Phương Pháp Giải Bài Tập Hóa Lớp 8

Phương Pháp Giải Bài Tập Hóa Lớp 8Phương Pháp Giải Bài Tập Hóa Lớp 8Phương Pháp Giải Bài Tập Hóa Lớp 8Phương Pháp Giải Bài Tập Hóa Lớp 8Phương Pháp Giải Bài Tập Hóa Lớp 8Phương Pháp Giải Bài Tập Hóa Lớp 8Phương Pháp Giải Bài Tập Hóa Lớp 8Phương Pháp Giải Bài Tập Hóa Lớp 8Phương Pháp Giải Bài Tập Hóa Lớp 8

HUYNH VAN UT

Giao viin bAi dlliJng h9e sinh gipi Giai thllllng sach hay VN

Phuong phap giai chi titt Lcii giai cac bai toan nang cao

� Nha xuat ban

�E)�i hQC Quoc gia Ha NQi

HUYNH VAN UT

Gicii thui:i11g sc,d, lwy Vi(?t Nam 2008

GV Roi d1iif"!! h f,J c sinh gioi

GV Trli<111g C/UO(' a: Vift - Uc

PHUONG PHAP GIAI BAI TAP

�

HOA HOC

Cung n'ii vi?c d6i m6i chLtung trlnh va sach giao khoa, vi�c d6i m6i phu·ong ph:1p hQc v6i h9c sinh Ii\ m(H trong nhung v�n d� co ban,

la 111()t trong nhu·ng l11l)C tieu phan dau cua nhieu th:1y co gi;10 va h9c sinh Nh0n thu\: du\1c di�u do rjc gi,1 xin ch?ln thi\nh gi{ji thi�u den

bc:tn GQC quyen s{teh: PHVONG PHAP GIAI BAI T �p HOA HOC 8

N◊i dung quytfn sftch gf>m sau chu·ong, tu.ling (tng vdi sau chuong trong sach gi,10 khoa HOA HQC 8 hi�n hanh tv1oi chu·ong gom:

A KIEN THUC CAN NHd

B PHVONG PHAP GIAI cAc D�NG TOAN

Cac hai t�p trong m6i d�ng toan du\Jc s:'ip xcp tt.i d� den kh6.Sau 1nbi c1� bai c6 loi giai chi tiet, ro rang nh�m giup (aC em doi chieu l�i ke'r qu,\ sau khi da tt! giai bai toan d6

Chung toi co g�ng khai thac tri�t di cac d0ng toan khac nhau v� Ii thuyet cCtng nhu bai t�p va c6 nhi�u G1ch giai kh:.k nhau nh�m lain phong pht1 kien thu·c cho c.k em

Voi kho�ng 200 trang s,kh, clnia OlJ di chung toi th� hi�n het au<)c y tu·Jng v?1 mong uoc cua m1nh va c6 th0 con m¢t so sai s6t Rat mong quytf n sach duc,.1c quy th�y co gic10 va c;k L'm h9c sinh d6n nh�n, g6p y

CHU'O'NGI

'" , ? "" ?

A. KIEN THUC CAN NHO

I. CHAT

(1} V(1t the': la nhung vc;it ton tc:i.i xung quanh ta Vi dy: quan ao,

gia'y dep, sach, but,

V(ll the' t1_t 11hie11: m�t trang, ng6i sao, cay xoai, d6i nui,

V(rt the' 11hu11 /�10: X(! m,iy ,ich v6, ti vi, tu lc:tnh,

bJ Clicft: v�t the · duqc t�o thanh tli chat

- Cha't tinh hhie't lit c/1ci't l<li611g co la11 cha't lduic hay d1lqc tgo thimh lit Cit/lg 1119/ loq.i µlia11 It! (nlio'C cat, muoi tinh, ).

- Hieu tinh cluit ctla cha't de: phan bi�t chat nay v&i chat khac,

tu'c la de· nh?n biet tu-ng chat.

- llai hay nhieu clui't tr911 Lan uao nhau g9i la hon hqp.

II NGUYEN TU'

1. Nguyen tu: la h�t v6 cung nho (vi mo) va trung hoa ve di�n.

2 C(iu tqo nguyen til

-+-) Laµ vo: duqc cau t�o bci'i m(>t hay nhieu electron mang di�n tich

am, chuye'n d()ng xung quanh h9t nhan va siip xep thanh tu-ng lap Khoi luqng cua h<;1t electron: 9,1.10 - 2 8 gam.

-+-) Hqt nhan: duqc cau t<;10 bci'i proton (p) va natron (n).

• l\1oi h <;1 t proton mang mQt di�n tich duang, cac nguyen tu' cung

lo<;1i c6 cung s6 proton trong h<;1t nhan Khoi luqng cua h�t

proton: 1 , 67 10 - 2 4 gam.

• Di�n tich cua proton c6 trj so bf!tng di�n tich cua electron nhung nguqc dau va nO'tron kht'ing mang di�n Khoi luqng cua h�t notron: 1,675 1 0- 2 4 gam.

- Trong m9t nguycn tu': so proton = so rlectron.

3 Nc;u h1nh dung nguyen tu' nhu m(>t qua c 1u, tam la h:::tt nhan, cac electron chuye'n dqng rat nhanh xung quanh h�t nhan, thl duang

4. Kho'i luqng nguyen tit bttng t6'ng hho'i luqng cua proton, natro11 ua electron Vi hho'i luqng electron rdt be so vai ldufi l1t<;1ng· c11a proton va natron nen ldio'i luqng nguyen til xdp xi bdng kho'i luqng cua proton va natron.

Ill NGUYEN TO HOA HQC

1. Nguyen to' h6a h9c la tq,p hqp nlulng nguyen ti£ ci'tng lo c;i i, co c1:ing s6 proton trong h c;i t nluln.

2. Moi nguyen to h6a h<;>c duqc bi�u <lien biing m(>t hay hai chu cai,trong d6 chu cai dciu duqc viet in hoa g<;>i la Id hi�u h6a h9c.

- Ki hif?U h6a h<;>c cho biet:

• Ten nguyen to

• Chi m(>t.nguyen tu' cua nguyen to d6

• Nguyen tu' khoi cua nguyen to

3. Nguyen til ldio'i (NTK) la hho'i lttqng cua 11ujt nguyen td tinh

bttng dan vi cacbon (dvCJ ldvC = _!_ khoi luong mot nguyen

cacbon

IV DdN CHAT VA HQP CHAT- PHAN Tl/

6

I Dan clui.t la nhitng chdt tc;w nen tit nz{)t nguyen to h6a lu;,c Co 2 loc;ii:

• Kim loCJ,i: hciu het a th� riin trong dieu ki�n thuong ( tru' thuyngan a th� long), c6 anh kim, dan di�n va dan nhi�t tot

• Phi kim,: ij dieu ki�n thu<1ng, c6 th� ton t�i ca ba tr�ng thai: riin(C, P, S); long (Br2); khi (Cl2, 02, N2 ) Phi kim khong c6 anh kim,khong dan di�n va dan nhi�t (tnJ than chi dan di�n va dan nhi�tduqc nhung kem)

2. Hqp chfit la nlulng chfit tc;io nen tit hai nguyen to' h6a h9c tro Len.

Gom hai loc;ii:

• Hqp cluit vo ca nhu: H20, H2S04, NaCl,

- Neu phan tit bf, chia nh6 han thi hhong con mang tinh chat c,111 clta't.

PHlJ<JNG PHAP GIA! OAI T,\P 1((1,-\ IIQC 8

V CONG THUC HOA HOC

- Cong tlu?c hoa h9c d(u,g de' biiu dien phan tit ct.la dan chc i t va /I(!!' chfi't.

·- Cong tlu?c hoa h9c cho biet: Phan tu' chat d6 gom nguyen tu' cua

nhUng nguyen to nao, moi nguyen to c6 bao nhieu nguyen ttl, phan

t1, khoi cua chat.

Chi gont hi hi¢u h6a h9c ctla m(Jt nguyen to va chl so iiguyen ltl trong m(Jt phan trt.

