, Huong din gid
3.Nhiing m6t thanh sat nang 100 gam vao 100ml dung dich h6n hop g6m Cu(N03), 0,2M va A g N O j 0,2M Sau m6t thoi gian láy thanh k i m loai ra, rua
sach lam kh6 can duoc 101,72 gam (gia thiét cac k i m loai tao thanh diu bam het
vao thanh sSt). Khoi luong sat da phan utig la
Ạ 1,40 gam. B. 0,84 gam. C. 2,16 gam. D. 1,72 gam.
Hu&ng dan gidi
Theo b^i ra: ncu(N03)2 = 0> 1 • 0,2 = 0,02 (mol);
" A g N O , = 0 , 1 . 0,2 = 0,02 (mol).
Truofc tifen: Fe + 2AgN03 > 2Ag i + Fe(N03)2
K h i 0,02 mol A g N O j phan u-ng het thi kh6'i luong thanh k i m loai se tang: 0,02. 108 - 0,01. 56 = 1,6 (gam) < 1,72 (gam).
=> A g N O j phan ling hd't, va m6t phSn Cu(N03)2 da phan iJng:
Fe + Cu(N03)2 > Cu i + Fe(N03)2
K h i 1 mol Fe > 1 mol Cu khd'i luong tang 8 gam K h i khd'i luong tang 1,72 - 1,6 = 0,12g
=> s6' mol Fe da phan ling la 0,12 / 8 = 0,015 (mol) < 0,02.
Vay t6ng kh6'i luong Fe da phan ling la mp^ = (0,01 + 0,015). 56 = 1,40 (gam).
Dap an dung la Ạ
4. Cho m gam bdt Fe vao dung djch A g6m A g N O j va Cu(N03)2, sau khi cac phan umg xay ra hoan toan thu duoc chat ran B va dung dich nude loc C. Cho nude loc
C tac dung v d i NaOH d u thu duoc 36,8 gam kd't tda ciia 2 hidroxit k i m loai Nung ket tua trong khong khi den khoi luong khCng doi thu duoc 32 gam h6n hcjp cac oxit k i m loaị Gia tri ciia m la
Ạ 14,6 gam. B. 16,8 gam. C. 17,4 gam. D. 18,9 gam.
Huong dan gidi
Nhan xet: K h i cho nude loc C tac dung v6i NaOH du thu duoc ket tiia cua 2
hidrioxit k i m loai, nung ket tua trong khdng khi d^n kh6'i luofng khong d6i thu duoc h6n hop cac oxit k i m loai (2 oxit k i m loai khac nhau) => 2 oxit d6 la Fe20.
va CuO => 2 hidroxit la Fe(OH)2, Cu(0H)2 ^ trong dung dich C c6 Fe(N03):. Cu(N03)2 d u => A g N O j phan ufng het; Cu(N03)2 c6 thé da phan iJng m6t phSn.
Goi X , y, z Ian luot la s6' mol AgN03; Cu(N03)2 da phan img v6i Fe; Cu(N03): con du sau phan ihig vdi sat.
^ly I i^ttTnvrmJWHKhana Viet
Fe +2AgN03 X — < - X (mol) -> Fe(N03)2 + 2Ag4. X 2 X (mol) Fe +Cu(N03)2 > Fe(N03)2 + C u i y ( m o l ) > y y ( m o l ) y
Cg6m: -^- + ymolFe(N03)2; z mol Cu(N03)2 dụ
Fe(N03)2 + 2NaOH > Fe(OH)2i + 2NaN03 X Cu(N03)2 + 2NaOH - z Taco: 9 0 ^ + y + 98z =36,8 - + y (mol) -> Cu(OH)2i + 2NaN03 z (mol) (1)
4Fe(OH)2 + O2 — ^ 2Fe203 + 4H2O
X
- + y (mol)
Cu(0H)2 — ^ CuO + H2O
z z (mol) T a c o : 160. {4 2 T f i f ( l , 2 ) t a c 6 : z = 0 , 1 + 80z = 3 2 => l + y + z = 0 , 4 (2) ^ + y = 0,3 V a y m = 5 6 . = 56.0,3 = 16,8 (gam). Dap an dung la B. (adsbygoogle = window.adsbygoogle || []).push({});
5. Cho 4,48 gam bdt F e v a 200 m l dung djch h6n hop g6m A g N 0 3 0,2M va
Cu(N03)2 I M , khua'y deu de'n phan ling xay ra hoan toan, thu duoc cha't r l n A va dung djch B. Khd'i luong cha't rSn A la
Ạ 8.16 gam. B. 6,84 gam. C . 9,48 gam. D. 7,32 gam.
Huong đn gidi -,„n(
T h e o b a i r a u p . = 0,08(mol). ' \
" A g N 0 3 = 0 , 2 . 0 , 2 = 0 , 0 4 (mol); '
/•//,/// loai vciplnfcmgphdp gidi n/niiih HI lí>< 12 -Cu Ifianh loan
Cac PTPLT xay ra theo thiif tu sau:
. Fe + ZAgNO, > Fe(NO,)2 + 2 A g i
0 , 0 2 ^ 0 , 0 4 > 0,02 > 0,04 (mol).
Fe + CuCNO,): > ¥e(NOj), + Cui
(0,08 - 0,02) 0,06 > 0,06 > 0,06 (mol). ^, .
^ ncu(N03)2 (P^^) = < 0,2 (mol) => CuiNO,), dụ
Vay chat rSn A gom 0,04 mol A g va 0,06 mol Cu: m^ = 0,04 . 108 + 0,06 .64 =8,16 (gam)
Dap an dung la Ạ
Dang 7; Bai tap ve hon h(yp kim loai sat, dong tac dung vol axit HNỌ,