c ) Goi hieu suat phan ufng cua C4H10 la h va cua CsHs la l,21i só mol da phan ufng: só mol da phan ufng:
cua C4H10 la 0,6h (mol)
cua C3H8 la l,2h x 0,4 = 0,48h (mol)
Só mol anken tao thanh:
0,6h + 0,48h = l,08h = 0,75 (mol) => h = 69,44% Phan trSm so mol da phan lifng cua:
C4H10: 69,44%; CsHg: l,2h = 83,33%
So mol cac ankan con lai (chiTa phan ufng) va so mol H 2 la:
C3H8 : 0,4 X 0,1667 = 0,0667 (mol)
C4H10: 0,6 X 0,3056 = 0,1833 (mol)
H 2 : 0,75 (mol)
Khi qua dung dich KMn04 dii thi C 4 H 8 va C3H6 hi hap thu het, coi
lai hon hop D gom 3 khi:
+ n„ =1 (mol)
Ho
CsHg: 0,0667 mol ^ 6,67% (the tich)
C4H10: 0,1833 mol -> 18,33% (the tich)
H 2 : 0,75 mol 75,0% (the tich)
d) Cac phiiong trinh phan (ing thiic hien siJ chuyen hoa tii X
* C3H8 T ^ C H 3 - C H 2 - C H 3 (X) * C4H10 T C H 3 - C H 2 - C H 2 - C H 3 (Y) TCr X ^ Y: C H 3 - C H 2 - C H 3 Cracking -> CH4 + C H 2 = C H 2 xt, t"- C H 2 = C H 2 C H ^ C H + CH^CH C H E ^ C H + H 2 xt, C H = C - C H= C H 2 ^"g' ^° > C H 3 - C H 2 - C H 2 - C I I 3 (Y) TCr Y ^ X: C H 3 - C H 2 - C H 2 - C H 3 (Y) Cracking -> CH4 + C H 3 - C H 2= C i ; CHa—CH2=CH3 + Ho xt, t'' l i l A -> CH3—CH2—CH3
Lfll GIAI THI HOC SINH GIOI HOA HC'
D E SO 31
OE THI HOC SIAIH GIQI KOA HOC 9, THAIVH PHO HO CKI MI1\IH NAM HOC ZOOS - 2007
1.1. Metan c6 Ida C2H2 vd CO2, b&ng phuang phdp hoa hoc lidy tiiih
die metan.
1.2. Viet phuang trinh phan icng de bieu dien sa do sau:
CaO > Că0H)2
CaCOs ^zr^ CaCl2 > CăN03)2 > CaCOa
CăHC03)2
1.3. Hay cJion cdc chat: X, Y, Z, E, T, G thich hap thoa man cdc dieu
kicn sau:
CO kill bay len X+Y Z + Y E + Y X+T z + r E + G
CO khi bay len
-> CO khi bay len
-> CO kct tua
CO ket tua
CO ket tila
Trong do X, Z, E, G Id cdc muol c6 goc axit khdc nhau, Y Id axit, T Id baza vd cdc phan iCng deu xdy ra trong dung dich. Viet cdc phuang trinh phan ling mlnh hgạ
Cau2
2.1. 0 30°C, 100 gam nuac hoa tan 5,59 gam barl hldroxlt, tao tlidnh
dung dich bdo hoa c6 kiwi lugng rleng Id 1,06 gam I ml. Hay tinh nSng do % vd nong do mol I lit cua dung dich a nlilet do dọ
2.2. Kill hoa tan cdc muol tan cua nhom hogc sdt vdo nUac thi dung
dich thu dugc luon luon bi due vd thuang tao thanh mot it ket tuạ NhUng kill them vdo mot it dung dich axit thi ket tua do tan het vd du&c dung dich tmng suot. Hdy gidi thich hlen tugng do vd vlet cdc phuang trinh phan ling mlnh hgạ Hdy du dodn do pH cua
dung dich tao thdnh (=7; >7; <7).
2.3. Nguyen tii cua nguyen to A c6 tong so hat Id 82, tong so hat niang
dlen nhleu han so hat khong mang dlen la 22.
a) Xdc dlnli nguyen to Ạ
b) Viet ba phuang trinh phan iaig cua A trong do A the hlen hoa trl 2.