Vi du:

• Cong thuc h6a h9c cua kim lo?i: kali, magie, la: K, Mg,

• Cong thuc h6a h9c cua phi kim: cacbon, luu huynh, la:

C, S,

nguyen to va chi soviet & chan cua ki hi�u h6a h9c: N 2 , _H 2 , 03,

2 Cong thuc hoa h Q C cua h<jp cha't

06m hi hi�u hoa h9c ctla nhilng nguyen to tg,o ra chat he,n theo chl so' a chan nu3i ki hi�u nhu: Ax B y ho�c A xB y C z ,

Trang do:

• A, B, C la ki hi�u nguyen to.

• x, y, z la nhung so bi€u thi so nguyen tu' cua nguyen to c6 trong m9t phan tt1 hqp chat, g9i la chl sc>'.

• Chi so l quy u'O'C khong ghi.

VI HOA TRI

1 Hoa trj, ctla 111. (Jt nguyen to (hay nhom nguyen tit) la con so biiu thj,

!did nang lien het ctla nguyen tii (hay nhom, nguyen tii) nay vai 1n(>t

so nhat dj,nh nguyen tii (hay nh6m nguyen tii) lduic.

2 Qui tdc hoa tri

N(>i dung: "Trang c6ng thuc hoa h9c, tich cua chl so va hoa trj,

cua nguyen to nay bling tich cua chl so va h6a trj, cua nguyen to

/,•ict"

B PHUdNG PHAP GIJ\I cAc D�NG TOAN

- 1 a lt:t- nip= , gwn; 1n 11 = ,o,, gwn; me= , u gcJJn

- Neu g9i m p , ,n il va 11l c ldn lt.tqt la llhoi lt.tqng cua proton, natro,n va electron.

::::::> mnguye11 t1l = 11lp + 1n,1 + 11le

Vt 11l c << ,n p , 11l n nen llt 11 guyc1 1 ft�= ,n p + ,n/1 (btuig ldwi lw;m.g Cllll hat nh6n).

- Klufi lt.tqng cila nguyen ti't dt1qc tinh theo gain gqi la hhoi lutq,,g tuy� t dtfi ( KLTD ).

⇒KLTD = M x 1,66.10 24 gam (um · M la nguyen til llh6i cua nguyen tfi)

- Khoi lt.tqng tuang doi (KLtdJ: la ldioi lt.tqng Lan gap N ldn lllh6i luqng tuy?t d6i, tinh bdng dan vj cacbon (dvC).

⇒KLtd = N x KLTD (N = 6,02.10 23: s6 Avogadro).

Bal I M{>t nguyen tu' nhom (Al) ·co 13 proton, 13 electron va 14 notiron

Hay xac dinh khoi lugng cua m{>t nguyen tu' nhom

+) 1 nguyen tu' cacbon 02 dvC) c6 khoi lugng la 1,9926.10-23 gann

1 nguyen tu' natri (23 dvC) c6 khoi luqng la a gam

2:J

a= 23 x 1,9926.10- = 3 82_10'_23 gam

12

+) 1 nguyen tu' cacbon (12 dvC) co khoi lugng la 1,9926 10-23 gann

1 nguyen ht sAt (56 dvC) co khoi luqng la b gam

+) l nguy&n tv cachon ( 12 dvC l c6 khoi lu\ing In 1,9926.10 2:3 gam.

1 nguyen tv nhom (27 c1vC) c6 khoi luqng la c gam

27 · · l ( C)26 10 n

c = · .::_J J = 4,48.10 2

:1 gam

12 +) 1 nguyen tv cacbon (12 dvCl c6 khoi luqng la 1,9926.l0- 2·3

gam

1 nguyen tv dong (64 dvC l c6 khoi lu<;Jng la d gam

d = 64 x 1, 9926 10 n =- 10 63.10 2:1 gam

1 2 +) 1 nguyen tv cacbon ( 12 dvC) c6 khoi lu<Jng la 1,9926.10 - 23 gam

i nguyen tu· kern ( 65 c.lvC > co khoi luqng la e gam

e = 65 y 1,9926.10 2:1 = 10, 79.10 23 gam.

12 Bai 3 Hay xac djnh khoi luqng cua m9t nguyen tu' OXl, luu huynh,

magie, bari ra dO'n V! gam?

- Trong m(}I nguycn f tl: t(Sng so' proton = t{/ng so' electroii.

- Ne'u g9i n, p · i1a c l611 l11c;t la so' natron, proton va electron

trong 11guye11 ttt fX) thi:

Vi p = e nen: T6'11g so' h c;i t trong X = 2µ + n (2)

Ket h e; µ (2) i1ai cclc da hifn d� cha de' lg,p h� phtlang trinh, gidi

It� ph ttang I ri II I, uti'a lg,p => p, n va e

Pl!L'O-.;(: PHAP <:!Al nAI TAI' 110,\ HOC 8 9

·- Neu dJ biii chi cho ttlng so' h [! t trong nguyen tit ,na llh6 , i g cho them dil ki�n nao thi ta rip d(Lng bat ddng thac: p s n s l 5p (8) (dlfqc phep <ip d,.1ng clui k'16ng clutng minh) Luc nay ta l<e't h c; p giila (2) va (3) di xur elf nh gia,· ,,,.�n cila p ⇒ p, n UCL e.

- Neu dJ cho ,n<)t loq,i h q. t llCLO do c/1,icm bao nhieu phdr tram

Vong to'ng s6 h q, t thi ta <ip di!-ng c6ng thllc tinh % ⇒ llet QilO.

- S6 kh6i (A) cila 1n<)t nguyen tit = nguyen tit ldio'i: t6'ng �o' hq,t proton + t6 ? ng s6 h f) t natron.

Bal I a) Biet t6ng so cac lo�i h�t (proton, natron va electron) trong

nguyen ta X la 28 va so h�t khong mang di�n la 10 Hay xac djnh so proton trong nguyen hi X?

b) Biet nguyen tu' Y co tong so cac lo�i h�t la 21, trong do ;,o hl;ltkhong rnang di�n chiem 33,33% Xac djnh cau t�o cua nguyen tu Y?

Bal 2 Biet t6ng so h�t trong nguyen tu' cua m(>t nguyen to h6a h� Z la

40; trong do so h�t mang di�n nhieu hon so h�t khong mang di�n la 12 Xac djnh cau t�o cua Z

Giai

G9i p, n, e lftn luqt la so proton, notron va electron cua (Z).

Theo de bai, ta co: p + n + e = 40 (1)

Vl nguyen tu- trung hoa di�n nen: p = e

Hay xac dinh ten va ki hi�u cua nguyen to Z?

b) M◊t nguyen tu' Y c6 t6ng so h�t (proton, notron va electnn) la28; so h�t khong 1nang di�n chiem xap xi 35,7% Hay xac dinhcau t�o cua Y

PHUONG PHAP GIA.I BAI TAP H<A HQ( 8

Gi<,i 5,:312 IO !I a) Ta c6 : NTK , z , = = :�2 <lvC: luu huynh (8)

Bii 4 al Cho ki hi�u h6a h9c sau: 1��Au Hay xac dinh so notron cua

nguyen tu' tren

b) Hay xac dinh cau t<;10 CUR h<;1t nhan uran 2 i�U.

Giai a) Ta c6: A = n + p (v6'i A la so khoi h<;1 t nhan)

⇒ n = A - p = 197 - 79 = 118

b) Tuong tlf a)

So proton = so electron = 92; so notron la: 238 - 92 = 146 Bili 5 a) Biet � nguyen tv X n�ng biing i nguyen tu' kali Xac dinh ten

va ki hi�u cua nguyen to X?

b) Nguyen to X c6 nguyen tu' khoi bAng 3,5 Ian nguyen tv khoi cua oxi Hay xac dinh ten va ki hi�u cua nguyen to X?

a) Theo de bai, ta co:

b) Theo de bai, ta c6: Mx = 3,5M 0 = 3,5 x 16 = 56: sAt (Fe).

Bai 8 Nguyen tu' silt (Fe) c6 di�n tich hc;it nhan la 26+ Trong nguyen tu'

srit, so hc;it mang di¢n nhieu han so h�t khong mang di�n la 22

Hay xac dinh nguyen tu' khoi cua scit

Giai

DiOn tich h�t nhan bilng 26+ ⇒ So proton trong h<;1t nhan la: 26

p!UtJN; !'HAP Git\! 13AI TAP 110 1 \ f!(,1C

11

D�NG TOAN 3:

PhucJng phap:

- De viet sa do electron Clla nguye11 til ta dung ccic duang t , roi

di bieu diJn M6i vong la ni<)t lap electron.

