C. Zn D Mg Ạ Al B Bạ
Al +H2O + NaOH NaA10 2+ H 2t (3)
H 2 O > H 2 T + Ozt (2)
0,05 -> 0,025 mol 1 12 1 12
n,i^ = - T — = 0,05 (mol) =>n(ô thoatrad anot trong (2)) = 0,025 mol ncu(catot) = 2. nCOz thoat ra & anot trong (1)) = 2.(0,05 - 0,025) = 0,05 mol. Vay: mcu(catot) = 0,05.64 = 3,2 gam.
Chon dap an B.
Bai 2: {BH B - 2009): Dien phan c6 mang ngSn 500 ml dung dich chufa
hon hop gom CuClz 0,1M va NaCl 0,5M (dien cifc trcJ, hieu suat dien
phan 100%) vdi cUdng do dong dien 5A trong 3860 giaỵ Dung dich
thu dufcJc sau dien phan c6 kha nang hoa tan m gam Al. Gia t r i Idn nhat cua m la . nhat cua m la .
Ạ 4,05. B. 2,70. C. 1,35. D. 5,40.
Hiiofng dan gidi
Cdch 1: n^M, = 0,5.0,1 = 0,05 (mol); nNaci = 0,25 (mol). Phifcfng trinh dien phan: Phifcfng trinh dien phan:
2NaCl + CuS04 "P'" > Cu + Na2S04 + Cht (D
0,1 <- 0,05 0,05mol Alt Alt
Ap dung cong thiJc Faraday: m =
96500n
rvux- • A-^ 1,- , 96500.2.(64.0,05)
Thai gian dien phan (1) la t, = = 1930s . 64.5 64.5
Sau phan ilng dtf NaCl: n(Nacidu) = 0,15 mol.
2NaCl + 2H2O "P"" ) 2NaOH + C l 2 t + H 2 t (2)
Al + H2O + NaOH NaA102 + - H 2 t (3) 2 2
Thdi gian dien phan (2) la ta = 3860 - 1930 = 1930 giay 5.1930 ,
Thdi gian dien phan (2) la ta = 3860 - 1930 = 1930 giay 5.1930 , Vay mAi = 0,1.27 = 2,7 gam.
Chon dap an B.
Cdch 2: nc„ci, = n^,^,. = 0,5.0,1 = 0,05 mol
nNaci = 0,25 mol
Y,n^y = 0,05.2 + 0,25 = 0,35 mol
So mol e trao đi of dien cUc la:
Ịt 5.3860 np = np = 96500 96500 d catot (-) = 0,2 mol Cu^^ + 2e Cu 0,05 0,1 0,05 2H2O + 2e -> 0,1 -> d anot (+) 2Cr ^ C I 2 + 2e 0,2 <r- 0,2 mol H 2 T + 20H- 0,05 0,1 mol
Sau phan iJng ta thu diTgfc 0,1 mol OH" nen ta c6 pt sau: Al + H 2 O + OH- AlO- + - H a t Al + H 2 O + OH- AlO- + - H a t
0,1 <- 0,1 mol Vay mAi = 0,1.27 = 2,7 (gam) Vay mAi = 0,1.27 = 2,7 (gam) Chon dap an B.
Bai 3: Dien phan hoa toan 2,22 gam muoi clorua kim loai 0 trang thai nong chay thu duoc 448 ml khi (d dktc) d anot. Kim loai trong muoi la: nong chay thu duoc 448 ml khi (d dktc) d anot. Kim loai trong muoi la: Ạ Na B. Mg C. K D. Ca
f
Hitdng dan gidi
MCỊ, 0,04 0,04 0,448 22,4 (Ipđ = 0,02 mol n n M -> M + - C I 2 <- 0,02 mol 2,22 2,22.n MCL 0>04 0,04 = 55,5.n n V(5i n = 2 thi M la Ca s Chon dap an D.