Fe203 + 3H2 ^—> 2Fe + 3H2O
Cau 3. Viet 10 phan ufng theo scf do:
Fe + 2HC1 —> FeCh (A) + Hat (1 >
2Fe + 3CI2 —> 2FeCl3 (B) (2i
2FeCl2 (A) + CI2 > 2FeCl3 (B) (3'
2FeCl3 (B) + Fe > 3FeCl2 (A) (4'
FeCla (A) + 2NaOH > Fe(OH)2i + 2NaCl (5' FeCl3 + 3NaOH > Fe(0H)3 ( D ) i + 3NaCl (6) 4Fe(0H)2 + O2 + 2H2O — > 4Fe(0H)3 (7)
Fe(0H)2 jrr. ^ T : . — T T T T —> FeO (E) + H2O (8)
(khong CO khong k h i )
2Fe(OH)3 — > FeaOs (F) + 3H2O (9)
4FeO + O2 —> 2Fe203 (F) (10)
Ị TInh hieu suat i ^ . t n ung.
Cho hon hop A: 2 mol => SO2: 1 mol; O2: 1 mol
P h a n ling xay ra: 2SO2 + O2 t ° , x t -> 2S03 (1) Neu (1) xay ra 1007c t h i SO2 het, O2 diị Goi x la so mol SO2 phan ufng: 0 < X < 1 2SO2 + O2 2 1 t O , x t Ban dau: Phan ufng: Sau: 1 X - > 1-x 2SO3 2 0 (1) X 2 1 - ^ 2 [ ' - 2 ) nB = ( l - x ) + | l - - | + x = 2 - - T a c6: Vay: "so, n„ x 35,29 100 X = 0,6 Hieu sua't (H) = = 60%
5. T i n h nong do mol chat tan trong dung dich Ạ
33 8
Trong 33,8 gam H2SO4.3SO3 c6 ^ - ^ m o l = 0,1 mol H2SO4 va
0 0 0
3 X 0,1 mol SO3.
Trong 800 ml dung dich HSO4 19,6% (d = 1,25 g/ml) c6:
n 800.1,25.19,6 - = 2 mol n •"2S0, 100.98 800.1,25.(100 -19,6) = 44,67 mol -> H2SO4 100.18
Ta CO phan ufng: SO3 + H2O —
Vi n„ y » ngQ^ do do sau (1) trong dung dich A c6:
Klidi li/ong chat tan la H2SO4 = (0,1 + 0,3 + 2)98 = 235,2g Klidi lugng dung dich A = 33,8 + 800.1,25 - 1033,8g N6ng do %H2S04 = X 100% = 22,75%
lUoo, o
6. T i n h so m o l ttrng chat t r o n g dung dich A .
T a CO p h a n urng:
2Fe + 6H2SO4 dac Fe2(S04)3 + 3SO2 + 6H2O (] 6 72
Vdri npe = = 0,12 m o l ; n^^g^^ = 0,3 m o l . V a y sau (1) t h i Fe d 0,02 m o l ; H2SO4 h e t ; Fe2(S04)3 s i n h r a 0,05 m o l . '
Fe d i i l a i tac d u n g l e n Fe2(S04)3. theo p h a n ufng:
Fe + Fe2(S04)3 > 3FeS04 (2 (adsbygoogle = window.adsbygoogle || []).push({});
Sau (2) t h i B^e h e t , Fe2(S04)3 dii 0,03 m o l ; FeS04 s i n h r a 0,06 m o l , V a y d u n g d i c h A chufa: 0,06 m o l FeS04 v a 0,03 m o l Fe2(S04)3.
7. a ) T i n h a M a ) T i n h a M P h a n ufng x a y r a : M g + H2SO4 > M g S 0 4 + H 2 t (1 Z n + H2SO4 > ZnS04 + H g t (1 T r o n g p h a n 2 l u g n g H2SO4 d u n g n h i e u hcfn p h a n 1 , d a n d e n lijgi,, H2 n h i e u h a n p h a n 1. V a y sau p h a n 1 liigfng H2SO4 h e t , lircfng kii, loai con d i i . "^""^ ^ P^^^" 1 - ^H.^ t h o a t r a a p h a n 1 = 0,15 m o l => a M = ^ = 0,75 0,2 b) T i n h k h o i lugng tUng k i m l o a i : " H ^ S O ^ ^ 6 m d u n g a p h a n 2 = 0,3.0,75 m o l = 0,225 m o l " H J S O, , P ^ ^ " ^^^S cf p h a n 2 = n,j^ a p h a n 2 = 0,2 m o l < 0,225 m o l V a y sau p h a n 2 k i m loai h e t , H2SO4 dạ
Goi X , y I a n liigt l a so m o l M g , Z n t r o n g h 6 n h g p dau, t a c6: Theo (1) v a (2), só m o l 2 k i m loai: x + y = 0,4
K l i o i l a g n g 2 k i m l o a i : 24x + 65y = 17,8 G i a i {*) v a (••:*) ^ kót quạ
f D E SO 26
HI CHON HOC SINH GIOI HOA HOC 9, TJNH KHANH HOA N A M HOC Z 0 0 4 - - ZOOS
(Jdi^ i . Cho cdc cap chat sau day, cap ndo c6 hogc khong the ton tgi trong
Cling mot dung dich? Giai thich vd viet phiiang trinh hoa hoc xay ra (neu cd):
a) NaOHvd KNO3 b) NaOH vd CuSÔ c) HCl vd Fe d) BaCls vd Na2S04 e) NaHCOs vd H2SO4. d) BaCls vd Na2S04 e) NaHCOs vd H2SO4.