- Qui uac s6 electron co trong nii)t lap:

• Lap tlul nhat chua to'i da 2 electron.

• Lap thu hai cluta toi da 8 electron.

• Lap tlul ba chlia 8 electron

• Lap tlllt tu cluta 8 electron.

Chu y: Lap thu ba, thu tu clula han 8 electron nhung trong gi,i lu;ui chliang trinh ta chi dt'tng l g, i 8 electron

- M6i electron bieu thj bling mqt dau chant tron d(un.

Bal I Nguyen tu'

Z c6 16 proton trong h�t nhan Hay ve cau tc;io · CUl

nguyi,n tU

Z

Gilli

�®� ·-t;'0_;.�)

Vl sop= so e = 16 ⇒ lop 1 c6 2 electron, _v

lop 2 c6 8 electron, lop 3 c6 6 electron

Sa d6 cau t�o (hinh ve hen)

a Cho biet Sc1 do mot so nguyen tu' sau:

l'IIUU!\G PIIAP Cl.-\1 BAJ T.-\1' Ill).-\ 1 1! ( 1( ';

D�NGTOAN 4:

L�P CONG THl.TC PHAN TU' CUI\ MQT cHAT DlfA vAo HOA TRJ

Pl1tt<1ng phap:

+) Kl, i bi( , ho(I tri Cl/(( Iii(_)/ ll/,?"llye11 to' ( fl h6lll 11guye11 I 1/) vi, c611g

t/11/c hoci h9c, ti111 hoa tri Cl/([ 11g11y{>11 to' COIi /qi

Ta g9i (J lu hod trj C(ICJ nguycn to' CCIII tim, dqt vao c611g tlurc hoa

h9c roi up d1_111g quy tdc hoa tri de' ti111 a

+) Kl, i bi i'>'t hoc, tri CIJCJ 2 11guye11 t o' ( ho(J,c lllQl 11guye11 to' va Ill Qt 11/,6111 11guye11 t1i'J, lq.p cung tlulc /u)(i l,9c c11a ch1111g G9i x, y ldn ltlql fo chi s6 Cl/G ch1111g, SCll/ d6 cip c/ 1_ 1.ng quy tdc hoa tri roi suy

Kl, i l�ip Ii so - la p!tdi d11a t•e phdn s6 t6i gidn r6i c/1911 x, y

y

Dl; l(,p nhanh c6ng tllltc huci h9c, ta lay lzo<i tri c11a nguyen to'

Sall do nit wm SC t/111 dli�1C c611g t/11/c hoci h9c,

Bai I Viet cong thuc h6a h9c va tinh phan tu' khoi cua cac hqp chat sau:

a) Canxi oxit (voi song), biet trong phiin tu' c6 1 Cava 1 0.

b) /\moniac, biet trong ph,1n hi co 1 N va 3 H.

b) Fe h6a trj Il, 0 h6a tri II

Ag h6a trj I, 0 h6a tr[ II

Si h6a tri IV, 0 h6a tri IL

PtlLJUNl; PHAP GIA! HAI T/\1' IIOA IIOC 8 13

Bil a a) Mc}t hqp chat X chua 94,118% h.tu huynh va con l�i la hidro

Hay xac djnh cong thuc nguyen cua hqp chat X

b) Hay tinh thanh phan phan tram theo khoi luqng cua cacnguyen to trong hqp chat CuSO4

Giai

a) Xac djnh cong thuc cua X

Theo d� bai: %S = 94,118% ⇒ %H = 100% - 94,118% = 5,88Zo/nXet 100 gam X ⇒ ms= 94,118 g�m va mH = 5,882 gam

G9i cong thuc tong quat cua X co d�ng: HxSy

Iii 4 a) Tinh h6a trj cua moi nguyen to trong cac hqp chat sau: ZnC12,

CuCl, AIC13 , biet Cl h6a trj I

b) Tinh h6a tri cua Fe trong hqp chat FeSO4

Giai

a) ZnCb, ta c6: 1 x x = 2 x I ⇒ x = II ➔ Hoa tri cua Zn la 11

CuCI, ta c6: 1 x y = 1 x I ⇒ y = I ➔ Hoa trj cua Cu la I

AlCb, ta c6: 1 x z = 1 x III ⇒ z = III ➔ H6a ttj cua Al la III.b) FeSO4, ta c6: 1 X k = 1 X II ⇒ k = II ➔ H6a tri cua Fe la II.Iii I a) Hqp chat Y chua 72,414% Fe va 27,586% 0 Hay xac djnh cong

thuc h6a h9c cua hqp chat Y Biet cong thlic nguyen cung chinh

la cong thlic phan tu-

14

b) Hqp chat Z chlia C va 0, trong do cacbon chiem 27 ,27% theokhoi luqng va phan tli khoi cua Z bAng 44 ·dvC Xac djnh songuyen tli cua oxi trong hqp chat Z

PH VONG PIIAP GIA! BAI T Ar l!(>A HQC 8

Giai

a) Xac d�nh c6ng thuc phan ht cua Y

Jm Ft · = 72, 41 4 gam Xet 100 gam Y ⇒ l m 0 = 27, 5 8 6 gam.

G9i cong thuc tong quat cua Y: Fe x O _y

Lap ti le: x: y = 72' 414 : 27'586 = 3: 4 => CTPT cua Y: Fe 3 0 1

Bai 8 a) Hqp chat Y co thanh phan phan tram theo khoi luqng cac

b) Xac djnh h6a tri cua sAt trong hqp chat Fe 3 0 4•

b) Fe 3 Q 4 gom 2 oxit: FeO Fe20:1

Bal 7 a) Cho cong thuc h6a h9c cua hqp chat X c6 d�ng Fe2(S04) a , phan

tu' khoi cua hqp chat X biing 400 dvC Xac djnh h6a trj cua silt trong hqp chat X?

b) Khi phan tich hqp chat Q chua: 27,273% cacbon va con l�i la oxi Xac djnh h6a trj cua cacbon trong hqp chat Q.

c) Cho hqp chat oxit cua nito c6 d�ng N a O b Hay xac djnh h6a trj cua nito trong hqp chat tren?

Git.ii

a) Theo de bai, ta c6: 56 x 2 + 96 x a = 400 ⇒ a = 3 Ap d l;l ng qui tAc hoa trj ⇒ Hoa trj cua siit: III

b) L � p A tl •• l" �: X : y = 27,273 72,727 l 212 : 16 = :

⇒ Cong thuc cua X: CO2 ⇒ H6a trj cua cacbon la IV.

c) Ap d l;l ng qui tAc hoa tri, ta co:

Nail ob ⇒ a.x = b.11 ⇒ x = - a

Bti a Oxit cua m{>t nguyen to R co h6a trj V chua 43,67% R v� khoi

luqng Hay de xuat cong thuc phan tu' cua hqp chat R

Giai

G9i cong thuc oxit t6ng quat co d�ng: R2Os

Th d� b' t , eo e a1, a co Cl 1cm1< = x2R 100° ic 1 = 43 , 67° 1 1 0

2R + 5 x 16

<=> 2R + 80 = 4,5798R ⇒ R = 31: photpho (P) ⇒ cong thuc: P:.O5

Bal • a) L�p cong thuc h6a h9c cua nhung hqp chat hai nguyen to SHU:

P (Ill) va H, C (IV) va S (II), Fe (III) va 0

b) L�p cong thuc h6a h9c cua nhung hqp chat t�o b<1i m◊t nguyen

to va nh6m nguyen tii' nhu sau: Na (I) va (OH) (I), Cu (Ill va(SO4) (II), Ca (II) va (NO3) (I).

