- VI, Fe co nd => AgN03,Cu(N03)2 phai phan ling hét Dodo dung dic hX chua A1(N03)3 va Fe(N03)
1. De hoatan hoan loan 2,32 gam h6n hop gom FeO,Fe304 va FeA (trong do so mol FeO bang s6' mol FeA) cdn dung vira du V lit dung dich H Q I M Gia tr
cua V la
Ạ 0,08. B.0,16. C.0,18. D . 0,23.
Huong đn gidi
, Vi s6 mol FeO bang s6' mol Fe^O,, do do c6 the quy d6i h6n hop thanh Fe304
(FeỌ F e P j ) :
56' mol Fe304 = 2,32 / 232 = 0,01 mol
PTPIT: Fe304 + 8HC1 > FeCU + IFeCl, + 4H2O
0,01 - > 0,08 (mol)
Suy ra: V, , H c i - 0,08 (1ft)
Dap an diing la Ạ 2, Cho 9,12 gam h6n hop g6m FeO, FêO,, Fe304 tac dung vdi dung djch H Q (du).
Sau khi cac phan ling xay ra hoan toan, duoc dung djch Y; c6 can Y thu duoc 7,62 gam FeCl. va m gam FeCl,.
Gia trj cua m la
Ạ 9,75. B. 8,75. C. 7,80. D. 6,50.
Hu&ng đn gidi , ^
Theo bai ra: So mol FeCl, = 0,06 mol. ^ j ^ j -u'Jr; ^
VI Fe304 = FeỌ Fe203, nen c6 thá coi h6n hop g6m FeO (x mol) va Fe203 (y mol). Theo bai ra ta c6: 72x + 160y = 9,12 (1) PTPLT: FeO + 2HC1 > FeCU + H^O
X X Fe203 + 6HC1 > 2FeC\, + 3H,0 Fe203 + 6HC1 > 2FeC\, + 3H,0 y 2y X = 0,06 (mol) (2) Tir ( 1 , 2) ta c6: x = 0,06; y = 0,03 (mol). vay m = 2. ỵ 162,5 = 2. 0,03. 162,5 = 9,75 (gam). Dap an dung la Ạ 3. Nung h6n hop ran gom a mol FeC03 va b mol FeS^ trong binh kin chua khong
khi (du). Sau khi cac phan \xng xay ra hoan toan dua binh ve nhiet d6 ban đu,
thu duoc chat ran duy nhat la Fe203 va h6n hop khị Biet ap suát khi trong binh trudc va sau phan irng bang nhau, mdi lien he giSa a va b la (biet sau cac phan
ung, luu huynh a mdc oxi hoa +4, the tich cac chat rdn la khdng dang ki)
Ạ a = 0,5b. B. a = b. C. a = 4b. D. a = 2b. jsiil
Hu&ng đn gidi
CacPTPirxayra: >r,afiiil>
4FeC03„, + O2,,, > 2Fc,0,,„ +
a (mol) -> — (mol) a (mol)
4
4FeS2,,, + 1102(k, > 2Fe203„, + 8S02(k) b ( m o l ) ^ — ( m o l ) 2b (mol)
4 »">
V i V , T = const, ma P (trudc) =P(sau) => njkh,-, = "(khf, sau)-
r^ a l i b Do do: - + = a + 2b
r : > a + l l b = 4a + 8b=>3a = 3 b=>a = b.
Dap an diing la B. 4. Chat phan ling vdri dung dich FeClj cho k6't tua la
Ạ C H , N H , . B.CH,COOH.
C C H 3 O H . D. C H 3 C O O C H ,
Hu&ng đn gidi
Dung djch F e Q j phan ung \6\, (metyl amin) tao ra kd't tua Fe(OH)3:
PeCl, + 3CH3NH, + 3H2O Fe(OH)3i + 3CH3NH3CI
Dap an dung la Ạ
Chu V. Cac ankyl amin bac I (RNH,, R la g6'c ankyl) c6 tinh bazo manh hon
amoniac (NH3):
R N H , + H O H RNH3* + O H "
NH3 + H O H N H / + 0 H "
5. M6t loai quang chura sat trong tir nhidn da duac loai bo tap chát. Hô tan quang nay trong dung djch HNO3 thay c6 khi bj hoa nau trong kh6ng khi bay ra, dung djch thu diroc cho tac dung vdi dung djch BaCl, tháy c6 kfít tua trSng. Loai quang sat do la
Ạ hematic. B. manhetit. C. xiderit. D. pirit sat.
Hu&ng đn gidi
Quang sat trong tu nhiSn, g6m c6: - Quang hematit do: Fe203 khan. - Quang hematit nau: F e j O j . nH^O - Quang manhetit: Fe304. - Quang xiderit: FeC03. - Quang pirit sat: FeS2.
Theo bai ra: \ + Quang sat — > khi bj hoa nau trong kh6ng khị
=> khi do la NO (2NO + O, 2NO2 nau)
=> phan ling giSa quang sat \6\3 la phan ihig oxi hoa - khu
=> sat trong quang t6n tai đri dang hop chat sat (II).
, *HNO, ^ + Ha CI 2 ^ I ,
+ Quang sat ' dung djch * trang. =::> ket tiia trang c6 thé la BaS04
=> quang sat la hop chát ciia sdt vdi liru huynh. Vay quang sat do la pirit sat (FeS,).
Dap an dung la D.
