bộ đề hóa học 9 ôn thi cho lớp 10

Moi de c6 kien thiic bao quat nhieu van de trong tam va nang cao cua chUdng trinh hoa hoc T H C S c6 do kho nhat dinh, doi hoi hoc sinh phai hieu biet, suy luan mcl rong, c6 kl nang va

5450-0 NHA GIAO LfU TU PHAM SY LL/U

THI/VIEN Vm

Nhiem vu cua cac trUdng chuyen, 16p chon clUdc nganh Giao due giao

nhiem vu la dao tao nguon nhan lUc, boi dudng nhan tai de gop ph^n day

manh sU nghiep cong nghiep hoa va hien dai hoa dat nudc

De dap ilng yeu cau do, Nha sach Khang Viet xin tran trong gidi thieu cuoh

sach: Bp de Hoa hoc 9 on thi vdo 10 cua Nha giao Uu tii Pham Sy Luu

Sach gom 3 noi dung chinh:

- Phan 1; Cac phUdng phap giai nhanh bai tap hoa hoc

- Phan 2: Tuyen chon gi6i thieu de thi tuyen sinh ciaa mot so'trUdng chuyen

- Phan 3 : Ba mUdI de thi thii

P h a n phiftfng phap : giiip cac em c6 each giai dung, hay va nhanh gon

nhat, thich hdp v6i tufng dang bai tap Moi phUdng phap dUdc minh hoa

bang nhieu vi du d l hieu de cac em c6 the van dung dUdc mot each nhanh

nhat De dam bao yeu cau chuan kien thiic, ki nang nen cac vi du da dUdc

can nhfic, Ixia chon chu yeu tii cac de thi DH & CD cua bo GD & DT cung

vdi Idi giai dUdc trinh bay chi tiet theo each giai tU luan phu hdp vcti kieu ra

de thi tuyen vao cac trUdng chuyen va 16p chon

P h a n giofi thieu de thi tuyen sinh cua cac trtfofng chuyen : gom 30

de cua nhieu trUdng chuyen tren ca nUdc trong 5 nam gdn nhat Moi de bao

gom nhieu kien thiic trong tam va nhieu ddn vi kien thiic nang cao, doi hoi

tinh ta duy sang tao, sU van dung Unh boat cac phUdng phap giai toan hoa

hoc hien dai Phan nay giup cac em trUcfc ki thi tuyen can nhdc, chuan bi

cho minh kien thiic va ki nang lam bai hieu qua de dat thanh tich tot nhat

P h a n de thi thii: gom 30 de Moi de c6 kien thiic bao quat nhieu van de

trong tam va nang cao cua chUdng trinh hoa hoc T H C S c6 do kho nhat

dinh, doi hoi hoc sinh phai hieu biet, suy luan mcl rong, c6 kl nang van dung

linh boat cac phiidng phap giai toan hoa hoc Nhieu van de dUdc lap lai ci

cac de khac nhau nhSm giiip cac em nim dUdc kien thiic va phUdng phap

giai tot nhat

Cuo'n sach nay la cong cu hiiu ich khong the thieu doi v6i hoc sinh trong

qua trinh hoc tap va on luyen mon hoa chuan bi cho ki thi tuyen vao 16p 10

chuyen, sach cung giiip cho cac bac phu huynh c6 dUdc tU lieu de c6 phUdng

hu6ng tU van, giiip dd cac em hoc tap dat ket qua tot trong ki thi

Mac du da rat c6 gkng va dau tU nhieu thbi gian cung v6i nhiing tich luy

trong nghe ciia tac gia nhUng chac chan khong tranh khoi nhiing thieu sot,

chiing toi rat mong nhan diidc nhiing dong gop cua ban doc cho noi dung

ciia cuoh sach hoan thien hdn trong Ian tai ban ;

Xin chan thanh cam dn!

O/ TNHHMTVDVVnKhan^ Vi6l

1 Q u y t a c thCr t a uu tien c u a p h a n Crng

(/) Phan ifng oxi hoa khu:

- Phan irng theo chi^u: chat o.xi hoa manh licfii oxi hoa chat khit nuiuh hc/ii tao thanh chat o.\i hoa yen hem vd chat khifyeh hon vu ,

- Khi xay ra phan irng ciia h6n hop chat oxi hoa vori h6n hap cha't khir: Phan

I'Oi^ mi tien la chat o.xi hoa manh nlid't o.xi hoa chat klu't nianli nhat tao thdnli

chat o.xi hoa yen nhat vd chat khi'f yen nhat

(2) Phan itng cua cac chat dien li trong dung dich theo thii tit uu tien sau:

- Phan ii-ng trung hoa: H"^ + O H " -> HgO

- Phan irng tao ket tiia hidroxit: M"^ + n O H " -> M(OH)„

- Phan irng hoa tan hidroxit lu5ng ti'nh trong k'lim dir:

M(OH)„ + ( 4 - n ) 0 H - > [ M ( 0 H ) 4 f Hay : M(OH)„ + (4 - n)OH -> MO^^ + 2 H 2 O

i'i

Vi du 1. Cho 29,8 gam h6n hop bot gom Zn va Fe vao 600ml dung dich C U S O 4 0,5M Sau khi cac phan ling xay ra hoan toan, thu diroc dung djch X va 30,4 gam h6n hofp kirn loai Phdn tram va khoi lucmg ciia Fe trong h6n hop ban ddula

CUSO4 het (2)

y ^ y - > y mol Nd'u chi CO Zn phan urng, Fe kh6ng phan ling thi: n^,, = n^u = 0,3 mol

Kh6'i luong kim loai sau phan urng giam so vai ban diJu: y;

=>mKL = m p , +mc., - 29,8 - Am = 29,8-0,3 x (65-64)-29,5 gam < 30,4 gam Neu Fe cung tham gia phan iing het thi kim loai chi c6 Cu Trai d^ ra

Vay Zn het, Fe con du Zn phan ung lam kh6'i luong kim loai sau phan ii:ng giam, Fe phan ling lam khd'i luong kim loai sau phan irng tang:

He PT:

ncuS04 =(x + y) = 0,3mol fx = 0,2 Jmz„ =0,2x65 = 13 gam

Am = (-lx + 8y) = 0,6gam ^ [ y = 0,l ^ [ m p ^ = 29,8-13 = 16,8 gam

>%mp^ = — x l 0 0 % =

// V( Ik• V; / / / / v',-;.' !0 - Phm S/Liru

Vi du 2 Cho 1,68 gam Fe va 0,36 gam b6t Mg tac dung v6i 375ml dung djch

C U S O 4 , khua'y nhe cho de'n khi dung djch mat mau xanh Nhan tha'y khd'i lugnig

kim loai thu duoc sau phan ling la 2,82 gam N6ng d6 mol cua dung dich

Sau phan irng dung dich mS't mau xanh chung to rang C U S O 4 da phan ung het,

C O the con du kim loai sau phan ling

Gia sij Fe phan iJng het khdng con du thl khoi luong Cu thu dugc se la :

mc„ = (0,03 + 0,015) X 64 = 2,88 gam > 2,82 gam (theo gia thie't)

Fe con dir sau phan ung

Goi X la so mol Fe phan ling Am : do tang khoi luong kim loai

Ap dung phuong phap TGKL ta c6:

Am =(64 - 24).0,015 + (64 - 56)x = 2,82 - (1,68+0,36) = 0,78gam

0,015 + 0,0225 0,375

>x = 0,0225mol => C M ( C U S O 4 ) = -—rrri = 0,1 M Chon B

Vidu 3 Cho m, gam A l vao 100ml dung djch gom Cu(N03)2 0,3M va AgNOj

0,3M Sau khi cac phan ling xay ra hoan toan thi thu duoc m2 gam chat rdn X

Neu cho m2 gam X tac dung v 6 i luong du dung dich HCl thi thu duofc 0,336 lit

khi (a dktc) Gia trj ciia m, va m2 Idn iuot la

A.8,10va5,43 B 1,08 va 5,43 C 0,54 va 5,16 D 1,08 va 5,16

Gidi

ncu(N03)2 " "AgNO, = 0,1-0,3 = 0,03 mol; n^^ = 0,015 mol

Phan ling khir AgNO^ va Cu(NO,)2bori Al theo thii tu uu tien, sau do chat rdn

2 tac dung v 6 i HCl tao khf H2=> Al duda khijf H C l : n^nu^, = - n ^ ^

Xet toan qua trinh : - Chat khir: A l

- Chat oxi hoa : A g N O , , Cu(N03)2, H C l

Vi du 4 (DHB 2009) Cho 2,24 gam b6t sdt vao 200ml dung dich chiia h6n hop

gom AgNOj 0,1M va Cu(N03)2 0,5M Sau khi cac phan ting xay ra hoan toan,

thu duoc dung djch X va m gam chat rdn Y Tinh gia trj cua m

A 5,08 gam B 4,08 gam C 3,72 gam D 6,24 gam

Gidi

npe =0,04 mol; nAgwo, = 0,02 m o l ; ^c^^no^,)^ =^,1 mol

Su khijf xay ra theo thii tu uu tien:

AgNO, tac dung het sau do Cu(N03)2 bi khii

Fe + 2AgN03 ^ 2 A g + Fe(N03)2 0,01<-0,02-) 0,02

Chon B

m,.„ = 0,02 X 108 + 0,03x64 = 4,08 gam

Vidu 5 Hoa tan 13,8 gam N a 2 C 0 , vao nude Vira khuay vifa them tirng giot dung

djch HCl I M cho t6i dii 180ml dung dich axit, thu duoc V lit khi Viet phuong trinh phan ufng xay ra va tmh V (dktc) '

"K2CO3 = 0,1 mo! =^ H H C K c d n d u n g ) = 0,2mol > H H C K C I C ^ r a ) = 0,18mol PTHH theo thii tu uu tien:

HCl + Na2C0, -> NaCl + NaHCO,

Vi du 6 Them tir tir dung djch NaOH 2,5M vao 400ml dung djch X chiia HQ I M va

AlCl, 0,5M Ti'nh the tich dung djch NaOH cin dung de thu duoc ket tiia 16n nhat

Gidi

Phan ling theo thii tu uu tien: ; •

NaOH + HCl ^ NaCl + H 2 O (1) '

Dp dS/l^>a hoc y on Iht y^o IP-Phaa 3/ MT

Vay the ti'ch dung dich NaOH can dung la : | V = 400 ml

Vidu 7 Dung dich D g6m cac chat N a A l O , 0.16 m o l ; Na,S04 0,56 m o l ; NaOH

0,66 m o l Can them bao nhieu ml dung dich HCl 2 M vao dung djch D de: (a)

Dugc khoi lirctng ket tua Idii nhat (b) Dugc ket tiia ma sau khi nung den khoi

lugng khong d 6 i thu dugc chat rdn can nang 5,1 gam

T H I : Khong xay ra pluin irng (3) : x = 0,05 x 2 = 0,1 mol ; y = 0 mol

S6 mol H C l Clin dung: n^ci = 0,66 + 0,1 = 0,76 mol

' V - i = — = 0,38 lit = 1380ml

T H 2 : Co phiin irng ( 3 ) : x = 0,16mol ; y = 0 , 1 6 - 0 , 1 = 0 , 0 6 mol

So mol HCl can dung: = 0,66 + 0,16 + 3.0,06 = 1 mol ' '

^HCi = - = 0 5 lit = I 500 mil

2 Phiiong phap bao toan khoi luong

Dinh ludt: Trong phdn ung hod hoc long khoi lumg cdc sdn phdin bdng long khoi luang cdc chat tham gia phdn dug

A + B ^ C + D ,

m ^ + mp = m ^ + m^j ( A , B : vira dii hoac con du)

m ^ , la khoi lugng ciia A , B tham gia phan ung

m ^ , m ^ : la khoi lugng ciia C, D tao thanh '

Apdung:

- Phan ung c6 n chat ma biet dirge khd'i lugng cua (n - 1) chat => khoi lugng chat con lai

- Trong cac bai toan X i i y ra nhieu phan I'rng, c6 the khong ciin viet day dii

cac phuong trinh phan itng, chi can lap so do phan irng de thay moi quan he ti le mol giOa cac chat can xac djnh va nh&ng chat ma de cho Sau do lip dung dinh luat de tim ket qua

- K h i CO can dung dich thi kh6'i lugng mu6'i thu dugc bang tong khoi lugng

cac cation k i m loai va anion goc axit

- Tfnh khoi lugng dung djch sau phan ihig:

' " ( d d s;ui pif) - "^(dd IriAK- pit)"'' ''"(clut l:m) '^(chiVl ki l l i i a ) ' ^ ( c h a l bay hai)

Vi du I Cho mot luong khf clo du tac dung vdri 9,2 gam k i m loai sinh ra 23,4 gam

muoi k i m loai hod trj I Hay xac djnh k i m loai hoa trj I va muoi k i m loai do

Vi du 2 Cho 7,8 gam h6n hgp k i m loai A l va M g tac diing vdi H C l thu dugc 8,96

li't H , (6 dktc) Hoi khi c6 can dung dich thu dugc bao nhieu gam muoi khan

Vi du 3 Hoa tan hoan toan 3,22 gam h6n hop X g6m Fe, Mg va Zn bang m6t luong

vCra du dung dich H2SO4 loang, thu diroc 1,344 1ft hidro (0 dktc) va dung djch

' chiia m gam mu6'i Tinh m?

Gidi

nH2S04 =(1-344:22,4) = 0,06 mol

P T H H : M + H 2 S O 4 M S O 4 + H2

DLBTKL: m^^g, = nix + " 1 H 2 S 0 4 - " I R Z =3,22 + 98 X 0,06 - 2 X 0,06 = |8.98 gam

Vidu 4 Hoa tan 10 gam h6n hop 2 mudi cacbonat cua cac kim loai hoa trj 11 va I I I

bang dung djch HCI du thu dugc dung djch A va 0,672 lit khf (dktc) Hoi c6 can

dung dich A thu dugc bao nhieu gam muoi khan?

Gidi

n^„^ = (0,672:22,4) = 0,03mol

Goi 2 kim loai hoa tri I I va I I I Ian lugt la X va Y ta c6 cac PTHH:

XCO, + 2HC1 -> XCl, + CO, + H , 0 (1) , Y,(C03), + 6HC1 2YC1., + 3CO, + SH.O (2) ' '

T i r ( l ) va(2) n H 2 0 = n c o 2 =^'^3 mol; = 2.0,03 = 0,06 mol

^ "iHCi(ph:min,g) =0,06x36,5 = 2,19gam

Ggi X la khoi lirgng mudi khan: XCK va YCl,

! Theo djnh luSt bao toan khdi lugng ta c6:

, 10 + 2,19 = x + 44.0,03 + 18.0,03 x = 10,33gam

Vidu 5 Cho 4,96 gam hdn hgp rdn gdm Ca va CaC, tac dung vdi nude dir thu dugc

2,24 1ft hdn hgp khf A (dktc) D3n A qua dng dung bdt Ni nung ndng mot thdi

gian thu dugc hdn hgp khf B Tiep tuc dfin hdn hgp B qua dung dich nude brom

du thl khdi lugng bhih dung dung dich brom tang m gam va cd 0,896 1ft hdn hgp

khf D (dktc) thoat ra, ti khdi ciia D so vdi H2 la 4,5 Tfnh m

Gidi

Cac PTHH : Ca + 2H2O -> Ca(0H)2 + Hj

CaC2 + 2H2O ^ Ca(0H)2 + C2H2

Theo cac PTHH ta cd : sd mol H2 va C2H2 liin lugt la x va y

>HePr: '40x + 64y = 4,96 _ [x = 0,06

x + y =0,1

=> =0,06x2 + 0,04x26 = 1,16 gam Theo DLBTKL: m ^ = mg = l,16gam

Va: m = m B - m ^ = l , 1 6 - 0 , 0 4 x 2 x 4 , 5 = [oj8jam

y = 0,04

Vi du 6 (DHA 2009) Xa phdng hda hoan toan 1,99 gam hdn hgp hai este bflng

dung djch NaOH thu dugc 2,05 gam mudi ciia mot axit cacboxylic va 0,94 gam

hdn hgp hai ancol la ddng dAng ke tiep nhau Cdng thuc ciia hai este dd la

A HCOOCH, va HCOOC.H, B C,H.,COOCH., va C2H5COOC2H5

C CHjCOOC^H, va CH^cboCH^. D CH,COOCH, va CH.COOC^H,

Gidi

DLBTKL : m^^oH = "imuoi + "lancoi - niesie - 2,05 + 0,94 -1,99 = 1 gam

=^nN.oH= (1:40) = 0,025 mol

Do tao mdt mudi va hdn hgp 2 ancol nen 2 este cd cung gdc axit vdi dang

RCOOR': RCOO R' + NaOH -> RCOONa + R^OH

^ R O H = Trbr =0,94 37,6 => 2 ancol la CH,OH va C^H^OH

0,025

2este: CH,COOCH, va CH.COOC^H, => Chgn D

Vidu 7 Cho 200 gam mot loai chii't beo c d chi s d axit la 7 tac dung vdi mdt lugng

