, Huong din gid
3. Cho 0,3 bot Cuva 0,6 b6t Fe(N03)2 vao dung dich chiia 0,9mol H2SO4 (loang) Sau khi cdc phan iJng xay ra hoan toan, thu duoc V lit khf NO ( san phdm khir
duy nhát, 6 dktc). Gia trj cua V la
Ạ 10,08 B. 4,48 C. 8,96 D. 6,72
Hu&ng dan gidi
Theo bai ra: n^^2+ = 0 , 6 ( m o l ) ; n _ = l,2(mol); n + = l,8(mol) . . 5
PTPIT: 3Cu + 8 H^ + 2 N O J >3Cu^++ 2 N O + 4 H 2 O
PIT: 0,3 ^ 0,8 ->• 0,2 ^ 0,2 (mol) Con: 0 1 1 3 F e 2 + +4H+ + N O 3 >3Fê+ + N O + 2 H 2 O ,1 P/u: 0,6 0,8->0,2 ^ 0,2 (mol) Con: 0 0,2 0,8 Vay V = (0,2 + 0,2). 22,4 = 8,96 (1ft). Dap an dung la C. 4. Cho a g i m Fe vao 100 ml dung djch h6n hop gom H N O 3 0,8M va Cu(N03)2 I M .
Sau khi cac phan lifng xay ra hoan toan, thu duoc 0,92a gam h6n hep kim loai va
D.5,6
+ 5
khi NO ^san phdm khií duy nhát ciia N ) . Gia tri ciia a la
Ạ 11,0 B. 11,2 C. 8,4
Hu&ng dan gidi
Theoba ra: nuNOj =0'08mol; ncu(N03)2 = 0 , l m o l
=>n^+ =:0,08mol; n^^2+ =0,lmol;n^^_ =0,18mol
V i sau plian ihig thu diroc h6n hop kim loai (=> h6n hop kim loai Cu, Fe du) => Fe du =i> Trong dung dich sau phan ling chiia Fe"* (Cu"* het, kh6ng c6 Fé*) PTHH: Fe + C u ^ * - > F e 2 + + C u i
0 , l < - 0 , l - > 0,1 (mol)
3Fe + 8H* + 2 N O 3 ->3Fê^ + 2 N 0 + 4 H 2 O
T a c o : 0,92a = [ a - (0,1 + 0 , 0 3 ) . 5 6 ] + 0,1.64 =>0,08a =0,88 =>a = 11,0
Dap an dung la A . 5. Cho h6n hap X gom 0,09 mol Fe va 0,05 mol FeCNO,),. VH^O vao 500 m l dung
dich H C l I M , k6't thiic phan urng thu duoc dung djch Y va khi N O (san phdm k h u duy nhat). H o i dung dich Y hoa tan toi da bao nhieu gam Củ
A . 3,84 B.4,48 C. 4,26 D . 7,04
Hu&ng đn gidi
Theo bai ra: n^Q = 0 , 5 . 1 = 0 , 5 ( m o l ) => n ^ ^ = n^^Q = 0 , 5 ( m o l )
n = 0 , 0 5 . 2 = 0,1 ( m o l ) , n , ^ = 0 , 0 5 ( m o l ) Fê+ + N O + 2 H 2 0 0,36 - > 0,09 (m ol ) 0,14 0,09 (mol) Fe + NO3 + 4H+ Phan li-ng: 0,09 0,09 Con: 0 0,01 (Tinh theo Fe v i Fe c6 kha nang het)
3Fê* + NO3 + 4H+ ^ 3Fé'^ + N O + 2H2O
Phan ling: 0,03 < - 0,01 ^ 0,04 0,03 => Trong dung djch Y c6: 0,03 + 0,09 = 0,12 mol Fé*
( 0 , 0 2 m o l F e 2 ^ ; 0 , l m o l H + ; 0 , 5 m o l C r )
2Fe-'^+ + Cu- -^2Fe2"+Cu2+
0,12 0,06 ( m o l ) Vay m c , = 0,06.64 = 3,84(g)
Dap an dung la A . 6. M o t dung dich chiia 0,02 mol Fe(N03)3 va 0,3 mol H C l . Dung dich nay c6 kha
nang hoa tan toi da s6' gam Cu la
A . 7,84 gam B. 7,2 gam C. 5,76 gam D . 6,4 gam
Hu&ng đn gidi
T h e o b a i r a : n = 0 , 0 2 m o l ; n = 0 , 0 6 m o l ; n , = 0 , 3 m o l .
Fé+ NO3 H+
PTPLT: 2Fe-^^ + Cu >2Fe2^+Cu2^
0,02 0,01 (mol)
3Cu + 2 N O J + 8 H ^ > 3Cu^+ + 2 N 0 + 4H2O
BandSu: 0,06 0,3 Phan ti-ng: 0,09 < - 0,06 0,24
=>ncu =0,01 + 0,09 = 0 , l ( m o l ) = > m c u = 6,4( g a m )
Dap aki diing la D.