6. Cho 19,3 gam h6n hgfp b6t g6m Zn va Cu c6 ti Id mol tuomg utig la 1: 2 vao dung djch chu-a 0,2 mol FcjCSÔ),. Sau khi cac phan ling xay ra hoan toan, thu duc»c m djch chu-a 0,2 mol FcjCSÔ),. Sau khi cac phan ling xay ra hoan toan, thu duc»c m gam kim loaị Gia trj cua m la ' T > ^
Ạ 12,80 B. 12,00 C.6,40 D. 16,53
Hu&ng đn gidi ^ '-'S
Dat n^n = X =:> n^u = 2 x ( m o l ) .
Taco: 65x + 64.2x = 19,3 => 193x = 19,3 =^ x - 0,1 PTPÚ (theo thii tu):
Zn + Fe2 ( 8 0 4 ) 3 0,1 -> 0,1 0,1 -> 0,1
->ZnS04 +2FeS04
Cu + Fej ( 8 0 4 ) 3 >>CuS04+2FeS04 0,1 <- ( 0 , 2 - 0 , 1 )
nQi(du)= 2x - 0,1 = 2. 0,1 - 0,1 = 0,1 (mol)
=>m = mcu(dí) =0,1 .64 = 6,40 (gam).
Dap an dung la C. 7. Cho m gam b6t Zn vao 500 ml dung djch Fe2(S04)3 0,24M. Sau khi cac phan ling
xay ra hoan toan, khd'i luong dung djch tang them 9,6 gam so vdi khd'i luong dung djch ban dSụ Gia trj cua m la
Ạ 20,80. B. 48,75. C. 32,50. D. 29,25.
Hu&ng đn gidi
Theo bai ra: Hp^^^sQ^j^ =0,5.0,24 = 0,12(mol)
Trudc tien: Zn + Fe2 (SO4 )^ (đ) -> ZnS04 (đ) + 2FeS04 (đ)
0,12 <-0,12 (mol) -> 0,24
Néu chi CO phan ihig nay thi nijj^,gj = m^n
TheoPTHH: m^^ =0,12.65 = 7 , 8 ( g ) < 9 , 6 ( g )
=> Phai CO them PTHH sau:
Zn + FeS04 ZnS04 + Fe i x -> X -> x ^ x ( m o l ) Taco: 6 5 x - 5 6 x = 9 , 6 - 7 , 8 = 1,8 => 9x = 1,8 => X = 0,2 (<0,24 => thoa man) v a y m = (0,12 + 0,2).65 = 20,80(g) D^p an dung la A . ' Of; it . 225
8. Co 32 gam h6n hop gom Fe304, ZnO, CuO tac dung viTa du vdi 300 ml dung djch H2SO4 2M. Khdi luong mu6'i thu duoc la
Ạ 60 gam. B. 80 gam. C. 85 gam. D. 90 gam.
Hu&ngđngidi
Theo bai ra: So mol axit H2SO4 = 0,3 . 2 = 0,6(mol).
So do phan ling: Oxit kim loai + axit H2SO4 (1) ^ muO'i + nirdc.
- Theo dinh luat bao toan nguyen t6' (ap dung cho nguydn t6' hidro)
=> nH 2 o = 0,6 (mol)
- Theo djnh luat bao toan khd'i luong: m„^j, + m„i, = m^^, + m,,^^ o 32 + 0,6.98 = m„„,i + 0,6.18 m„„,i = 80 (gam)
i *5' Dap an diing la B.
9. Khi nung h6n hop cac chát Fe(N03)2, Fe(OH)3 va FeCOj trong khong khi den khdi lirong khdng d6i, thu duoc m6t chát rSn la
A.Fe304. B. F e O . C. Fẹ D. FẹOạ ,
Huong dan giai
PTHH ciia phan iJng xay ra khi nung Fe(N03)2, Fe(0H)3, FeCOj la: 2Fe(N03)2 FêOj + 4NO2T + 1/202!
§ 2Fe(OH)3 Fe,03 + SH^Ot
2FeC03 + 1/20, (kk) > FêOj + 2 0 0 , t Vay cha't ran thu duoc la FcjOj (nau do).
Dap an dung la D. 10. Cho 100 ml dung dich A g N O j 2a mol/1 vao 100 ml dung djch Fe(N03)2 a mol/1.
Sau khi phan ting ke't thiic thu duoc 8,64 gam chát ran va dung djch X . Cho dung djch HCl du vao X thu duoc m gam kd't tiiạ Gia tri ciia m la
Ạ 11,48. B. 22,96. C. 17,22. D. 14,35.
4 Huong dan giai
Theo bai ra: n^gNOj =0.2a(mol); np^(N03)2 =0.1a(mol)
Chát ran thu duoc la Ag: n^g = 8,64/108 = 0,08(mol) AgNOj + Fe (NO3 \> Fe(NO3 )3 + Ag i Banđu: 0,2a 0,1a
Phan iJng: 0,1a < - 0,1a "m«ort 0,1a
Con: 0,1a 0 0,1a => Dung djch X chiia 0,1a mol AgN03; Fe(N03)3
0,1a = 0,08 a = 0,8 226
y^iy 1111111 ivii V uvvH KhanJVTet
AgNOj + HCl -> AgCl i +HNO3
(0,1.0,8) (0,1.0,8) ,
=>m = '"Agca=(0'10-8)-143,5 = l l , 4 8 ( g ) .''^-H^^ "
Dap an dung la Ạ