Giai

a) Cong thuc h6a h9c cua: P (III) va H la PH3, C (IV) va S (11) la

CS2, Fe (III) va O la Fe20a

b) Na (I) va (OH) (I) la NaOH, Cu (II) va (SO4) (II) la CuSO4, Ca {ll)

va (NOa> (I) la

Ca(NOah-Bal 10 M9t hqp chat 410 b<1i 2 nguyen to S va 0, biet ti l� theo khoi lu<J11g

cua S doi v6'i O la 2 : 3

a) Xac djnh ti so giua so nguyen tu' S va so nguyen tu' 0 c6 trongmc)t phan tii' hqp chat

b) Xac djnh phan tti' khoi cua hqp chat tren, biet trong m◊t phantll' hgp chat c6 1 nguyen tti' S

Giai

a) Tinh ti so giua so nguyen tti' S va 0

G9i X, y Ian lugt la so nguyen tll' S va O trong m(>t h<,1p chat

Th d�b�· t ,, ms 2eO e al, a CO - = - C> 32.x 2= - C> -x 2 16 1 =

-X - =

b) Tinh phan tu' khoi:

Phan tll' khoi cua hgp chat tren la: 32 x 1 + 16 x 3 = 80 dvC

Bal 11 M{>t hgp chat oxit X co dC;1ng R.2O3 • Biet phan tti' khoi cua X Ja 102

dvC va thanh phftn phfin tram theo khoi luqng cua oxi trong X

bAng 4 7,06% Hay xac djnh ten R va cong thuc oxit cua X.

Tu-121 >a= 3, thay vao phuong tdnh (1), ta c6: 2.l\1H + 1(-i.�3 = 102

� I\1 11 = 27: nh<">m (All

-V,:1 v I{ Iii kim loc:1i nh6m va cong thu·c Oxit: Al 2 O 1

Bai 1a I3i<",t X chlia � nguyen to C va H, trong do cacbon chie'm 85,71 o/c

theo khoi Iuqng va phan tu' khoi cua X nh� hon 7._ Jan phan tu'

8 khoi cua 0� Xt1c djnh cong thlic h6a h9c cua X

Giai

G9i cong thu'c tong qwit cua X: C x l I�­

Thco de bl1i, ta co:

Mx = � M 0 _ ⇒ Mx = f 32 = 28 dvC c:> 12x + y = 28 (1)

V,'.l <¼Mc =- > l�� -·100 = 85,71'7c 28 ⇒ x = 2, the' vao (1), ta duO'c:

y = 28 - l 2x == 28 - 12.2 = -1 V�y cong thvc h6a h9c cua X: C2H4

Bai 13 Tien hanh phan tfch m9t hgp chat X chva: 32�C; 6,67%H; 18,67%N

va 42,66'/r O theo khoi lugng Hay xac dinh cong thvc h6a h9c cua

h9'p chat X, bict trong µhan hi chi c6 m9t nguycn tti nitci

⇒ Cong thvc nguyen tu' cua X la: (C 2 HsOzN) ll

Vi trong phan tu' X chu'a m()t nguyen tu' nit<1 nen n == 1

V,}y cong thu'c h6a h9c cua X la: C2 H 5 0 2 N

Bal 14 Khi µhi1n tich djnh luqng m◊t chat huu co A, ta c6 ket qua sau:

cv 4 phan khoi luqng cacbon thi c6 m()t phan khoi luqng hidro v£:1 2,67 phan khoi luqng oxi Biet phan tu' khoi cua A gap 11,5

DAI HOC 4.::::JUOC G'tA nM ','t.'I

1•11 L ' l1�c t'IIAP CL\t BAI TAP Hu:\ 11n t �N_ � TAM THONG TIN T H U _ '!I Ef\ j

Giiii

G9i cong thuc t6ng quat cua A: C x HyO z

Va me : mH : mo = 12x : y : 16z = 4 : 1 : 2,67

4 2,67

�x :y :z=-:1: =2 :6:1

⇒ Cong thuc nguyen cua A: (C2H6O)11 •

Ma: MA= (2 X 12 + 6 +16)n = 46 ⇒ n = 1 ⇒ CTHH cua A: C 2 H 6 O Iii II De dot chay 1 mol chat X can 6,5 mol oxi, thu duqc 4 mol CO 1 va

5 mol H20 Hay xac djnh cong thuc phan tu- cua X.

Giai h� phlldng trinh, ta duqc: x = 4; y = 10; z = 0

V�y cong thuc phan tti' cua X: C 4

H10-lil 18 Biet m<)t hqp chat cua nguyen to A h6a trj II vdi nguyen to oxi,

trong do nguyen to oxi chiem 20% ve khoi luqng Hay xac djnh ten nguyen to A?

18

Giiii

A c6 h6a trj II ⇒ Cong thuc chung cua A vdi oxi la: AO

Theo de bai, ta co:

%0 = 16 > 100% = 20% <:::- A + 16 = 30 ⇒ A = 64.

A+ 16 V�y A la kim lo�i d6ng (Cu )

PHUONG PHAP <;fAJ BAI T \P 11n ; H<ic �

Bai 17 I ,(1 p CT ll 11 cua c.ic hqµ chftt vdi oxi cua cac nguyen lo sau day:

c) Ab03 g) SO:i

Bai 18 Vie ' t cong lhlic h6a h9c cua cac hqp chat t�o b6i c,ic thanh phan

C,lU t<;to sau V,1 ti nh phan tu' khoi cua cac hqp chat <l6-:

nl Pb (ll) v;.1 N0 : 1 bl Ca Ya PO 4 cl Fe (III) va Cl

ell Ag va S0 1 el Bava CO 3 g) :Mg va HSO4

(Pb= 207; Fe= 56; Ca= 40; Cl= 35,5; P = 31; l\lg = 24; Ag= 108;

1•1 19 Tinh so nguyen tL.i' hidro lien ket duqc v6'i nguyen tu', nhom

nguyen tL.i' sau Viet cong thlic h6a h9c cua c,k hqµ chat do

S(ll); l3r(I); N(Illl; SO 4 (1IJ; NO 3 (1); CO 3 (lll; PO.,(IIIJ; C(IV); SiO;i(IIJ; P(IIJ L

Giai

S: 2 (11:.$l; Br: 1 (Ill3r); N: 3 (NH 3 ); SO4 : 2 (I-bS04 ); NO 3: 1 (HNO 3);C0:3: 2 (I l�CO:1l; PO , : 3 (H 3P01 ); C: 4 (C8➔ ); SiO3 : 2 nI�SiO 3 ); P: 3 <PH3 )

1•1 20 Cho hqp chi{t Z cua nguyen to A h6a trj 11 v6'i nguyen to oxi,

biet nguycn to oxi chiem 20% ve khoi luqng Hay cho biet t�n nguyen to A la gi?

=>A= 80 - 16 = 64 dvC: dong (Cu)

Vq.y cong thlic µhan tu' cua Z: CuO

PHL'Cli':<; Pllr\P Cl,\l TL\l Tt\P I!< 1.\ HOC 8 19

CHU'C1NG II

A KIEN THUC CJ\N NHO

_,,, , ,

I St) BIEN DOI CHAT

1 Hifn tllqng trong do chfit bi bien d6'i (trgng tluii, hinh d<mg,

kich tlutac ) ma vein girt 11guye11 la clui't ban dciu duqc g9i la hi�n ltl<Jng t·(tt ly.

2. Hi{!n tt1911g !rung do co Si;l bien d6'i clui't na.y than/, c!ui't !duic, tile La co s-lf sinh f'(I cltd't mai dtl<JC g9i la hi�n ll1911g hoa '19c.

II PHAN UNG HOA HOC

-Phii.n lfllg hoa IU,JC la qua lrinl, bi�n do'i cha�t nay tlu'mh ch(it ldujc.

- Trong phii11 1/11g hoa h9c chi co Lien ldit giila ccic "l.Jl(Ven tli thay d6'i Ldm cho phan t11 nay bien d6'i thanh pluin t11 lduic.

- Clui't tham gia (ha.y chdt phdn 1i'ng) ghi ben trrii.

- Chtit t<;w thr111h (hay sdn plufmJ ghi bcn pluii.

Sa d6: Ten c,ic chli't than1 gia - -➔ Ten cac chi1t san µhftm.