214
CtyTNHHMTV DWHKfiang Viet Chu V. - K i m loai Cu chi c6 the khir NO3 trong m 6 i trucmg ve N O .
- K h i khu NO3 ve N O se hoa tan duoc nhi6u kirn loai Cu hon khi khir NO3 ve NỘ 7. Hoa tan 19,2 gam Cu vao 500 ml dung djch NaNO, I M , sau do them vao 500ml dung
djch HC l ? M . Ket thuc phan ling thu duoc dung djch X va khi N O duy nhSit, phai them bao nhieu m l dung djch NaOH 1M vao X de ket tua het ion Cu"* ?
A . 600 B.800 C.400 D. 120 v
Hu&ng đn gidi
Theo bai ra: n^^ = 19,2/64 = 0 , 3 ( m o l )
nNaN03 = 0 , 5 . 1 = 0 , 5 ( m o l) ; n H c , = 0 , 5 . 2 = l ( m o l ) = > n , = l m o l ; n = 0 , 5 ( m o l )
NO3 ^
3Cu + 8 H ^ + 2 N O ; - > 3Cu^+ + 2 N 0 + 4H2O iurf
0 , 3 - > 0 , 8 - ^ 0,2 ^ 0,3 ( m o l ) n ^ (con du) = 1 - 0,8 = 0,2 (mol)
H H + + 0 H - - ^ H 2 0 H + + 0 H - - ^ H 2 0 0 , 2 ^ 0 , 2 ( m o l ) C u 2^ + 2 0 H ~ - > C u ( O H ) 2 i 0,3 - > 0 , 6 ( m o l ) = > n ^ ^ _ = 0 , 2 + 0,6 = 0,8(1) = 8 0 0 ( m l ) Dap an dung la B. 8. Cho 0,87 gam h6n hop g6m Fe, Cu va A l vao binh dung 300 m l dung djch H2SO4
0 , 1 M . Sau k h i cac phan ling xay ra hoan toan, thu duoc 0,32 gam chat ran va c6 448 m l k h i (dktc) thoat rạ Them tiép vao binh 0,425 gam N a N O j , k h i cac phan
ling ket thiic thi the tích k h i N O (dktc, san ph^m khiJr duy nhát) tao thanh va khd'i lugng muoi trong dung djch la
A . 0,224 l i t va 3,750 gam. B. 0,112 l i t va 3,750 gam. • ] C. 0,112 l i t va 3,865 gam. D . 0,224 l i t va 3,865 gam.
Hu&ng đn gidi
Theo bai ra: nH2S04 = 0 ' 0 3 " i o ' ; "NaN03 = 0 , 0 0 5 m o l
n^^ = 0 , 4 4 8 / 2 2 , 4 = 0 , 0 2 ( m o l )
V i nH2 < nH2S04 => F e , A l bj tan het; chát rSn thu duoc la Cu j • ,
' 1 M U M l u u i vu jviaui]^ jjiiuir }(iui niiunn ui iiuu iiyi lị —K.UI iiumi t uun
n ^ (con dir) =(0,03-0,02).2 = 0,02(mol) H
Fe + H2SO4 (l)->FeS04 + H2 T
X - > X - > X
2A1 + 3H2SO4 (1)^AI2 (504)3 + ^ f >H (!>!;
y l.5y
^ , r56x + 27y = 0,87-0,32 = 0,55 ^
Taco: =:>x =0,005; y = 0,01
| x + l,5y = 0,02 Khi phan 'ing vdi NaNO, c6:
n + =0,02mol;n _ =0,005mol;ncu =0,005mol
H
n 2+ =0,005mol
Thii tu trong day dien hoa: Cu^^ /Cu;Fế^ /Fê""
3Cu + 8H+ + 2NO3 ^ 3Cu^++2NOt+4H20
P/iing: 0, 0 0 5-^0, 0 4/ 3 ^ 0,01/3-> 0,01/3(mol)
Con: 0 0,02/3 0,005/3
3Fe2* + 4H+ +NO3 ^ 3 F e ' ^ + N O T + 2 H 2 0
0,02/4<-0,02/3-^0,02/12-> 0,02/12
(Cac chat phan utig vira hd't \6\)
Vay VNO=(0,01/3 + 0,02/12).22,4 = 0,112(1)
^(...0= nipẹca,Ai +-^^a- ^""sô- = 0,87 + 0,005.23 + 0,03.96 = 3,865(g)
Dap an dung la C.
Chu v.- Cu,Fê^ (FeS04) diu bi oxi hoa boi NaNOj trong m6i trucmg axit H2SO4.