NaOH vua du thu dugc 207,55 gam hdn hgp mudi khan Tfnh khdi lugng NaOH

da tham gia pluin ung

Gidi

Chi so axit la \<'>'niii KOH de tiling hoa axit heo co tron}> / i ; chat heo

Sd mol KOH trung hda axit beo : l^!:!^—-^T^ _ Q Q25,,^Q|

56 PTHH: RCOOH + K O H ^ RCOOK + H O ( I )

C , H , ( 0 C 0 R) 3 + 3NaOH ^ C3Hs(OH)3 + 3R'COONa (2)

Ta cd : nM.,oH drujig hoa axil beo) = " K O H (Irung hoa axit beo) = " H i t J ^ ^ ' ^ ^ ^

Goi X la sd mol triglixerit cua chii't beo=> nNaOH(.huy phA,> t r i g i i x c i ) = 3x

" N a O H ( p h ; i i i linj.) = iNaOHCIhiiy phiii Iriglixeril) + "Na()H(lrunp hoa axil boo) ~ ( ^ X +0,25) DLBTKL : mc,,,, + mN,OH(phan m.g) = ^ 6 1 + "Iplixerol + mH20

200 + (3x + 0,025) X 40 = 207,55 + 92x + 18 x 0,025 => x = 0,25 mol

niNaOH = (3 X 0,25 + 0,025) x 40 = |31gam

bg de H6a kx; 9 on Ihi viio lO- Pham S/ltSu

3 Phaong phap tang, giam khoi luong

Nguyen tdc:

So siinh k h o i l u g i i g cua chii't can xac d i n h v 6 i l u o i i g san p h a m cua no m a g i a

thiet c h o biet, de tir k h o i lirong tang hay g i a m nay, ket h o p \d\n he t i le m o l

giiJa 2 chat de t l m ra iiroiig cha't can xac d j n h ( c 6 the la so n h o m chuc, so mol, )

Pham vi su dung:

D o i v 6 i cac bai toan phiin irng x a y ra thu6c phan l i n g phfin h u y , phan u'ng

g i u a k i m loai m a n h , k h o n g t a n t r o n g nirdc diiy k i m loai y e u ra k h o i d u n g d j c h

m u o i phan u n g , phan l i n g t r u n g hoa axit c h o biet krgng m u o i tao thanh, Dac

biet k l i i chira biet r o phan u'ng x a y ra la hoan toan hay k h o n g t h l p h i r o n g phap

nay t o ra ra't h i e u q u a

Vi du I N h u n g m o t thanh sdt naiig 8 g a m vao 5 0 0 m l d u n g d j c h C U S O 4 2 M Sau

m 6 t thofi g i a n UYy la sdt ra can hi thay nang 8,8 g a m X e m the tfch d i u i g d j c h

k h o n g thay d o i t h i n o n g d o mol/li't cua C U S O 4 t r o n g d u n g d j c h sau phan ii"ng la

T h e o ( I ) va ( 2 ) ta nhiui thay cur I m o l C O , bay ra t u c la c 6 1 m o l m u o i cacbonat

chuyd'n thanh m u o i c l o r u a va k h o i l u o n g tang t h e m 11 g a m (goc C O ^ " la 6 0

g a m c l u i y e n thanh goc 2C1 c 6 k h o i l u o n g 71 g a m )

V a y CO 0,2 m o l k h i bay ra thi k h o i krgng m u o i tang la: 0,2.1 1 = 2,2 g a m

V a y t d n g k h o i l u g n g m u o i c l o r u a khan t h u dirge l a :

m (muoi ;i khan) = 2 0 + 2 , 2 = 22,2 gam

ci/ imnmvuwnKhang vie/

Vidu 3 N h u n g m o t thanh k i m loai M hoa trj I I vao 0,5 li't d u n g d j c h C U S O 4 0 , 2 M

Sau m o t thcfi g i a n phan i t n g , k h o i l u g n g thanh M tang l e n 0 , 4 0 g a m t r o n g k h i

n o n g d 6 C U S O 4 c o n l a i la 0 , 1 M

(a) X a c d j n h k i m loai M

(b) L a y m g a m k i m loai M c h o v a o 1 li't d u n g d j c h chira A g N O , va C u ( N O , ) 2 , nong

d o m 6 i m u o i la 0 , 1 M Sau phan urng ta t h u dugc eha't rdn A c 6 k h o i l u g n g 15,28

g a m va d u n g d i c h B T i n h m g a m ?

Gidi '•'}^''•

(a) ncuso4(b<i) = 0 , 5 0 , 2 = 0,1 m o l ; n c u s o 4 ( d ) = 0 , 5 0 , 1 = 0 , 0 5 m o l •> > • : ncu.s()4(|«) = 0 , 1 0 - 0 , 0 5 = 0,05 m o l

P T H H : M + C U S O 4 - > M S O 4 + C u

A m = 0 , 0 5 ( 6 4 - M ) = 0 , 4 = 5 6 = > M la Fe (b) T a c h i biet so m o l eija A g N O , va so m o l cua C u ( N O , ) , -

N h u n g k h o n g biet so m o l ciia Fe > '

Day H D H H : P T H H theo tM tir iru tidn :

Vi du 4 N h u n g m o t thanh sdt va m o t thanh k e m vao c u n g m o t coc chi'ra 5 0 0 m l

d u n g d j c h C U S O 4 Sau m o t thtJi gian lay hai thanh k i m loai ra k h o i coc t h i m 6 i

thanh c 6 t h e m C u b a m vao, k h o i l u g n g d u n g d j c h t r o n g coc bj g i a m mat 0,22

g a m T r o n g d u n g d j c h sau phan u n g , n o n g d o m o l ciia Z n S 0 4 g a p 2,5 liin n 6 n g

d o m o l cua F e S O j T h e m d u n g d i c h N a O H d u vao c6c, Igc lay ket tua r o i i i u n g ngoai k h o n g k h i d e n k h o i l u g n g k h o n g d 6 i , t h u dugc 14,5 g a m chdt rdn So g a m

C u b a m tren m 6 i thanh k i m loai va n 6 n g d o m o l c u a d u n g d j c h C U S O 4 ban dau

la bao nhieu?

• Gidi ••••'^

P T H H : Fe + C U S O 4 - > F C S O 4 + C u ( 1 )

Z n + C U S O 4 - > Z n S 0 4 + C u (2)

Goi a la so mol cua FeS04

Trong cuiig mot dung djch nen tl le v6 nong do mol ciia cac chat trong dung

dich cung chmh la ti le ve so mol

Theo bai ra: CM/z„s(i4 = 2.5 x CM/CUS04 => nznS04 = 2,5 x n^^^^^

Khoi luong thanh sat tang : (64 - 56)a = 8a gam

Khoi luong thanh kem giam: (65 - 64)2,5a = 2,5a gam

Khoi luong ciia hai thanh kim loai tang: 8a - 2,5a = 5,5a gam

Mii thirc te bai cho la : 0,22g

5,5a = 0,22 ^ a = 0,04 mol

Vay khoi luong Cu bam tren thanh sdt la: 64.0,04 = 2,56 gam

va khoi luong Cu bam tren thanh kem la: 64 x 2,5 x 0,04 = 6,4 gam

Dung djch sau phan ling 1 va 2 c6 : FeS04, ZnS04 va CUSO4 (neu c6)

TIFCJO., = 160.0,5a = 160.0,5.0,04 = 3,2gam

IncuS(34(b;m L) = a + 2,5a + b = 0,28125 mol

0,28125

-CUSO4

0,5 0,5625M

Vi du 5 Nung 6,58 gam Cu(N0,)2 trong blnh kin khong chua khong khi, sau mot

thori gian thu dugc 4,96 gam chat rfui va h6n hop khf X Hap thu hoan toan X

vao nude de' duoc 300ml dung djch Y Dung djch Y c6 pH bang

->H^ +NO3

0,1M pH = l

Chon D

Cl/ mmMiyDVVnKhang K/c,

Vidu 6 (f)HA 2 0 1 1 ) Trung hoa 3,88 gam h6n hop X g6m hai axit cacboxylic no,

don chufc, mach ho bang dung dich NaOH, c6 can toan bo dung djch sau phan iJng thu dugc 5,2 gam muoi khan Neu d6't chay hoan toan 3,88 gam X thl the tich oxi (dktc) cdn dung la

A 4,48 lit B 3,36 lit C 2,24 lit D 1,12 lit

-^COONa

22 = 0,06 mol Axit no, don chiic, mach ho => 2 dong dang lien tie'p: C-H2-O2

1 Khdi niein gid tri trung binh :

Co cac so duong A, B, a, b > 0 Trong do A < B

Ta C O : A.a < B.a

A.b < B.b (A.a + A.b) < (A.a + B.b) < (B.a + B.b) Chia tat cii cho (a+b), ta duoc: A < i ^ ^ - ^ i i ^ l ^ ^ <B <; •

(a + b)

A = -—'• 5^ ( A : la gia tri trung binh ciia A va B)

(a+b)

2 Cac gid tri trung binh trong bai tap hoa hoc:

Nguyen tu khoi trung binh : nguyen tir khoi ciia nguyen to c6 iihiSu d6ng vj la

nguyen tu khoi trung binh ciia h6n hgp cac dong vj c6 tinh deii ti le phan tram

so nguyen tu ciia m6i d6ng vj

Cong thuc : - _ X i A | +X2.A2 + - + x„.A„

Bp d: Hda hoc 9 on Ihi vAo lO - Pham S/ Ltfu

tir ( n|,n2,n,.-)cua m6i dong vj :

— _ n | A | + 112^2 + n^.Ậ

n,+112+113

Khoi Imng mol trung binh : 'i! ' J

Khoi lugiig mol trung blnh ciia h6n hop la khoi luoiig ciia 1 mol h6n horp c6

tinh den % so mol ciia m6i chat trong h6n hop

- Xet h6n hop X gom 3 chat A, B, C:

M x : khoi luong mol trung blnh ciia h6n hop X

Neu h6n hop gom i chát ta c6 cong thiic tong quat tinh M x :

- _ _ ^ I i i i M i _ Z x , M i

100 Thucnig gap h6n hop gdm 2 chat: A va B

Taco: % A + %B = 1 0 0 % B = ( 1 0 0 - % A )

- a M ^ + b M , X ^ M ^ + ( 1 0 0 - X ^ ) M B

(a + b) 100

- Trong cac h6n hop k h i : % the tich = % so mol

Cdc gid tri trung binh khdc:

S6 nguyen tu trung binh ( C , H , O , N ) , d6 kh6ng no trung binh ( k ), so

nhom chiJc trung binh, phan til khoi trung blnh ( M )

(ạMA+b.MB)

H6n hop

X

A : C ^ H y O , N , (amol) B:CpHqO,N,(bmol)

O/ nun MTV DVVn Khang Viet

Vi du 1 Trong tir nhidn, nguyen to dong c6 2 d6ng vj la 29Cu va 29Cụ nguyen tijr khoi trung binh ciia d6iig la 63,54 Thanh phan philn tram tong so nguyen tir ciia

Vi du 2 Trong tu nhien nguyen to brom c6 2 dong vj trong do dong vj 33 Br chiem

54,5% ve so lugng Biet nguyen tu khoi ciia brom la 79,91 Xac djnh s6' khoi ciia dong vj con laị

Gidi

Goi s6' kh6'i ciia dong vj con lai la Ạ

Phan tram so nguyen tii ciia dong vj c6 so khoi A la: 100% - 54,5% = 45,5%

„ , - 54,5X79 + 4 5 5 X A ^ ^,

Ta co: A = = 79,91 => A = 81

100 Vay dong vj con lai c6 so khoi |A = 81

Vi du 3 Cho 6,2 gam h6n hop 2 kim loai kiem thuoc 2 chu ky lien tiep trong bang

tudn hoan hoa hoc phan ling vdri HiO du, thu dugc 2,24 lit khi (dktc) va dung

djch Ạ Tính thanh phdn % \i khoi lugng tirng kim loai trong h6n hgp ban đụ

Gidi

= (2,24:22,4) = 0,1 mol =>n,,, =2nH2 =0,2 mol

= > M( A , B ) = ^ = 31g/mol Gia su: A < B => A < 31 < B ^ A = 23(Na);B = 39(K)

Vidu 4 Cho 28,1 gam quang d6l6mit g6m MgCOj va BaCOj trong do %MgC03 la

a% vao dung djch HCl du thu dugc V (lit) CO, (o dktc) Xac djnh V (lit)

Bp dSH6a hoc 9 on Uii v6o lO- Phm Sy Lmi

Vay: |3,21it < V c o , <7,49Iit

Vi du 5 Hoa tan 115,3 gam h6n hap gom MgCO, va RCO, bang 500ml dung djch

H2SO4 loang ta thu duoc dung djch A, chat rdn B va 4,48 lit CO (dktc) C6 can

dung djch A thi thu diroc 12 gam muoi khan Mat khac dem nung chat rdn B tori

khdi luong kh6ng doi thi thu duac 11,2 1ft CO (dktc) va chat rdn B, Tinh nong

do mol/lit ciia dung dich H2SO4 loang da dung, khdi luong ciia B, B, va nguydn

tir khdi ciia R Biet trong hdn hap ddu sd mol ciia RCO, ga'p 2,5 Idn sd mol cua

MgCO,

Gidi

Cdng thiic trung binh ciia hdn hop: M C O 3

Tdng sd mol khi COj = ^'^^^^ ' ' ' ^ = 0,7mol 22,4

M = 1 1 ^ - 6 0 = 104,71 1.24 + 2,5.R

0,7 3,5

Mudi khan khi cd can dung djch A la MgS04

nMgSO4=(12:120) = 0,lmol

Chat rdn B gom mudi cacbonat con dir va BaS04

Theo phuang phap TGKL:

= 104,71 =^R = 137 =^RlaBa

V'« Co 0,2 mol gdc C O 3 ' da chuyen thanh gdc SO '4 • 2-

=> Khdi luong tang: 0,2.(96-60) = 7,2 gam

Do CO 12g MgS04 tach ra ntn khdi luong chat rdn B la:

me = 115,3 + 7 , 2 - 1 2 = 110,5gam

0\^X rdn B, gom cac oxit CaO, BaO va BaS04

11 2 m„ = 1 1 0 , 5 - — 4 4 = 'Bl 22,4 88,5 gam

Vidu 6 A la hdn hap gom hai chat ddng ddng lien tia'p cua ancol etylic Cho 7 gam

hdn hap A tac dung het vdi Na thu dugc 1,12 lit khi H , (dktc) Tinh % sd mol va

% khdi lugng mdi chat trong hdn hap A

Gidi "•

Goi 2 chat ddng ddng lien tiep A, B => M R = M ^ + 14

C r P T cua A, B la C„H2n+,0H (a mol) va C„+,H2n+30H (b mol)

CT trung binh cua hdn hop C-H2-^,0H (a + b) mol

24,86%

75,14%

Vi du 7 A va B la hai chat trong day ddng dang ciia etylen Ca hai d^u la chat kh

trong dieu kien thudng 5 gam hdn hap A, B chid'm th^ ti'ch bang thi tich ciia 4,4 gam khi CO trong cung ditu kien ve nhidt dd va ap sua't Xac djnh % v^ thd' tich

ciia mdi chd't trong hdn hc»p

Gidi

Cong tht(c: CT ciia A, B la : C„\{.„ (a mol) va C ^ H , , (b mol)

CT trung binh ciia hdn hop C-H2- (a + b) mol Gia sir n < m va A, B deu la chat khi trong dilu kien thudng

• n nhan 2 gia tri la n = 2 va n = 3 CTPT ciia A la C 2 H 4 hoac C j H ^

• m nhan 1 gia tri la m = 4 CTPT ciia B la C4Hg

vay CTPT ciia A , B l a : C3H4vaC4H8 ( T H I )

Vi du 8 Ddt chay hoan toan 2,76 gam hdn hap X gdm C^H^COOH

C^H^COOCH,, CH,OH thu duoc 2,688 lit C02(dktc) va 1,8 gam H , 0 Mai khac, cho 2,76 gam X phan ung vto dii vdi 30ml dung dich NaOH I M , thi dugc 0,96 gam CH,OH Cdng thiic ciia C^H COOH la

A.C,H,COOH , D CILCOQH C C^HaCOOH D C^H^COOH

T H U V I E N T I W H I I N M THU m 17

B6 dSOda hga^dalhivio lO-Phm3ylHtu

S o d 6 : X + Y + Z + HNO3 XCNOj )3 + YCNOj )2 + ZNO3 + NO2 + N O

Bao toan nguyen to nito :

