Cciu 2. a ) Co cac p h a n ufng: C5H10 + H2 C5H12 t ° , N i CsHiz (Xi) t ° . P, xt -> CH4 + C4II8 (X2) 1500°C 2CH4 u u K 1 1 K 1. K—> C H ^ C H + 3 H , * tach n h a n h , l a m l a n h n h a n h S C H ^ C H 600° C b) CeHe + H2 S + O2 Cacbon t O. N i CeHg (X4) —> C6H12 Xiclohexan -> S O , 450° C 2SO3 iau 3. 2SO2 + O2 ^ 45(
SO3 + H2O > H2SO4
SO2 + 2H2O + Br2 > H2SO4 + 2 H B r
H2SO4 + NaaSOs > S O z t + H2O + Na2S04 P h a n ufng cua M vori H2SO4 loang:
2 M + nH2S04 > M2(S04)„ + nHaT
T L r ( l ) = > an = a => n 2 P h a n ijfng vdi H2SO4 dac nong:
2 M + 2 m H2SO4 > M2(S04),u + mS02T + 2mH20
T C r ( 2 ) = . « ' ^ - ™ . ^ ' ° ' -
' 2 M 22,4 =0 M = 18,66m
m 2 3 1
M 37,3 56 18,66
Ket qtia Loai D i i n g loai
a) K i m loai can tim l a F e .
b) Dung dich A l a F e S 0 4 . C 6 cac phan ufng:
F e S 0 4 + 2 N a O H > F e ( 0 H) 2 i + Na2S04 4Fe(OH)2 + 2II2O + O2 > 4 F e ( O H) 3 i
2Fe(OH)3 —^—^ FeaOa + SHaO
V a y B l a FeaOạ
Cdu 4. Theo de b^i: d^^^^^ M = 18,5 => M = 37
Co cac phiiong trinh phan uTng: ( C J l ^ la cong thijfc phan tuf trung
binh cua 2 chat)
CnHm + \ + —j02 3H2O + P2O5 - m -> nC02 + — HaO 2 CO2 + Că0H)2 12,6 2H3PO4 —> CaCOai + H2O b ) Sau (3): B 1^ Fe304 c6 dụ
Fe304 + 4CO - 3Fe + 4CO2T mpe = 0,3 X 56 = 16,8 gam , 'au 6. n n C a C 0 3 ~ ^ h o n hcJp ~ 18 50 " 100 7,4 37 = 0,7 mol = 0,5 mol = 0,2 mol 2 chat
truang hap 1: 1 chat 1^ CH4 (M = 16 < 37) 0,1 X 16 + 0,1 X M = 7,4 => M = 58 => C„H2„+2 la CTTQ cua chat 2.
=> 14n + 2 = 5 8 = > n = 4r:> Cong thijfc phan ttf la (C4H10)
Truang hap 2: 1 chat la CgHe (M = 30 < 37) 0,1 X 30 + 0,1 X M = 7,4 M = 44
=> 14n + 2 = 4 4= > n = 3=i> Cong thijfc phan tijf la (CsHs)
Ket luan: Hon hop c6 2 cap nghiem: Cap 1: CH4 ; C4H10
Cap 2: C2H6 ; Calh
Cdu 5.
