* Kim loai manh + HNO3 loang > san ph^m Ichii' N.O, N j , NH4NO3:
4Zn + lOHNOj,,, > 4Zn(N03), + NH4NO3 + SH.O
* Dac thii phuong phap giai bai tap dang nay la phuong phap bao toan electron.
Ngoai ra c6 the sir dung ket hop cac phuong phap bao toan kh6'i luong, bao toan didn tich, bao toan nguyen to.
* Thi du: Cho 2,16 gam Mg lac dung vo^ dung dich HNO, (du). Sau khi phan ung xay ra hoan toan thu duoc 0,896 lit khi NO (6 dktc) va dung dich X. Kh6'i luong muoi khan thu duoc khi lam bay hoi dung dich X la
A. 8,88 gam. B. 13,92 gam. C. 6,52 gam. D. 13,32 gam.
(Trich De thi tuyen sink DH - CD khoi B) Phantich: . ^ . . . c , „ , „ u . . i •
- Dau hieu de' nhan ra bai nay c6 kha nang c6 san ph£m khir NH4NO3: , ,
+ Kim loai manh (Mg) + HNO3
+ Di bai khong khing dinh NO (chat khi) la san phdm khir duy nha't.
- NH4NO3 la mu6'i tan trong dung djch ntn trong khi tinh khdi luong mu6'i thu duoc se bao gom muoi nitrat cua kim loai va muoi amoni nitrat NH4NO3.
- Gia sir chi c6 san phdm khir la NO, dua vao dinh luat bao toan electron ta thay s6' mol electron cho (Mg cho) > s6' mol electron nhan, nen bu6c phai c6 them san phdm khir khac, turc la c6 san phdm NH4NO3.
Theo bai ra: nwg = ~~- = (mol); =
24 = 0,04 (mol)
Xay ra cac qua trinh:
- 2e
fed,*/
+2
Mg Mg
0,09 0,18 (mol) 0,09 (mol)
=>2:n„M,cho, =0,09. 2 = 0,18 (mol).
+5
N + 3e 0,12 (mol)
+2
N 0,04
+5
=> En, ( N nhan) = 0,04. 3 = 0,12 (mol).
Suy ra: En, > Sn, „,hj„, => c6 san phdm NH4NO3
N + 8e > N £ ;
' ' ' (0,18-0,12) 0,0075 (mol) cM'^-iiU,-'mi-^.,r^
Vay khdi luong chat rdn khan thu duoc:
m = mMg,N03)2 + ' " N H 4 N 0 , =0,09.148 +0,0075.80= 13,92 (gam).
Dap an diing la B.
C6ng thirc de giai nhanh cac bai tap dang nay:
C6ng thu-c tinh s6 mol HNO3 cdn dung de' hoa tan m6t h6n hop cac kim loai:
nHN03 = 4 n N o+ 2 n N 0 2 + 1 0 n N 2 o + • 2 n N 2 + ' ^ " N H4N0 3
2, Cac thi du minh hoa:
ffii du 1: Hoa tan hoan toan m gam Al bang dung dich HNO3 loang, thu duoc 5,376 lit (dktc) h6n hop khi X gom N,, N.O va dung djch chiJa 8m gam muoi.