V I du 3 Khir het m gam Fe304 bang CO thu diroc h6n hop A gom FeO va Fe

H6n hop A tan viTa du trong 0,3 lit dung dich H2SO4 I M giai phong 4 , 4 8 lit khf

(dktc) Tinh gia tri cua m

Vi du 4 (DHB 2012) D6't chay hoan toan 20ml hoi hop chat hihi co X (chi g6m C,

H , O) cdn vira du 110ml khi O, thu duoc 160ml hdn hop Y g6m khi va hoi D i n

Y qua dung djch H2SO4 dac (dir), con lai 80ml khi Z Biet cac the tich khi va hoi

do o Cling didu kiSn C6ng thiic phan tix ciia X la

Vay CTPT : C4H8O => Chon D

Vidu 5. ( C D A B 2010) H6n hop Z g6m hai este X va Y tao boi cung m6t ancol

va hai axit cacboxylic ke tiep nhau trong day dong dang ( M ^ < M^) Dot

chay hoan toan m gam Z cdn dimg 6,16 lit khi 02(dktc), thu dugc 5,6 lit khi

CO2 (dktc) va 4,5 gam H , 0 C6ng thiic este X va gia t n ciia m tuong ling la

A (HCOO),C2H, va 6,6 B HCOOCH, va 6,7

C CH3COOCH, va 6,7 D HCOOC^Hj va 9,5

-n c o 2 = 0,25 mol; -n H 2 0 = 0,25 mol; -n o 2 = 0,275 mol

nco2 = " H 2 0 =^ Z gom 2 este no, don chiic: ^ H ^ p ,

S^d6 : C,.H2„02 - » ^ " ' - ^ ^ - ^ - > > { n H ^ o = n e o 2 =0,25 mol

Bao toan so nguyen tijf o x i : no(esie) + " 0 ( 0 2 p/u) = no(C02) " O C H J O )

=> no(es.c) + 2no2 (p/u) = 2nco2 + n H 2 0 = > "o(estc) =(2nco2 + "H2o)-2no2(p/u)

Vidu 6. H6n hop X g6m 2 chat hiru co A , B (chi co chua C, H , O, N trong phan tir), trong do ti le m , , : m^, = 80 : 21 Thanh phdn % khoi luong cua nito trong h6n

hop X la 10,966% Dot chay hoan toan 3,83 gam h6n hop X can 3,192 lit O, (dktc), thu diroc cac san phdm chay gom khi va hoi la CO,, HjO va N , Toan b6

san phdm chay trdn dem sue vao binh dung dung djch nuoc v6i trong du Tinh khoi luong ket tiia thu duoc

A 13 gam B 20 gam C 15 gam D 10 gam

I .i vi., Gidi

m^ = 3,83X 10,97% = 0,42 gam mo = — x0,42 = 1,6 gam

=> m x = m c + m „ + m o +mf^ => " 1 ^ = 3,83-(0,42 +1,6) = 1,81 gam Goi a la so mol C va b la s6' mol H => 12a + b = 1,81 gam (1)

' [CO2 (a mol)

Xet phan ling chay ciia X : 3,83 gam X—"•''^^-'''""'"2 > H2O — mol

Bao toan kh6'i luong o x i : 32a + 8b = 1,6 + 0,1425 x 32 = 6,16 gam (2)

Co mot so bai toan ta cho dudi dang gia trj tong quat nhu a gam, V lit, n mol

hoac cho ti le the tich hoac ti \t s6 mol cac chat

Nhu vay ket qua giai bai toan khong phu thu6c vao luong chtft tong quat da

Bd dSnde hgc96nthivAo lO-Pham S/lM

— ^ j - gam C O 2 va gam N6'u thfem vao h6n hop X m6t nira luomg A

C O trong h6n hop X r6i d6't chay hoan toan thi thu duoc " ~ ~ g^rn C O j va

^ ^ ^ ^ gam H j O Bigt A , B khdng lam mat mau nirdc Br, Xac dinh CTPT

cua A va B

Gidi

Chon gid tri cho thdng so : Chon a = 41 gam

Dot X: n 'CO2 = ^ = 3'"o' " H 2 0 = — = 2,5mol

v mafA TNHHMTVDVVHKhang Vict

J Phuong phap quy doi hdn hpp nhieu chdt thanh so chat it hon

• Khi quy doi hOn hop X g6m nhieu chat (tuf ba chat tro len) thanh h6n hop hai chat hay chi con mOt chat ta phai dam bao nghi6m diing djnh luat bao toan (s6'mo! nguyfen tir, khoi luong) ' : " "

• Co thi quy doi h6n hop X v^ ba't ky cap chat nho, tham chi quy d6i \i mot

chat Tuy nhien ta nen chon cap chat nao don gian c6 it phan iJng nhat de don gian viec tinh toan

• Khi quy doi h6n hop X ve mot chat thi chat tim duroc c6 the la gia djnh khOng C O thifc

Vidu 1 H6n hop X gom cac oxit sat sau: FeO, Fe^O, va Fe304 trong do so mol cua

FeO bang s6' mol ciia Fe^O, Cho 2,32 gam h6n hop X tac dung vira du vdi Vml dung dich HCl I M T i m V

Gidi

Dung phep quy d6i:

X :

FeO (a) Fe20,(a) Fe304(b)

232 = 0,01 mol PTHH: Fe,04 + 8HC1 FeCl2 + 2FeCl3 + 4 H 2 O

mol 0 , 1 0 ^ 0,80

Z ^ O H C I = 0 , 0 1 x 8 = 0,08 m o l = > V H c i I M = — = 0,08 lit=|80ml

1

Vi du 2 Cho 11,6 gam h6n hop Fe^O, va FeO c6 ti le mol 1:1 vao 3(X)ml dung djch

HCl 2M duoc dung dich A (a) Tinh nong do mol cac chat trong dung djch sau phan ufng (th^ tich dung djch thay doi kh6ng dang ke)

(b) Tinh the tich dung dich NaOH 1,5M di tac dung hd't vdi dung dich A

Gidi

(a) Quy doi h6n hop:

^ JFeO (a) 1 ,.y.6, Fe,o^(a)|n = a = - ^ = 0,05 mol

Fe203(a)J n^ci = 0 , 3 x 2 = 0,6 mol PTHH: Fe304 + 8HC1 -> FeCl2 + 2FeCl3 + 4 H 2 O mol 0,05-^ 0,6 '')<JJ|

Vi du 2: Hoa tan 200 gam S O , vao m gam dung djch H 2 S O 4 49% ta diroc dung

dich H,S04 78,4% Ti'nh gia tri cua m,

=> irij = — X 200 = 300 gam

Vidu 3: Phiii them bao nhieu gam nude vao 200 gam dung djeh KOH 20% dd dugc

dung djeh KOH 16%

Gidi Dung moi nude dugc quy doi thanh dung djch c6 nong do chflit tan: 0,0%

20 16

m ll2()

"^KOH 20% 16 • m )l70

= 2 0 0 x - = 50 gam

Vi du 4 Dc didu che 560 gam dung dich C U S O 4 16% can phai lA'y bao nhidu gam

dung dich C U S O 4 8% va bao nhieu gam linh the CuSO^-SHp

Giai

Cach 1: phuoiig phap duonig eheo

160 Quy doi: CuS0 5H20=>%Cf.„c(> = 100% = 64%

,ir,> ijfro'ub <'h xy-' ,

' ioa iiii'o <i-j'i

= — = - ::ib vtlu i'.n k

560

'Cu.sO4.5H2O ~ j X I = 80 gam va Wf 'CUSO4 m 480gam

Vi du 5: Can bao nhieu lit axit n,S04 (D = 1,84) va bao nhieu lit nuoc cat dc pha

thanh 9 lit dung dich H.SO^ c6 D = 1,28 gam/ml? "v - i' ^

' H 2 0

V U , O = 2 V H2SO^ ( * )

3 lit

Dgng 2 Bdi toan xdc djnh thanh phdn hSn h^p

Vidu I Nguyen tir kh6'i trung binh ciia brom la 79,319 Brom c6 hai d6ng vi b^n:

35Br va jsBr Tinh thanh phSn % sd'nguyen tif cua 3^Br

Vi du 2 Mot h6n hgp gom O., O, b didu kien tieu chudn c6 ti kh6'i hcri vdri hidro la

18 Ti'nh thanh phfin % si the tich cua O, trong h6n hgp

C O , + N a O H - > N a H C O j (1)

[ C O 2 + 2 N a O H ^ Na2C03 + H 2 O ( 2 ) ^

T i n h s 6 - m o l m u 6 ' i : ^^^^ ^^^^^

_^ j n C 0 2 = "NaHCO, + "NajCO^ _ "Na2C03 = ("NaOH " "CO2 )

"NaOH "^"NaHCOj '^^^Na2COi "NaHC03 - ( 2 " C 0 2 "NaOH )

• "NaHCOj =(2nc02 " " N a O n ) (*) nNa2C03 = ( " N a O H " "cOj )

- Gia thi^'t iru tidn tao thanh mu6'i axit va sau do kifem d u chuy^n mu6'i axii

thanh muoi trung hoa :

| C 0 2 + N a O H - > N a H C O j ( l )

N a H C O , + N a O H ^ Na2C03 + H 2 O (3)

Tinh Sd'mol mu6'i : nNa2C03 =nNaOH(3) =("NaOH - " C O 2 ) ( * * )

^ (**)

Bao toan cacbon: n^aHcOj = " c o 2 -"Na2C03 2"C02 "NaOH

- Bai toan xay ra trong tinh huong sue khi CO, v a dung djch kidm nen uu tidn tao

ra mu6'i trung hoa va sau do la khi CO, d u chuyen mu6'i trung hoa thanh mu6'i axit

C O 2 + 2NaOH ^ Na2C03 + H 2 O ( 2 )

C O 2 + H 2 O + Na2C03 ->• 2NaHC03 ( 4 )

T i n h sd'mol m u 6 i :

(4)=>"NaHC03 =2nco2(d.r) = 2 ( n c 0 2 -O-^nNaOH ) = (2"C02 " " N a O n ) (*)

Bao toan cacbon:

"Na2C03 = " C 0 2 ~"NaHC03 = "CO2 ~ ( 2 " c » 2 ~ "NaOH ) ("NaOH ~ " C O 2 )

Nhdn xet: Sir tao thanh san ph^m chi phu thu6c s6' mol chat tham gia phan ihig

khdng phu thuoc qua trinh phan ling

Ne'u: n CO2 < "NaOH < 2nco2 ta lu6n c6 :

"NaHC03 - ( 2 " C 0 2 " " N a O n ) (*) va Na2C03 NaOH " C O 2 (**)

1 BAI T O A N THUAN

Biet so mol CO2 va so mol kiem Tim so mol muoi

Sir dung cap phan ung (1) va (2) de xet gi6i han t i le m o l ;

V J du 1 Nung 13,4 gam h6n hop 2 muoi cacbonat ciia 2 k i m loai hoa trj 2, thu

duoc 6,8 gam chat rdn va khi X Lucmg khi X sinh ra cho ha'p thu vao 75ml dung

djch N a O H I M , kh6'i iuong mud'i khan thu dirac sau phan limg la (cho H = 1,

Vidu 2 Hap thu het 2,688 l i t CO, (dktc) bang 200ml dd N a O H I M Sau th6i gian

phan ung, c6 can dung dich thu duoc m gam chat rdn T i m m

Gidi i»

2 688

n c o 2 = ^ ^ = 0,12 mol < n^aOH = 0 , 2 x 1 = 0,2 mol

Tao thanh h6n hop 2 muoi N a H C O , va Na,CO,

Vr J « 3 ( C J ) A B 2012) Ha'p thu hoan toan 0,336 i.'t khf CO, (dktc) vao 200ml dung

dich gom N a O H 0 , 1 M va K O H 0 , I M thu dirge dung djch X Co can toan bo dung djch X thu duoc bao nhidu gam chat riin khan?

A 2,44 gam B 2,22 gam C 2,31 gam D 2,58 gam

33

Gidi

Quy d6i dung djch h6n hop NaOH 0,1M va K O H 0,1M thanh 1 chat la M O H 0,2M

" M O H = 0-2.0,2 = 0,04 mol va nco^ = 0,336 : 22,4 = 0,015 mol

^ JlMOH ^ 2:2^ = - > 2 =^ Tao M , C O , va M O H con d u • '

Biet so mol mud'i Tim so mol CO2

Vidu Tinh the tfch C O , a dktc cdn cho vao 100ml dung dich K O H 2 M de thu diroc

Dgng 2 Todn C O , td c di/ng vdi dung djch CalOH) , ha y BoIOH),

Cac phan i/ng c6 the xay ra :

CO2 + Ca(OH)2 - > C a C O , + H2O ( 1 )

T i i f ( l ) v a ( 3 ) t a c o : • • " C 0 2 ( l ) ~ " C a ( O H ) 2

* " C 0 2 ( 3 ) - " C O 2 ~ " C 0 2 ( 1 ) - " C O 2 ~ " C a ( O H ) 2

" C a ( H C 0 3) 2 = " C 0 2 ( 3 ) = ( " C 0 2 - " C a ( 0 H ) 2 ) ( * ) Bao t o a n sd n g u y d n tijT c a c b o n : n c a C 0 3 = " 0 0 2 - 2 n c a ( H C 0 3 ) 2

Biet so mol CO2 vd so mol kiem Tim so mol mud'i

Six dung cap phan ling (1) va (2) d6' xet gidi han t i 16 m o l ; Bien l u i n theo cac phan umg (1) va (2): Dat T = " " ^ ^ = -

n C a ( H C 0 3) 2 = "COj - " C a ( O H ) 2

n C a C 0 3 = 2 n c a ( O H ) 2 - " C O 2

n C a C 0 3 = 0 Tuong tir khi thay C a ( O H) 2 bang Ba(OH)2 •

35

Cac PTHH

H o a c :

Vidu 1 Hap thu hoan toan V h't khf C02(DKTC) vao 600ml dung djch Ca(OH)2

0,5M thu duoc 10 gam ket tua Tinh gia tri cua V

"caC03 ~ "ca(OH)2 ~ ^'-^^ ""^^ C6 2 trucrtig hop :

T H l : chi xay ra phan ling (1) Ca(OH)2 du

^ "CO2 = "caCOj = 0,10 mol => |V= 2,24 lit

TH2: xay ra 2 phan ling (1, 2) hoac (1, 3):

^ "CaC03 =2nca(OH)2 " C O 2

(1) : o H , •A '

(2) ',r,«,,

(1) a (3)

.i4V

Cac PTHH

Hoac :

=> n c o 2 =2nca(OH )2 -ncaC03 - 0 3 x 2 - 0 , 1 =0.5 m o l ^ | V = 11.2 lit|

Vi du 2 Ha'p thu hoan toan x mol C O 2 vao dung dich chira y mol Ba(0H)2 thu duoc

5,91 gam k6't tua Neu ha'p thu hoan toan 2x mol C02vao dung dich chiia y mol

Ba(0H)2 thi luang kd't tiia thu duoc la 7,88 gam Xac dinh cac gia trj cua x va y

X€i thi nghidm 2:

Ne'u Ba(0H)2 du thi luomg BaC03 ga'p d6i luang BaC03 cua TN1 Trai v6i d^ ra

la 7,88 gam Vay Ba(0H)2 thieu va m6t phdn BaCOjda tan trong COj

=^ "caco., = 2nB,(OH)2 " "coa => 2y - 2x = 0,04 (1)

Xet thi nghiem 1:

Nd'u Ba(OH)2 cung thia'u nhu TN 1 thi trong TN2 khd'i luomg BaC03< 5,91

gam Trai v6i de ra la 7,88 gam Vay Ba(0H )2 ^^"g nghiem 1

^ " & 1 C O 3 - " C O 2 ~ ^ ~ ^ ' ^ - ^ Tir (1) => y = 0,05 mol

VrrfM3 (CDAB2010)Ha'pthu hoan toan 3,36 lit khf C02(dktc) vao 125 ml

dung djch Ba(OH), I M , thu duoc dung dich X Coi the tich dung dich khdng

thay d6i, n6ng do mol cua chat tan trong dung djch X la

A 0 , 6 M B.0,2M C.0,1M D 0,4M

36

Cty mnn MTV D Wn Khan^ Vict

Gidi

n^y^ =(3,36:22,4) = 0,15 m o l ; nB,(OH)2 = 0,125x 1 = 0,125 mol

, = 1,2 < 2 ^ Co 2 muoi: BaCO, va BaCHCOj), ' ' ' ^ ' •

' 0,125 • ' ' ' -y^^^ir

Tif cac PTHH, ta c6:

nBa(HCO,)2 - " C 0 2 - " B a , O H)2 =0.150-0.125 = 0.025 mol • nBaC03 =2nB,(OH)2 - " C 0 2 =2x0,125-0,150 = 0,10 mol :£ ^H-> ,

