- Hoatan het phd n2 bang dung dich HNO3 dac, nong (du) tháy c6 11,76 lit khi bay ra (dktc).
7. Ch om gam h6n ho pX gom Ba va Al (trong do Al chiem 37,156% ve khoi luong) tac dung vdi H O d u thu dugc V lit khi H (or dktc) Mat khac, neu
cho m gam h6n hop X tren tac dung vdri dung djch NaOH du thi thu dugc 12,32 lit H , (o dktc). Gia tri ciia m va V Idn lugt la
Ạ 19,1 va 10,08. ' B. 21,8 va 8,96. C. 19,1 va 9,408. D. 21,8 va 10,08.
Huong đn gidi
Theo bai ra: nH2 = 12,32/22,4 =0,55 (mol)
* PTPl/: X + dung djch NaOH du:
Ba + 2H20 -> Bă0H)2 + H ^ t
62,844. m 62,844.m 100.137 100.27
A l + NaOH + 3H.O NaAl(OH)4 + - H ,
2 ^ 37,156.m 1,5.37,156m 100.27 100.27 „ . 62,844m 1,5.37,156m „ Taco: + =0,55 ' 13700 2700 = > m = 21,8
Suy ra: n^^ = ^ = 0 , l ( m o l ) ; n ^ , = i l^ ^ M ^ = 0 , 3 (mol)
* PTPU": X + H , 0 du: i^ / ^ i ' :
Ba + 2A1 + 8H2O -> B a [ A l ( 0 H ) 4 ] , + 4 H , t >E11
0,1 0,3
0,1 0,2 0,4 (mol)
Vay V = 0,4 . 22,4 = 8,96 (lit) ' ' '
Dap an diing la B. 8. Cho 0,5 gam h6n hgp X gom L i , Na, K vao nu6c thu dugc 2 lit dung djch c6 pH = 12.
Tr6n 8 gam h6n hgp X va 5,4 gam bgt A l roi tha vao nudfc den phan li-ng hoan toan C O V lit khi thoat ra (dktc). Gia tri cua V la ^ ' *
Phdn loai 1 <i /hhniin.; •>li,in ỵi<ii nhanh BT Hoa hoc 12 —Cu Thanh Tocin
Huong dan gidi
C a c k i m loai L i , N a , K la cac kim loai kidm, ki hieu chung la M .
M + HOH -> + OH" + 1/2 H , t p H = 12 ^[H'] = IQ-'-M ^ [OH"] = 10"-M. V i V = 2 lit =^ n ^2.10"^ = 0,02 (mol) ' OH" (1) = > V H ^ C I ) =0,02. 1/2 =0,01 (mol)
Ta c6: 0,5 gam X 0,02 mol OH" va 0,01 mol H , 8 gam X -> a mol OH" va bmol H ,
^ a = 8. 0,02/0,5 = 0,32 mol b = 8.0,01 / 0,5 = 0,16 mol.
Al + OH" + 3H2O ^ [A1(0H)4]" + 3/2H2 *
0,2 0,32
0 , 2 - > 0,2 0,3 (mol)
Vay = (0,16 + 0,3) . 22, 4 = 10,304 (lit) " '
Dap an dung la D. 9. Hoa tan hokn loan 8,1 gam A l bang dung djch chiia m gam NaOH thu duoc
dung djch X. Cho 900ml dung dich HCl I M vao X thu duoc 15,6 gam k6't tuạ Gia tri 16n nhát ciia m la
Ạ 32 B. 60 C.40 D. 24
Hu&ng đn gidi
Theo bai ra: n Al = 8,1 / 27 = 0,3mol;
H H C I =0,9.1 = 0 , 9 ( m o l ) ;
két tua thu duoc la A l ( O H) 3: n^^oHj^ = 15,6/78 = 0,2mol
2A1 + 2NaOH + 6 H 2 O -> 2Na
0,3 0,3 0,3 (mol)
Al(OH)^ + 3H,
=>Xchiira N a [ A l ( O H ) ^ ] ( 0 , 3 m o l ) vaNaOHdu(x mol) NaOH + HCl ^ NaCl + H j O NaAl(OH)4 + HCl -> A1 ( 0H) 3 + NaCl + H 2 O 0,2 < - 0,2 <- 0,2 = ^ x + 0 , 2 =0,9 = > x = 0 , 7 Vay m „ , , = (x + 0,7).40 = (0,3+ 0,7).40 = 40(g) Dap an dung la C.
Chu y: Gia trj nho nhát ciia m khi: NaOH + HCl -> NaCl + H 2 O N a A l( O H ). + HCl ^ A l( O H ) . i +NaCl + H 2 O ' -I:; 0,3 -> 0,3 -> 0,3(>0,2) A1( 0 H) 3 + 3HC1^A1C13+3H20 ( 0 , 3 - 0 , 2 ) - > 0,3 = ^ x + 0 , 3 + 0,3 = 0,9=>x = 0,3 =>ni„in = ( x + 0,3).40 = (0,3 + 0,3).40 = 24(g) ^ 10. H6n hop X g6m hai kim loai K va Al c6 khoi luong 10,5 gam. Hoa tan hoan
toan h6n hop X trong nude duoc dung djch Ạ Thdm tir tCr dung djch HCl I M vao dung djch A: luc đu khdng c6 kfít tua, khi th^m 100ml dung djch HCl I M thi bat dSu CO ket tiiạ Tinh thanh phdn % s6 mol ciia cac kim loai trong X.
Hu&ng đn gidi
Dat X, y Idn luot la s6' mol K, A l trong h6n hop.
2K + 2 H 2 O > 2 K 0 H + H2 T ( 1 )
2A1 + 2 K 0 H + 2 H 2 O > 2 K A I O 2 + 3H2 t (2) y ^ y
Do X tan het => A l hd't, KOH du sau (2) Khi them HCl, ban đu chua c6 k6't tua vi:
HCl + KOH(du) > KCl + H 2 O ( 3 )
(x-y) < - (x-y) m o l
Khi HCl trung hoa het KOH thi bat diu c6 két ttia tao r a .
K A I O 2 + HCl + H j O > A l ( O H) 3 i + KCl (4)
Vay dé trung hoa KOH du c^n 100ml dung djch HCl I M
Ta c6: nHci = nKOH(3) = x - y = 0,1.1 = 0,1 ( m o l ) (a) ; ., j . , . ^
Mat k h a c: 3 9 x + 27y= 10,5 (b)
T i r ( a, b ) g i a i r a: x = 0,2;y = 0,l
Vay % s6' m o l c a c k i m loai trong h6n hop X:
^ 0,2.100% , : ' %n^^^-^ = 66,67% > * ^ 0,2 + 0,1 %nA, =100-66,67 = 3 3 , 3 3 % . .<rj .f. 113
Dang 7: Bai tap ve tinh luong tinh cua nhom oxit, nhom hidroxit