ^ ' ^ " ^ ^ = 4 0 % ' ^ •

„ , J, ^ 0,05 ,., ^

Hieu suát tao este cua Y : H,ỵ = " ' ^ ' " ^ ^ ^ ^ = 2 0 % ' „

( Y ) - . H .

^FomVKS; • Dap an diing la Ạ

CdM27: * D u n g d i c h ( l ) : H 2 N C H 2 C O O H c6 p H « 7 t S i l : ; | | , ^

* Dung djch (2): C H j C O O H ^ i ^ + CH3COO" 4 - «i ,

C O p H < 7: (moi trucmg axit)

* Dung djch (3): C H 3 C H 2 N H 2 + H 2 O ^ C H 3 C H 2 N H 3 + O H "

CO pH > 7 (moi truong bazo)

Vay p H ( 2 ) < p H ( l ) < p H( 3 ) * ,„„,;-}, Dap an diing la Ạ Cdu 22: PTHH: -1 -2 +7 3 C 6 H 3- C H- C H 2+ 1 0 K M n O 4 > a p/ +3 +4 + 4 ' , 3 C 6 H 5 C O O K + 3 K 2 C O 3 + 1 0 K M n O 2 + K O H + 4 H 2 O 3x - I -2 +3 44 C - C ^ C + C + lOe +7 +4 M n + 3e -> M n lOx

vay tong hfe s6': 3 + 10+3 + 3 + 1 0 + 1 + 4 = 3 4

Dap an dung la B.

Cdu 23: + T^r 1,2 => X la andehit

+ Tir 1,2 => Y | , Y , CO cung so nguyen tir cacbon trong phan tiJr.

Tir (3) va CTPT Y , => Y , , Y2 deu c6 3 nguyen tir cacbon trong phan tir (bang m6t nixa cua phan tir Y 3 ) => X c6 3 nguyen tu cacbon trong phan tir.

+ Phan t u Y , C O 2 lien ket n => Phan tir X c6 2 lien két 71 (C = C,C = O ) . '

vay X la C H 2 = C H - CHO (andehit acrylic).

Dap an diing la D.

Chu v: C?c P T H H xay ra:

C H 2 = C H - C H O + - 0 2 ) C H 2= C H - C O O H

f

C H 2 = C H - CHO + 2H2 ) C H 3 - C H 2 - C H 2 O H

CH2 = CH - COOH + C H 3 C H 2 C H 2 O H —^ ^ y^ C H 2 = CH - COOCH2CH2CH3 + H j O

I ( Y , ) ( Y 2 ) ( Y 3 )

" Cdu 24: * Xac dinh ti le s6' mol O, (x mol), O 3 (y mol). ^ _ j,| -

32x + 48y

• = 22.2 = 44 I , .

x + y

T - > J O A A A A A t'^ O 'T" ¥ S " i - X + >' +• X ''•„•'.;•.

=>32x + 48y = 44x + 44y => 4y = 12x = > y = 3x

Gia sir trong Vj lit X c6 1 mol OT thi c6 3 mol O3 (trong V2 c6 4 mol khO

= > C O 11 mol O . .

. +. uO<r-Oo:J +' (M

* Xac djnh ti le s6' mol C H 3 N H 2 (a mol) va C2H5NH2 ( b m o l ) . Ta c6:

^ ' - ^ ^ ^ ^ = 17,833.2 = 35,666 a + b

=> 31a + 45b = 35,666a + 35,666b => 9,334b = 4,666a ^ 2b = a

2 C H , N H 2 + 9 0 2 C O 2 + 5 H 2 O + N ,

2b ^ 9 b )}..,:, ,

2 C 2 H 5 N H 2 +150 ^ 4 C O 2 + 7 H 2 O + N2 ; ^

b 7,5b - ,<

Taco: 9b + 7,5b = l l => 16,56 = 11 b = 2/3 Trong V, lit Y c6 3b mol (= 2 mol)

v a y V, :V2 = n Y : n x = 2 : 4 = 1:2 it (XH >; ' "

I j i 1 ,> Dap an diing la D.

Cdu 25: Co can bang: 2SO2 + 0 2 ? = ^ 2SO3, A H < 0

Jkj Cac bien phap lam can bang chuyén djch theo chieu thuan:

~ (2): Tang ap suát chung ciia he phan iJng. (3): Ha nhiet do

(5): Giam nong do SO3

>' Dap an dung la D.

Chii y: Phan ling thuan c6: 1 (' ,v * /

+ So mol khi giam (3 ^ 2) => tang ap suát ; ... K + A H < 0 (toa nhiet) => giam nhiet do

+ Th6m chat tham gia d^u lam cSn bang chuyen djch theo chieu thuan.

