Chi dugc dung th6m mQt thu6c thri hdy trinh bdy phucmg ph6p h6a hoc c6 minh hoa bdng phin ung nh?n bitit c6c dung dfch ttpg-riOng U4t-trong c6c lg m6t nhdn sau: BaOH2 ;BaCl2; HzSO+; HCI;
Trang 1PHONG GD &ET
CUM THI TAM HgP
ri, rnr cHeN Gv DAY crOr cAp rnUoNc
NAvr Hoc 2ort-zorz
lr6N rru, H6a Hgc Thdi gian ldm bdi: 120 phrlt (Khdng kO thdi gian giao dA)
ThAy (CO ) H6y hufng d6n HS lirm c6c bii tgp sau:
CAu I: ( 1.5 di6m)
1 Chi dugc dung th6m mQt thu6c thri hdy trinh bdy phucmg ph6p h6a hoc (c6 minh hoa bdng phin ung) nh?n bitit c6c dung dfch ttpg-riOng U4t-trong c6c lg m6t nhdn sau:
Ba(OH)2 ;BaCl2; HzSO+; HCI; NaCl ; NaOH Cffu II (2.s diCm)
1.cho h6n hqrp vun cdc kim 1o4i: Ag, cu, Al, Fe Hdy nou phucrng ph6p h6a hgc (c6 minh hoa cu th€ bing phuong trinh phin rmg) t6ch ri6ng biQt c6c kim^loai ndy
C A u I I I : ( 3 t t i 6 m )
_ , chia 4,04 gamhSn hqp vun ddng nh6t gdm 2 kim lopi ld zn vdMg thdnh 2 phdn bang nhau vd ti6n hdnh 2 thi nghi6m
Thi nghiOm 1: cho phdn 1 vdo ctic dlrng 200 ml dd HCl sau ph6n img dun n6ng cho bay hoi htit nudc, ta thu dugc 4,86 gamchAt rin
Thi nghidm 2: cho phAn 2 vi,o c6c drmg 400m1 dd HCI (c6 cr)ng nti.tg dQ v6i dd HCI d thf nghiQm 1) Sau ph6n ring dun n6ng cho bay hoi htit nu6c, ta thu dugc 5,57 gam chSt rfin
1 Tinh ktrOi luqng mdi kim 1o4i trong hdn hq'p ban clAu
2 Tinhthd tich khi da bay rao thf nghiem r Gh6 tich do o Dktc)
3 Tinh n6ng d0 mol/lit cria dd HCI dA ding trong 2 thi nghiQm
CAu IV :( 3 di6m)
Hdn hqp A gdm cScl<himetan, etylen vd axetylen Ngudi ta ti6n hdnh cdcthi nghiQm sau:
TNl: DAn 2,8 lft h5n hqp A (do d dktc) qua binh dung drng dich nu6c Brom th6y binh bi nhgt mdu di mQt ph-An vd c6 20g biom dA phan rin! a
TN2: oOt chay hoin toan 5,6 lit A (do o clktc) rdi cho-todn bQ s6n phAm chiry qua binh clWrg 175,2 gam dung dich NaOH20% sau thf nghiOm thu cluqc dung dfcir chria I,57yo NaOH
Tfnh Yotheo th6 tich cua m5i khi c6 trong hdn hqp A /
I3:.-T:1_O=tU, Zn-65,M9=24,Na=23, C=I2,CI=35.5, S =32, Br = 80 ).
;;;;;.1"**
Trang 2pHoNG GD &DT xY rnr cHeN GV DAy cr6r cAp rnUoNc
oAp AN vA srEu DIEM
I 1
Hoc sinh ldp duoc so dd nhan bidt vd sau d6 nOu duoc cdcbu6c co bin sau
Trich m6u thrl vh dine qui tim ph0n biOt duoc thanh 3 nh6m
Nh6m lhm qui tfm ho6 xanh lh NaOH vd Ba(OH)" (Nl)
0,25
Nh6m ldm qui tfm ho6 d6 le HCI vi H,SO" (N2)
Nh6m khOns lhm qui tim thav d6i mdu ld NaCl vd Ba Cl" (N3)
Ding 2 chdt A Nl) ldn luot cho t6c duns vdi2 chdt trone (N2)
ChAt nho trons nh6m 1 tao kdt tfia mdu trFr;rs, vdi chAt & N2) ld Ba(OH)" 0,25
Chdt cbn lai li NaOH
Chdt duoc Ba(OH), tao kdt ttra mdu tr6np,ld H.,SO, chat cdn lai ld HCI 0.25
Dtne H,SO, vila mdi nhAn ra cho t6c duns vdi2 chAt trone (N3)
0,25
ChAt tao k€t tira mdu tr6ng vdi H,SO, ld Ba Cl, chAt cbn lai ld NaCl
Phuong trinh ho6 hoc
Luu y trudc m6i thi nghiOm phai trich m6u thrl