Cha^ tan trong dung dich la BaCHCOj),: , ^,,(1 ,|))fj ; / i

0,025 ' Qj^jj

N 6 n g d 6 : C^^^,^^^^^^^ ^5^^ = "'^^ =^ C h o n B

1 B A I T O A N N G H I C H

Biet so mol Ca(OH)2 vd so mol ket tua Tim so mol CO2

De giai quyet bai toan ta so sanh so mol Ca(0H)2 vdi so mol CaCOj

- Neu ncaco3 - "ca(OH )2 • ''^y ""^ P^^"' if'ig ( ' ) , C O 2 va kiem tac dung viia

- Neu Hcaco, < nca(Ofi)2 • 2 truong hop xay ra

TH1: C O 2 thieu, chi xay ra phan ung (1), Ca(0H)2 du i=> n^Q^ = ric^coj

TH2: Ca(0H)2 thieu, Ca(0H)2 tao ket tiia hoan toan theo phan ling (1),

luong C O 2 con lai hoa tan mot phan CaCO, theo phan ung (3)

"C02 =2n^^^.,((j„,^-n(-.,,co3 =»j^j^ ncaC03

( G i i i i tuong tu khi thay Ca(OH )2 bfmg BaCOH)^)

Vidu 1 Thoi V lit (dktc) khf SO, vao 200ml dung djch Ba(OH), 1,25M thu duoc

32,55 gam ket tiia Ti'nh V

Gidi ; ,,;•„, ,

Ba(OH), + SO3 > BaSOj + H.O (1) ,, / , :

Ba(0H)3 + 2SO3 — > Ba(HSO,): (2) ' • ^ ^ryx

D6 dSnda hoc 9 6nUuvio lO- Pham S/LiTu

nso2 =2iiBăOH)2 -"ft-cọ, = 2 x 0 , 2 5 - 0 , 1 5 = 0 , 3 5 mol

|V = 0,35 • 22.4 = 7,84 lit

Vidu 2 Hap thu hoan toan 2,688 lit khf CO, (đktc) vao 2,5 1ft dung djch BăOH)-,

nong do a mol/1, thu duoc 15,76 gam k^'t tiiạ Gia trj cua a la (cho C = 12, O =

16, Ba= 137)

Ạ 0,032 B 0,048 C 0,06 D 0,04

(jiai

nco, = (2,688:22,4) = 0,12 mol > nR^co, =(15,76:197) = 0,08 mol

V i khf CO, bj hap thu hoan toan nen da xay ra cac phan irng :

Dgng 3 To6n CO, tdc dyng vdi dung djch NaOH vd BăOH)j

Xet bai toan theo 2 buoc:

- Bifoc 1: xet sir tao thanh s6' mol cac ion CỐ va HCÔ dira vao ty le so

mol CO, va OH

- Budc 2: tinh so mol kd't tiia CaCOj theo s6' mol CỐ va s6' mol mu6'i tan

theo so mol HCOJ

" c e o , =(2"c;,(OH)2 " " ^ - 0 2 ) = ( " O H - - " C O 2 ,

Tircnig tir khi lhay CăOH),bAng Bă0H)2va NaOH bang KOH

Vi du 1 (f)HB 2012) Sue 4,48 lit khf CO, (dktc) vao 1 lit dung dich h6n hop

BăOH), 0,12M va NaOH 0,06M Sau khi cac phan ling xay ra hoan toan thu

dugc m gam kct tiiạ Gia trj cua m la

Ạ 19,70 B 23,64 C 7,88 D 13,79

Gidi

Quy d6i BăOH), 0,12M thanh NaOH 0,24M => Ta c6 NaOH 0,30M

nco, = 0 2 ] nN,oH 0,3 ^ ^ , ^ , [NaHCO,

2 lz::>l<-Jli!i2H =_L_ = |,5<2=::.C6 2mu6i <^

Tircac PTHHtatinh diroc:

nNa2CỌ, =nNaOH - " c o j = 0 , 3 - 0 , 2 = 0 , 1 mol

nNaHC03 = 2 n c o 2 -"NaOH = 0 , 4 - 0 , 3 = 0,1 mol

Do lofp 9 chua hoc s u dien li ntn ta dung phuong phan quy doi h6n hop NaOH

va BăOH), thanh chat tirong duong la NaOH

Thuc te trong dung djch c6: ion C O 3 " (0,1 mol); ion H C O 3 (0,1 mol);

ion Nâ (0,06 mol) va ion Ba-* (0,12 mol)

Ion Ba"* con du sau phan ling tao ket tiia: 'r;

Vidu 2 Cho 0,448 1ft khf CO, (0 dktc) htfp thu het vao 100 ml dung djch chíra h6n

hop NaOH 0,06M va BăOH), 0,12M, thu duoc m gam ket tiiạ Tính gia tri ciia m

Ạ 3,940 B 1,182 C 2,364 D 1,970

Gidi

Hoan toan tuong tu vf du 1 Ta c6 :

nco2 =0,02 m o l ; n^2+ =nBăOH)2 =0,012 mol

" O H - ="NaOH +2nBăOH)2 = 0,1(0,6 + 2 X 0,12) = 0,03 mol

1< OH"

^C02

0,03 0,02 = l , 5 < 2 = > C 6 2muoi

^' du 3 ( » H A 2 0 1 1 ) Háp thu hoan toan 0,672 1ft khf CO, (dktc) vao 1 1ft dung

djch g6m NaOH 0,025M va CăOH), 0,0125M, thu duoc x gam ket tiiạ Gia tri

Tir cac PTHH ta tfnh duoc: iiNajCO, = "NaOH - "coj = ^.05 - 0,03 = 0,02 mol

nN,HCỌ, = 2 n c o 2 - " N a O H = 0 ' 0 6 - 0 , 0 5 - 0 , 0 1 mol • , v >i,á<

Thuc tétroiig dung djch c6 : ion CO3"(0,02 m o l ) ; ion HCO3 (0,01 mol); '

ion Na* (0,025 mol) va ion Ca"" (0,0125 mol)

Ion CỖ con du sau phan úng tao ket tua :

Ca-^ + COJ- >CaCO, i fx-0,0125x 100= 1,25 gam

hay : NaOH + A K O H ) , NaAlOj + 2H2O (2)

PTHH dang ion thu gon: A l ' " + 3 0 H -> Al(OH),

Al(OH)3 + OH ^ AIO2 +2 H2O

Hay : 4NaOH + A I C I 3 ^ NaAlÔ + 3NaCl + 2H2O (3 )

^ "NaOH(l) - ^ " A 1 C 1 3 ' "Nă)H(2) ^ ' ^ " A I C I ^ ' "NaAI02 = " A I C I 3 ( 2 )

= > H e P T :

^" • " — "NaAI02 - ( " N a O H -3n^lQ^ j (*)

»AI(OH)3 = ( 4 n A I C l 3 """NaOn) ( * * ) Vay neu c6 : 3 n A i c i , <nNaOH < 4 n A i a 3 ^''^ ^"A,-^+ < " O „ - ^ ' ^ " A I - ' *

"A1(OH)3 +"NaAI02 " "AICI3

3"AI<OH)3 "*"^"NaA102 = "NaOH

Oy mm MTVDVVn Khang Viet

Thi CO 2 san ph^m Al(OH), va NaA102 vdi cac lien he so mol voi cac chat ban

(jau tham gia phan ling nhu sau :

nNaAi02 - ( " N a O H - 3" A i c i 3 ) ^ay n^^^ =(n 'AIO2 \ " O H " " ' A | - ' + ) _ - 3 n ( * )

; ^ A I ( O H ) 3 i = ( ' ^ " A l C l 3 - " N a O H ) | ^ay "Al(OH)3i " ( ^ " A I - ^ ^ ~ " O H " ) ( * * )

N6'u mudi tham gia phan úng la Al2(S04),: cong thuc dang phan tu diroc viet

lai nhu sau :

" A I ( 0 H ) 3 i ~ ^ " A l 2 ( S 0 4 b ~"NaOH

"NaA102 - "NaOH - ' ' ^ " A I 2 ( S 0 4 ) 3

(**)

(*)

• Trong dung dich gdm h6n hop cac ion (axit), M""^ (tao hidroxit k6't tua) va

Ar^^chap nhan c6 3 nhom phan ling xay ra theo thu tu nghiem ngat uu tien

v^ thcri gian la:

Trung hoa - tao cac ket tua hidroxit - hda tan A l ( O H ) ,

X n = n + n , + n

O H H* O H (taocac>lhidroxil) O H (hoa tan iAI(()n)3)

• Ki^m yd'u: dung djch NH3 thi AI(0H)3 khong tan trong ki^m du:

3NH3 + 3H2O + AICI3 A1(0H)3 + 3NH4CI

Dang ion thu gon: Al^" + 3 NH3 +3 H2O ^ AI(OH)3+3 NH4 ; iiifv"

liSdSH6ohpc9Sn Uil vAo 10 - Pham S/ UT

\idu 1 Tron dung djch chira a mol A l C l , \6\g djch chiia b mol NaOH De thu

duoc ket tua thi cdn c6 ti' le

A a : b = 1 : 4 B a : b < 1 : 4 C a : b = 1 : 5

Gidi

3NaOH + AICI3 -> Al(OH)3 + 3NaCl

4NaOH + AICI3 ^ Na[ A1(0H)4 ] + 3NaCl

' H a y : 4 N a O H + A I C I 3 N a A I 0 2 + 3 N a C l + 2H2O (3")

Ket tua cue dai khi: n^aOH = 3nA|ci^ => b = 3a

: K€i tua tan bet khi: n^aOH - ' * " A I C I 3 b = 4a

1 C O ket tua Al(OH)3 :

n N a O H < ' * n A i c i , => b < 4 a = > l < 4 - = > - > - = > C h o n D

b b 4

Vidu 2 Nho tir tir 0,25 lit dung djch NaOH 1,04M vao dung djch gom 0,024 mol

j FeCl,; 0,016 mol Al2(S04X, va 0,04 mol H2SO4 thu diroc m gam ket tua Gia tri

cija m la

A. 4,128 B 2,568 C 1,560 D 5,064

Gidi

C a c P T H H : 2NaOH + H2SO4-> Na^SOj + 2H2O (1)

3NaOH + FeCl3 -> 3NaCI + Fe(0H)3 ^ - (2) 6NaOH + AI2 (SO4 )3 -> 3Na2S04 + 2A1(0H)3 (3) 8NaOH + AI2(804)3 -> 2NaA102 + 3Na2S04 + 4H2O (4)

"NaOH = 0'26 (mol), nFeci3 = 0,024 (mol)

Ket tua A1(0H)3 tao thanh sau khi trung hoa va ket tua het Fe'^

AI(OH)3 tan mot phiin trong NaOH (hay O H " ) : '

Tinh lugng A l ( O H ) , va NaAlO, theo cac c6ng thiic (*) vh (**)

• Ne'u dung djch ban dau g6m AlCI, (hay Ar^"") va HCI (hay ) thi: >

AI(OH), tao thanh sau khi da trung hoa het a x i t H * Do do :

( " N a O H ~ " H C | ) - ( 4 n A I C l 3 " " A K O I D I I )

hay ^ ) H " V^ = ( ' * - " A - ^ - - " A I ( O H ) ,

Vidu 1 Cho 200ml dung djch AlCI, 1,5M tac dung vdi V lit dung djch NaOH

0,5M, lucfng ket tua thu duoc la 15,6 gam Gia tri Idn nha't ciJa V la (cho H = 1,

«A1(0H)3 tan m6t phdn trong N a O H

P T H H : 3NaOH + AICI3 -> A1(0H)3 + 3NaCI ( I )

4NaOH + AICI3 ^ Na[AI(OH)4 ] + 3NaCl (2)

^ ' ^ ' ^ ( 2 ) ^ % < O H ) 3 = ' ^ " A C l 3 - n N a O H

"NaOH =4nAici3 ~ " A I ( O H ) 3 = 4 X0,3 - 0 , 2 = 1 mol

= > V = i i ^ = 2 1 i t = ^ C h o n D 0,5

^' du 2 Cho V lit dung dich NaOH 2 M vao dung djch chiJa 0,1 mol A l j (SO4 )j va

0,1 mol H2SO4 den khi phan ling hoan toan, thu duoc 7,8 gam ket tua Gia trj

I6n nha't cua V de thu duoc luong ket tiia tren la

A 0,45 B.0,35.' C 0,25 D 0,05

Gidi • ^

"NaOH = 2V (mol) , , , V " •

" A I( O H ) 3 " ^ " A ' 2 < - ^ 4 > 3 =0'2 mol

Lircmg NaOH da dung nhieu nha't => AKOH), da tan m6t pMn

Vay C O cac phan irng : 5, ,ji>^

6NaOH + AI2(804)3 3Na2S04 + 2A1(0H)3 (2)

8NaOH + Al2(S04)3 ^2NaA102 +4Na2S04 + 4 H 2 O (3)

Ta C O : (n^aOH - 2nH2S04) = (4nAici3 - " A K O H ) , )

=> ("NaOH - 2nH2SO4) = ( 8 n A i 2 ( S 0 4 - " A K O H ) , ) O 2V - 2 X 0,1 = 8 X 0,1 - 0,1

2 V - 0 , 9 m o l V - — = 0,45 1ft 0 9 Chon A

Vidu 3 Them m gam kali vao 300ml dung djch chura Ba(OH), 0,1M va NaOH 0 , i M

thu duoc dung djch X Cho tir tu dung dich X vao 200ml dung dich AU(SO^),

0,1M thu duoc ket tua Y De thu duoc luofiig ket tiia Y Ion nha't thi gia tri ciia m la

Ket tiia gom BaS04 va A1(0H)3

Kd't tua lorn nha't => Al(OH)3 khong bi hoa tan ^

Vi du 4 (J)HA 2012) Cho 500ml dung djch Ba(OH), 0,1M vao Vml dung dich

Al2(SO4),0,lM, sau khi cac phiin ung ket thiic thu duoc 12,045 gam k6't tiia Gia

NS'u chi C O ket tua BaS04, AKOH), tan het theo phan ling (2)

T a c o : mBa so4 = 0,05x233 = 11,65 gam <12,045 gam Trai de ra

vay : C O 2 phan li-ng ( I ) va (2): kd't tua c6 BaS04 va Al(OH)3

m | = n i B a S 0 4 + " 1 A I ( O H ) 3 ( 1 ) ~ ' " A 1 ( O H ) 3 ( 2 )

^ 233 X 0,3V+ 7 8 ( 0 , 2 V - ( 0 , 1 - 0 , 6 V ) ) = 12,045

|V^,0,15 1ft = 150ml| Chon D

Dgng 2 Mu5i k§m tdc dyng vdi dung djch ki^m mqnh

Phuffng phdp gidi hodn loan tuang tu mud'i nhom tdc dung v&i dung dich kiem manh

Bao toin s6' nguyen tu kem : nz„(OH)2 = "Z"Ci2 " "Na2Zn02

=> nZn(OH)2 ""ZnCl2 "(O.SnNaOH " "Z11CI2 ) = (2nZnCl2 -O.SnfjaOH) ( * * ) Ket luan:

• ZnCl2(hoac Zn^"") du (hoac vira dii), NaOH thie'u

Néu : 2 < < 4 => Co 2 san phdm : Z n ( O H ) , va NâZnO,, c h i t tham gia

"ZnCl2

N a O H va ZnCU d^u hét

"Zn(OH)2 - 2 n y „ Q 2 ~^>5'lNa NaOH

"Na2Zn02 " ^ ' ^ " N a O H ~ " Z n C l 2

(**) (*) hay

nz.i(OH)2 = 2 " z „ 2 + - 0 ' 5 n O H "

n 1 - 0,5n

V.nỐ O H Zn 2+

• NaOH thieu :

Vi du Ị T m h the tich dung djch NaOH I M can cho vao 200ml dung djch Z n C l ,

2 M dc dugc 29,7 gam ket tuạ

J "/,nci2 > "zn(OM)2 " ^ ' " ^ => 2 tru6ng hofp :

"zn(OH)2 ~ 0 ' 5 " N a ( ) H

" N a O H = 2 n z , „ ( ) „ ) 2 = 2 X 0,3 = 0,6 V = 0,6 lit

•NaOH hoa tan I phfin hidroxit luong tinh : ny^^^Q^^^^ - ^^yA\C\i ""^-^"NaOH

"NaOH = ' ^ " / M a 2 -2'izn(OH)2 = 1,6 - 0 , 6 = 1 mol ^ V = 1 lit

Vidti 2 ( J ) H A 2009) Hoa tan het m gam ZnS04 vao nuoc duoc dung djch X Cho

110 m l dung dich K O H 2 M vao X duoc a gam ket tuạ M a t khac neu cho 140ml

dung dich K O H 2 M vao X thl cung duoc a gam ket tuạ Gia tri m la :