Ti khoi ciia X so v6i H2 bang 18. Gia tri ciia m la
A. 21,60 B. 18,90 C. 17,28 D. 19,44
(Trich de thi tuyen sink DH khoi A nam 2013) Hu&ng ddn giai
Nhdn thdy Al Id mot kim loai manh plidn I'fng vcfi dung dich HNO^ lodng; de hid khong noi ro ngodi A',, N2O khong c6 sdn phdm khu ndo khdc nen chung ta du dodn trong sdn phdm khuc6 NH^NO,. , ; , i t i.: : ,, -jy,, j^,,^, Sefmol h6n hop khi: n^hkhf = 5,376/22,4 = 0,24 (moj) ^ s : ' ^ } ' "
- Theo bai ra ta c6:
n N 2 + " N 2 O= 0 , 2 4 ' ^ ^ t f / W^ l v , ,
_ 4 4 - l 8 . 2 _ 8 _ l ^ ^ n ^ , ^n^^o =0,12 (molj^'J';'^^^^^^^^
18.2-28 8 1
N 2 0 - - ,
o +1
=> T6ng s6' mol electron do N nhan tao ra N j . N j O la:
0,12.10+ 0,12.8 = 2,16 (mol)
* Tru&ng hap 1: Co san phdm khu NH4NO3
=> Muoi gom CO A1(N03)3 va NH4NO3
+-') -.1
N+ 8 e- ằ N( N H 4 N O , )
8x < - X
Theo bao toan electron ta c6:
3m
H ô
27 = 2,16 + 8x
Mat khac: —.213 + 80x = 8m 27
(1) (2) T u ( l , 2 ) = > m = 21,6(gam)
Dap an diing la A. ' ' Tru&ng hap 2: Kh6ng c6 san phdm khiJr NH4NO3
Theo dinh luat bao toan electron, ta c6:
^ ' " = 2 , 1 6 = ^ m = 19,44(gam) 27
157
Trong dung dich chi c6 mu6'i A1(N03)3
Kh6'i lirong A1(N03)3 = ^ ^ ^ ^ ^ = 153,36(g)
=> Loai trucmg hop nay (vi khO'i luong muO'i ^ 8m).
Thi du 2: Cho 29 gam h6n hop g6m A l , Cu va A g tac dung vijra du vdi 950 ml dung dich HNO3 1,5M, thu duoc dung dich chiia m gam mu6'i va 5,6 lit h6n hop khi X (dktc) g6m N O va N.O. Ti kh6'i ciia X so vod H j l a 16,4. Gia tri ciia m la
A . 98,20. B. 97,20. C. 98,75. D. 91,00.
(Trich de thi tuyen sink DH khoi B) Hu&ng ddn gidi ,,,.1
Dat sd mol HNO3 la X. i w d
S6' mol NO3 tao mu6'i vori ion kim loai: • n„^_ „aomu6ivôi„„Mm,oai, = 3n(NO) + 8n(N20) + 8n(NH4N03)
NO3
= 3.0,2 + 8.0,05 + 8x = 1 + 8x ^.
V i nguyen t6' nito duoc bao toan ndn: > ,v 1,425.1 = 2.0,05 + 1.0,2 + (1 + 8x) + 2x
Suy rax = 0,0125 (mol).
Vay kh6'i luong mu6'i khan thu duoc:
m = m(kim loai) + mCNO," tao mu6'i vori ion kim loai) + m(NH4N03) m = 29 + (1 + 8. 0,0125). 62 + 0,0125. 80 = 98,20 (gam).
Dap an diing la A.
Thi du 3: Hoa tan hoan toan 8,9 gam h6n hop gdm M g va Zn bang luong viifa dii I 500 ml dung dich JlNOj I M . Sau khi cac phan umg ke't thiic, thu duoc 1,008 lit khi N2O (dktc) duy nha't va dung dich X chiJa m gam mu6'i. Gia trj ciia m la A . 34,10. , B. 31,22. C. 33,70. D . 34,32.
(Trich dS tuyen sinh Cao dang khoi A)J Hu&ng ddn gidi
Nhan tha'y M g , Zn la cac kim loai manh phan ihig vdi dung dich HNO3 loang;
d^ bai khdng noi ro ngoai N j O kh6ng c6 san phim khix nao khac n6n chiing i^' du doan trong san ph^m khir c6 NH4NO3. •
Theo bai ra: n^^j^Q^ =0,5 (mol); Uf^^Q =0,045 (mol) ' "
Goi X , y, z ISn luot la s6'mol M g ( N 0 3 ) 2 ; Z n ( N 0 3 ) 2 ; NH4NO3.
Taco: 24x + 65y = 8,9 ' M ; !
V i nguyen t6' nito duoc bao toan ntn ta c6:
2x + 2y + 2z + 0,045.2 = 0,5.1
^ x + y + z = 0 , 2 0 5 (2) eo nguydn tac bao toan electron, ta c6:
2x + 2y = 0,045.8+ 8z
2x + 2 y - 8 z = 0,36 (3)
•Tiy(l,2,3) =^x = 0,l;y = 0,l;z = 0,005
Vay m = m^,g(N03)2 " ' Z n ( N 0 3 ) 2 + ' " N H 4 N O 3 (iv .• f
m = 0,1.148 + 0,1.189 + 0,005.80 = 34,10(g) ' Ddpan dung la A .