a) Dat a, b la so mol AI2O3, FexOy Phan ufng vdi HCl:
AI2O3 + 6HC1 > 2AICI3 + 3H2O FexOy + 2yHCl > xFeClgy/x + yHgO
TU (1), (2) c6: 6a + 2yb = 0,5.4 = > H H ^ O = 3a + yb = 1 Phan ufng vai NaOH:
AI2O3 + 2NaOH > 2NaA102 + H2O
T L T (3) ^ 2a = 0,2.2 a = 0,2 mol (
^ yb = 0,4
TCr (I), (II) c6: 0,2 x 102 + b(56x + 16y) = 43,6 va xb = 0,3
xb 03 X 3
4
a) Dat X , y, z, t la so mol cua C3H8, C2H4, C2H2 va H2 trong 0,6 mol Ạ
T a c6: x + y + z + t = 0,6 (]
44x + 28y + 26z + 2t = 13 (I]
Cho A qua dung dich Bra dU c6 phan ijfng:
C2H4 + B r 2 > C2H4Br2 (]
C2II2 + 2Br2 > C2H2Br4
b ) Sau (1, 2) trong B gom C3H8 va II2. 6,76 6,76 X + t = 22,4 = 0,3 mol ^ 8,33% = TO (III), (IV) 2t .100% o 2x = t TCr (I), (II) va X , ta c6: 44x + 2t X = 0,1 mol; t = 0,2 mol 28y + 26z = 8,2 y + z = 0,3 y = 0,2 mol; z = 0,1 mol Vay: %C3H8 = M 0,6 100% = 16,6(6)% = %C2H2 %C2H4 = %H2 = 33,3(33)% c) T L T (1), (2) c6: m = 28 X 0,2 + 26 x 0,1 = 8,2 gam Cdu 7. a) Co phan ufng:
2I<[Mn04 + 16IIC1 - > 2KC1 + 2MnCl2 + SClgt + 8II2O CI2 + 2NaOII > NaCl + NaClO + II2O
TO(l)^n,,^=- X n .
TCr (2) => nNaOH phan img = 2 n ^ , = 0,02 mol
0,16 x 0,2 = 0,01 mol
r:> dung dich A c6: [NaCl] = [NaClO] = 0,01
Ta CO t i le:
yb 0,4 y
C 6 n g thufc p h a n tiJf c u a oxit sat: F e 3 0 4 .
[NaOM dul = 0,2 0,2.0,2-0,02 0,2 = 0,05M = 0,1M
104 Ldl CIAI Oi THI HOC SINH Cl5l HOA HOC ^
b ) C h o A p h a n tifng v6i ( N H 4 ) 2 S 0 4 c6 p h a n ufng:
2NaOII + (NIl4)2S04 > N a 2 S 0 4 + 2NI-l3t + 2H2O
^ n(NH,),so, = I " N a O H dii = 0,01 m o l ^ V = M i = 0,1 (lit). Kill n£ T u i u n r Giuu nihi u n i unr a
D E SO 4 1
DE THI HOC SINK GIOI HOA L^P 9 , TP. HO CHI MiNH N A M HOC 2 0 0 8 - 2 0 0 9
Cau 1. (4 diem)
1.1. Hay chon 6 dung dich muoi Aj, As, A3, A4, A5, A g ling vai 6 gdc axif
khdc nhau thoa man cdc dieu kien sau: a) Ai + As > Co khi bay len b) Ai + A3 > CO kit tua
c) As + A3 > CO ket tua vd c6 khi bay len
d) A4 + As > CO ket tua
e) As + A g > CO ket tila
1.2. Viet cong thiCc cdu too c6 the cua hap chat c6 cong thiCc phdn td CJigClo 1.3. Co 5 goi bgt trdng Id KNO3, K2CO3, K2SO4, BaC03, BaSÔ chi duw,
diing them nuac, khi CO2. Hay trinh bay each nhdn biet tiing chut bgt trdng noi tren vd viet phuang trinh hoa hoc de minli hgạ
Cdu 2. (4 diem)
2.1. Cho 5 gam CaO tdc dung het vai lOOud nUdc cat trong mot chicc cdc, khudy deu hon hap de phdn dng xdy ra hoan loan, de yen cue trong mot thai gian ngdn, thdy ket tua trdng Idng xudng đy cue phdn tren la dung dich trong. De cdc ra ngodi trai sau vdi ngay thuy
tren mat dung dich trong cdc c6 mot lap vdng trdng. Hay gidi thick hien tugng vd viet phitang trinh phdn dng xdy rạ Biet do tan cua CăOH)s a nhiet do phong (25°C) la 0,153g/100g nude vd khdi lugiii> rieng cua H2O Id Ig/ml.
