cua hai mud'i bang nhaụ Neu them vai giot quy tim vao dung dich sau khi dien phan thi m a u cua quy tim se thay doi nhif the naỏ
^ i du 8. CAn bao nhieu tan quang manhetit chufa 80% Fe304 de san xuat
diTofc 800 tan gang c6 ham liiOng sat 95%. Biet rang trong qua trinh san xuát lufgng sat bi hao hut la 1 % .
V i du 9. H o a t a n 10 gam FeS04 c6 Ifin tap chát la Fe2(S04)3 trong nude
diicfc 200cm^ dung dich, 20cm^ dung dich nay dtfOc axit hoa b^ng
H 2 S O 4 loang da l a m m a t m a u t i m cua 25cm^ dung dich K M n 0 4 0,03M.
a) V i e t phLfOng trinh hoa hoc dang ion riit gon va cho b i e t vai tro ciaa
ion Fê"" va ion M n 0 4 ?
b ) Co bao n h i e u m o l Fê* tac d u n g v d i 1 m o l M n 0 4 ?
c) Co bao nhieu mol Fê^ tac dung vcfi 25cm^ dung dich KMn04 0,03M?
d ) Co bao nhieu gam ion Fê"^ trong 200cm^ dung dich b a n daủ e) T i n h p h a n t r a m theo k h d i laong ciia FeS04 t i n h k h i e t . e) T i n h p h a n t r a m theo k h d i laong ciia FeS04 t i n h k h i e t .
V i du 10. De tac d u n g viTa dij v d i 2 0 m l dung dich FeS04 (c6 pha v a i giot
dung dich H 2 S O 4 loang) can 15,1ml dung dich K M n 0 4 0,02M. Xac d i n h
n o n g do mdri cua dung dich FeS04 va cho biet m u d n pha 1 l i t dung dich do t h i can d u n g bao nhieu gam F e S 0 4. 7 H 2 0 ?
V i du 11. H d n hcfp A chufa F e va k i m loai M c6 hoa tri khdng đi trong m o i hcfp chát. T i le so m o l ciia M va F e trong h d n hcfp A la 1 : 3. Cho 19,2 gam h d n hcfp A t a n het vao dung dich H C l t h u duTcfc 8,96 l i t k h i
H2. Cho 19,2 gam h d n hcfp A tac dung het v d i k h i C I 2 t h i can dung 12,32 l i t k h i CI2. X a c , d i n h k i m loai M va p h a n tram k h d i Itfcfng cac k i m l o a i trong h o n hgfp A . Cac the tich k h i do d dktc.
V i du 12. N u n g n o n g 16,8 gam hot sat ngoai khdng k h i , sau m o t t h d i gian t h u di/gc m gam h d n hcfp X gdm cac oxit sat va sat dif. H o a t a n het h d n hqp X bang H 2 S O 4 dac n o n g t h u dirqc 5,6 l i t S O 2 (dktc). H a y xac
d i n h khd'i l u g n g ciia h d n hcfp X?
V i du 13. Cho 4,15 h d n h c ^ Fe, A l p h a n ijfng v d i 200 ml dung dich CUSO4
0,525M. K h u a y ki hdii'hgp de p h a n ufng xay ra hoan toan. D e m loc ket tua (A) g d m h a i k i m loai n a n g 7,84 gam va dung dich nadc loc B. De hda t a n k e t t u a (A) can it nhát bao n h i e u ml dung dich H N O 3 2 M , biet
p h a n ijfng tao ra N O .
V i du 14. Cho h d n hop (Y) gdm 2,8 gam Fe va 0,81 gani A l vao 200 ml dung dich (C) chufa AgNOg va Cu(N03)2. K e t thuc p h a n ufng t h u duac
dung dich D va 8,12 gam chat r a n (E) gdm ba k i m loaị Cho (E) tac
d u n g vdri dung dich H C l dU, ta t h u duac 0,672 l i t H 2 (dktc). T i n h nong
du 15. H 6 a t a n m o t l u g n g o x i t sat FexOy b a n g d u n g d i c h H N O 3 t h u
dugc 2,464 h't k h i N O (or 27,3°C; l a t m ) . Co c a n d u n g d i c h t h u diiqc
72,6 g a m m u o i k h a n .
a) T i m cong thufc hoa hoc ciia o x i t s ^ t da dung.
b ) D S n 4,48 K t k h i CO (dktc) qua 10,8 gam oxit sit t r e n n u n g nong.