Thi du 4: Hoa tan hoan toan m6t luong b6t Zn vao m6t dung djch axit X. Sau phan ling thu duoc dung djch Y va khi Z. Nho tir tir dung dich NaOH (du) vao Y, dun nong thu duoc khi khdng mau T. Axit X la , .
A. H2SO4 loang B. HNO3 C. H2SO4 A&c D. H3PO4
(Trich de tuyen sinh Cao dang - Khoi A) Hu&ng ddn gidi
V i NaOH + ddY — ° - > T T (khdng mau) + ...
=> Trong dung dich Y c6 NH4NO3 ^
NH4NO3 + N a O H — ^ N H 3 T +3H2O *
Axit X la HNO3 . ' M . i , i
4Zn + IOHNO3 -> 4 Z n ( N 0 3 ) 2 + NH4NO3 + 3H2O Dap an dving la B.
Thi du 5: Cho h6n hop gom 6,72 gam M g va 0,8 gam M g O tac dung hd't vdi luong du dung dich HNO3. Sau khi cac phan ung xay ra hoan toan, thu duoc 0,896 lit mot khi X (dktc) va dung dich Y. Lam bay hoi dung djch Y thu duoc 46 gam mud'i khan. K h i X la
A. N O , B.N3O C N , D . N O (Trich de tuyen sinh Cao dang - Khoi A) . . . J Hu&ngddngidi
S6' mol cac cha't: nj^g = 0,28(mol); n^go = 0,02(mol); Ox = 0,04(mol)
" Mg(N03)2 = " M g + " M g O = 0,3(mol) (bao toan nguyen t6' magie)
=^'"Mg(NO3)2=0'3-148 = 4 4 , 4 ( g ) < 4 6 g . Suy ra trohg dung djch Y c6 mu6'i NH4NO3 (amoni nitrat), kh6'i luong amoni nitrat tao ra la: mNH4N03 = 46 - 44,4 = l,6(gam)
159
1,6 = 0,02 (mol)
" N H 4 N 0 , -
Cac qua trinh nhuong - nhan electron xay ra:
M g M g + 2 e +2
0,28 -> 0,56(mol)
+ 5 N + 8e - 3 N
- 3
NH4NO, v
I , u I 1
2 N ^ 2ne ->
0,08n <- 0,04(mol)
, Theo nguyen tdc bao toan electron, ta c6:
j : 0,56 = 0,16 + 0,08n =>n = 5
0,16 <- 0,02(mol)
f i
+5 0
V a y k h i X l a N , : 2 N + 1 0 e^ N 2 Dap an dung la C.
Thidu 6: Hoa tan hoan toan 13,00 gam Zn trong dung djch H N O , loang, du thu dirge dung djch X va 0,448 lit khi N , (dktc). Kh6'i luong muoi trong dung dich X la
A. 37,80 gam B. 18,90 gam C. 28,35 gam D. 39,80 gam (Trich de tuyen sink Cao dang khoi A)
Nluhi tlui'y Zn la mot kim loai manh phdn Ung vai dung dich HNOj, loang; de hdi khong noi ro ngoai khong c6 sdn phd'm khi( ndo khdc nen chimg ta di(
dodn trong sdn phdm khifco NH^NO,.
Theobaira: n^„ = — = 0,2(mol); nN2 = ^ ^ = 0,02(mol) Cac qua trinh xay ra:
+2 + 5
Zn -> Zn + 2e 2 N + lOe N2 0,2 0,2 ^ 0,4(mol) 0,2 0,02(mol) V i 0,4 > 0,2 => CO qua trinh sau xay ra:
+5
N + 8e - 3 N (NH4NO3) ( 0 , 4 - 0 , 2 ) -)>0,2/8 = 0,025(mol)
Trong X C O mu6'i Zn(N03) 2 (0,2 m o l ) ; NH4NO3 (0,025 mol) Vay m = 0,2.189 + 0,025.80 = 39,08(gam)
Dap an diing la D. , 160
C t y TNHH MTV DWH Khana Vi^t
fhidu 7: Hoa tan hoan toan 12,42 gam A l bang dung djch HNO3 loang (du), thu duofc dung djch X 1,344 lit (or dktc) h6n hop khi Y g6m hai khi \k N^O
^f2. Ti khd'i cua h6n hop khi Y so vdi khi H j la 18. C6 can dung djch X, thu dugc m gam chat rSn khan. Gid trj cua m la