2.2.
a) Hay tun 4 phdn dng hoa hgc trong do khi 2 chat khdc nhau tdc duíb
vdi nhau, thu dugc dung dich NaOH. Viet phuang trinh hoa hgc ciic cdc phdn dng dọ
b) Co mot nueng Na kim loai de ngodi khong khi sau mot thai giÓ'
ngdn, no bien thdnh hdn hgp A, hoa tan het A vdo nUdc, thu di/ó^' dung dich B. Cho vdo dung dich Bads vd mot gigt chat chi thi ma" phenolphtalein. Hay mo id hien tugng xdy ra vd viet phuang triíl'
hoa hgc cua cdc phdn dng dọ
Cdu 3. (4 diem)
3.1. Cho hon hgp A gom C2H2 vd Hs- Cho 10,08 lit A di qua dng dUng
chat xdc tdc Ni dun nong, thu dUgc 6,944 lit hon hgp khi B gom 4 chat. Ddn B di chgm qua blnh dUng du nUdc brom cho phdn dng xdy
I ra hodn todn, thu dugc 4,48 lit hdn hgp khi C. Biet rdng Imol A c6 khdi lugng 10 gam vd cdc the tich khi deu do d dieu kien tieu chudn. I Hay viet cdc phuang trinh phdn dng xdy ra vd tinh thdnh phdn phdn
tram theo the tich cua cdc khi trong hon hgp A, B, C.
I 3.2. Cho 4,48 lit (dktc) hon hgp A gom 2 hidrocacbon no mgch ha c6 khdi lugng 4,88 gam. Ddt chdy hodn todn hon hgp A do, cho todn bg
L cdc sdn phdm chdy hap thu het vdo binh dUng lugng du dung dich
I Bă0H)2, sau thi nghiem thdy khdi lugng binh tang niig vd trong \ tgo thdnh nisg ket tua trdng. Viet cdc phuang trinh hoa hgc xdy
I ra vd tinh mi, nis.
'du 4. (4 diem)
4.1. Cho td td dung dich B chda x mol HCl vdo dung dich C clida y mol Na2C03. Sau khi cho het B vdo C ta dUgc dung dich D. Hdy xdc dinh cdc chat tg,o thdnh vd sd mol cdc chat trong dung dich D ( theo x, y) 4.2. De xdc dinh nguyen td khdi cua do ( x dvC) vd kali (y dvC), ngudi
ta nhiet phdn migam kali clotrat d nhiet do cao den khdi lugng khong dot thu dugc nisgam kali cloruạ Hoa tan sdn phdm vdo nUdc rSi cho tdc dung vdi bgc nitrat dU thi thu dugc nis gam ket tuạ Clio nguyen td khdi cua oxi Id A dvC vd bgc Id B dvC.
a) Hdy thiet lap cong thdc tinh x vd y theo nii, nis, nis, A vd B.
b) Tinh x, y khi nii = 3,0642; nis = 1,8648; nis = 3,5838; A = 15,964;
B = 107,868. du 5. (4 diem) du 5. (4 diem)
5.1. Ddn 2,24 lit khi CO (dktc) di chgm qua dng sd dung 7,2 gam hon hgp Xgom CuO vd C'u den khi phdn dng xdy ra hodn todn thu dugc chat rdn Y vd khi D c6 tl khdi hai so vdi hidro bdng 18. Hoa tan Y trong dung dich HNO3 vda du, can dung 0,5 lit dung dich HNO3 vd CO khi-NO2 thodt rạ
a) Tinh khdi lugng moi chdt trong X.
b) Tinh nong do mol/lit cua dung dich HNO3 vd the tich khi NO, thu dugc (dktc). thu dugc (dktc).
5.2. Cho 43 gain hon hap bari clorua vd canxi clorua vdo 1,5 lit dung
dich NagCOa 0,4M. Sau khi phdn dug xdy ra hodn todn thu dugc 39,7 gam ket tiia A vd dung dich B.
a) Tinh % khoi lugng cdc chat trong Ạ
b) Tinh tong khoi lugng muoi trong dung dich B.
Biet: H= 1;C = 12; N =14; Na =23; Mg = 24; Al = 27; S = 32; CI = 35,5; K = 39; Ca = 40; Fe = 56; Zn = 65; Ag = 108; Ba = 137.
Hoc sinh khong dugc sii dung bang tinh tan vd bang he thong tuui,
hodn cdc nguyen to hoa hoc.