Ạ 20,125 B 12,375 C 22,54 D 17,71

Gidi

Z n S 0 4+ 2 K a O H ^ Z n ( O H ) 2 + K 2 S O 4 (1)

Z n ( O H ) 2 + 2 K O H - > K 2 Z n 0 2 + 2 H 2 0 (2)

- Trong thf nghiem 1: neu ZnS04 het, K O H con d u thi trong thf nghiem 2 voi

luong K O H nhi^u han luong ket tija se be hon a gam Vay trong thf nghiem 1,

K O H thieu va ZnS04 dụ

nzn(0H)2 =0,5nKOH=^ ^ = 0,5x0,22 = 0,11

' * Trong t h i nghiem 2: Neu K O H cung thieu va ZnCU tiep tuc d u nhu trong tin

nghiem (1) thl luong ket tiia da nhi^u hon cua t h i nghiem (1) Vay trong thf nghiem

(2), ZnS04 thieu va K O H da hoa tan mot phdn Z n ( O H ) ,

L0AI3:UmG mCH AXITTAC DUNG raiC/Vdl DUNG DjCH MU6l ALUMINAT

(1) V 6 i cac axit manh, A K O H ) ^ tan trong luong axit dụ

"A1CI_, - : ^ ^ H C \ ( 2 ) = ' ^ ( " H C I ~"NaAI()2

Bao toan só nguyen t u nhom : n ^noH), = nN,Ai02 " I A I C I ,

(2) K h i C O , tac dung voi dung djch N a A l O , tao A l ( O H)3 nhung Ai(OH)3

khong tan trong luong CO, dụ

NaA102 + CO2 + 2H2O Al(OH)3 + NaHC03

47

Bid't H H C I (hay ) va n^aAiOz (hay n^|^_ ) tim n

- Neu 1 < T < 4 hay b < a < 4b hoa tan m6t phdn A K O H ) , :

Tinh lirqmg saii philm theo cac lien h6 (*) va (**)

• Neu dung djch ban dfiu c6 N a A l O (hay A I O 2 ) va NaOH (hay O H ' ) thi:

Ket tiia A I ( O H ) , chi duoc tao thanh sau khi da trung hoa het O H "

Hay ( V - " 0 H - ) = 4n _ - 3n AIO2 Al(OH), y

Vidu 1- Mot dung djch A c6 chiia NaOH va 0,3 mol NaA102 - Cho 1 mol HCI vao

A thu duc«c 15,6 gam ket tua Khoi luong NaOH c6 trong dung dich A la

A 32 gam hoac 16 gam B 32 gam hoac 28 gam

C 32 gam hoac 8 gam D 32 gam hoac 14 gam

Gidi

C a c P T H H : NaOH + HCI ^ NaCl + H j O (1)

NaAI02 + H C I + H 2 0 ^ A K O H ) , + NaCl (2)

A l ( O H ) 3 + 3 H C l - > A l C l , + 3 H 2 O (3)

" A K O H ) , = '7^^ = (^'2 mol <0,3 mol = nMaAi02 • hai truong hop:

+ T H I axit thieu: chi CO phan irng (1), (2) xay ra:

("HCI - "NaOH AKOH), • (I - "NaOH) = (^>2 n ^ ^ H = 0,8 => m = 32gam + T H 2 axit dir: phan irng (3) c6 xiiy ra nhung khong hoan toan

HCI - n NaOH 4 " N a A I 0 2 - 3 " ^ , ^ , 3 „ , ^ < = > ( ' - n N a O H ) = 4 x O , 3 - 3 x O , 2

=> n|vj|()j, = 0 , 4 => m = I6gam => Chon A

Vi du 2 (WHA 2012) Hoa tan hoan toan m gam h6n hop gom N a , 0 va AKO, vao

nucfc thu duoc dung dich X trong suot Them tir tir dung dich HCI I M vao X, khi het lOOmI thi biit dau xuA't hien ket tiia; khi het 300ml hoac 700ml thi deu thu duoc a gam ket tiia Gia tri ciia a va m Ian luot la

I Them 100ml HCI (0,1 mol) bat ddu c6 ket tua: phan ling (3) hoan toan

- HHCI """NaOH = ( 0 , 3 - 0 , 1 ) = 0,2 a = 7 8X 0 , 2 = [ l 5 , 6 g a m AI(OH)3

T h 6 m 7 0 0 m l ( 0 , 7 m o l ) cQng t h u duoc a g a m k e t tua: A K O H ) , tan m o t phai

t r o n g H C l

( " H C I ~ " N O O H )

-( 0 , 7 - 0,1) = (4nN,A,02 - 3 x 0 , 2 ) : ^ nN,Ai02 = 0 - 3 m o l

Bao l o a n s 6 ' n g u y ^ n tir N a va A l : n^j^o^ = - n N a A i 0 2 = 0 1 5 m o l

"NazO = |(nNaOH + nNaAI02 ) = j c " ' ' + ^,3) = 0,2 m o l

m = 62 n N„20 + ' 02 n A i = 62 X 0,20 + 102 X 0,15 = |27,7 g a m C h o n A

LOAI 4 : DUNG DjCH AXIT TAC DUNG TCT TC/ V 6| DUNG DjCH MU(5l CACBONAl

(HOAC MU6l SUNFIT)

• So sanh lire a x i t : H , C O , la axit m a n h hon HCO3

=> N a j C O , CO t i n h bazo m a n h h o n N a H C O ,

{I) Trinh tu uu tien cua phdn itiig giua muoi cacbonat vd axit khi them tCc tit

axit vao dung dich muoi Id:

- H o a c ( I ) + ( 2 ) :

d u N a H C O , (hay

"CO2 ~ " H C I "Na2C03 hay: " C 0 2 = " H - " „ „ 2 -

Cac phan lirng x a y ra hoan toan; the' ti'ch k h f la'y o d k t c Ti'nh V va V , theo a, b

Gidi

H C l + N a 2 C 0 3 ^ N a H C O , + N a C l (1)

N a H C 0 3 + H C l N a C l + C O j + H2O ( 2 )

V i sau phan u n g t h u d u o c k h i nen phan l i n g ( 2 ) da x a y ra

Suy ra N a 2 C 0 , da phan ung het or ( 1 ) S6' m o l H C I c o n lai sau phan u'ng( 1) la : (a-b) m o l

"Naiico3 = b ( m o l ) > "HCI (du) = (^1" b) ( m o l )

= ( a - b ) :

(a) P T H H :

"CO2 ~ " H C I (dir) V = 2 2 , 4 ( a - b ) l i t (b) N a 2 C 0 3 + 2 H C i - > 2 N a C l + H 2 0 + C 0 2

G i a sir N a C O , phan urng het t h i so m o l H C l can d u n g l i i 2 b m o l M a theo d^ bai

t h i a < 2 b vay H C l phan u n g het, N a C O , con d u

n^o^ = 0 , 5 a ( m o l ) = > V, = 1 l , 2 a ( l i t )

du 2 C h o tir t u d u n g djch chiia a m o l H C l vao d u n g d j c h chiJa b m o l N a ^ C O ,

d o n g t h d i k h u a y deu, t h u dugc V l i t k h f ( d d k t c ) va d u n g djch X K h i c h o d u nuoc v o i t r o n g v a o d u n g djch X thay c 6 xuQ't hien ket tiia Bieu thuc lien he giua

Bo dS' Hda hoc 9 on Uii vAo lO - Pham Sf Im

"NaHCO., = b ("lol) > nHci (du) = " b) (mol)

nco2 = " H C I (du) = (a - b) => | V = 22,4(a - b)lft Chon A

Vidu 3 Nho tir tir tirng giot den het 30ml dung djch HCl I M vao 100ml dung dich

I chu-a NajCOj 0,2M va NaHCO., 0,2M, sau phan ling thu diroc so mol C O , la

NaHCO, + HCl -> NaCl + C O j + H 2 O (2)

" C O 2 = " H C I - " N i n c o , = 0 , 0 3 - 0 , 0 2 = 0,01 mol => Chon I)

LOA/5: KIM LOAI SAT T A C D U N G V6l DUNG DICH AXIT S U N F U R I C

Fe tac dung voi H 2 S O 4 c6 cac trirong h(.rp sau day xay ra :

H,S04 dac nguoi : thu dong hoa hoc siit (khong phan u"ng)

H,S04 dac va nong : 2Fe + 6H ,S04 -> Fe2 (SO4), + 3 SO2 + 6 H2O (2)

Neu Fe con du: Fe + Fe2 ( S O 4 ) , -> 3FeS04

(2) + ( 2 ) ^ Fe + 2H,S04 -> FeS04 + SO2 + 2 H 2 O

Lien he nn^s(.)4 vc'rf np^ va san pham khi c6 2 muoi

2Fe + 6H2SO4 -> Fe2(504)3 + 3SO2 + 3 H 2 O

Fe + 2H2SO4 -> FeS04 + SO2 + 2 H 2 O

(2') (3)

"FC2(.S()4)., - 2'"''''

(2) Fe504 + 5 O 2 1

"FC.S()4 - " F C (2) va ( 2 ) hoac (2) va (3)

H,504 dac, nong, thieu

H25O4 dac, nong

Fe va H:504 deu het

Vi du 1 Cho 6,72 gam Fe vao dung dich chua 0,3 mol H,504 dac, nong (gia thict

S O , la san phdm khir duy nha't) thu duoc dung djch B Tinh so mol muoi sdt thu duoc sau phan ling

A 0,04 mol Fe2 (SO4 )3 va 0,04 mol Fe504 B 0,06 mol Fe2 (5O4 )3

C 0,03 mol Fe2 (SO4 )3 va 0,06 mol FeS04 D 0,12 mol FeS04

Gidi

" F e = 0,12 mol nH2S04 = 0.3 mol <|V'J ' ' '

2 < X = ! ! M 2 S O 4 _ ^ 0 3 0 ^ 2 =>c6 2mu6'i Fe2(S04), va Fe504

npe 0,12 ' ' •

J2Fe + 6H25O4 Fe2(504)3 + 35O2 + 3 H 2 O ,

Fe + 2H2SO4 ^ FeS04 + SO2 + 2 H 2 O

nFe2(S04)3 = 2 " H 2 S 0 4 " "Pe ^ | n F e 2 ( S 0 4 ) , = 0,12 - 0,03 = 0,030101

nFc-so4 = 3 n F e -nH2S04 [nFc.so4 =3.0,12-0,30 = 0,06mol

Vi du 2 Mot dung djch co chu-a b mol H,S04 hoa tan het a mol Fe thu dirge khi A

va 42,8 gam mu6'i khan Nung lugng muoi khan do a nhiet do cao trong diau

kien khong eo khdng khi den khdi lugng khong doi thu duge h6n hop khi B

(a) Tinh gia tri eiia a, b (biet ty s6' - = — ) ' "

• H2SO4 dae va nong, khi A la SO^

•Tao muoi Fe,(S04), va FeSOj

P T H H : 2Fe + 6H2SO4 Fe2(S04), + 3SO2 + 3H2O

Fe + 2H2SO4 -4 FeS04 + SO2 + 2H2O

nFe,so4 = (3a - b) = 3a - 2,4a = 0,6a (mol)

> 400X 0,2a + 152 X0,6a = 42,8 gam ^ a = 0,25 mol

nso2 "^"f-V2(.s(i4), +nFc.sc)4 =3x0,05 + 0,15 = 0,3 mol

"02 =>'5nFe2,s()^,^ +0,25np,so4 =1,5x0,05 + 0,25x0,15 = 0.1125 mol

Ti kho-i doi voi khong khf: d , ^ = ">3x64 + 0,l 125x32 ^

1 niinat C O the giai bang phuong phap sir dung gidi han ti le mol Ngoai phirong

^[llip tren thl cae bai toan nay cung co the giai bang phuong phap do thi

^ Dgng 1. Khf CO, tdc di/ng vdi dung djch Ca(OH)2 hay BaCOH),

PTHH: CO2 + Ca(0H)2 - > CaCO, + H2O

CO2 + CaCO, + H2O - > Ca(HC0,)2

(1) (2)

Do thi:

nco2 = X nCa(OH)2= ^

nCaC03 = y

0 x i a X2 2a ^ "^^^

Tr6n true y chgn diem y = a, trdn true x chon 2 diem x = a x = 2a Tir diem a eiia true y va a eiia true x, ke vuong goe ehiing giao nhau tai di^m P Tir P noi vdti tga do O va 2a ta dugc tam giae

Vdri mot gia tri ciia s6' mol kfi't tiia tren true y, ke ducmg song song vdi true x cat tam giae tai 2 diem Q va R, tif Q va R ke vu6ng goe v6i true x ta co cac gia

trj X, va Xi ciia so mol CO2

fy = x vai 0 < x <a

Ta eo : y = (2a - x ) vdi a < x < 2a i j

y = 0 vdi x > 2a Tir do thj suy ra :

Oiig vdi m6t gia trj eiia x (O < x < 2a) thi luon cd 1 gia trj eiia y

Neu cd mot gia trj eiia y (O < y < a) thi lu6n cd 2 gia trj eiia x

du Ha'p thu hoan toan 2,688 Ift khf CO, (dkte) vao 2,5 1ft dung djch Ba(OH),

nong do a mol/Ift thu dirge 15,76 gam ket tiia Gia trj ciia a la

A 0,032 B.0,06 C 0,04 D 0,048

Gidi

nco2 =(2,688:22,4) = 0,12 mol;

"BaC03 =('5,76:197) = 0,08 mol

'^AI(OH)7 ^

'OH = x

diem a eiia triie y va 3a eiia true x, ke vuong goe cliilng giao nhau tai diem P Tit

P noi vdi toa do O va 4a ta duoe tam giac

Vdfi mot giii tri eiia so mol kct tiia tren true y, ke duong song song vdi true x

eit tam giac tai 2 diem Q va R, tir Q va R ke vuong goe v6\e x ta c6 cae gia

Tir d6 thj suy ra:

• Uiig voi mot gia trj cua x (O < x < 4a) thi luon c6 1 gia tri ciia y

Nd'u CO mot gia trj eiia y (O < y < a) thi luon c6 2 gia tn eua x

56

yidu- Cho 200ml dung dich A l C l , 1,5M tac dung v6i V lit dung djch NaOH 0,5M,

lucmg ket tua thu duoc la 15,6 gam

Gia tri lorn nha't ciia V la (cho H = 1, O = 16, A l = 27)

V CO gia trj tri Idn nha't vSy : V = 2,0 lit => Chon D

Dgng 3. Mud'i aluminat tdc di/ng vdi dung djch axit mqnh

•^Ifm a ciia true y va a ciia true x, ke vuong goc chung giao nhau tai d i e m j V T & T

57

noi voi toa d6 O va 4a ta duoc tam giac

V d i mot gia trj cua so mol ket tua tren true y, ke ducmg song song vdri true x

cat tam giac tai 2 die'm Q va R, tir Q vii R ke vu6ng goc vdi true x ta c6 cac gia

tri X | va X j ciia so mol H*

' Uiig vori mot gia trj cua x (O < x < 4a) thl lu6n c6 1 gia t r i cua y

> Neu CO mot gia t r i ciia y (O < y < a) thi lu6n c6 2 gia t r i ciia x

Vi du Cho 200ml dung djch H C l a M vao 200 m l dung dich N a A l O , 2 M thu duoc

15,6 gam ket tiia Gia trj ciia a la

A 1,0 hoac 2,0 B 2,0 hoac 3,0 C 1,0 hoac 5,0 D 2,0 hoac 4,0

0 0,2 0,4 1,0

T C r d d t h j v d i nAi(OH)3 = y = 0,2

r.x, = y = 0 , 2 m o l = : > a = (0,2:0,2) = l , 0 M

I | X 2 = ( 4 a - 3 y ) = ( 4 x 0 , 4 - 3 x 0 , 2 ) = l , 0 m o l = > t = ( l , 0 : 0 , 2 ) = 5,0M

I Vay : a = 1,0M hoac 5,0M => Chon C

11 Phuong phap giai bai tap xac djnh thanh phdn h6n hpp tCf cac

phdn h6n hop khong deu nhau

Cdch nhdn dang:

H6n hap duoc chia thanh nhieu phiin nhung khdng cho bid't t i le giffa cac phdn

thuomg gap nha't la loai bai tap c6 s6' lidu of cac phdn c6 don vj khac nhau : phdn 1

dan vj gam, phdn 2 dcm vj lit (hoac m o l )