Sau k h i d i i n g p.han ufng t h u dLfgc k h i A c6 t i kho'i so vdri H 2 b a n g 18. T i n h h i e u suat cua qua t r i n h khuf oxit sSt.
du 16. H o a t a n h 6 n hgp X gom 11,2 gam k i m loai R va 69,6 g a m oxit
RxOy cua k i m l o a i do b a n g dung d i c h H C l d i i t h u dtfcfc 4,48 h t k h i H 2
(dktc). N e u cung h6a t a n h o n hgp X n h i i t r e n b a n g d u n g d i c h H N O 3
idii) t h i t h u dtrgc 6,72 l i t k h i N O (dktc). Xac d i n h cong thufc cua RxOy
du 17. H o a t a n h o a n t o a n 46,4 gam m o t k i m loai o x i t b a n g d u n g dich H 2 S O 4 dac, n o n g (vi^a d i i ) t h u dirgrc 2,24 l i t k h i SO2 (dktc) va 120 gam
m u o i . XAc d i n h cong thufc ciia k i m loai oxit.
HlfdNG DAN GIAI
du 1. P h a n ufng:
(1) 3Fe + 2O3 ^—> FeaOi
(2) Fe304 + 4 C 0 ^ — > 3Fe + 4 C 0 2 t (3) Fe + 4 H N 0 3 > Fe(N03) 3 + N O T + 2 H 2 O (4) Fe + H 2 O 1° > 570"C y YeO + (5) FeO + 2 H C l > FeClg + H 2 O (6) 2FeCl2 + C I 2 > 2FeCl3 (7) FeClg + 3 N a O H > Fe(OH)34 + 3 N a C l (8) 2 F e + 6 H 2 S O 4 d — > Fe2(S04)3 + 3SO2 + 6 H 2 O (9) Fe2(S04)3 + Fe > 3FeS04 ( 1 0 ) FeS04 + 2 N a O H > F e( 0 H ) 2 i + Na2S04 (11) 4Fe(OH)2 + O2 + 2 H 2 O > 4 F e ( O H) 3 i du 2. P h a n ufng d a n g p h a n tuf va i o n
FeS04 + BăN03)2 > Fe(N03)2 + B a S 0 4 i ^ S O f + Bâ" > BaS04
FeS04 + 2 N a O H — > Fe(OH)24 + Na2S04 => Fê* + 2 0 H • > Fe(0H) 2
1 B6| DUONG HdA HOC 12
3 F e S 0 4 + 4 H N O 3 > Fe(N03) 3 + Fe2(S04)3 + N O t + 2 H 2 O
^ 3Fê^ + 4ir + N O 3 > 3Fê* + N O T + 2 H 2 O
B a ( N 0 3) 3 + K2SO4 > BaS04 i + 2 K N O 3
=> Bâ^ + SÔ- > BaS04
Fe(N03) 3 + 4 N a O H du > SNaNOg + N a [ A l ( 0 H) 4 ]
Af^ + 4 0 H - > [A1(0H) 4 ] "
N a O H + H N O 3 > NaNOg + HgO H ^ + O H - > H2O
V i du 3. Cac phiforng t r i n h p h a n ufng:
CO + FeO ^ — > Fe + CO2 (1
0
SCO + Fe203 ^—> 2Fe + 3CO2 (2
2 24
T a c6: n^o = ^ = 0,1 ( m o l ) .
Goi X va y la só m o l ciia FeO va Fe203 c6 t r o n g h o n hcfp.
= X + 3y = 0,1
mpgo + mpg^o, = 72x + 160y = 6,64
=:> X = 0,07 va y = 0,01.