Cdu 2. Hay xdc dinh cdc hap chat A, B, C, D vd viet phuang trinh Jioa hoc
bieu dien bien hoa do:
A B C D Cu
Cdu 3- Mot lion hap goin: đ voi, voi song, thqch cao, niudi an. Chi diCgc
phep dung nhiet do vd cdc hoa chat Id niCac, axit HCl vd Na2C03, hay tdcli rieng ra tUng chat nguyen chat.
Cdu 4. Dot lion hap goni cacbon vd lUu huynh trong O2 du, thu dugc hon
hap kill Ạ •
Cho —A Igi qua dung dich NaOH thu dUgc dung dich B vd khi C. Cho
2
khi C qua hon hgp chíca CuO vd MgO nung nong thu dugc chat rdn D vd khi Ẹ Cho khi E Igi qua dung dich Că0H)2 thu dUgc ket tiia F vd dung dich G. Them dung dich KOH vdo dung dich G Igi thdy c6 ket tua F xudt hicn. Dun nong G cung thdy cd ket tua F.
Cho —A cdn Igi qua xiic tdc nong thu dugc khi M. Ddn M qua dung dich
2
BaCl2 thdy cd ket tila N.
Hay xdc dinh thdnh phan A, B, C, D, F, G, M, N vd viet tat cd cdc phuang trinh phan ling xay rạ
5. Cho 80 gam bgt dong vdo 200 ml dung dich AgNOa, sau mot thai gian phan dug, dem Igc thu dugc dung dich A vd 95,2 gam chat rdn B. ' Cho tiep 80 gam bgt Pb vdo dung dich A, phan dug xong dem Igc thi
•
tdch dugc dung dich D chi chda mot mudi duy nhdt vd 67,05 gam clidt rdn Ẹ Cho 40 gam bgt kim logi R (cd hoa tri 2) vdo 1/10 dung dich D, jksau plidn dug hodn todn dem Igc thi tdch dugc 44,575 gam chat rdn F. mTinh nong do mol cila dung dich AgNOa vd xdc dinh kim logi R.
LCil GIAI
Cdu 1. Xet tiVng cap chat da cho ta thay:
a) Kliong phan lifng.
b) Co phan ilng: 2NaOH + C U S O 4 > Na2S04 + Cu(OH)2i (adsbygoogle = window.adsbygoogle || []).push({});
c) Co phan ufng: Fe + HCl > YaCh + H2T .
d) Co phan ufng: BaCla + Na2S04 > BaS04i + 2NaCl
e) Co phan uTng: 'iNaHCOg + H2SO4 > Na2S04 + 2C02t + 2H,0
Ket luaii: Chi c6 cap (a) ton tai trong dung dich, cac cap con 1^;
khong ton tai vi da phan ufng tao thanh chat khac.
Cau 2. Tim D: D Cu: D c6 the la CuO neu chat tac dung la H2, CỌ
D CO the la dung dich muoi dong: CuCl2, C U S O 4 , v.v... neu chat tac dung la kim loai hoat dong manh han Cu (Fe, Zn, Mg...). dung la kim loai hoat dong manh han Cu (Fe, Zn, Mg...).
Tim C: Neu iii C tao ra CuO thi C la Cu(0H)2 con neu C la CuSÔ
thi C la Cu(0H)2. Do do B la muoi dong tan nhuT Cu(N03)2,... A la
muoi dong tan CuCl2.
Vay CO the c6 hai day bien hoa sau:
CUCI2 Cu(N03)2 '^Q" ) Cu(0H)2 ) C U S O 4 — ^ ^ C u
hoac: C U S O 4 — ^ - 5 2 1 ^ CuCls ^"Q" > Cu(0H)2 — ^ CuO "'^ > Cu
Cau 3. Hoa tan trong nUdc: CaO + H2O > Că0H)2
Rifa nhieu Ian thu diigc chat rfin A (c6 CaCOs va CaS04) va niiox loc B (c6 NaCl va Că0H)2). loc B (c6 NaCl va Că0H)2).