<;8

Ca s0 vd phuang phap gidi:

Cac phdn la tur mot h6n hop ntn thanh phdn h6n hop kh6ng thay d6i do do ta

^at lucnig chdt trong phdn nay bdng k Idn lucnig chat trong phdn kia

Sau do theo cac gia thiet di ra ta viet diing cac PTHH va qua do lap cac

hiTOiig trinh toan hoc Giai he phuong tr'inh de di den ket qua

yi du 1: Cho 5,6 lit h6n hop khf A g6m metan, etilen va axetilen loi qua dung

dich nUdc brom d u thi c6 52 gam brom tham gia phan Ceng

E)6't chay hoan toan 2,6 gam h6n hop khf A thl cdn vira dii 30,24 li't khong khf

Xac dinh thanh phdn phdn tram theo the tfcii cac khf trong h6n hap A, biet cac khf

j o ^ d i e u kien tieu chua'n, trong khong khf oxi chiem 20% va the tfch con lai la nito

^idu 2 Cho h6n hap A c6 khoi luong m gam gom bot A I va sdt oxit F e , 0 ^ • Tifi'n

hanh phan ung nhiet nhom h6n hop A trong di6u kien khong c6 khong k h i , duoc h6ii hap B Nghien iiho, tr6n deu B r6i chia thanh hai phdn

- Phdn 1: Co khoi luong I4,49g duoc hoa tan het trong dung djch HNO3 dun

"ong, duoc dung djch C va 3,696 lit kiif N O duy nha't (dktc)

• Phdn 2: Tac dung vai luong d u dung djch N a O H dun nong thay giai phong

0.336 1ft khf H2(dktc) va con lai 2,52g chat rdn Cac phan iJng deu xay ra hoan

^oan Xac djnh CTPT ciia oxit sdt va khoi luong m ^ r

Gidi

Phan ung nhiat nhom: 2 y A l + 3 Fe.Oy - > y A!20,+3xFe (1)

59

H6n hap B thu dugc sau pu (1) tac dung v6\H cho , phan ufng lai

hoan toan => Trong B: A l con du Fe^Oy tac dung het

=> Vay B gom : A l O , , Fe, A l (du)

- Phdn 1 + dd H N O , dun nong : dat n ^ i j c ^ , " ^' "AI~ ^' "FC =

y-(2) (3)

i - Phdn 2 + dd NaOH du : dat n^i^o^ = kz ; n ^ , ^ kx ; np^ = k y

AlA+ 2NaOH -> 2 NaA102 + H : 0

(*) (**) (III)

Xac djnh Fe_jO^ va tinh m:

Ta c6: 6- " F e ^ " F e ^ Y _ 0,135 3

3nA,203 3z 3.0,06 4 = —=>oxitsdt: F e , 0 4 Kh6'i luomg h6n hop A (m):

m = m, + m 2 = m| + k x m | = 14,49 + - x 14,49 = [ 19,32 g

H I

Ct/ ninnMTV D VVH Khang Viet

• 2 phuong phap xac djnh iupng chdt tan can them vac tioqc tachi

\Q khi thay ddi ntii^t do dung djchi bao hioa

-—cin"^"^ vung khai niem dung dich bao hoa, do tan va cong thuc tinh nong

(Jo c6ng thiJc lien he n6ng do % cua dung dich bao hoa v6\o tan

• BU(5c I : Tinh kh6'i lugng chat tan va khoi lugng dung moi c6 trong dung djch bao hoaat,("C)

• Bifdc 2: Dat a gam la khoi lugng chat tan A can them hay da tach ra khoi dung djch ban ddu, sau khi thay doi nhiet do tir t, ("C) sang t, ("C) (t, # t,)

• Birdrc 3: Tinh kh6'i lugng chat tan va khoi lugng dung moi c6 trong dung djch bao hoa b UCQ-

• Budrc 4: Ap dung c6ng thiirc tinh do tan hay nong do % dung dich bao hoa

de tim a

Luu y: Ne'u 66 ydu cdu tinh lugng tinh the' ngam nude tach ra hay can them

vao do thay doi nhi6t d6 dung djch bao hoa cho sin, 6 bu6c 2 ta phai dat an so la s6' mol (n)

Vidu 1 6 12''C c6 1335g dung djch CUSO4 bao hoa Dun nong dung dich len

den 90°C Hoi phai thdm vao dung dich bao nhieu gam CUSO4 de dugc dung dich

bao hoa a nhidt do nay Biet a 12°C, do tan cua CUSO4 la 33,5 va a 90"C la 80

Gidi

'"cuS04(iroi,gdddiiu) = > 3 3 5 x - ^ = 335g=> m u p = 1335-335 = lOOOg Trong qua trinh dun nong, gia thiet nude khong bay hoi, khoi lugng dung djch

dugc bao toan Dat x la khoi lugng CUSO4 can them vao Ta c6:

DLBTKL dung djch =^ m^^sau) = (1335 + x)

DL BTKL chift tan => mc^so^ = (33,5 + x)

Ap dung cong thiic d6 tan => = — |x=465g

'"cuso4(,rong dd ddu, = > 877 X ± ^ = 877g mH^o = 1877 - 877 = lOOOg

^at x la khoi lugng CUSO4.5H2O tach ra khoi dung djch:

^ ' " C u S f J 4 ( , a c h r a ) = ^ X - ^ = 0 6 4 x g ; mH^Odce,linh) ^ ^ ^ " " ' ^ ^ " ^

-§au khi ha nhiet do:

.1)1

Bo dc ttda hoc 9 6a (hi vio lO - Pham 3/LUu

Kh6'i lirgng dung djch: m j j = (1877 - x)g

Khoi lucmgchat tan: m^,ha,,a„ = (877-0,64x)g

Kh6'i lircrtig nu6c con lai trong dung moi: m^^ = (1000-0,36x)

Ap dung c6ng thirc ti'nh d6 tan: mchauan _(877 - 0,64x)^ 40

(1877-x) 140

m

x=961,7g

dd

Vidu 3 Cho 0,2 mol CuO tan trong luong vira du dung djch H 2 S O 4 20% dun nong, s;,,^

do lam nguoi dung djch den lO^C Ti'nh khdi luong tinh the C U S O 4 5 H 3 O da tach

r-I khoi dung dich, biet ring do tan ciia C U S O 4 b 10°C la 17,4g trong lOOg H.O

Khoi lugng dung djch: m^j = ( l 14-x)gam

Khtfi luong chat tan: mj.ha,,a„ = (32 -0,64x)gam

Ap dung cong thiic ti'nh d6 tan:

"ichauan (32-0,64x) 17,4

m dd ( 1 1 4 - x ) 117,4

x = 30,71 gam

13 Phaong phap bao toan mol electron

* Khi CO iiliieu chat oxi hod, chat khtf trong mot hSn h(/i) phan i(n}> (nhieu pluin

ling hoac phan I'oig qua nhieu giai doan) thi tcfng so electron nui cdc chat khi(

nhudng ra hang long so electron nid cdc chat oxi hod nhdn vdo

Hay Id : Tong so mol electron nhu&ng = Tong so mol electron nhdn

• Luu y: Ta chi cdn nh^n djnh dung trang thai dau va trang thai cu6'i ciia cdc chai

oxi hoa hoac chat khif, c6 the bo qua viec can bang cac phuong trinh phan ihig ^

Vi du 1 Cho m gam h6n hop g6m hai kim loai Mg va A l tac dung vofi dung did'

HCl dir thu diroc 0,672 lit khi H, (6 dktc) Mat khac, cQng cho m gam h6n h"T

hai kim loai tren tac dung vdfi dung dich HiS04 dac, nong du thi thu duoc V 1''

khi SO, (dktc) Xac djnh V

Gidi

Goi a, b, x Idn luot la so mol cua Mg, Al va SOi

O/ mmMiyDVVtiKhang Viet

Trong phan u-ng vdi dung dich HCl: M g va A l la chat khij (nhirang electron);

HCl (thirc chat la ion H*) la chat oxi hoa (nhan electron) ,, Qua trinh nhuong electron Qua trinh nhan electron

Trong phan ung vdi H 2 S O 4 (dac): vai tro chat khir khong thay ddi chat cung nhu s6' luong nen t6ng so mol electron nhircmg khong thay d6i

Chat oxi hoa la H 2 S O 4 dac (thuc chat la nguydn tif S*^ cua g6c axit S O 4 " ) nhan electron

Qua trinh nhan + 2e -> S*"*

2x - > x Bao toan so mol electron: (2a + 3b) = 2x (**) Tir (*) va (**) =^ X = 0,03 mol =i> V = 0,03.22,4 = 0,672 lit S I '>

Vidu 2 Oxi hoa 0,728 gam Fe ta thu duoc 1,016 gam chat rdn X Hoa tan hoan

toan luong chat X tren vao dung djch H 2 S O 4 dac, nong du thi thu duoc V lit khi' SO^od'ktc Xacdinh V

Gidi

= 2 l Z ^ = 0 , 0 1 3 m o l ^ 5 ^ c a c oxit sdt ""^^^dac ^ p ^ 3

Chat khu la Fe va chat oxi hoa la OT va HiS04 dac, nong

Ta C O cac qua trinh:

Nhuong electron Nhan electron

Fe -> Fe''^+3e 0,013- -^0,039

0,288

= 0,009 mol

^2 32 Bao toan so mol electron: 2x + 0,036 = 0,039 => x =0,0015 mol Vay: V = 0,0015.22,4= |0,0336 Iff

Bg de nda hoc 9 on thi i '< < I >him 3/Ltfu

Vi du 3 Hoa tan hoan loan 20,88 gam m6t oxit sSt bang dung djch H2SO4 dac,

nong thu duoc dung dich X va 3,248 h't khi SO, (san phdm khir duy nhát, o

dktc) Co can dung dich X , thu duoc m gam muoi sunfat khan Gia t r i ciia m la

A 52,2 B 4 8 , 4 C 54,0 D 58,0

20 88

nso2 = ( 3 , 2 4 8 : 2 2 4 ) = 0 , 1 4 5 m o i ; H p , n

56x + 16y m o l Sir oxi hoa: F e , 0 + 2 y H ^ ^ xFê^ + y H O + (3x - 2y)e

20,88

Su khir:

(56x + 16y) SỐ + 2e- + 4 H * - > SO2 + 2H2O 0,29 < 0,145

•>(3x - 2 y ) x 20,88

(56x + 16y)

Bao toan so m o l electron cho-nhan: (3x - 2y) x

= > X = y => CTPT cua oxit sdt: FeO

20,88 (56x + 16y)

= 0,29

n F c 2 ( S ( ) 4 ) , _ " F c O _

20,88 72.2 = 0,145 mol =>mpg2(so4), = p 8 gam

^ Chon I)

Vi du 4. Tron 0,81 gam bot nhom vdi bot FêO, va CuO roi dot nong de tien hanli

phaii mig niiict niiom thu dirge h6n hop A Hoa tan hoan toan A trong dung dicli

H N O , dun nong thu dugc V lit khi N O (san pham k h u duy nhát) a dktc Gia tri

^ h6n hop A + duiig dich H N O 3

hoa Ian hoan loan

Thuc chat trong bai toan nay chi c6 qua trinh cho va nhan electron ciia nguyeii

yidu 5- T^""?" S"'" bot liru huynh r6i nung nong (trong dieu

Jcien khong c6 khong khQ, thu dugc h6n hop ran M Cho M tac dung v6i lugng

du dung djch H C i , giai phong h6n hgp khi X va con lai m6t phdn kh6ng tan G

\Di d6't chay hoan toan X va G cdn vCra dij V l i t khi O, (o dktc) Gia trj cua V la:

Fe 0,1 ^ 0,2

S 0,075

j + 4

+ 4e 0,30

• Dua vao cac dai lugng c6 gidi han, chSng han :

- Nguyen tir khoi trung binh, phan t u kh6'i trung binh K L P T T B ( M ), hoa trj trung binh, so nguyen t i j trung b i n h ,

- Hieusuat: 0 ( % ) < H < 100(%)

- So mol chat tham gia ( n ^ g j ) : 0 < n(,g) < ng,,„ (s6' mol chat ban dau),

• De suy ra quan he vdi dai lugng ciin t l m Bang each:

- T i m sir thay doi a gia trj nho nhat va Ion nhát cua 1 dai lugng nao do de dSn

giai han cdn t i m

- Gia sir thanh phdn h6n hgp ( X , Y ) chi chua X hay Y de suy ra gia trj nho nhát j^gjon nhát ciia dai lucmg can t i m

^^du 1 Cho 6,2 gym h6n hgp 2 k i m loai kiem thu6c 2 chu ky l i ^ n tiép trong bang

tu5n hoan phan ling vdi H, 0 du, thu dugc 2,24 Ift khi (dktc) va dung djch Ạ Tinh thanh phdn % ve kh6'i lugng tung k i m loai trong h6n hop ban đụ

Gidi

nH2 = ( 2 , 2 4 : 22,4) = 0,1 mol => n,,, =2nH2 = 0 , 2 m o l

% N a = 1 0 0 - 6 2 , 9 = 3 7 , 1 % " K - " N a = 0 , l m o l

Vi du 2 C h o 13,8 g a m ( A ) l a mu6'i cacbonat cua k i m loai k i e m v a o 110ml dung

d j c h H C l 2 M Sau phan u n g tha'y c o n axit t r o n g d u n g d i c h t h u ducfc v a the tich

k h i thoat ra V , virgt qua 2 0 1 6 m l V i e t p h u o n g t r i n h phan l i n g , t i m ( A ) v a t i n h V,

2HC1 + B a C O , ^ BaCl2 + CO2 + H2O

0.143 = ^ < n 197 - ••CO2 - "BaCO., + "MgCO j ^ — < — = 0,3345 m o l 28 I

0,143.22,4 = 3 , 2 l i t < Vco^ < 0 , 3 3 4 5 2 2 , 4 = 7,491it

Vay: 3 , 2 1 i t < V c o 2 < 7 , 4 9 l i t

Vi^V ^- •''^^ ^^"^ ^^"^ ^y^^S v o i 5 0 0 m l d u n g d i c h H Q I M

(a) C h i i t i g m i n h r a n g s a u p h a n ufng v 6 i M g v a A l thi a x i t v 5 n c o n d u ?