Cdu 3. Ngdy nay ngudi ta dieu che axetilen bdng each nhiet phdn metan Q nhiet do cao theo phuang truth phdn iCng:
2CH4 > C2H2 + 3H2
Khi lay CH4 thuc hien phdn ling chuyen hoa tren, thu dugc hon hap kin
A gom metan, axetilen vd H2. Dot chdy hodn todn m (gam) hon hap thu dicac 17,6 gam CO2.
a) Viet cdc phUang trinh phdn ting xdy rạ
b) Tinh khoi lugng m (gam) hon hap A đ dem dot.
Cdu 4. Cho bgt sdt tdc dung hodn todn vai lugng du dung dich H2S()^
đc, nong nong do 78,4%, thu dugc dung dich A, trong do nong do
cua ¥62(80^)3 vd H2SO4 du bdng nhau, gidi phdng khi SO2. a) Tinh nong do % cua mud'i sdt vd H2SO4 dU trong dung dich Ạ
b) Tinh the tich dung dich NaOH 2M can de tdc dung het vai 50 gam
dung dich Ạ
Cdu 5. Cho hon hap A dang bgt gom Mg vd Al. Lay 12,6 gam A cho tuc
dung viCa du vai 300 ml dung dich hon hap HCl Ci moll lit vd H2S(),
lodng C2 moll lit. Biet C ; - 2C2. Sau phdn ling thu dugc dung dich B vu
13,44 lit khi H2 (dktc).
a) Viet cdc phuang trinh phdn ling xdy rạ
b) Xdc dinh Ci, C2, % khoi lugng moi kim logi trong hon hap Ạ Cho biet: H = 1, C = 12, O = 16, Na - 23, Mg = 24, Al = 27, S = 32, CI = 35. Cho biet: H = 1, C = 12, O = 16, Na - 23, Mg = 24, Al = 27, S = 32, CI = 35.
Fe = 56, Cu ^ 64, Zn = 65, Ba = 137.
Hoc sink CO the sii dung bdng he thong tudn hodn vd bdng tinh tan. Ldl GIAI
Cdu 1.
1.1. Dan hon hop k h i di qua dung dich Că0H)2, se thay c6 ket tu;' trang tao th^nh. trang tao th^nh.
Că0H)2 + CO2 > CaCOgi + H2O
Tiep tuc dan hon hcfp k h i c5n lai qua dung dich brom -> mat maU dung dich brom v a k h i khong phan lifng bay r a 1^ metan (CH4).
C2H2 + 2Br2 > C2H2Br4
l A i G U I THi Hnr: SINH nini HOA HOC 9
1.2. 1000°c Cau 2. 1.1. 2.2. CaO + CO2 > CăOH)2 > C a C O a i + H2O CaCOa CaO + H2O CăOH)2 + CO2
CaCOs + 2HC1 > CaCla + H2O + COzt CaClz + 2AgN03 > CăN03)2 + 2 A g C l i CăN03)2 + Na2C03 > CaC03i + 2NaN03
CaCOa + CO2 + H2O > C a ( H C 0 3) 2
CăHC03)2 '° ) CaCOgi + H2O + COat X: M g ( H C 0 3) 2 Y: H2SO4
T: Bă0H)2 E: Ba Hoc sinh tiT viét phan tfng.
Z: BăHS03)2
G: CuCl2
a 30°C, 100 gam H2O + 5,59 gam Bă0H)2 ^ dung dich bao hoa
nidung dich bao hoa = 105,59 gam 5,59 C% Bă0H)2 = 105,59 100 = 5,29% 5,59 171 =5- Bă0H)2 Vduug dich 0,0327 = 0,0327 mol CM - = = 99,613 (ml) = 0,099613 (Ht) 1,06 = 0,328M 0,099613
K h i hoa tan muoi tan cua A l hoac Fe se xay ra qua t r i n h thuy phan cua cac ion Al^\* hoac Fé^*.