LCil GlAl
C a u 1. 1.1.
a ) 3K2HPO4 + SNaHCOg > Na3P04 + 2K3PO4 + SCOgT + 3H2O
b ) K2HPO4 + Ba(HS04)2 > B a S 0 4 i + K2SO4 + H3PO4
c) 2NaHC03 + BăHS04)2 > BaS04^ + Na2S04 + 2CO2T + 2H2O d ) K2SO3 + BaCl2 > B a S O g i + 2KC1 d ) K2SO3 + BaCl2 > B a S O g i + 2KC1
e ) BaCl2 + 2AgN03 > BăN03)2 + 2 A g C U
A l : K2HPO4 A2 : N a H C O a A 3 : BăHS04)2
A 4 : K2SO3 As : BaCl2 A« : AgN03
Chii y: Bdi todn nay c6 nhieu chat de liia chgn. Cdc ban c6 the hia
chgn chat khdc cho phii hap vdi de bai!
1.2. Cac cong thufc cau tao (dong phan) cua C4H8CI2:
CH3-CH2-CH2-CHCI2 CH3-CH2-CCI2-CH3
C H 3 - C H 2 - C H C I - C H 2 C I C H 3 - C H C I - C H 2 - C I I 2 C I
C H 2 C I - C H 2 - C H 2 - C H 2 C I . C H 3 - C H C I - C H C I - C H 3
H 3 C- C H - C H C I 2 CIH2C-CH—CH2CI H3C-CCI—CH2CI
C H 3 CH3 C H 3
1.3. Trich m5i goi mot i t hot lam mau thijf : Cho 5 mau thijf M n luot vao H2O
Nhom m&u tha tan Id: KNO3, K2CO3, K2SO4 (I)
Nhom mau thii khong tan (ket tua): BaCOa, BaS04 (II)
I
D a n tii tii khi CO2 vao hai ket tiia or nhom ( I I ) , thay ket tua nko tan
ra BaCOa, con lai la BaS04
Phan iJng: BaCOg + CO2 + H2O > Ba(HC03)2
Lay san pham tao thanh cf tren BăHC03)2 cho Ian liigt vao cac mau
thijf d nhom I .
+) M S u thuf nao khong c6 hien ti/cfng la KNO3
+) Mau thtf nao xuat hien ket tiia la K2CO3 va K2SO4, phan ufng sau: BăHC03)2 + K2CO3 > BaC03i + 2KHCO3
BăHC03)2 + K2SO4 > BaS04l + 2KHCO3
Loc ket tua tren dan ttf tii k h i CO2 vao, ket tua nao tan ra chat
ban dau l a K 2 C O 3 . Ket tua khong tan => chat ban dau l a K2SO4
Phan ufng: BaC03 + CO2 + H2O > BăHC03)2
C a u 2.
2.1. Theo de bai, CaO tac dung het vdi H2O va phan ufng xay ra hoan toan => san pham la Că0H)2
Phan ang: CaO + H2O > Că0H)2
+) d nhiet do phong (25°C): m Ma: m,. C a ( 0 H ) 2 do 5 g a m C a O tao t h a n h 185 X 74 = (gam) 56 28 gam > 0,153 gam Că0H)2
C a ( D H ) 2 - 28
=> Că0H)2 khong tan het.^do do ket tua trSng l^ng xuong Ih Că0H)2
K h i de coc ra ngoai trcfi t h i CO2 trong khong k h i se tac dung v d i
dung dich Că0H)2 tao thanh vang trSng la CaCOs
CO2 + Că0H)2 > CaCOs^ + H2O
2.2.
a ) Phan ufng: Na2C03 + Ca(0H)2
Na2S04 + Ba(0H)2
-> CaC03i + 2NaOH -> BaS04i + 2NaOII , Na + H2O -> NaOH + - H 2 t
b) K h i de inieng Na ngoai khong k h i t h i Na t^c dung vdi O2 tron^, khong k h i tao th^nh Na20. khong k h i tao th^nh Na20.
4Na + O2 > 2Na20
Sau do Na20 trong khong k h i se tAc dung vdi CO2 tao th^nh Na2C0,
NasO + CO2 > Na2C03 Vay hon hop A gom: Na du, N a 2 0 va NagCOs. Vay hon hop A gom: Na du, N a 2 0 va NagCOs.