(U) p h a n l i n g t r e n l a m thoat r a 4,368 lit k h i H i ( d k t c ) H a y t i n h s 6 g a m M g v a

H H C I (tacdung) = 2 n M g = 0 , 1 6 1 X 2 = 0 , 3 2 2 m o l < 0,50 m o l Ne'u h 6 n hofp c h i CO A I : nHci(,.icdung) = ^ H A I = 0 , 1 4 3 x 3 = 0 , 4 2 9 m o l < 0 , 5 0 m o l

=> v a y H C l vSn c o n d u sau phan u'ng

Vidu 5 X l a h 6 n h o p h a i k i m loai M g v a Z n Y l a d u n g d j c h H 2 S 0 4 c h u a r o n o n g do

Thi n g h i e m 1: C h o 24,3 g a m X v a o 2 lit Y s i n h r a 8,96 lit k h i H , ( d k t c ) Thi n g h i e m 2: C h o 24,3 g a m X v a o 3 lit Y s i n h r a 11,2 lit k h i H , ( d k t c )

(a) C h i i n g t o r i n g t r o n g thi n g h i e m 1 thi X c h u a tan het, t r o n g thi n g h i e m 2 thi X tan hat

3 : 2 = 1,5 Idn m ^ kh6'i l u o n g H , giai p h o n g c h i tang 0 , 5 : 0 , 4 < 1,5 Idn

mau thu5n g i u a d 6 tang the' t i c h cua axit v a t h e t i c h cua k h i H j c h u n g t o

t r o n g T N 1 c o n d u k i m l o a i , t r o n g T N 2 con d u a x i t ( b ) G o i X l a so m o l M g , y l a s6' m o l Z n T r o n g T N 2 ta c 6 : ' >'

K h i g i a i cac bai l o a n hoa hoc theo p h u o i i g phap dai s6', ne'ii s6' p h u o n g trinh

l o a n hoc thiet lap diroc it hom s6' dn so chua biet can t i m t h i phai h'l&n l u a n Ta

c h o n 1 dn so l a m c h u d n r o i tach cac an so con l a i N e n difa \6 p h u o n g t r i n h toan

hoc 2 d n , t r o n g d o it nha't c 6 1 dn c 6 g i a i han Sau d o c 6 thd' thid't lap bang bieri

thi6n hay dua vao cac d i ^ u k i d n khac de chon cac gia trj h o p l i

Vi du 1 H o a tan 3,06 g a m o x i t M , O y bang d u n g d i c h H N O , d u sau do c 6 can thi

t h u d u o c 5,22 g a m mu6'i k h a n H a y xac d i n h k i m loai M biet n o c h i c 6 m 6 t hoa

trj d u y nha't

Gidi

x M ( N 0 3 ) 2 y / + y H 2 0 + 2 y H N 0 3 ^

Vi du 2 A , B la 2 chat k h i a diiu ki&n thucmg, A l a hop cha't ciia n g u y d n t6 X \d\

o x i ( t r o n g d o o x i chid'm 5 0 % khd'i l u o n g ) , c o n B la h o p cha't cua n g u y d n t6' Y vdi

h i d r o ( t r o n g d o h i d r o c h i e m 2 5 % kh6'i l u o n g ) T i k h o i ciia A so vori B bang 4

X a c d j n h c 6 n g thiJc phan tir A , B Biet t r o n g 1 phan tir A c h i c 6 m 6 t n g u y d n tii

d u n g de phan l i n g het h 6 n hop A

0 , 3 2 - 0 , 0 8 ^

=i> m = m p Fe203(du) 0 , 1 - - 160 = 3,2 g a m 3,2 g a m < m < 4,8 g a m

Phan II: BA MirOfl DE THI VAO LQTP 10 0 CAC TINH

S d G D & D T

T / P D A N A N G

K Y T H I T U Y ^ N S I N H V A O L C J P I O T R L ; O N G THPT CHUVeN le QUtDON rinP

NAM HOC: 2 0 1 2 - 2 0 1 3

Di C H I N H THirC

K Y T H I T U Y ^ N S I N H V A O L C J P I O T R L ; O N G THPT CHUVeN le QUtDON rinP

NAM HOC: 2 0 1 2 - 2 0 1 3 MON: HOA HOC

Th()1 gian lam bai: 150 phut (khong keth&i gian giao de)

Cau 1 (3,0 diem)

(1) Viet cac phudig trinh hoa hoc ciJa phan ling xay ra trong qua trinh san xuat

thiiy tinh tir cat trdng, s6da, da v6i

( 2 ) Tron 50 gam dung djch muoi sunfat ciia m6t kirn loai kiem (dung dich A) n6iig

d6 26,4% v6i 50 gam dung djch NaHCO, thu dixgc dung djch X c6 khdi lirgng

nho hon 1 0 0 gam Cho 0,1 mol BaQ, vao dung dich X thay vSn con du muoi

sunfat Them tiep vao do 0,02 mol BaQ thi dung djch thu duoc vSn con Baa2 dir

Bie't cac phan ling xay ra hoan toan

(a) Xac djnh cong thiirc hoa hoc ciia mu6'i sunfat ban ddu

(b) Viet phuong trinh hoa hoc cua phan ling xay ra (neu c6) khi cho ISn lugt cac

chat sau tac dung v6i dung djch A: Fe, Fc(OH),, Fe,04, Ag, NaAlO,

Ckull {4,50 diem)

(1) Khong dung thuoc thir, biing phuong phiip hoa hoc, hay phftn biet 6 dung dich

kh6ng mau: MgCl., NaHCO,, H,S04, BaCI,, NaCl va NaOH dung rieng biet

trong cac binh mat nhan Viet cac phuong trinh hoa hoc minh hoa

( 2 ) Dun nong 10,8 gam b6t A l trong O, mot thofi gian, thu diroc m gam h6n hop

chat rdn A Hoa tan het A bang mot lirong vua du dung djch h6n hop HCl I M va

diroc (m + 44,34) gam muoi khan Tinh m, V

(3) Hoa tan hoan toan 8,4 gam kirn loai M trong dung dich H 3 S O 4 dac, nong (vira

dii) tha'y thoat ra 5,04 lit khi SO, (dktc) Dung djch sau phan ung dem c6 can thi

thu duoc 42,15 gam chat rdn X Xac dinh M va X

cau I I I {3,0 diem)

(1) Trinh bay phuong phap hoa hoc de' tach rieng biet

cac khi tir h6n hop sau: HCl, O,, CO,

(2) Cho mot binh km dung ti'ch khong doi chu-a 80ml

nude va 4 lit khdng khf (xem hinh ve) Phdn khong

khi chi cMa N , va O, theo ti le 4 : 1 v^ the' tich

Bom 0,04 mol h6n hop khf B bao g6m NO, va NO c6 ti kh6'i so vdi H , bang 19

vao binh va lac kT binh tdi khi cac phan ung sau xay ra hoan toan:

-a d U c K dung dich X Tinh n6ng d 6 % dung dich X

Gia sir ap sua't hoi liudc trong binh khong dang ke'; cac the ti'ch khi deu do dktc; khdi luong rieng cua nude Dn^o = ' g/m'-

cau IV (^,<>

(1) '^'^"0 Ve hinh minh hoa each tien hanh thi nghiem dieu che va thu khf

axetilen an toan trong phong thi nghiem Viet phuong trinh hoa hoc minh hoa

(2) ( 2 diem) X la hdn hop gdm axetilen va hidro cd ti khdi so vdi heli la l,*-)

Cho toan bo X qua dng sii dung N i , dun nong mot thdi gian, thu duoc hdn hop

Y Cho Y tac dung vdi luong du dung djch AgNO, trong N H , thay tao thanh 7,2 gam ke't tiia va hdn hop khi Z D3n Z qua nude brom du thu duoc hdn hop khi T, ddng thdi thay cd 4,8 gam brom da tham gia phan tag Dot chay hoan hoan hdn hop T thu duoc 0,896 lit CO, (dktc)

(a) Xac djnh thanh phdn hdn hop Y, Z, T

(b) Tinh ti khdi cua Y so vdi heli ^

Cau V {2,50 diem)

(1) Neu hien tuong vii giai thi'ch:

(a) Nhd vai giot iot vao mat mdi cdt cua cii khoai lang

(b) Cho vai giot chanh vao cdc sua bd

(c) Cho mot mdu cao su tu nhien vao xang

(2) Len men giam 115ml dung djch rugu etylic 10° mdt thdi gian thu duoc dung djch

A Neu cho toan bd dung djch A tac dung vdi luong vua dii Na den khi phan ung

xay ra hoan toan thu duoc 251,7 gam cha't rdn khan Tinh hieu sudt qua trinh len men giam, biet khdi luong rieng DC^H^OH = g/"^'- ^ ^H o = ' g/f"'- •

Cau VI {3,0 diem)

Ddt chay hoan toan 3,74 gam hdn hop X gdm CH,COOH, CH,COOC,H,, C,H,OH thu dugc 3,584 lit CO, (dktc) va 3,42 gam H , 0 Mat khac, cho 3,74 gam X phan ung het vdi 40ml dung djch NaOH I M , thu duoc dung djch Y va 0,05 mol C.HyOH Cd can dung djch Y, thu duoc 2,86 gam cha't rdn khan

(a) Xac djnh c6ng thiJc phan tir ciia ancol C^H^OH

(b) Tinh % theo khdi luong cac chat trong X

Phuong trinh hda hoc: SiO, + Na,CO, > Na,SiO, + CO,

SiO, + CaCO., - 1 ^ - ^ CaSiO, + CO3

(2) 2,0 diem

(a) m + m > m mudi sunfat kirn loai kiem la mud'i axit

(MHSO4) do cd S U thoat khi CO, nen lam giam khdi luong dung djch sau khi pha trdn

-71

2MHS04 + ZNaHCOj -> M^SO^ + Na,S04 + 2CO, + 2H,0

M2SO4 + B a C l , - > B a S 0 4 + 2 MCI

Dat MHSO4: X mol

( M 9 7 ) x = ^ = 1 3 2 ^ x = ' "

100 M+97

T h e o d a O , l < x < 0 , 1 2 : ^ 13 < M < 35 => M = 23 (Na) la phu hop

Vay cong thirc cua muoi sunfat la NaHS04

(b) 2NaHS04 + F e - * Na3S04 + FeS04+ H ,

8NaHS04 + Fe304 ^ 4Na.S04 + Fe,(S04), + FeS04 + 4 H , 0

2NaHS04 + 2Fe(OH)2 -> Na2S04 + 2FeS04 + SH.O

NaHS04 + NaAlO, + H.O-* Na.S04 + Al(OH),

Co the': 6NaHS04 + 2 A 1 ( 0 H ) 3 A l 3 ( S 0 4 ) 3 + 3Na,S04 + 6H.O

Cdu U (4,50 diem)

Tn'ch cac mSu thu va danh dtfu Ldn luot cho cac m5u thir tac dung v6i nhau

tirng doi m6t, ta c6 bang ket qua sau:

MgCK NaHCO, H.SO4 BaCl NaCl NaOH

MSu thir nao khi cho vao cac mSu thu kia tao 1 chat khi thoat ra: NaHCO,

MSu thu nao khi cho vao cac mSu thir kia tao duoc 1 chat khi va 1 ket tiia la

H2SO4, chat tuong urng la NaHCOj va BaCl,

Kh6ng C O hien tuong nao la NaCl

2NaOH + M g C l 2 - > Mg(OH), i + 2Naa

; H:S04 + BaCl, -> BaS04 i + 2HC1

; 2NaHC03 + H,S04 -> Na,S04 + 2H.O +2C0,

Phan bidt NaOH, MgCl,:

C6 can va nung nong 1 mSu NaHCOj, hoa muoi thu duoc vao 2 m3u thir nay,

neu C O ket tua tao duoc la MgCU, kh6ng c6 hien tuong gl la NaOH » 1

* K : t 2NaHC03 — ^ Na^CO, + CO, + H^O

MgCU + Na3C03 -> MgCOj + 2NaCl

(2) (2,0 diem)

Dun nong A l trong O,: 4A1 + 3 0 , - -> 2AKO3

72?

S5'niol A! = 10,8:27 = 0,4 mol phuong trinh hoa hoc: A l + 3HC1 AICI3 + ^ H , t

2A1 + 3H,S04 -> Al2(S04)3 + 3 H , t ' AI2O3 +6HC1 ->2AlCl3 + 3 H 2 O AI3O3 +3H2SO4 ^ A 1 : ( S 0 4 ) 3 + 3 H , 0 oat s6' mol H:S04 la x => s6' mol HCl: 2x '•'0)^:)

Iv1u6'i thu duoc la h6n hop: A1:(S04)3: — mol va AICI3: — mol

3 3 2x 2x

Bao loan nguyen to Al ta c6: —- + — = 0,4 => x = 0,3 Kh6'i luong muoi khan:

mAia, + '"Ai2(S04 )3 = " I A I + nig^caxi. = 10,8 + 0,3x96 + 0,3x2x35,5= 60,9 Theo d^: m + 44,34 = 60,9 => |m = 16,56 gam|

m-10,8 16,56-10,8 ^ , „ ,

Bao toan khoi luong ta co: no2 = —— — = — = 0,18 mol

.S6'mol A l d u t r o n g A l a : 0 , 4 - - 0 1 8 = 0,16 mol

> S6' mol khi H , thoat ra: 3/2 0,16 = 0,24 mo!

Vay the tich khi H , 6 dktc la: |V= 0,24 x 22,4 = 5,376 lit

=> Chat rdn thu duoc la tinh the ngam nu6c: Fe3(S04)3.xH20

— = 0,075 X = 9 vay chat rdn la: |Fe,(S04),.9H,^

18x

f^^ni. (3,0 diem)

^^^(i,50die)n)

^^n h6n hop khi qua binh dung dung dich Ca(OH)2 du, khi O, kh6ng tham gia

phan ling thoat ra, thu la'y O,; HCl va CO, bj hap thu het:

7-*

2HC1 + Ca(OH)2 -> CaCl, + 2H2O

CO, + Ca(OH), -> CaCOj + H, 0

- Lpc lay ket tua, hoa tan het trong dung djch HCl du, thu khi CO, thoat ra

CaCO, + 2HC1 CaCl, + H^O + CO^t

- Co can dung djch chira CaCU, CaCOH), den khan, sau do cho tac dung v6

H2SO4 dun nong, thu khf HCl thoat ra '

no2 (ban ddu) = = 0,036 (mol) > n o , (p/iing) = 0,03 (mol)

' Y / H e - 4 , 4 6

Cau V (2,50 diem) (1) (0,75 diem)

(a) Ch6 nho iot tha'y xufi't hien mau xanh dac trung do trong khoai lang c6 chiJa tinh

b6t, tao phan ihig mau v6i iot

(b) Co hidn tucmg dong von siJa do trong sOa c6 1 so protein d6ng tu khi ti6'p xiic v6i moi truong axit

(c) M5u cao su tan tao thanh dung dich nhdt do cao su la polime tan dupe trong dung m6i hiru co: xang, benzen, ddu hoa

2H2O + 2Na 2NaOH + H2

0,20

Cau V I (3,0 diem)

(a) CTPT cua ancol :

Ta CO so do chay X: X + O, -> CO, + H , 0

nco2 = (3,584 :22,4) - 0,16mol; nn,,) = (3,42 :18) = 0,19 mol

TCr cac PTHH ciia phan ling chay:

Ta c6: C H , C O O H va este C H ^ C O O Q H ^ trong do goc Q H ^ boa trj 1 va no

khi chay tao thanh so m o l H , 0 b i n g so m o l C O ,

Theo de ra: n^^o > "002 C^HyOH la ancol no don chiic: y = 2x + 1

T C r d o t a c o : n c ^ n ^ ^ ^ j O H = 0 , 1 9 - 0 , 1 6 = 0,03 mol

Sau khi thuy phan thu dugc long s6' mol ancol la 0,05 mol

Suy ra : so mol este CH3COOC^H2,+| la : 0,05-0,03 = 0,02 mol

V a y X g 6 m : CH,COOH (a mol), CH3COOC,H2,+, (0,02 mol)

va C,H2^+,0H (0,03 mol)

PTHH ciJa phan ung v6i dung djch NaOH ciia X:

CH ,COOH + NaOH -> CH ,COONa + H^O

a -> a -> a (mol)

CH,COOC,Hy + NaOH ->• CH^COONa + C H y O H

0,02-> 0,02-> 0 , 0 2 ^ 0,02 (mol)

- Neu NaOH phan ling vira dii vdri X, chat rdn la CH ^COONa

nCH3COONa = " NaOH = 0,04 mol => mcH3CO()Na = "'O^ X 82 = 3,28g > 2,86g

- Vay NaOH con du

KV THI TUY^N SINN VAO idP 10

TRLTONG THPT CHUYeN NGUYEN TRAI

NAM HOC: 2012-2013

MOH: HOA HOC

Thoi gian lam bai: 120 phut (khong kethdigian giao de)

C^l (2,0 diem)

(1) Cho h6n hop g6m AKO,, Cu, Fe^O, vao dung djch H,S04 loang du thu duoc

dung djch X va chat ran Y Cho X\x tir dung djch NaOH toi du vao dung djch X

thu diroc dung djch Z va ket tiia M Nung ke't tiia M ngoai kh6ng khf tdi kh6'i lircmg kh6ng doi thu duoc chat ran N Cho khi H , du di qua N nung nong thu

ducK; chat ran P Sue khi CO2 toi du vao dung djch Z thu duoc ket tiia Q

(a) Xac djnh thanh phdn cac chat c6 trong X, Y, Z , M , N , P, Q Bia't cac phan ling

xay ra hoan toan ; ' i >

(b) Vid't cac phuong trinh phan utig hoa hoc xay ra

(2) Cho h6n hop kim loai M g , Fe vao dung djch chiia h6n hop mu6'i Cu(N03)2,

AgNOj Phan ung xay ra hoan toan, thu dugc h6n hofp chat rdn A g6m 3 kim loai va dung djch B chu-a 2 muoi Trinh bay phuong phap tach riang tirng kim loai ra khoi h6n hop A Viet phuong trinh hoa hoc

cau 2 (2,0 diem)

Cho hai hop chat hffu co X , Y chii-a (C,H,0) chi chffa mot loai nhom churc da

hoc va c6 khd'i lugng mol phan tir deu bang 46 gam

(1) Xac djnh c6ng thffc ca'u tao cua X, Y Biet X, Y d6u phan ling vdri Na, dung djch

cOa Y lam quy tim hoa do

(2) Tir X Viet cac phuong trinh hoa hoc dieu che' Polyvinylclorua (PVC) va

Polyetylen (PE) , ; >

cau 1.(2,0 diim)

(1) Hay chon cac chat thich hop va viet cac phuong trinh phan irng hoan thanh so

<J6 bie'n h6a sau:

Cac chat A , B, D la hop chat cua Na;

Cac cha't M va N la hop chat cua A l ;

Cac chat P, Q, R la hop chat cua Ba;

Cac chat N , Q, R khong tan trong nu6c

X la chat khf kh6ng miii, lam due dung djch nu6c v6i trong;

muO'i Na, dung dich Y lam do qui tfm ^

(2) Tir 9 kg t i n h bot c6 the' didu che dirge bao nhiSu h't rugu (ancol) etylic 46°? Biq

hieu suat ciia ca qua trinh didu che la 72%, kh6'i luong r i e n g cua ruou etylj^,

nguyen chat la 0,8g/ml

Cau 4. (2,0 diem)

Nung 9,28 gam h5n hop A g6m FeCO, va mot oxit sit trong khong khi den k h d j

lugng khong doi Sau khi phan lirng xay ra hoan toan, thu dugc 8 gam mot oxj(

sdt duy nha't va khf CO, Ha'p thu het luong khi CO, vao 300ml dung dic|i

Ba(OH)2 0,1M, ket thiic phan ling thu dugc 3,94 gam k€i tiia

(1) Tim cong thiic hoa hoc ciia oxit sat

(2) Cho 9,28 gam h6n hop A tac dung vofi dung djch HCl du, sau khi phan lifng ket

thiic thu dugc dung dich B DSn 448ml khi CU (dktc) vao B thu dugc dung dic|,

D Hoi D hoa tan tdi da bao nhieu gam Cu?