tr Hda tan h6n hqtp trong dd NaOH chi c6 Al tan, loc chAt khdng tan, dtng dd nudc
loc dd t6ch nhOm
0,25
2Al + 2 NaOH +2HrO -)
Na AlOr + CO, + 2HrO +
2Na AlOr,oo, + 3H,
2AI(OH)3 'o-) Al2O3 + 3HrO
2Al P3 DiQnPhan n6ng chaY t 4AL + 3.,,
Ti€p tuc cho chdt rilntan trong dd HCI chi c6 Fe loc chdt kh6ng tan thu duoc h6n
lrqp Ag vd Cu DDng dd nudc loc dd t6ch Fe
0,25
Fe + 2HCl + FeClr+ H,
FeCl +2NaQH+ Fe(OH), +2NaCl
0,5
Fe(OH), chdn kh6ng
FeO + CO 'o-)
to -) FeO + HrO
Fe +CO, Ddt n6ng h6n hqtp trong oxi sau d6 hod tan h6n hgp trong dd HCI cbn lai ch6t
CuO + zHA, -+ CuCl, + H"O
CuClr+ 2NaOH + Cu(OH)2 + 2NaCl
A(QH), ,o _) CuO + H"O
Trang 3Zn +2HCl + FeCl,* H, (1) 0,25
Mg + zHC' + MgClr* H, (2)
Theo (1) vd (2) ctl 2 mol HCI tham gia phin trng thi ch{t r6n thu dugc khi cO can
duns dich tlne 7L s,am
0,25
Theo bii ra kh6i tuong chat r5n ting 4,86 -2,02 =2,84 sam 0,25
J trucr dd tham gia ph6n fng trong thf nghiOm (1) le 4#= 0,08 mol
X6t thi nghiOm (2), ndu HCI phin ttng hdt thi theo kdt quA thf nghiem(l)khdi
luong chdt rin sd tdng gdp doi so vdi thf nghiOm(l) tric ld chdt r6n sau khi c0 can
dd sE
ld, : 2,02 + 2,84x2 ='1,'7 gam l6n hon khdi luong dd bei cho ld 5,57 gam Do vAy
trong thf nghiOm (2) HCI phin rlng chua hdt c6 nstria ld kim loai dd bi hod tan hdt
0,5
GQi x ,y (mol) ldn lugt ld sd mol cfia ZryME trons 2,02 samh6n hop 0,25
Ta c6 phuong trinh 65x + 24 y -2,92
Theo (I) n r.^.,r= fl7, = x mol
0,25
Theo (2) tt
".,r= 1 Mo - y mol
Ta c6 phuong trinh 136 x + 95y = 5,57
Ta c6 h€ phuong trinh
165x L l 3 6 + 24 y =2,02 Gi6i ra ta du-o c f x = 0,02 mol
x + 9 5 y = 5 , 5 7 L y = 0 , 0 3 m o l
0,5
ffi Mn= 0,03 x24x2 - t,44 (g)
Theo (1) vh (2) cf 2 mol HCI tham gia phin ring thi ei6i ph6nF,22,4lir khf
Theo tinh to6n d tI94 4ncr = 0,08 J rHz= 0,04 -) V"r= 0,04 x22,4 - 8,96lit 0,25
cr'q=
N Ggi x, y, zldnluqt ld c6c s6 mol cria CH4, CzHqua C
2.8
I l h h :
f f i : 0 , 1 2 5 ( m o l ) - ) x * y + z:0,125 (I)
0,5
Khi cho 2,8lithh A di qua binh dlrng nudc Brom chi c6 CzHqvirC2H2phan rmg
Phucrng trinh phan img:
CzHq + Brz -> C2HaBr2 (1)
CzHz* 2Brz -> C2H2Br (2)
Theo (l) (2) vd bdi rcTac6: nBr2 = y + 2t= ,20^:0,125 (mol)-
160 K6t ha,p v6i (I) -) x: z
0,5
Trang 4ot5t chay 5,6lith6n hqp
cll4 + 2oz -> co2 + 2H2o (3)
czH+ + 3Oz-> 2CO2+ 2H2O (4)
2Czkba oz -> 4 coz+ 2 H2o (5)
Theo (3) (4) (5) Ta c6: n COz:2x+ 4y + 4z : 0,375 + y ( v i : 2 x + 4 y + 4 z : 6 2 * 4 y = 3 ( y + 2 z ) + y : 0 , 1 2 5 * 3 + y )
0,5
Theo bdi ra: n NaOH : 0,876 mol
COr + 2NaOH -> Na2CO3 + H2O (6)
Theo (6) n NaOH phin img : 2n COz: 0,75 -r 2y
n NaOH du : 0, 876 - 0,75 - 2y :0,126 - 2y
0,5
Ta c6 hQ phuong trinh
x + y + z = 0 , 1 2 5
y +22 = 0,125 40.(0,126 -2y)
(0,375 + y).44 +175,2 1 0 0 = 1 , 5 7
0,5
Giei hQ ta dugc: y:0,025
x : z : 0 , 0 5
Yo CH4:40o/o
o/o Czfu:20o/o
Yo CzHz : 40Yo
0,5
GV hufns din HS llrm c6ch kh6c mir drins vffn cho tli6m t6i da