Phcin ung xi.iy ra khi cac chat phai tiep xuc Pai nha.u {di�n tith

tiep xuc cling l6'n thi phan ung ci1ng de xay ra), dun /IOllg ho�lC

can co chat Xtic tac'

+) Ditu hi�u nh(in biet phan ung xay ra:

Phan ung xay ra khi co chat moi t�o thanh qua cac d,1u hi�u:

• Thay d6'i 11u1u sdc

• T q., o chu't bay Jun

• T q., o .cha't /<.et tu.a

• Toa nhi�t ho(f.c pluit scing.

Ill DINH LU� T BAO TOAN KHOI LlJQNG

20

Dinh lu(u: Trong 1119! phan ting hoa lu;>c, to'ng llhoi lt1q11g c11a ccic chat Sall plufm bilng tiSng ld16i lttqng Cl/Q ccic ch(it tham gia phew dng

Giai thich ,ljnh lu(lt: Trong m(>t ph,in Ling hoa h9c, chi c6 �l/

thay doi lien ket giua ecic nguyen tu', con so nguyen tLi' cua c,ic

nguyen to van giu nguyen nen khoi luqng duqc brio to,'tn

A. p d�ng: Trong m(>t ph,in ting co n chat, n<:-'u biet kh6i luqng cua

(n -1) chat thi tinh duqc khoi luqng cua chftt con l�ii

PIIU<JNC Pll,\P CIA! BAI Tr\P H6A I tn,· 8

IV PHVdNG TRINH HOA HQC

J S1_1 bi/11 <li,�11 JJluin 1/11/l Iii)(/ lt9c b611g c<LC c611g t/111c h6a h9c dll<7c

,�1 .>i ht p/111a11g lrinlt !ti)(/ lt9c.

2 Ph 1lu11g I ri II It hou h9c cho bi{,{·

- Ch11t phrin ung va S[ln phflm.

- Ti I� so nguyen tu', phan tLi' cac chat trong phan ung Ti l� nay

Vf dy: CaC0: 3 r+ 21 ICl - ► CaCl 2 + CO2 + H20

Trong ph,in ung tr<'n, ti I� µhan tl.i la:

CaC0:1 : IICI = 1 : 2

CaC0 1 : IICI : CaCl 2 : CO 2 : H 2 0 = 1 : 2 : 1 1 : 1

B PHUdNG PHAP GIAI cAc D�NG TOAN

D�NG TOAN 1:

AP Dt,JNG l>JNH LU]:\T BAO TOAN KHOI LU'QNG

Phuong phap: S/1 rli,111g cong Llui'c:

'\� m 1 = ' m 1 1 ( ' : doc la t6'11g)

L.,, \,H· ,t11pldlll L.,, CaC'C)Alll,llll)-!1� L.,, •

Nf:'u trong pluin 1111g co 11 cl1ti't, ld1i biet hho'i luqng cua (n - 1)

cha't tlti tinh dt(�JC ld1o·i illqng Clta clt{r't COil l c;i i

1•1 I LMu huynh ch,ly theo Sd do phan ung sau:

Neu dot chay 48 gam luu huynh va thu duqc 96 gam khf sunfuro thi khoi l119ng cua oxi tham gia phan Ll'ng la bao nhieu?

Giai

ffi1t111 huynh + ffioxi = ffikhi sunfuro ffioxi = ffikhi sunfuro - ffiht11 huynh = 96 - 48 = 48 gam

chat ban dau elem dot?

Bil 3 Cho 65 gam kem (Zn) tac dl,lng v<li dung djch HCl thu du'Q'c 136.

gam ZnC12 va 22,4 lit H:2 a dktc Hay tinh khoi luqng HCI da tham gia phan u'ng

Giai

Zn+ 2HC1 ➔ ZnCh + H2i

Ap dl;lng djnh lu�t bao toan khoi h.tQ'ng, ta co:

ffiHCI = 136 + 22,22,4 4 x 2 - 65 = 73 gam

■ii 4 Cho hon hQ'p 2 muoi X2S04 va YS04 c6 khoi luqng 22, 1 gam tac

dl;lng vua du v<li dung djch chl!a 31,2 gam BaC 11, thu duqc 34 ,95 gam ket tua BaS04 Tinh khoi luqng cua hai muoi dem dung

Giai

• X:2S04 + BaCl2 ➔ BaS04-l, + 2XC1

YS04 + BaCh ➔ BaS04t + YCh

Ap dl;lng djnh lu�t bao toan khoi luqng, ta c6:

⇒ mh iu 1111101 rau = Ill I., V :, S() YS(' , 4 • • ) � I + ffin.,c, 1>11 , - m,, "O ,it.'a �

= 22,1 + 31,2 - 34,95 = 18,35 gam , Dot chay hoan toan 12,8 gam dong (Cu) trong binh chaa khi oxi

(02), thu duqc 1'6 gam dong (II) oxit (CuO) Tinh khoi luQ'ng md tham gia phan ung

Giai

2Cu + 0 2 ➔' 2Cu0

Ap d1:1ng djnh lu�t bao toan khoi luqng ta c6:

mcu + mo2 = mcuo ⇒ moi = mcuo - 'mcu = 16 - 12,8 = 3,2 (gam) ail I Cho 16,25 gam Zn tac dl;,lng vai dung djch axit sunfuric (H2S04),

thu qugc dung djch chua 40,25 gam ZnS04 va 5,6 lit H2 (dktc) Xac dinh khoi lugng axit H2S04 cAn dung?

22

Giai

So do: Ke·m + axit sunfuric ➔ Kem sunfat + hidro

· Ap dl;lng djnh lu�t bao toan khoi luqng, ta c6:

ffikern + ffiaxit sunfuric ➔ ffikem slmfat + ffitudro

PHUONG PHAP Glr\l BAI TA.P HOA JIOC 8

I

= 40,25 + � x 2 - 16,25 = 24,5 (garn)

22,4

(NaOl - 1), thu duqc 10,7 gam s<'it (Ill) hidroxit Fe(OI-1)3 va 21,3

gia phan 11ng'?

Giai

Sci do:

Sih (Ill) sunfat + natri hidroxit -� SAt (III) hidroxit + natri sunfat

Ap dl;lng dinh lw)t bao toan khoi luqng, ta co:

ffis rit till) suufat + ffin:itn h1drox1t = ffis,h till> h1droxit + ffinatri sunfat

⇒ ffinatri h1droxit = ffisiit !II11 hidroxit + ffinatri sunfat - ffisiit (III> sunfat

= 10,7 + 21,3 - 20 = 12 gam

Bal 8 Dan 36 gam hon hqp khi gom (CO, H 2 ) di tu tu qua 139,2 gam

b<)t Fe3O4 dun n6ng cJ rihi�t de) cao, thu dvqc m gam sAt va 7 4,4

gam hon hqp (CO 2 , H 2 O) Tinh khoi luqng sAt thu duqc sau phan ling, biet phan U'ng vua du?

Iii I De dot chay het a gam hgp chat X can 10,24 gam oxi, thu dugc

thi thay khoi lugng binh tang them 15,96 gam, (biet binh dt!ng

nuac voi trong hap th11 cd CO 2 va H:t:)J Tinh khoi lugng chat X dem dung

Ap dl;)ng djnh lu�t bao toan khoi luqng, ta co:

Iii 10 Nung hon h<Jp X g6m CaCO 3 va MgCO 3 theo phan Llng:

CaCO 3 t• CaO + CO 2

Neu dem nung 31,8 gam hon h(!p X thi thu duqc 7,84 it CO2

<dktc} Xac djnh khoi lu(!ng hon h<Jp oxit thu duqc sau phan Llng •

vua du vai 93,5 gam dung djch AgNO3 Khi ket thuc phani vng thu du(!c 64,95 gam hon h<Jp hai muoi NaNO 3 va Ba(NO3}i va a gam ket tua trrlng (AgCl) Tinh khoi luqng ket tua thu dt:qc.:

Giiii

(NaCl+ BaCb) + 3 AgNO 3 ➔ 3AgCl-!-+ (NaNO3 + Ba(>J(J 3 )J

Ap d1:1ng DLBTKL, ta c6:

= 50,375 + 93,5 - 64,95 = 78,925 (gam) Iii 12 Cho 22,2 gam hon hqp X gom Fe, Al, Zn tac <ll;)ng vua cfo i vdi

dung djch c6 chlla 38,325 gam HCI, thu duqc m gam hon1 hqµ muoi AlCb, FeCl 2 , ZnCl 2 va 1,05 gam hidro Tinh gici tri cuia m?