Cau 5. (2,0 diem)

Thuy phan hoan toan 19 gam hcfp chat hCru co A (mach his, phan ling dugc vo,

Na) thu dugc m, gam cha't B va m , gam chat D chiia hai loai nhom churc

- Dot chay m, gam chat B cin 9,6 gam khi O, thu dugc 4,48 lit khi CO2 va 5,4

gam nu6c

- Dot chay m , gam chat D can 19,2 gam khi O, thu dugc 13,44 lit khi CO, va 10,X

gam nuorc

(1) T i m cong thiic phSn tir A, B, D

(2) Xac djnh cong thiic ca'u tao ciia A , B, D

C h o : F e = 56; B a = 1 3 7 ; C = 1 2 ; 0 = 1 6 ; H = 1; Na = 23; CI = 35,5; Cu = 64

.5^ Hl/CJNG DAN GIAI

Cau 1. (2,0 diem)

1. (1,0 diem)

(a) Dung dich X : A1,(S04)3, CUSO4, FeS04, H,S04du

Chat rdn N : CuO, Fe,03

Chat ran Y : Cu - ; r

Chat ran P : C u , Fe

Dung djch Z : NaAlO,, Na,S04, NaOH

Ket tua Q : AKOH)^ '

Ket tua M : Cu(OH),, Fe(OH), '

Cu + H , 0

- > 2Fe + 3 H , 0 NaHC03

Cha't ran A : A g , Cu, Fe

Dung djch B : MgCNO,),, FeCNO,),

Cho chat ran A vao dung dich HCl du Thu dugc pMn cha't ran la kim loai Cu,

Ag va phdn dung dich FeCl, va HCl

Fe + 2HC1 ) FeCl, + H , t ,^; Cho NaOH d u vao phdn dung dich, thu dugc ket tiia la Fe(OH), ^

N a O H + HCl > N a C l + H , 0 2NaOH + FeCl, > F e ( O H ) , i + NaCl Nung ket tua ngoai khong khi den khoi lirgng khong doi Cho luong khi H , d u

di qua cha't rdn, nung nong, thu dugc Fe

-> MgCNOj), + Cu )-Fe(N03), + 2 A g -> FeCNOj), + Cu

4Fe(OH), + O, Fe,03 + 3 H ,

->2Fe,03 + 4 H , 0 -> 2Fe + 3 H , 0

^ " n g 2 k i m loai trong kh6ng khi toi khoi lugng khong d6i thu dugc cha't rdn 8<5m CuO va A g Cho cha't rdn sau phan ling vao dung dich HCl du thu dugc k i m

J o a i A g

2Cu + 0 , CuO + 2HC1

-^2CuO -»CuCl, + H , 0 Cho dung djch NaOH du vao dung dich sau phan iJng thu dugc kd't tua, Igc la'y

•^^'t tua dem nung ngoai khdng khi t6i khoi lugng khdng d6i, cho ludng H , du di 9Ua chat rdn, nung nong thu dugc Cu tinh khia't

Vay c6ng thiic phan tir cua X, Y c6 the la QH^O, CH,0,

Vi Y phan irng \6\, lam do quy tfm, Y c6 nhom -COOH

Dieu che |x)lyvinylclorualcn (PVC)

CjH^OH + O2 >CH,COOH + H^O

CH,COOH + NaOH > CH,COONa + H,0

CH.,COONa + NaOH - > Na.CO, + CH4

jxTHH: 2NaOH + CO, ->Na,C03 + H,0

Na,C03 + CO2 + H,0 -> 2 NaHC03 2NaOH + 2A1 + 2H,0-> 2NaA10, + 3H, NaHC03 + NaOH-> NajCOj + H,0

3Na,C03 + 2AICI3 + 3H ,0 -> 2Al(OH)3 i+ 6NaCl + 3CO,

2NaHC03 + Ba(OH), du-> BaCOji + Na,C03+ 2H,0

BaC03 + 2NaHS04 -> BaS04 i+ Na,S04 + CO, + H,0 ,

Ba(HC03)2 + 2NaHS04-> BaS04 i+Na,S04 + 2CO,+2H,0 Ba( HC03)2 + Na2S04 -> BaS04 i+ 2NaHC03

PTHH: 4FeC03 + Oj —L-^2Fe.O, + 4COj (1)

2Fe,0, + ( ^ I Z M ) O , — i l ^ x F e ^ O j (2)

8 3 94

"Fe203 =7^5 = ^ ' ^ ^ ™ ' = "Ba(OH)2 =0,3x0,1 =0,03 mol; n e a c o j = - J ^ = 0,02mol

PTHH: CO, + Ba(OH), j-BaCO, (3)

2CO, + Ba(OH), >Ba(HC03), (4)

Cho 9,28 gam h6n horp A vao d u n g d j c h HCl d u

FeCOj + 2HC1 FeClj + COj + H j O (5) 0,04 -> 0,04 mol

Fe304 + 8HC1 -> FeCl2 + 2FeCl3 + 4 H 2 O (6) 0,02-> 0 , 0 2 ^ 0,04 mol

Dung djch B g6m: FeCl, 0,06 mol; FeCl, 0,04 mol; HCl du Cho khi CI2 = 0,02 (mol) vao dung djch B

2FeCl2 + CI2 0,04 <- 0,02 0,04 mol

Dung djch DCOchiia: npgc^ =0,08 mol; np^Q^ =0,02 mol

2FeCl3 + Cu - » 2FeCl2 + CUCI2 (8) 0,08 0,04 mol

=> mcu =0,04x64 = 2,56 gam

Cfiu 5. (2,0 diem)

Goi c6ng thiJc tong quat cua B la C.H^O, (x, y, z € N*)

no2 = 0,3 m o l ; nco2 = 0,2 m o l ; n H 2 0 = 0,3 mol

A p dung djnh luftt bao loan khd'i luong:

• " i + = m c o 2 + m H 2 0 =>m, =4,6gam

= > m o ( B ) = 4 , 6 - ( 0 , 2 1 2 + 0,3.2) = l , 6 g a m = ^ n o ( B ) = 0 , l mol

=> x : y : z = nc : n H :no =0,2:0,6:0,1 = 2:6:1

=> CTPT CO dang: (C^H^O),,

Theo quy luat hoa tri ta c6: 6n < 2 x 2n + 2 n < I => n = 1

=> B CO cong thiJc phan tir: C,H(,0

Do B la san ph.4m ciia phan ling thuy phan n d n B c6 CTCT: CHjCH^OH Goi c6ng thurc t6ng quat ciia D la C H A (a, b, c e N*)

m H20 = m, + m j - m ^ = 4,6+ 18-19 = 3,6 gam => n^^o =0,2 mol

Bao toan cac nguyfin t6' cacbon va hidro: <

niccA) = " i c ( B ) + "10(0) = 0,2.12 + 0,6.12 = 9,6 gam => =0,8 mol

mH(A) = mH(B) + mH(D) - mnjn^O) = 0,6 + 1,2- 2.0,2 = 1,4 gam => = 1,4 mol

A CO 2 nhom chifc este, khi thuy phan cho 1 phan tijf C2H,0H

=> D c6 2 loai nhom chirc va c6 CTPT dang (CH20)q va D la san ph^m cua

phan ling thuy phan: CgH,405 + 2H2O ^CjH^OH + 2(CH20)q

Bao toan cac nguyen to => q = 3 => D c6 CTPT : CjH^Oa

2 (0,5diem)

B CO cong thirc ca'u tao CHj-CH^-OH

D c o C r C T : CH3-CH(OH)-COOH hoac HO-CHj-CH.-COOH

KY THI TUY^N SINH L 6 P 10 THPT CHUYEN

NAMHOC 2011-2012 MONIHGAHQC ,y,n;f Ngay thi 04 thang 07 nam 2011 [ g c H I N H T H i r c

KY THI TUY^N SINH L 6 P 10 THPT CHUYEN

NAMHOC 2011-2012 MONIHGAHQC ,y,n;f Ngay thi 04 thang 07 nam 2011

Thin g i a n l a m b a i : 150 p h i i t (khong ke thai gian giao de)

(1) Viet cac phuong trinh hoa hoc theo so do phan ling sau:

FeS04 — ^ FeCl2 — ^ FeClj — ^ Fe2 (SO4 \^ Na2S04

(2) Cho Na20 vao nirdc du, duoc dung djch X Chia X thanh hai phdn bang nhau

Sue khi COi du vao phdn 1 duoc dung dich Y, cho het phan 2 vao Y duoc dung

djch Z, cho Z tac dung \6\g dich Ca(N0,)2 Hay viet phuong trinh hoa hoc

ciia cac phan ufng trong thi nghiem trdn

(3) Cho cac dung djch sau, m6i dung djch duoc dung trong m6t lo rieng mat nhan:

K H S O 4 , KHCO,, K.CO,, Mg(HCO,)2, BaCHCO,), Hay neu phuong phap nhan

?i6't cac chat tren neu chi duoc dung them each dun nong Viet phuong trinh hoa hoc minh hoa

,4) Co bao nhieu gam phen chua KA1(S04)2.12H20 duoc tach ra khi lam bay hai

160 gam nu6c tir 320 gam dung djch KAI(S04)2 bao hoa a 20"C Biet rang a

20°C, dung djch KA1(S04)2 bao hoa chiJa 5,5% KA1(S04), ve khoi luong

Cau 2 (4,0 diem)

(1) Cho dung djch H2SO4 loang, du Idn luot vao cac 6'ng nghidm, m6i 6'ng ehiia mot chat trong so cac chat sau: xenlulozo, chat beo, natri axetat, saccarozo r6i dun nong Viet phuong trinh hoa hoc ciia cac phan ung

(2) Hidrocacbon A tham gia phan ling c6ng v6i CU theo ti Id mol 1:1 thu duoc dSn

xua't diclo B Dot chay hoan toan a gam B thu duoc ^'^gam COi, ^a/^^gam H2O va ^ ^ / ^ g gam HCl Tim e6ng thiie phan tu, viet c6ng thurc ca'u tao, goi ten

A v a B • '

(3) Tliuc hien qua trinh len men gia'm tir 57,5ml dung djch ruou etylic lO", sau mot

thai gian thu duoc dung djch A Dem dung djch A tac dung \6i Na (vira dii) dd'n

khi phan ting xay ra hoan toan thu duoc 125,85 gam chat rdn khan Tfnh hieu suat qua trinh len men gia'm, bid't rang: dt^^H^OH =0,8g/ml va dH20 = '.Og/ml

Cau 3 (4,0 diem)

(1) Cho 68,5 gam kim loai bari vao 500 gam dung djch H2SO4 loang, sau khi phan

"•ng hoan toan thu duoc h6n hop A va V 1ft H2 (dktc) Nd'u c6 can A duoc 97,9

gam chat ran khan

(a) Viet phuong trinh hoa hoc cua cac phan iJng va tinh V r (b) Ti'nh n6ng d6 % ciia dung djch tru6c va sau phan iJng, coi nuoc bay hoi khong

dangkd' ' ' l*- 85

(c) Cho dung djch Ba(HCO,): du vao h6n hop A Ti'nh kh6'i lirong ket tua tao thanh

(2) Hoa tan hoan loan a gam h6n hop A gom Na va m6t kim loai R c6 hoa trj II vao

nude, sau phan urng thu duoc dung djch B va V lit khf • Ne'u cho dung dich B

tac dung vCra dii vdri 300ml dung djch HCl 0,25 M tao thanh m6t dung djch chi

chua hai chat tan Mat khac, khi h&p thu vira het 1,008 lit khi COi vao dung dich

B, thu dugc 1,485 gam m6t chat ket tua va dung djch nude loc chi chiia chat tan

NaHCO, Bict cac phan urng xay ra hoan toan va the tich cac khi do a d\6u kien

tieu chuan Hay xiic djnh ten kim loai R

Cau 4 (4,0 diem)

(1) Nhiing mot thanh Mg vao 200m! dung djch h6n hop chiia Fe(NO,)3 2M vu

Cu(NO,): 0,15M Sau m6t thori gian nha'c thanh kim loai ra tha'y kh6'i luong kini

loai tang them 2,8 gam so vdri ban dau Gia thid't ring ihi ti'ch dung djch khong

thay ddi sau phan urng va toan bo kim loai tao thanh diu bam het vao thanh Mg,

(a) Vict cac phuong trinh hoa hoc c6 the xay ra

(b) Tinh nong do mol ciia cac chat trong dung djch thu duoc sau phan irng

(2) Khu hoan toan 34,8 gam m6t oxit kim loai 6 nhiet d6 cao, ngudi ta cSn diJng

13,44 li't CO La'y toan b6 luong kim loai tao thanh cho vao dung djch chiJa h6ii

hcifp hai axit HCl va H2SO4 loang, du Sau khi phan iJng xay ra hoan toan thu

diroc 10,08 li't H , Xac djnh cong thiic oxit kim loai do, biet ring th^ tich cac khi

•"f do a dktc

Cau 5 (4,0 diem)

(1) Diln 5,6 lit (dktc) h6n hop khi gom etilen va axetilen vao 300ml dung djch broni

I M (trong dung moi CCIJ tha'y khoi luong binh dung dung djch brom tang them

6,7 gam, dong thdi kh6ng tha'y khi thoat ra Ti'nh khoi lugng tirng san phdm tao

thanh, biet rang cac phan (xng xay ra hoan toan

(2) Chia 44,8 gam h6n hop X g6m axit axetic, glixerol va etyl axetat lam ba phan

(ti le so mol ciia cac cha't trong m6i phan la nhu nhau)

- Phan 1 tac dung het voi Na thu duoc 1,344 lit (dktc) khi H,

- Phiin 2 tac dung vira du voi 500ml dung djch NaOH 0,4M khi dun nong

- PMn 3 (c6 khoi luong bang khd'i luong phdn 2) tac dung vdri NaHCO, du thi c6

2,688 lit (dktc) khi bay ra

j< Ti'nh khoi luong m6i chat c6 trong h6n hop X, biet rang hieu suat cac phan fog

Lay miu ciia cac dung djch can nhan bi6t vao cac 6'ng nghidm kh6 sach va danh so Dun nong tirng m5u, ta thay :

+ Cac m3u kh6ng tha'y c6 kd't tua la K H S O 4 , KHCO, va K^CO, (Nhom I)

2KHCO3 ^ K^COj + CO + H,0

+ Cac m^u xua't hidn ket tua trlng la dung djch Ba(HC03 )2 va Mg(HC03), (Nh6m II)

Ba(HC03), ^ BaCOji + CO.t + H , 0

Mg(HC03)2 MgC034 + CO.t + H,0

- La'y tirng mSu of nhom I (vdi lucmg du) cho tac dung Idn luot vori tirng mSu 0

nhom I I + Mdu nao 6 nhom I deu khdng phan fog la KHCO,

+ MSu nao 0 nhom I deu tao ket tiia trdng la K.CO,

K3CO3 + Ba(HC03)3 BaCOji + 2 KHCO3

K,C03 + Mg(HC03), ^ M g C O j i + 2 KHCO3

+ MSu con lai 6 nhom I c6 m6t Idn vira tao ket tiia trSng, vilra tao khi; m6t Idn

chi tao khi la KHSO4

2 KHSO4 + Ba(HC03), ^ BaS04i + K,S04 + 2C0,T + 2 H , 0

+ Mdu nao thu6c nhom I I chi tao khi v6i KHSO4 la Mg(HC03)„ mdu con lai la Ba(HC03),

(4) (1,0 d)

Goi kh6'i luong ciia tinh th^ phen chua tach ra x (gam)

- Khoi luong KA1(S04), trong 320 gam dung djch bao hoa la: "^^^^^'^ = 17,6g