Cdu 26. Theo bai ra: = 15,68/22,4 = 0 , 7 ( m o l )

n ^ o = 8 , 9 6 / 2 2, 4 = 0,4( m o l ) . ^ ' '^s''-?^^,'' H j O + C - ^ C O + H j • f , . - l i ' J - - . . h - : ' : . ; u - V ' X x ( m o l ) : ; , : : , : „ : . . , , , > , / _ . u ' " ' ' 2 H 2 0 + C ^ C 0 2 + 2 H 2 ;<):Obfn y ^ 2 y ( m o l ) l^ls:-:? i r > c 6 : x + y + x + 2y = 0,7 ^ 2 x + 3y = 0,7 ( l ) . .^^^ , C O ^ C u O - . C u + C02 , X ^ x m o l ,-. , , , , + CuO ^ Cu + H , 0 ;,; x + 2y - > ( x + 2 y ) m o l

3Cu + 8 H N O 3- > 3Cu ( N O 3 )2 + 2NO + 4H2O

2x + 2y 4( x + y ) / 3 T a c o : 4( x + y ) / 3 = 0,4 = > x + y = 0 , 3 ( 2 ) T i r ( l , 2 ) = ^ x = 0 , 2 ; y = 0 , l ^r. « x.100% 0,2.100% Vay % V c o / x = % n c o / x = = Q ^ = 28,57% Dap an dung la A .

Cdu 27: K h o i luong nguydn 16' n i t a trong 14,16 gam X : 14,66.11,864 I T I M = ^ 100 • = l , 6 8 ( g ) = > n N = 0 , 1 2 ( m o l ) = n =0,12.62 = 7 , 4 4 ( g ) ^ N O J N O 3 = > m k i m i o a i = m x = m ^ ^ _ = 1 4 , 1 6 - 7 , 4 4 = 6,72(gam) Dap an dung la 6.

Cdu 28: Theo bai ra: H J ^ Q H = 0 , 1 4 4 m o l ; ng^^Q^j^ = 0 , 0 1 2 m o l

= : > n ^ ^ _ = 0 , 1 4 4 + 0,012.2 = 0,168mol;n^2+ = 0 , 0 1 2 m o l Theo dinh luat bao toan didn tich, ta c6: 0,1.1 + z.3 = t . l + 0 , 0 2 . 2

= > t - 3 z = 0,06 (1)

2+ ^ "so2- ^ " B a S 0 4 4 = ^^2^ = 0 , 0 1 2 m o l V i n 9 ^ < n

Ba

Cty TNHH MTV DWH Khang Viet

> = 0,012.233 = 2 , 7 9 6 ( g ) < 3,732(g) Trong kd't tua c6 A l ( O H ) ^ : ' " A , ( O H ) 3 = 3 , 7 3 2 - 2 , 7 9 6 = 0,936(g) ' n A i ( O H) 3 = 0 , 9 3 6 / 7 8 = 0 , 0 1 2 ( m o l ) + O H' - j- H j O 0,1 - > 0,1 ( m o l ) A l ' + + 3 0 H " ^ A l ( O H) 3 i 0 , 0 1 2 < - 0 , 0 3 6 < - 0 , 0 1 2 ( m o l ) O Al-''+ + 4 0 H ~ - > A l ( O H ) ^ 0,008 < - ( 0 , 1 6 8 - 0 , 1 - 0 , 0 3 6 ) ^ 2 = 0,012 + 0,008 = 0,020 ( 2 ) T i r ( l , 2 ) = 0 , 0 6 + 3.0,02 = 0,120 Vay z, t iSn lucrt la 0,020 va 0,120

, 0

Dap an dung la C .

Chu y: M o t each gSn dung, coi khong do nuoic phan l i ra (bo qua sir dien l i ciJa nuoc).

Cdu 29: Theo bai ra: n^g = 0 , 1 7 ( m o l ) ; nco2 = 0 , 0 3 5 ( m o l ) V I sinh ra C O , => Y la andehit fomic

H C H O + 4 A g N 0 3 + 6 N H 3 + 2H2O ^ 4 A g + 4 N H 4 N O 3 + ( N H 4 \3 x ( m o l ) ^ 4x - > X (1^ (NH4)2C03 + 2 H C 1 - > 2 N H 4 C 1 + C02 t + H 2 0 ,\046 , ' ' f> ^ ^ He-! i f f ' n V v - X = 0,035 (mol) m y = 0,035.30 = l , 0 5 ( g ) = > m z = 1 , 8 9 - 1 , 0 5 = 0 , 8 4 ( g ) Dat Z la R C H O (y mol) R C H O + 2 A g N 0 3 + 3NH3 + H2O ^ RCOONH4 + 2 A g i + 2 N H 4 N O 3 y - > 2y T a c o : 4 x + 2y = 0,17 = > 2 y = 0 , 1 7 - 4 . 0 , 0 3 5 = 0,03 = > y = 0,015

'TTi7iTi loaT \ J>]},'ii> i;uii n/uiiili til Hoa Hoc IZ -Cii I liaiili Iodn

Suyra: = = SeiC.W^CHO)

: ^ 0,015 ^ ^ ' ' Vay Z la CH2 = CH - C H O (andehit acrylic)

• Dap an diing la D.

Cdu 30: Xet cac phucfng an:

Ạ Sai, V I este khong tan trong niroc.

Một phần của tài liệu phân loại và phương pháp giải nhanh bài tập hóa học vô cơ 12 (Trang 169)

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