Al^" + 3 I I 0 I I > AKOIDsi + 3H^ Fê" + 2 H 0 I I > Fe(0H)2i + 2H*- =^ 4Fe(OH)2 + O2 + 2H2O > 4Fe(OH)3i
Fê" + 3 H 0 H Fe(0H)3^ + 3ir
K h i cho axit vao dung dich, se xay ra phan ijfng trung hoa theo phan'ijfng:
i r + OH- > H2O
Xciy ra theo cac phiiong t r i n h vdri H ^ axit manh. AUOlDa + 3H^ > Al^^ + 3H2O
Fe(0H)3 + 3H" > Fê" + 3H2O
Fe(0H)2 + 2 H ' > Fê" + 2H2O
Do trong dung dich van con c6 ion H'' ban dau moi tri/cmg axit
2.3. T o n g so h a t t r o n g A la 82. => p + n + e = 82 => p + n + e = 82 Do n g u y e n ttf t r u n g hoa ve d i e n => so p = so e (1) o 2p + n = 82 T o n g só h a t m a n g d i e n n h i e u h o n só h a t k h o n g m a n g d i e n => 2p - n = 22 p = 26 G i a i (*), (**) t a duoc: n = 30 A l a Fẹ Cdu 3.
Goi a la so m o l CII4 b a n daụ b l a só m o l CH4 p h a n ufng.
2 C H 4 '" > C2H2 + 3H2
(mol) b -> 0,5b ^ 1,5b CH4 + 2O2 ' ^" > CO2 + 21120 (mol) (a-b) ^ ( a - b )
2C2H2 + 5O2 > 4CO2 + 2H2O
(mol) 0,5b b H 2 + ^ 0 2 -> H2O n C O , = a - b + b = a=: - 0,4 m o l 44 ^ ™ C H , banđu = X 0,4 = 6,4 g a m A p d u n g d i n h l u a t bao t o a n k h o i iLfcrng, t a c6: ™ C H , bandau = khi A = 6,4 g a m => m = 6,4 gam
Cau 4. G o i a la só m o l Fe ban daụ
b la só m o l H2SO4 ban daụ
2Fe + 6H2SO4 d — ^ Fe2(S04)3 + SSOzt + 6H2O (mol) a ^ 3a 0,5a -> 1,5a
m , , ^ At = 98b g a m
H 2 S O 4 b a n d a u "
m đ W.^SÔ 78,4% 98b
X 100
78,4 = 125b g a m
n i d u i i g dicii sau phan iing = 56a + 125b - 96a = 125b - 40a (gam)
m H.^SÔ da = 98b - 294a (gam)
"Ve,(so,)^^ = 200a gam
I Theo de b a i : C% H2SO4 d u = C% Fe2(S04)3
^ ( 9 8 b - 2 9 4 a ) X 100 _ 200a x 100
1 2 5 b - 4 0 a ~ 125b - 40a
^ 98b - 294a = 200a « 98b = 494a b * 5,0408a C% H2SO4 = C% Fe2(S04)3 = ^^^^ * 33,89% )) 50 g a m dung dich A . T a c6: m „ ụ^ > = m _ 16,945 = ^ " ^ H . S O ^ - . g g 16,945 => n . 125b - 4 0 a 33,89 x 50 ,aaAK H,so, d . = = 16,945 g a m 100 = 0,1729 m o l = 0,0423625 m o l 5. P h a n ijfng:
H2SO4 + 2 N a O I I > Na2S04 + 2H2O (mol) 0,1729 "> 0,3458 Fe2(S04)3 + 6 N a O H > 3Na2S04 + 2 F e( O n) 3 i (mol) 0,0423625 ^ 0,254175 => I^ N a O H = 0,599975 m o l Vd„.g dich N a O H 2 M * 0,3 l i t = 300 m l M g + 2HC1 > MgCl2 + H a t M g + H2SO4 > MgS04 + I l 2 t . 2A1 + 6HC1 > 2 A I C I 3 + 3 H 2 t 2A1 + 3II2SO4 > Al2(S04)3 + 3 H 2 t
I n F T H i Hnr. KIUM mm HOA H O C 9 (1) (2) (3) (4) 151
H C l : C j M
300 m l dung dich < M
300 m l dung dich <
HgSO, : C, M
=^ Z ^ H C l = 0,3Ci mol
Z i^H,so, = 0'3C2 mol
Theo (1), (2), (3), (4), ta c6: Z " H C I = ^Z"H2(1),(3) Z"H2S04 = Z"H2(2),(4) H 2 ( l ) . ( 3 ) - 2 z - H C l Z v = 13,44 22,4 M^: Ci = 2C2 => Ci - 2C2 = 0 Giai (*), (**) ta dUcfc: Ci = 2 va C2 = 1 Mg : a mol = 0,6mol r:> 0,15 Ci + 0,3C2 = 0,6
H5n hofp ban dau
[Al : b mol Co: Z ^ M g = Z % 2 ( i ) , ( 2 ) I ' ^ Z ^ A l = Z"H2(3).(4) => a + 1,5b = 0,6 Giai (***), (****) ^ a = 0,3 ^ b = 0,2 24a + 27b = 12,6 m^g =7,2 gam mAi =5,4 gam %Mg = 57,14% %A1 = 42,86% DE SO 32 (TLT LUYEN)
f)E THI HOC SINH GIOI HOA HOC 9, HUYEN CU CHI, TP. HCM NAM HOC Z006 - Z007
Can 1: (4 diem)
1. Neu phuang phdp tdch hon hap gom 3 khi CI2, H2, CO2 thdnJi cdc chai nguyen chat nguyen chat
2. Co 4 chat man trdng tilang tU nhau Id : NaCl, AICI3, MgCOs, BaCO chl dung tiilac, nong, trinh bay cdch nhgn biet tCctig chat tren.
