Cho bi~t cong thirc mot muoi X va vhh phuong trinh hoa hoc trong cac tnrong hop sau:.. X tac dung voi dd HC} va dd NaOH d~u cho khi.[r]
Trang 1DE THI CHQN DQI TUYEN ~QC SINH GIOI LOP 9
MON THI: HOA HOC Ngay thi: Thu Bay 04/01/2020
Tho; gian lamwbai: 120 phut (khong kJ phat i1~)
NAM HQC: 2019-2020
CaD 1: (4.0 d)
1.1 Cho sa db di€u che nhtr sau :
Vi€t cacphirong trinh hoa hoc xay ra ?
Cho bi~t thanh phfin h6n hQ"Pkhi X va Y
Neu each phan biet khi trong h6n hQ"PX?
1.2. Mu6i an co lfin tap chdt : Na2S04, NaBr,
CaCh, CaS04 trinh bay each tinh ch~ d~ co muoi an tinh khiet
H61l bqp
kbiX
csco
CllSO,;
Nooc
brom(du)
CaD 2: (4.0 d)
2.1 Cho bi~t cong thirc mot muoi X va vhh phuong trinh hoa hoc trong cac tnrong hop sau:
X tac dung voi dd HC} va dd NaOH d~u cho khi
X tac dung dd HCI co khi va tac dung voi dd NaOH cho k~t tua
2.2 Duoc dung them mot thuoc thu, tim each nhan bi~t cac dung dich trong cac lo mdt nhan : Ba(OH)2, Ba(N03)2, KC1, H2S04, HN03
Call 3: ( 4.0 d):
Hoa tan h~t 24,16 gam h6n hQ"PX g6m Cu va Fe304 trong dung dich HCl loang du thay con lai 6,4 gam Cu khong tan M~t khac hoa tan h~t 24,16 gam h6n hQ"Ptren trong 240 gam dung dich HN03 31,5% (dung du) thu duQ'c dung dich Y (khong ch(ra NH4N03) Cho 600
ml dung dich NaOH 2M vao dung dich Y LQc b6 k~t tlia, co ci;lndung dich nuac lQc sau
do nung t6"i kh6i luqng khong dbi thu duQ'c 78,16 gam chdt rin khan Bi€t cac phim (rng xay ra hoan toano
3.1 Tinh kh6i luqng m6i chAttrong X
3.2 Tinh n6ng dQ % clia CU(N03)2 co trong dung dich Y
Call 4: ( 4.0 d):
4.1 TrQn 10,8 gam bQt nhom vai 34,8g bQt Fe304 rbi ti~n hanh phan (rng nhi~t nhom trong di€u ki~n khong co khong khi Hoa tan h6n hQ'p r~n sau phan (rng b~ng dung dich H2S041oang
~u thu duQ'c 10,752 lit H2(dktc) Tinh hi~u sudt phan (rng nhi~t nhom? (gia su Fe304 chi bi khu thanh Fe)
4.2 D6t chay hoan toim m gam mQt mfiu than chua 8% t(.lPch~t b~ng oxi thu dUQ"c2,24 lit h6n hQ"PA gbm 2 khi hQ'PchAt (dktc) Svc tir tir.A vao 40 ml dd ch(ra Ba(OH)2 1M va KOH 0,5M sau phan (rng thu duQ'c 5,91 gam k~t tlia Tinh m va th~ tich oxi tham gia phan Ung (dktc)
Trang 2C,D 5 : ( 4.0 d):
5.1 H6n hop X g6m CH4,C2H4,C2H2
_ D6t chay 21,6 gam X thu duoc m g C02 va 32,4 gam H20
_ 11, 2 lit X (dktc) tac dung vrri dd B 2 dirthico 0,35 mol Brz phan irng
Xac dinh m va % th~ tich cac khi t ing X
5.2 H6n hQ'PX g6m hydrocacbon (A) it oxi (cy l~ mol ttrong img lit 1:7) D6t chay hoan toan
X thu duoc Y (g6m khi va hoi) 0 ngung tu h~t hoi mroc trong Y thi duqc h6n hopZ
g6m 2 khi co tY l~ moll: 1 Xac di cong thirc phan tft va vi~t cong thirc c~u tao cua A.