24

Giiii

So'do:

PHU<NG l'HAP GIAI BAI T:\P 1 10:' • \ ll(>C 8

/\p <l1._.ing DLl3TKL ta c6:

= 22,2 + 38,325 - 1,05 = 59,'175 gam a•i 13 a) De· dot chay het 3,36 gam Fe can 1,44 gam khi oxi Xac djnh

khoi luqng oxit sc'it (Fe2O3) thu duqc?

b) Khi nung da voi c6 phan ling: CaCO3 - CaO + CO2

Nf;u nung 1 tan da voi thl thu duO"c 560kg CaO va a kg CO2

Xac di nh gia tri cua a

Giai

mFep, == m0! + m Fe = 1, 44 + 3, 36 = 4, 8 (gam)

Ap dyng dinh lu�t bao toan khoi luqng, ta c6:

mcaco.1 = mcao + mco� ⇒ mco� = 1000 - 560 = 440 (kg) Bii 14 Cho 32,4 gam sc1t (II) oxit (FeO) tac dyng vl.fa du v&i 12,6 gam

cacbon oxit (CO) d nhi�t d◊ cao, thu duqc si:it (Fe) va 19,8 gam khi cacbonic (CO 2 ).

a) Viet sc:1 do phan ling bllng chu va ki hi�u h6a h9c.

b) Tinh khoi luqng silt thu duqc sau phan ling.

Giai

a) So do bAng chu:

SAt (II) oxit + khi cacbon oxit

So do bAng ki hi�u h6a h9c: FeO + CO

b) Tinh khoi luqng sat thu duqc:

Fe+ CO2

Ap dyng djnh lw}t b<:io toan khoi luqng (DLBTKL), ta c6:

= 32,4 + 12,6 - 19,8 = 25,2 (gam) Bal 15 Dot chay hoan toan 33,6 gam sAt (Fe) trong binh chu'a oxi (vl.fa

du) De ngu◊i blnh, thu duqc 46,4 gam oxit sc'it tu (Fe 3 O 4 ) Hay tinh khoi luqng oxi dem dung cho phan ling tren

Giai

So do: siit + oxi oxit sc'it tu'

Ap dl;lng dinh lu�t bao toan khoi luqng, ta co:

ffistit + ffioxi : ffiox1t sat tit

⇒ ffioxi = ffiox1t slit ti.I' - ffistit = 46,4 - 33,6 = 12,8 (gam)Bti 11 Cho 22,2 gam hon hqp b9t gom Al, Fe, Zn tac dl;lng vu'a <tu voi

dung djch HCl co chu'a 38,325 gam HCI, thu duqc 1,05 gam lh

va dung djch chua cac muoi: AIC13, FeC12, ZnC12

a) Viet cac So' df> phan u'ng tren

b) Co clµ} dung djch sau phiin ung, thu duqc a gam mu6i khan

Ap dl;lng djnh lu�t bao toan khoi lugng, ta c6:

ffih6n hqp kim lo,i + ffiHCI = ffihl,n hqp mu6i + ffiludro

⇒ ffihou hqp muoi = a = ffihon hqp kim lo1,1i + ffiHCI - ffihidro

Bil 17 De dieu che nhom sunfua, nguai ta dem nung 6,75 gam nhom voi

15 gam htu huynh Sau khi phan &1ng xong, thu dt1<;1c 18,75 gain nhom sunfua (Al2S3) • Dieu do co mau thuAn vai djnh lu�t bao toan khoi lu<;tng khong?

Nhu v�y, theo (1) Cu' 6,75 gam nhom phan u'ng hoan toan thi can

12 gam S d� t�o ra 18,75 gam san pha'm la hoan toan phu h<;1p DLBTKL

PHUONG PI-IAP GIAI BAI Tt\P HOA HOC 8

Bal 18 : Jeu dem <lot ch,iy hoan toan 14,08kg qu�ng pirit sAt (FeS2) c§.n

Jung 1,28kg oxi thu duqc 12,8kg siit (Ill) oxit (Fe203 ) va khi ,unfur0 <SO�)

:l) Hay viet Sei c16 chv va S0 c16 ki hi�µ h6a h9c

1) Tinh khoi luqng khi sunfurO' thu duqc

clohidric, phan vng h6a h9c xay ra theo sd do sau:

Al + HCl AlCb + H2 i

a) L�p phVdng trlnh h6a h9c cua phan u'ng tren.

b) Tinh the· tich hiclro thu dugc (dktc).

c) Tinh khoi lugng muoi AlCh t�o thanh sau phan u'ng.

Giai

Ta c6: nA1 = 5· 4 = 0,2 (mol)

27 a) Phudng trinh phan Ctng:

V�y: mA1 c1� = 0,2 x 133,5 = 26,7 (gam)

Bil 20 Khi nung da voi (CaCO 3 ) thu duqc canx1 oxit (CaO) vii khi

a) Hay l�p phudng trinh h6a h9c.

b> Tinh khoi lugng khi C0 1 sinh ra khi nung 1,75kg CaCO 3 va thu · dugc 0,98kg CaO.

c) Sau khi phan ling ket thuc, thu dugc 50,4kg canxi oxit va 39,6kg khi cacbonic thi khoi lugng CaCO 3 dem nung la bao nhieu?

Giai

a) Phudng trinh: CaCO 3 I' Cao + CO2 i ( 1)

b) Ap dyng DLBTKL cho phan vng < 1 ), ta c6:

=> mc02 = mc,.<·o3 - fficao = 1,75 - 0,98 = 0,77 (kg)

c) Tinh m('a( ' O : dem dung:

Tudng tl!, ap dl;Jng DLBTKL cho phan vng ( 1 ), ta co;

mc H co 3 = fficao + ffico = , 50,4 + 39,6 = 90 (kg)

D�NGTOAN 2:

28

CAN BANG pllJUf UNG HOA HQC

a) Phuung phap: De can biing mc)t phudng trinh hoa .'l<_>C, ta thlfc hi�n cac bu6'c sau:

Bu/Jc 1: Viet Sd d6 phan ung gom dAy du chat tham gia va san pham.

Builc 2: Can hllng so nguyen tu' cua moi nguyen to iJ 2 ve.

b) Cac phuung phap c� the":

cua phudng trinh bAng nhau.

- Builc 2: Giu nguyen phan so ho�c khu' mau d6' duqc phu'dng trinh hoan chinh.

Vi d� 1: Can bllng phan vng hoa h9c sau:

Fe2O 3 + H2 1°

Fe + l-12O

-Fe 2 0 3 + 3H 2 1

0

2Fe + 3II 2 0 PlltfUN(; l'IIAI' (;f \I 13;\J TAI' 11)-'\ ll')C' 8

Vi dl;l 2: Ccin L>hng phan Ling ho,i h9c sau:

P + KCl03 - � -► P205 + KCI

Dua h� s6 viw tru6'c P 2 0 :-)t h� s6 5 vao tn.tdc KCI0 : i va h¢ s6 6

, ;_\o tn16'c P, ta duqc phuong trinh hoan ch.inh

6P + 5KCl0 3 _ t _ "

-(1 2) Plutong pllCip "chdn - le"

nguyen tu' cua cung m<)t nguyen to trong m<)t so cong thlfc hoa h9c la s6 ch,'ln, con a c6ng thlfc khac l,:ti la so le thl phai d�t h¢

�6 2 truck cong thlfc c6 so nguyen tu' la so le, sau d6 tim cac h¢

�6 con l<;1i

Vi d i!, 1: Can biing phuong trlnh hoa h9c sau:

KC10 3 t" KCl + 0 2

Ta thay so nguyen tu' oxi trong 0 2 la so chan va trong KCl0 3 la

�6 le nen d<(lt h¢ so 2 tru6'c cong thlfc KC10 3

2KC10a t 0 KCl + 02

- Tiep theo can bllng so nguyen tii' K va Cl, di[it h� so 2 truac KCl

t 0 2KC1 + 02

- Cuoi cung can bAng so nguyen tii' oxi nen di;it h¢ so 3 truck 0 2

Vi di.' 2: C{1n bAng phan vng hoa h9c sau:

AI + 02 - Ab0a

- So nguyen tu' oxi trong Al,0 3 la so le nen them h� so 2 vao 1ru6'c n6.