+) K h i h5a tan vao H2O => c6 k h i H 2 t khong mau thoat ra Na + H2O > NaOH + | H 2 t Na + H2O > NaOH + | H 2 t
N a 2 0 + H2O > 2NaOH
Cho dung dich B vao dung dich BaCl2 xuat hien ket tua trdng BaCOg I Na2C03 + BaCla > BaCOg^ + 2NaCl I Na2C03 + BaCla > BaCOg^ + 2NaCl
Do trong dung dich B vSn chufa dung dich NaOH nen dung dich phenoltalein se hoa do (c6 the hong la tuy theo nong dp). phenoltalein se hoa do (c6 the hong la tuy theo nong dp).
C a u 3. 3.1. Phan ufng: 3.1. Phan ufng: C 2 H 2 + H2 > C2H4 (1) (mol) X X X C2H2 + 2H2 > C2H6 (2) (mol) y 2y y
Do hon hop B gom 4 chat => nhSn hgp A = ^^'^^ = 0,45 mol 22,4
=> Hieu suat phan ufng khong hoan toan.
=> Hon hop k h i B gom: C2H2 dtf, Hg di/, C2H4 va CgHẹ 6 944
Va nusn hap kill B = ^'^ ^ = 0,31 mol
Phan ufng: C2H2 + 2Br2 > C2H2Br4 (3)
C2H4 + Bra — ^ C2H4Br2 (4)
4,48
=> nh6n h<?p khi c = -^rr— = 0,2 mol
Goi X, y Ian lugt la so mol cua C2H2 tham gla trong phan ijfng (1) va (2) Va a, b Ian luot 1^ so mol cua C2H2 va H2 trong hon hop A ban daụ Va a, b Ian luot 1^ so mol cua C2H2 va H2 trong hon hop A ban daụ
=> a + b = 0,45 (I) Theo de hki, ciJ 1 mol A c6 khoi iLTcrng 10 g a m Theo de hki, ciJ 1 mol A c6 khoi iLTcrng 10 g a m
0,45 mol A C O khol lugng x gam => m A , n > u g o , « „ . o . =4,5(gam) <^ 26a + 2b = 4,5 => m A , n > u g o , « „ . o . =4,5(gam) <^ 26a + 2b = 4,5
Giai he phaong t r i n h (I) va (II), ta diicfc: a = 0,15 va b = 0,3
(II
Ma: n . , . + n,.„ + n ^ „ = 0 , 3 1 <::> a - x - y + b - x - 2 y + x + y = 0,31 <::> a - x - y + b - x - 2 y + x + y = 0,31
cị a + b - x - 2 y = 0,31 o - x - 2y = - 0,14 => x + 2y = 0,14 (Hi Do H2 du va C2H6 khong phan ufng vdi niidc brom => k h i C gom: H^ Do H2 du va C2H6 khong phan ufng vdi niidc brom => k h i C gom: H^ du va C 2 H 6 .
=^ ^ho^^imC = " H , d u + " c , H e = 0,2
« b - x - 2 y + y = 0,2 < = > - x - y = - 0 , l < = i > x + y = 0,l (IV Giai he phifcfng t r i n h (III) va (IV), ta di^pc: x = 0,06 va y = 0,04 Giai he phifcfng t r i n h (III) va (IV), ta di^pc: x = 0,06 va y = 0,04
Do la chat k h i => %V = % n +; H6n hap A: = %n^^,,^ = ^ ^ ^^^^^ = ^^'^^"^^ +; H6n hap A: = %n^^,,^ = ^ ^ ^^^^^ = ^^'^^"^^ %V„ = %n„ = 66,67% .) H6n hap B: %V,^„^ = % n , ^ , , ^ = x 100% = 16,12% b - X - 2y 0,31 X 100% = 51,61% %V„ „ = % n 0,31 X 100% = 19,35% %V^ „ = %nc H = 100^^ = 12,92% +) Hdn hap C: %V,^„^ = %n,^„^ = ^ x 100% = 20% %V„, d . = d u = 80%