Cho H=l; C=12 ; N=14 ;0= 16; =23 ; S=32 ;K=39 ;C=40 ;Fe=56 ; Cu=64 ;Ba=137
Trang 3
-H8t-~ I' HU'O'NG DAN CHAM MON HOA HOC QUAN GO VAp NAM HOC 2019-2020
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2d CaC03 + 2HCl + CaCh + S02 + H2O
1.2 Muoi an co 1fin tap chat : Na2S04, NaBr, CaCl«, CaS04:
2d Cho mu6i an co lfin tap chelt vao dung dich BaCh dir: 0,75d
Na2S04 + BaCh ~ BaS04 + 2NaCl CaS04 + BaCh ~ BaS04 + CaCh Loc bok~t tua, dung dich mroc loc g6m: NaCl, CaCh, NaBr va 0,75d BaCh duo Cho tac dung voi Na2C03 dir
CaCh + Na2C03 ~ CaC03 + 2NaCl BaCh + Na2C03 ~ BaC03 + 2NaCl Dung dich con lai gom NaCl, NaBr, va Na2C03 du, cho tac 0,25d dung voi dung dich HCl dir :
Na2C03 + 2HCl ~ 2 NaCl + C02 + H2O Sue khi Clo du vao dung dich ,r6i co can dung dich thu duoc 0,25d NaCI khan
Cli + 2NaBr -+ 2NaCl + Br2
2 2.1 X co th~ la NH4HC03
NH4HC03 + NaOH +NaHC03 + NH3 + H2O
X co th~ la Ca(HC03)2
Ca(HC03)2 +2NaOH +CaC03 +Na2C03 +2H20 2.2 Nh~n biet : Ba(OH)2, Ba(N03)2, KCl, H2S04, HN03.
(2d) - Dung quy tim nhan ra Ba(OH)z (hoa xanh) , dung dich con lai 0,5 0
chia lam 2 nhom :
- Nhom 1 : HN03, H2S04 lam quy tim hoa do
- Nhom 2 : Ba(N03)2, KCl Khong lam dbi mau quy
-Dung Ba(OH)2 cho vao mfiu eua nhom 1, H2S04 t~o ket tua 0,750 tr~ng , con l;;tila HN03
Ba(OH)2 + H2S04 + BaS04t + 2H20
- Dung H2S04 dtinh~n bi~t cho vao mfiu nh6m 2 chMtl;loket tila 0,75d
la Ba(N03)2 , mftu con l;;tila KCl
H2S04 + Ba(N03)2 +BaS041 + 2HN03
Trang 4-D~t a s6 mol Cu phan img, b hi s6 mol Fe304 ban dau
=> 64a+ 232b = 24,16 - 6,4 = 17,76 (I)
1,5d
0,25d "
PhU(Jl1gtrinh phan irng
0,5d
Cu + 2FeCb ~ CuCh + 2Fe~h (2)
b f -2b b 2b
0,25d V~y a=b; theo 1~ a= b =0 06
V~y trong 24,16 gam X co: ( ,16 mol Cu; 0,06 mol Fe304
~ mcu=0,16.64= 10,24 (gam; mFeP4= 0,06.232= 13,92 (gam) 0,5d 3.2 Tile dung voi dung dich ID~03: nHNO] (bd) = 1,2 mol
2,5d Sa db:
+ 1~2(mol)NaOH t
dd NaN03 [
dd Y X+HNO~
Khi
0,5d
Ta co: N~u NaOH h~t chat rAn chi co: NaN02 = 1,2.69 =82,8 gam> 78,16
