X ^ X X (mol)
=> Trong dung djch X c6 x mol HNO3 va (0,2 - x) mol AgNOạ Gia sir kim loai Fe h6't thi chat rSn la Ag Gia sir kim loai Fe h6't thi chat rSn la Ag
= > n A g =22,7/108 = 0,21 mol > n^gNOj => v6 lỵ Vay kim loai Fe con du
Dodo: Fe + 2AgN03 -> Fe(N03)2 + 2Ag 4. 0,5.(0,2-x)<-(0,2-x)-> (0,2-x)mol 0,5.(0,2-x)<-(0,2-x)-> (0,2-x)mol 3Fe + 8HN03-^3Fe(N03)2+2NO + 4 H 2 O
3 x / 8 « - x (mol) " Chat ran g6m: (0,2-x)mol Ag va 0,3-0,5.(0,2-x)-3x/8=(l,6 + x)/8mol Fe Chat ran g6m: (0,2-x)mol Ag va 0,3-0,5.(0,2-x)-3x/8=(l,6 + x)/8mol Fe
Theo bai ra, ta c6: (0,2 - x ) . 108 +56.(1,6+ x)/8 = 22,7 ^Ufm
=> 21,6 - 108X + 11,2+ 7x = 22,7 => lOlx = 10,1 x =0,1
v , y, = i ; ^ = 5 i i^ = , , o o( h ) " * '
1.3600 2,68.3600
Cdu 96: Dung dich c6 pH = 13 =; =>n _ =3.10"'=0,3(mol) =>n _ =3.10"'=0,3(mol) OH ' V y = 10-'^M: OH" I D a p a n d u n g l a B . = I O - ' M ....'^ M+ H 2 0 M ^ + O H" + - H 2 t 2 ^ 0,3(mol) 0,3(mol) <- =>M = 10,1/0,3 = 33,67
vay hai kim loai kiem la Na (M = 23) va K (M = 39)
Cdu 97: Ta c6: n c = - ^ ^ = - = l,5 "hh 2 "hh 2
Trong '.i6n hop khi c6 CH4 (Y la CH4)
Dat X la C„ H 2 n - 2( x m o l ) va C H 4( y mol) ^ 3 n - n ^ 3 n - n D a p a n d u n g l a C. C„H n " 2 n - 2 + O2 ^ n C 0 2+ ( n - l) H 2 0 \> nx (mol) C H 4 + 2O2 - > C O 2 + 2 H 2 O y -> 2y -> y (mol) x + y = 2 nx + y = 3 l,5nx-0,5x + 2y = 4,5 =>x = l;y = l;n = 2 ( C 2 H 2 )
vay c6ng thiic cira X, Y la CjH^ va CH4
Theo b^i ra, ta c6 he:
Cdu 98: Phuomg trinh didn phan:
CUSO4 + 2NaCl C u i + CI2 t + Na2S0 4 ,,„ .
q( m o l ) 2a(niol) catot anot
A n o t ( + ) : 2 C r ^ C l 2 T+2.1e .-.v^!\,,,
C a t o t( - ) : C u 2 * + 2 e ^ C u i
K h i Cu-* hfét (CUSO4 hat => NaCl cung het) thi phuang trinh dien phan xay ra:
' 2 H 2 O 2 H 2 T + O2 t
•' ' ' " catot anot •
Vay khi thu duoc a anot la CI2 va O, (dáu vet) Dap an diing la B.
Cdu 99: Theobai ra: n^.^ =0,6{mo\)\nQQ^ = 0 , 3 ( m o l )
* M + NaHCOj (du): Chi c6 CH,COOH phan ung.
C H 3 C O O H + NaHCOj ^ CHjCOONa + CO2 T +H2O
0,3 <r- 0,3 ( m o l )
* M + Na:
2CH,COOH + 2Na ^ 2CH3COONa + H2 T
0,3 -> 0,3 0,3 0,15(mol)
mcH3COONa =0.3.82 - 24,6(g)
Theo dinh luat bao toan khoi lucmg ta c6: m^^ + m^^ = m,, ^ + nij^^
= ^ m n , = 2 7 , 4 + 13,8-40,65 = 0,55(g) =>nH2 = 0 , 2 7 5 ( m o l ) V i 2.np = 0,55 <nNa = 0 , 6 =>Na dụ
Dat cong thiic chung cua hai ancol ROH 2 R O H + 2Na ^ 2RONa + H2 t
0,25 <- ( 0 , 2 7 5 - 0 , 1 5 ) ( m o l )
mROH = 2 7 , 4 - 0 , 3 . 6 0 = 9,4(g) =>MROH =9,4/0,25 = 37,6 vay hai ancol la CH3OH ( M = 32) va CjH^OH ( M = 46).
Dap an dung la B.
Cdu 100: K h i tSng nhiSt d6 thi t i khdi ciJa h6n hofp khi giam
=> t6ng so mol k h i tang Idn (vi khoi luong h6n hofp kh6ng thay d6i)
can bang trdn chuyen djch theo chi6u nghich.
vay phan u-ng nghjch thu nhiet (=> phan iJng thuan toa nhidt), can bang chuyen djch theo chi6u nghich khi tang nhifet dọ
.^k ^, , . «w ' " t t , Dap an dung la Ạ
M W C L U C
Chuomg 5. Dai cuomg ve kim loai
Ạ Li thuyet cff ban 3
B. Phuang phdp gidi cdc dang bdi tap 9
Chuomg 6. K i m loai kiem - kim loai kiem tho va nhom
Ạ Li thuyet ca ban 74
B. Phumg phdp gidi cdc dang bai tap 84
Chuong 7. C r o m - sit - dong va mot so kim loai khac
Ạ Li thuyet ca bdn 168
B. Phuang phdp gidi cac dang bdi tap 174
Chuong 8: Nhan biet mot so chat v6 co va chuan do dung dich t^ m / v
Ạ Li thuyet ca bdn 168
B. Phuong phdp gidi cdc dang bdi tap 174
Chuomg 9: Hoa hoc va van de phat trien kinh te, hoi va moi truomg
Ạ Li thuyet ca bdn
B. Phuang phdp gidi cdc dang bdi tap ••
168 174