- Hoatan het phd n2 bang dung dich HNO3 dac, nong (du) tháy c6 11,76 lit khi bay ra (dktc).
8. Hoatan hoan toa nm gam h6n hgp g6m NajO va AUOj vao nu6c thu dugc dung djch X trong su6't Thdm tCr tir dung dich HCl I M vao X, khi het 100 ml thi bat
đu xu^t hidn kd't tiia; khi het 300 ml hoac 700 ml thi d^u thu dugc a gam ket tiiạ Gia trj ciia a va m Mn lugt la
Ạ 23,4 va 56,3. B. 23,4 va 35,9. C. 15,6 vk 27,7. D. 15,6 va 55,4.
Hu&ng dan gidi
Theo bai ra: n^Q = 0 , l m o l;nHQ =0,3mol; n^Q = 0 , 7 m o l . Dat s6' mol NajO; AI2O3 Mn lugt la x, y (mol).
fhdn loai vdphuan^plúip i^idi nhanh HI Hoa hoc 11 — Cii I nan/i i oan
N a j O + H 2 O X -
-> 2NaOH
2NaOH + A I 2 O , + 3 H 2 O - > 2 N a r A l( O H ) ^
=> Dung dich X chi c6 Na Al(OH)^ hoac c6 dong thai Na A l ( O H)^J va NaOHdụ
Theo bai ra, khi cho HCl vao X chua tao ra ket tiia ngay = > X c 6 N a O H ( v a N a r A l ( O H ) 1 ) .
Trong X c6 2y mol N a r A l( O H ) ^ l va (2x - 2y) mol Nạ * De thu duoc ket tiia can trung hoa het NaOH trong Y:
NaOH (du) + HCl ^ NaCI + H.O 0,1 <- 0,1
= > 2 x - 2 y = 0,1 = > x - y = 0,05 ( l ) (S6' mol NaOH trong X bang 0,1 mol)
* De thu dugc a (gam) kát tiia vdi lugng HCl ft nhat; NaOH + HCl ^ NaCI + H j O
0,1 ^ 0,1
HCl + NaAl(OH)^ -> Al(OH)^ i + N a C l + H j O (0,3-0,1) > 0,2
=>a = 0,2 molAl(OH)3 i
* De thu dugc a (g) ket tua \(s\g HCl Idni nhat: NaOH + HCl-> NaCI+ H 2 O
0,1 -> 0,1
NaAl(OH)^ + HCI ^ Al(OH)3 i + N a C l + H 2 O
• ' 0,2 <- 0,2 <- 0,2
NaAl(OH)^ + 4 H C I -> NaCI + AICI3 + 4 H 2 O
0,1 <- ( 0 , 7 - 0 , 1 - 0 , 2 ) => 2y = 0,2+ 0,1 =0,3 =>y = 0,15 (2) T i r ( l , 2 ) = > x = 0 , 2 ; y = 0,05 Vay a = 0,2.78 = l5,6(g) m=0,2.62 + 0,15.l02 = 27,7(g) Dap an dung la C
p_ Hoa tan het m gam A1,(S04)3 vao nude dugc dung dich A . Cho 300 ml dung dich NaOH 1M vao A , thu dugc a gam ket tiiạ Mat khac, neu cho 400 ml dung dich NaOH I M vao A , cung thu dugc a gam ket tuạ Gia trj ciia m la
Ạ 21,375 B. 42,75 C. 17,1 D. 22,8
Hu&ng đn gidi
* Thinghiem 1: UNaOH = ^ ' 3 1 = 0,3(mol)
Trong thf nghiem nay: NaOH hd't, AI2 ( 8 0 4 ) ^ (dir). '' ' 6NaOH + AI2 ( 5 0 4 ) 3 ^ 2Al(OH)3 i +3Na2S04 -
0,3 0,1 (mol) J / ' =>a = 0 , l . 7 8 = 7,8(g)
* Thi nghiem 2: n^aOH =0,4.1 =0,4(mol)
Trong thi nghiem nay: NaOH, Al2( S 0 4 ) deu h e t ; A l( 0 H ) 3 bi hoa tan mot phSn.
6NaOH + AI2 ( 5 0 4 ) 3 2 A I ( 0 H)3 i +3Na2S04 6x < - X -> 2x(mol) A l ( O H) 3+ N a O H ^ N a [ A l ( O H ) / y -> y(mol) Ta c6: j ^ ^ ^ ^ " ^'"^ x = 0,0625 (y = 0,025) [ 2 x - y = 0,l ' \^ ' J vay m = 0,0625.342 = 2l,375(g) Dap an dung la A .
10. Nho tir tir 0,25 lit dung djch NaOH 1,04M vao dung djch g6m 0,024 mol FeClj; 0,016 mol Al2(S04)3 va 0,04 mol H 2 S O 4 thu dugc m gam ket tiiạ Gia tri cua m la 0,016 mol Al2(S04)3 va 0,04 mol H 2 S O 4 thu dugc m gam ket tiiạ Gia tri cua m la Ạ 4,128 B. 1,560 C. 5,064 D. 2,568
Huong đn gidi
Theo bai ra: HNaOH =0,25.1,04 =0,26(mol) . •
H2SO4 + 2 N a O H -> Na2S04 + 2 H 2 O '
0,04 0,08
FeCl3 + 3NaOH Fe(0H)3 i + 3NaCl - - ' ^
0 , 0 2 4 ^ 0,072 0,024 ,< ^ -
Al2( S 0 4)3 + 6NaOH -> 2A I( 0H) 3 i + 3Na2S04 , 0,016 ^ 0,096 - ) . 0,032
Than locii / Jin-óín: iTfiapfildi nluin/i tSl Hoa Hoc IZ -Cu I hanh I ogrT Al(OH)3 + NaOH 0,012 N a AI(OH)^ '0,26 - (0,06 + 0,072 + 0,096)" ^ = • " F e( O H) 3 i + ' " A , ( O H) 3 4 (sau) = 0-024.107 + (0,032 - 0,012).78 = 4,128(g) Dap an dung la A .
11. Hoa tan hoan toan 47,4 gam phen chua KAl(S04)2.12H2O vao nudrc, thu duoc dung djch X. Cho toan b6 X tac dung vdi 200 ml dung djch BăOH)2 I M ,