~ NaOH phai dir: theo So' d4~tren ta co:
O,5d x+y= 1,2; 40x+69y =78, 6 => x= 0,16; y = 1,04
0,25d kl2
k
k= sf, mol HN03 phan irng v ;iX;
nNaNO = 0,16.2 + 0.18.3 + nUN03 du = 1,04
2
O,5d
~ -~ -+~~~
O,25d
nHNO, pu (*) = 1,2 - 0,1~ = 1,02 (mol) = k
Theo bao toan khoi hrong:I
I i
~ ~ · -~I -~ ~
Trang 5-mkhi=24,16 + 63.1,02 -( 0,18.242+ 0,16.188+ 18.1,02/2) = 5,6g 0,25d
=>C% (CU(N03)2)= 0,16.188 =11,634%
240 + 24,16 - 5,6
R~n sau chua: Ah03 ; Fe304; Al (0,4-8x); Fe(9x) 0,25d Cac phan img voi H2SO4
Fe304 + 4H2S04 ~ FeS04 +Fe2(S04)3 + 4H20
2Al + 3H2S04 ~ Ah(S04)3 + 3H2
Fe + H2S04 ~ FeS04 + H2
Tilso mol H2 ta co:
Ta t ayh;' 80,4=-3- ; nen0,15 ~ H ' h htm t eo c at nao cungh;' , ? diroc
0,4 4.2 Cac phan irng :
C + 02 ~ 2CO H6n hop A g6m hai khi hQ'Pchdt nen 02 h~t
m = 12.0,1.100/92 = 1,304(gam)
Mol Ba(OHh = 0,04; mol KOH = 0,02; mol BaC03 = 0,03
Ta thdy molt < mol Ba(OH)2
Co 2 tnrong hQ'P: TH 1: k~t tua chua tan:
Mol C02 = 0,03; mol CO=0,07 Mol 02 = mol C02 + Y2mol CO
~ mol 02 = 0,065 (mol) V02 = 1,456 (1)
TH2: k~t tua tan 1 phful
Trang 65 5.1 X g6m CRt; C2Rt; C2H2;
C02 +KOH ~ KHC03 + Hfl
0,02 0,02 C02 +BaC03 +H20 - BaObC03)2 0,01 0,01
Mol C02 =0,07 Mol CO=O,I 3 Mol 02 =mol C02 +Y2mol ~O
=>mol 02 = 0,085 (mol) V b2 = 1,904 (1)
O,5d
O,5d
O,25d O,25d O,5d
,,
Giai (1),(2),(3),(4) => a=O,2 ; b=0,15; c=O,I; k=2
5.2 H6n hop X: CnH2n+2-2k(k 18 (, lien k~t7t)
2d Phanirngchay:
3n+1-k
CnH2n+2-2k+ 02 -~ n C02 +(n +1 -k) H20
2
1 (l,5n +0,5 - 0,5k) n (n + 1 - k)
co tY l~ moll: 1 => n= 6 ~ +0,5k - 1,5n
=> 4n - I =l3 => k=3; n=4
O,25d
0,5d
0,5d h6n hop Z gomC02 (n); va b2 dir (7 - (l,5n +0,5 - 0,5k»; 0,25d
2d Phan irngvoi Br2
C2H4 +Br2 ~ C2H4EIto2 C2H2 +2Br2 ~ C2H2tlr4 f)~t a.b,c la so mol CH4;C2H ; C2H2;trong 11,2 lit hh X
a+ b +C =0,5 (It
b +2c = 0,35 2) f)~t ka,kb,kc 18.so mol CH4; ~2H4; C2H2; trong 21,6 ghh X
16ka+28kb +26kc = 21,~ (3) 2ka + 2kb +kc = 1,8 (4)
%CH4 = 50%; %C2H4=30~~; %C2H2=20%_
m = 66 (gam)
0,5d