=J> n p e = X + 2y - 0,07 + 0,01 x 2 = 0,09 (mol)
==> m p e = 0,09 X 56 = 5,04 (gam)
V i du 4. P h a n ufng:
Fe^Oy + yHa ^—> xFe + y H 2 0 (1
Fe + H 2 S O 4 > FeS04 + H2 (2)
^ T a c6: n p e = n^,^^^^ = = 0,06 (mol)
M a n„ = 0,09 (mol)
H2(i) 22,4
TCr t i le p h a n ufng (1) t a c6: - = = - C o n g thufc oxit: Fe203. y 0,09 3
k^kfiil DUdNR H h A H n n 17 205
i du 5.
a) P h a n ufng: FegOg + SHg ^—> 2Fe + SHaO (1)
FeO + H 2 ^—> Fe + H 2 O (2)
G o i X , y I a n li/cft l a s6' m o l Fe203 v a FeỌ
Theo de b a i , t a c6 he phi/cfng t r i n h :
leOx + 72y = 9,6
3x + y = 2 ^ = 0,16 18
G i a i he phiforng t r i n h , t a diTOc: x = 0,034; y = 0,057
Vay: %Fe203 = ^ ^'^^"^ x 100% = 56,66% v a %FeO = 4 3 , 3 4 %
9,6
b) I B H ^ = (3x + y) X 2 = 0,159 X 2 = 0,318 (gam) 7 6 0 _ x 0 , 1 5 9 ^ 725V ^ , p ^ , , , ^ . t ) 7 6 0 _ x 0 , 1 5 9 ^ 725V ^ , p ^ , , , ^ . t )
273 273 + 17
du 6. Cac p h a n ufng c6 t h e xay ra t r o n g qua t r i n h d i e n p h a n : T a c6: nc^sô = 0.2 x 0,2 = 0,04 (mol)
2CUSO4 + 2H2O P ' ' ^ "—> 2Cu + O2 + 2 H 2 S O 4 (1)
+) N e u CUSO4 h e t t h i :
2H2O '^^^"P'^'" > O2 + 2H2 . (2) +) N e u CUSO4 b i d i e n p h a n h e t
^ ^o, (p. 1) = I " c u s o , = I -ÓÔ = 0,02 (mol)
M a theo gia t h i e t : k h i t h o a t r a d anot (la O2) c6 so m o l : = 0,1 (mol)
22400
=> CUSO4 b i d i e n p h a n chiia het.
V a y t h d i g i a n de d i e n p h a n l a :
t = ^''^Q-^^-Q'^^ = 386 giay hay 6 p h u t 26 g i a y
Vi du 7. T a c6: n^^.so, ban dAu = " K B r ban d^u = a (mol)
Thuf t\i cac p h a n ufng xay r a k h i d i e n p h a n :
CUSO4 + 2 K B r — C u + Br2 + K2SO4 (1) (mol) a/2 a
Sau (1), CUSO4 dii hay dung d i c h sau p h a n ufng (1) chufa K2SO4 v a CUSO4. K h i d i e n p h a n t i e p t h i :
2CUSO4 + 2H2O ^'^"^ P^^" > 2Cu + O2 + 2H2SO4 (2) P h a n ufng (2) tao H2SO4: m o i triJcfng a x i t n e n l a m quy t i m h 6 a dọ
Vi du 8. K h d i Itfong sat c6 t r o n g 800 t a n gang chufa 9 5 % sSt l a :
8 0 0 X 95 ,
mpe = — — = 760 (tan)
L i r g n g sdt thirc te can p h a i c6 l a : ^ = 767,68 ( t a n ) , So do chuygn hoa: Fe304 > 3Fe
( t a n ) 232 3 x 5 6
M u d n CO 767,68 t a n s^t, can k h d i liJOng Fe304 l a : 767,68 X 2 3 ^ ^
168
K h d i l u g n g q u a n g m a n h e t i t chiJa 8 0 % Fe304 can d u n g 1^:
1060,13 X 100 ,
= 1325,1625 (tan). 80
Vi du 9. a) P h u o n g t r i n h hoa hoc d a n g i o n r i i t goh:
2 M n 0 4 + lOFê^ + 1 6 H " > 2Mn^^ + lOFê^ + 8H2O Fê^ l a c h a t khuf; M n 0 4 l a c h a t o x i hdạ
ị b) Theo p h u o n g p h a p h o a hoc c6 5 m o l i o n Fê"" p h a n ufng v d i 1 m o l
i o n M n 0 4 .
c ) Ltfcfng M n 0 4 c6 t r o n g 25cm^ d u n g d i c h K M n 0 4 0 , 0 3 M la: 0,03 X 0,025 = 0,00075 (mol) 0,03 X 0,025 = 0,00075 (mol)
L u g n g Fê* tac d u n g h e t v d i lircfng K M n 0 4 t r e n l a :
0,00075 X 5 = 0,00375 (mol)