Al + 02 - 2A}z03 fiep theo d�t h� so 4 vao truck Al: 4Al + 02 2Ali0J

- Cuoi cung can bllng so nguyen tii' oxi ncn them ht) so 3 tru6'c 0 2

4Al + 302 - 2Ab03

o.3J Phuang µhap "d {l i so"

- Builc 1: Dua cac h¢ so hqp thU'c a, b, C, d, C, f, ]an luqt vao

aic cong thlfc a hai ve cua phue1ng trlnh phan Ling.

- Builc 2: C,ln bking so nguyen tii' d 2 vc cua phue1ng trinh bl!ng n<)t h� phucing trlnh chlfa cac ftn: a, b, c, d, c, f,

- Buiic 3: Gic1i h¢ phue1ng trlnh vu'a l�p dti tim cac h◊ so.

- Buiic 4: Dua c�ic h� so vua tim vao phue1ng trinh phi.in U'ng.

30

Vi d11, 1: Can bAng phan ling sau:

Cu + H 2 S0 4 1° CuS0 4 + S0 2 + H 2 0 ( 1)

Bribe 1: Dien cac h� so hc;,p thuc a, b, c, d, e, f, vao cac <hat

tntac va sau phan ling ( 1)

0

Bribe 2: Tiep theo l�p h� phuong trinh dva vao moi quan hi ve

khoi luc;,ng giua cac chat truac va sau phan ling, (khoi lt.11ng nguyen tii' cua moi nguyen to e1 2 ve phai bAng nhau).

Cu : a = c ( 1)

S: b = c + d (2) H: 2b = 2e (3)

0 : 4 b = 4c + 2d + e ( 4)

Bribe 3: Giai h� phuong trinh bAng each:

Tu' pt ( 3 ), ch9n e = b = 1 Tu'pt(2),(4)va(l) ⇒ c=a=d= !_

2

Briae 4: Dua cac h� so vu'a tim vao phuong trinh phan u1ng ta

duc;,c phuong trinh hoan chinh:

3

PlllJONG PHAP GIAI UAI TAP 1!(11A l(>C 8

Builc 4: l:)ua h◊ so vlia tim vao phuO'ng trinh phan ling:

3� Al+ 2HN0:i - ! AHN0:1h + N01 + H:>.0

:3 flo(1c: Al+ 6HN0 3 - Al(N0J): 3 + 3N0T + 311 2 0

so' llguyen Ill trong ccic c:611g tlui'c lwci hoc

a•t I Can bang c:-ic phrin vng h6a h9c sau:

L�p pht1<1ng trinh h6a h9c va cho biet ti l� so nguyen tu', so

ph{1n tu' cua cac chat trong moi phiin ling.

TI 1�: so phan w Fc<OH J 3: so phan tli Fe 2 0 3 : so phan tli H 2 0 = 2 : 1 : 3.

Bal 3 Thanh phan cua thuoc no den gom: 75% KN0 3 <ka1i nitrat 1 ; 10%

8 (]uu huynh) va 15% C (than) Sau khi dot ch,iy thu c.11 Nc khi cacbonic (CO 2 ) va khi nite1 <N2 ) Dan san ph{im khi viw dung <l!th

cacbonat (CaCOa) va nuck Hay viet phudng trinh chu va fhtWng trinh h6a h9c cua cac hi�n tuqng mo ta a tren

Giai

• Phu<1ng trinh chu:

Kali nitrat + luu huynh + than ,··- kali sunfua + khi cacbmic

• Phu<1ng trinh h6a h9c:

· Khi cacbonic + <ld nu<lc voi trong ➔ canxi cacbonat + ·1Lfo'c

• Phudng trinh h6a h9c:

CO2+ Ca(OI02 -► CaC03i + H20

Bil 4 Hoan thanh cac phuong trinh h6a h9c sau:

f) Ba(N03);.? + AliS0 4 h -► BaSO a ! + Al(NO j h

Hay l�p phuong trinh h6a h9c va cho bict ti I� so ngu:en tu', phan t11 cua cac chat ttong moi phan Ctng.

Plll.'<1:-,;C PIIAP l;l:\1 13,\I Tr\P ll):\ IHJC 8

Bal 8 'Jhay cac chi so X, y trong cac c[rng thvc hoa h9c roi can bl-lng cac

phuong tr,nh ph,in 1.J'ng Hay x,ic dinh ti 1� so nguyen hi phan tu' giua c,ic ch,'tl tnftk \· 'i sau phrin ling

7 Hoan thanh cac phi.long trinh h6a h<;>c sau, ghi them difu ki�n

phan un.g (ne'u c6):

a) FeaO• + ? ➔ FeCla + FeCl2 + ?

b) NaAI0 2 + HCI ➔ NaCl + ? + ?

a) Fea0 4 + 8HC1 ➔ 2FeC1 3 + FeCl 2 + 4H 2 0

b) NaA10 2 + 4HCI -➔ NaCl + AlCla + 2H 2 0

CHU'aNG Ill

A KIEN THuc cAN NHd

I MOL

1 - Mot la lw;nig chat cluJ.a 6 la2 3 nguyen tit ho<J,c ph{uz til cila chat do.

· · So ' 6.1 (i2 3 g9i la so' Avogadro, fly hi�u N.

2 Kh6i hiqng Mol la gi?

.• K!to'i luqng nwl (fly hi¢u MJ ctla 1n9t cluit la llho'i luqng tinh bting ga,n ctla N nguyen til ho<J,c phdn til cha't d6

Kho·; l1tqng nwl ctla chat c6 cung so' tri vai nguyen tii ldio'i ho<J.,c pluJ.n ltl kho'i ctla clui't, co dan vi kho'i luqng la gani.

3 Th� tich mol ctia cha't khi

- The tich mol ctla chat hhi la the tich chiem bai N phan til cua chdt hhi d6.

- M9t mol ctla bat ki, chat khi nao, trong cung dieu lli�n ve nhi�t d9 vii <ip suat, dJu chiem nhilng the tich biing nhau.

- 0 dieu ki�n tieu chudn (dktcJ: o ° C vii 1 atm the tich mo/ cua c<ic cha't khi deu biing 22,4 lit

- 0 dieu ki�n thuiJng (nhi�t dq 20 ° C, <ip sua't lat,nJ, 1 nwl chat ldii co the tich la 24 lit

11 CHUYEN 061 GIUA KH61 LllQNG, THE TicH vA uJQNG CHAT

1 S\i chuy�n doi giua lu'9ng chflt (so mol chflt) va kh<>i htcjng ctillt.

Cong thac: n = m (mol) ⇒ m = n x M (gam) ⇒ M = m (g/mol)

2 SI/ chuyln doi giii'a hiqng cha't va th� tich cha't khi.

Cong thuc: v(d dktcl = n x 22,4 (lit.) ⇒ n = _:! _ ( mol)

�2.4

3 SI/ chuy�n doi giua lu'(.1ng cha't va so' nguyen tti', phan tti'.

- S6 nguyen tu' A = N x so mol nguyen tu'

- S6 phan tu' A = N A so mol phan tu'

V&i N la so Avogadro: 6.1023

., _,, , ,, , ,,

Ill Tl KHOI CUA CHAT KHI

I Ti kho'i hdi cua khi A do'i vdi khi B ( d.'I Ii):

Chu y: Neu biet hho'i ltt<;1ng rieng ciia hhi A (D.,J a dlltc, ta tinh

dt1<;1c ldi6'i lt1<;1ng mol ldii A: M A = 22,4 D A

IV TiNH THEO CONG THVC HOA HQC

36

nguyen to trong hqp chat, gom cac bu6'c sau:

Bu"ilc 1: Tim khoi lugng mol cua hgp chat.

Builc 2: Tim so mol nguyen tit cua moi nguyen to trong 1 mol h <;p 1 chat.­

Buac 3: Chuyen doi so mol nguyen tu' thanh khoi luqng.