3. Nung agani CaCOs roi cho todn bo CO2 vdo Vml dung dich KOJ^ 0,5M. Sau phdn ihig, c6 can dung dich thu dugc 5,52 gam K2CO3 r •
2 gam KHCO3. Tinh gid tri cua a ud V.
1A9 Ldl GIAI Oi THI HOC SINH GIOI HOA HOC 9
dich H2SO4 lodng thi thu dugc 10,08 lit khi H2 (dktc), dung dich (Y) vd tgo ra 13,98 gam ket tuạ
a) Tinh khoi lugng mdi kim logi trong hon hgp (X).
b) Cho til tic dung dich KOH 2M vdo dung dich (Y). Tinh the tich dung
dich KOH can dung de tgo ra mot ket tua duy nhdt.
C&u 3: (4 diem)
Cho 31,8 gam hon hgp (X) gom 2 muoi MgCOs vd CaCOs vdo 0,8 lit dung dich HCl IM thu dugc dung dich (Z)
a) Hoi dung dich (Z) c6 da axit khong? b) Lugng CO2 c6 the thu dugc boo nhieụ b) Lugng CO2 c6 the thu dugc boo nhieụ
c) Cho vdo dung dich (Z) mot lugng dung dich NaHCOs du thi the tich CO2 thu dugc Id 2,24 lit khi (dktc). Tinh khoi lugng mdi muoi trong CO2 thu dugc Id 2,24 lit khi (dktc). Tinh khoi lugng mdi muoi trong hSn hgp (X)
Cdu 4: (4,5 diem)
Trgn 50ml dung dich AgNOs 0,44M vai 50ml dung dich Pb(N0s)2 0,36M thu dugc dung dich Ạ Them 0,828 gam bgt Al vdo dung dich A dugc chat rdn B vd dung dich C. .
1. Tinh khoi lugng cua B.
2. Cho chat rdn B vdo dung dich Cu(N0s)2- Sau khi phdn ling ket thuc thu dugc 6,046 gam chat rdn D. Tinh thdnh phdn phdn tram theo thu dugc 6,046 gam chat rdn D. Tinh thdnh phdn phdn tram theo khoi lugng cdc chat trong D.
Cdu 5: (3,5 diem)
1. O 85°C CO 1877 gam dung dich bdo hoa CUSO4. Ldm Ignh dung dich
xuong con 25°C. Hoi cd. bao nhieu gam CUSO4.5H2O tdch khoi dung \ Biet do tan cua CUSO4 d 85°C Id 87,7 gam vd d 25°C Id 40 gam.
%.2. Co 5 Ig mat nhdn dUng 5 Ig dung dich NaOH, KCl, MgCh.CuCls-
I AlCls. Hay nhgn biet titng dung dich md khong can them hoa chat ndo khdc vd viet phdn ling xdy rạ
Cho: H = 1; Cu = 64; S = 32; O = 16; Ag = 107; Pb = 208; Al = 27, Mg = 24; Ca = 40; C = 12; K = 39; Ba = 137; CI = 35,5; N = 14.