Them Na2C03 vao nU6c loc B:
NaaCOa + Că0H)2 > CaCOsi + 2NaOH
LOG ket tua thu diioc niidfc loc C. Dem nung ket tua:
CaCOg —> CaO + CO2T
Trung h6a niidrc loc C roi c6 can thu ducfc NaCl: NaOII + HCl > NaCl + H2O NaOII + HCl > NaCl + H2O
Them Na2C03 vao niJcJc loc D de thu dtfoc lai CaCOs:
CaCl2 + NaiCOs > CaCOsi + 2NaCl
i o n Ifll RIAI THI Hfin SINH r.lli] HflA HOC 8
4 . 20 + O2 ' > 2CO
C + O2 — — — > CO2
S + O —> SO2
Vay khi A gom: CO2, SO2, CO, O2 dif. — A di qua dung dich NaOH: 2 — A di qua dung dich NaOH: 2
CO2 + 2NaOH > NasCOs + H2O SO2 + 2NaOH > Na2SOs + H2O SO2 + 2NaOH > Na2SOs + H2O
502 + Na2COs > NazSOs + CO2
Vay dung dich B chufa NazSOs, Na2C03; khi C chufa Cp2, CO, O2. (adsbygoogle = window.adsbygoogle || []).push({});
Khi C di qua CuO, MgO nung nong:
CuO + CO —> Cu + COat
Vay chat rdn D gom MgO, Cu, khi E c6 CO2, O2, CO diT.
Khi E loi qua Că0H)2:
Că0H)2 + CO2 > CaCOsi + H2O
Că0H)2 + 2CO2 > CăHC03)2
Vay ket tua F la CaCOs, dung dich G c6 CăHC03)2.
Them KOH vao G va dun nong G:
2K0H + CăHC03)2 > CaCOsi + K2CO3 + 2H2O
Ca(HC03)2 —> CaCOsi + C02t + H2O
~ A xuc tac nung nong:
2SO2 + O2 ^=^^ ^ 2SO3
450° C
Vay khi M la SO3.
Khi M qua dung dich BaCl2:
503 + H2O + BaCl2 > BaS04i + 2HC1
Cdu 5. Phtrcfng t r i n h phan tirng: Cu + 2AgN03 > Cu(N03)2 + 2 A g l (1 ( m o l ) X 2x X 2x rv' u 9 5 , 2 - 8 0 ,, Tinh x: x = — ^ 0,1 (mol) 2 1 6 - 6 4 Pb + Cu(N03)2 > Pb(N03)2 + C u i (•• (mol) 0,1 0,1 0,1 0,1
Theo (2) thi do giam khoi lugng kim loai (do mat Pb = 207 ta Cu = 64) se la:
(207 - 64).0,1 = 14,3 (gam) > 80 - 67,05 = 12,95 (gam). Dieu nay chufng to trong dung dich A vSn c5n muol AgNOa da d
CO phan dng:
Pb + 2AgN03 > Pb(N03)2 + 2Agi (:' (mol) y 2y y 2y '
Phan ufng (3) lam tSng mot liigfng:
(216 - 207)y = 14,3 - 12,95 = 1,35 =^ y = 0,15 TCf (1) va (3) ta CO => nong do mol/1 cua dung dich AgNOa la:
M = 2 , 5 M 0,2 ' 0,2 '
Dung dich D chufa Pb(N03)2: 0,1 + 0,15 = 0,25 (mol)
R + Pb(N03)2 > R(N03)2 + P b i ('i (mol) 0,025 0,025 0,025 0,025 (mol) 0,025 0,025 0,025 0,025
Theo (4) thi do tSng khoi liTcfng kim loai la:
(207 - R).0,025 = 44,575 - 40 = 4,575 (gam) => R = 24. Vay kim loai R la Mg.
DE SO 27
D E THI CHON HOC SINH GIOI HOA HOC 9 TJNH QUANG NGAI NAM HOC 2 0 0 4 - 2 0 0 5 (adsbygoogle = window.adsbygoogle || []).push({});
Cdu 1. NgUai ta thUc hien cdc thi nghiem sau:
Thi nghiem 1: Cho Ca Idn lugt vdo cdc lo chiCa cdc dung dich: KOH, CuClọ
Thi nghiem 2: Cho vdi giot quy tim vdo dung dich anioniac, ta dUo
dung dich X.
Cho vdo X mot lUgng dung dich axit clohidric chiia so mol HCl bhng mol NH3 CO trong X. mol NH3 CO trong X.
Cho til tit dung dich AICI3 den du vdo dung dich X.
Thi nghiem 3: Ngdm thanh kern vdo dung dich H2SO4 96%.