BuiJc 4: Tinh thanh phan phan tran1 cac nguyen to trong hqp chtat.

Vi d ll, : Tinh phan tram cac nguyen to trong hqp chat: K2 S0

4-Buac 1: Khoi luqng mol cua hgp chat: M i.; tso4 = 17 4 (gam)

Buac 2: Trang 1 mol hgp chat cht1a: 2 mol nguyen tu' K; 1 rnol

nguyen tu' S va 4 mol nguyen tu' 0

Bu'ilc 3: Chuyen doi lugng chat thanh khoi luqng

mi == 2 x 39 = 78 (gam); ms = 1 x 32 = 32 (gam)

V8 01 0 = 4 X 16 = 64 (gam).

PlltJ0: IG PIIAP GIA! BAI L\P HC\ !HQC 8

Bu&c 4: Th:'tnh phan phan tram enc nguyen t6 trong h9·p chat la:

Ru&c 2: Tim so mol nguycn tlf cua moi nguyen to trong 1 mol hqp

chc''lt

Bu'itc 3: Quy ra so nguycn tlf cua moi nguyen to trong mqt phan

tu' hqp chat

Bu·&c 4: Viet cong thu'c hoa h9c CUR hqp chat

Vi d1_1: l\19t hqp chAt X c6 thanh phan khoi luqng cua cac nguyen to la: 32,;VitNa; 22,54 < 1,S vit 45,1':-1<0 Khoi · 1uqng mol cua hqp chat bl:'ing 142 gam Hay xac dinh cong thu'c hoa h9c cua hqp chat X

BuiJc 3: So nguyen tll' cua moi nguyen to trong 1 phan tll' hc,1p chat (X):

M◊t phan tu' X chu'a: 2 nguyen tu' Na, 1 nguyen tu' S va 4 nguyen tlf 0

V TINH THEO PHlJdNG TRiNH HOA HQC

1 Y nghia cua phudng trinh hoa hQc:

Phuang tdnh hoa h9c cho biet:

Ti 14- so phan tlf chill tham gia va tc;lO thanh.

2 Tinh s6 mol hotc kh6i hif/ng hay thl tich cac chflt tham gia va cac chflt sin phlm:

- Vie't phu'<1ng trinh hoa h9c.

- Chuy�n d6i khoi lugng chat ho�c the tich thanh so mol chat.

- DVa vao phu'<1ng trinh hoa h9c d� tim so mol chat tham gia hoJc chat �o thanh.

- Chuy�n d6i so mol chat thanh khoi lu'gng (m = n x M) ho�c th� tich khi d dktc (V = n x 22,4).

B PHlJdNG PHAP GIAI CAC D�NG TOAN

D�NGTOAN 1:

MOL, cil.CH Thm sb MOL cilc cllAT,

sb HGUYDf TU'VJ\ sb PHAM TU'

• Sfi nguyen tii = n nguyln til X 6.1 ()2 3 .

• Sfi phan tii = n p ha n til X 6.1 ()2 3

mol nguyen tii ho(l,c sd inol piian tii

V=·22,4 xn lit ⇒n = � mol 22,4

I a) Hay tinh so nguyen tu' hidro chu'a trong 1 mol nude

38

b) Hay xac djnh so nguyen tu- sAt c6 trong 280 gam sAt.

c) Hay tinh so mol phan tlt N 2 c6 trong 280 gam nit<1.

d) Phai lay bao nhieu mol phan tu- CO 2 d� c6 1,5.10 23

phan tit CO 2 ?

Gidi

a) 1 mol H 2 O c6 6.10 2 3 phan tlt H 2 O.

1 phan ta nuoc c6 2 nguyen ta hidro

V�y 1 mol nuoc chu'a so nguyen tu- hidro la:

6.10 23 X 2 = 12.10 23

PHVONG PHAP GIA! BAI Tt\P H6A H<;>c 8

280

b) Ta co: nFP = - 56 = 5 (mol)

⇒ So nguyen hi sAt c6 trong 280 gam silt la: 5 X 6.10 23 = 3Q.10 23

280 c) Ta c6: nN, = - 28 = 10 (mol)

, 1,5 10 :t:l

d) Ta co: nl.0 2 = 6.10 2J = 0,25 (mo))

••• a �ay cho biet so nguyen tu' ho?C phan tu' c6 trong moi h.tqng chat

sau:

a) 1,5 mol nguyen tu' Al;

c) 0,25 mol phan tv NaCl;

Bai gidi

b) 0,5 mol phan tu' H2;

d) 0,05 mol phan ttl H20.

a) 1,5 X 6 10 23 = 9.10 23 nguyen tu' Al.

b) O,b X 6 10 23 = 3.10 23 phan tu' H 2

c) 0,25 x 6.10 23 = 1,5.10 23 phan tv NaCl.

d) 0,05 X 6 10 23 = 0,3 1023 phan tu' H 2 O.

1•1 a a) Phai lay bao nhieu gam magie d� c6 so nguyen tv bhng so

phan tu' c6 trong 1,2 gam H 2 ?

b) Phai lay bao nhieu gam NaOH d� c6 so phan tu' bAng so phan tu' c6 trong 49 gam H2SO 4 ?

c) C6 bao nhieu phan tu' H 2 chua trong 1 mililit khi H2 cJ dktc?

4 a) Tinh so phan ht H2O c6 trong m<)t gi9t rn.f<ic (0,05 gam )':

b) M(>t lit hon hqp gom C3H8 va CH4 co the' tich b5.ng nh:iu do ddktc Hay xac djnh khoi luqng cua hon hqp tren?

Bil I Trong m<)t binh tr<)n khi SO2 vdi SO:-i- Khi phan tich ngubi ta

thay c6 2,4 gam luu hu)'·nh vi1 2,8 garn OX} Xac djnh ti S<J0 mol

SO2 va SO3 trong binh sau khi trt)n

Giai

GQ.i so mol SO2 la X thi ns la X va no lit 2x

G9i so mol cua SO3 la y thi ns la y va no la 3y

Theo de ba1, ta co hf? phuang tnnh: � · , , , , {(x + y) x 32 = 2,4

(2X + 3y) X 16 = 2,8Giai hf$ phu<1ng trinh, ta duqc: x = 0,05; y = 0,025

Bil 1 Hay tinh:

y = 0,025 = -1

a) So mol cua: 28 gam Fe; 64 gam Cu; 5,4 gam Al

b) Th� tich khi (dktc) cua: 0,175 mol CO2; 1,25 mol H2; 3 nol N2.c) So mol va the tich cua hon hgp khi (dktc) gom co: 0,44 gam CO2 ;

b) Yeo� = Ilcoi 22,1 = 0, 175 x 22,4 = 3,92 (lit);

V N, = nNi .22,·1 = 3 x 2'2,4 = 67 ,2 (lit)

m,.1, = -0 4'1 ' = 0,01 ( mol) l\l('(,, 12 t- (2 x 16)

a•i 7 Co 100 gam khi oxi V,1 100 gam khi cacbon dioxit, ca 2 khi c1eu d

20 ° C va 1 atm Bie·t rking the" tich mol khi d nhung dieu ki�n nay

la 24 lf t Neu tn)n 2 kho'i luqng khi tren v6'i nhau (khong c6 phan ling xay ra) thi hon hqp khi thu duqc c6 the· tich la bao nhieu?

V ( '.() 1 = n('U1 X 24 : 25 X 24 ::;: 54,55 (lit)

11 Th� tich cua hl>n hqp:

vhh = V o l + v (' Ol = 75 + 54,55 = 129,55 (lit)

a.i 8 Tinh th� tich cua cac hon hQ'p khi sau a dieu ki�n tieu chuan

(dktc) va dieu ki�n thuang:

a) 0, 15 mol CO2 ; 0,2 mol N02 ; 0,02 mol S02 va 0,03 mol N2

b) 0,04 mol N20; 0,015 mol NH3 ; 0,06 mol H2 ; 0,08 mol H2S

G{fi y: Aµ dtplg c6ng thlic:

The tich ldii a ( dhtc) = so' mol x 22,4 (lit)

The tich ldii a di�u hi�n thllang = so nwl x 24 (lit)