A. 97,98. B. 106,38. C. 38,34. D. 34,08.
(Trich Dithi tuyin sink DH khd'i A) Hu&ngddngidi . ,
Nhan tha'y A l la m6t kim loai manh phan ung voi dung dich HNO3 loang; dg bai kh6ng noi ro ngo^i N2, NjO kh6ng c6 san ph^m khiJ n^o khdc ndn chung ta dir doan trong san ph^m khu c6 NH4NO3.
Theob^i ra:
12,42 1,344 HAI = — r r - = 0,46 (mol); Uy =
27 22,4 Goi X, y \in lirot la s6' mol NjO, Nj.
x + y=:0,06
= 0,06 (mol)
Tac6: 44.x + 2 8 y ^ j ^ 2 = ^ ^ x + y
Q c qua trinh cho - nhan electron:
A l - 3e > A l ' "
0,46 > 1,38 mol
x = 0 , 0 3 y = 0,03
I n. „ H o ) = 1,38 (mol)
2N"
+ 5
2 N
lOe 0,3 8e 0,24 <r
0,03 (mol) 2 N +1
0,03 . 2 (mol)
I. H
=> Sn, = 0,3 + 0,24 = 0,54 < 1,38
Suy ra, CO qua trinh: _ ; • ? N + 8e > N (NH4NO,) "
(1,38-0,54) > 0,105 (mol) ' ' Chat ran g6m c6 A1(N03)3 (0,46 mol) va NH4NO3 (0,105 mol)
% m = 0,46. 213 + 0,105. 80= 106,38 (gam).
^ap an dung la B. ' . ' A ' . '
3. C a c bai tap tu luyen:
cau 1: Cho 7,8 gam Z n tan h6t trong dung dich HNO3 thu duoc V h't k h i duy nha\
la N2O (dktc). Lay dung dich con lai lam bay hoi lis tic thu duoc 23,8 gam chj, ran khan. Gia t r i cua V la (cho N = 14. O = 16, H = 1, Z n = 65)
A . 0,7168 B. 0,672 C. 0,5 D . 0,3584 Hu&ng ddn gidi
S6 mol kern: = 0 , 1 2 ( m o l ) ' ' ' ^ P T H H : 4 Z n + IOHNO3 - > 4Zn(N03)2 + NjO t + 5H2O l ' Gia sir chat ran khan thu duoc chi c6 Zn(N03)2 ' '
^ ' " z n( N 0 3 ) 2 = 0 ' 1 2 . 1 8 9 = 22,68(g) ; . 3 2 , 8 ( g )
=> Trong chat rSn c6 san ph^m k h u la NH4NO3:
m N H 4 N 0 3 = 2 3 , 8 - 22,68 = l , 1 2 ( g ) =^ n^^^^o, = 0 , 0 1 4 ( m o l ) '^J' P T H H : 4 Z n + IOHNO3 ^ 4Zn(N03)2 + NH4NO3 + 3H2O
, Theo djnh luat bao toan electron ta c6: i 0,12 . 2 = — . 8 + 0 , 0 1 4 . 8 = > V - 0 , 3 5 8 4 ( l i t )
22,4 Dap an dung la D .
C a u 2: Cho h6n hop g6m 5,6 gam Fe va 7,8 gam Z n vao dung dich HNO3 loang dir, sau k h i cac phan iJng xay ra hoan toan thu duoc 3,36 l i t N O duy nha't (a dktc) va dung dich X chiia m gam mudi. Gia t r i ciia m la
A . 47,78 B. 46,88 C.41,3 D . 41,58.
Hiring ddn gidi
S6'mol cac chat: np^ = 0 , l ( m o l ) ; nzn = 0 , 1 2 ( m o l ) ; n^o = 0 . ' 5 ( m o l ) - Qua trinh nhudng - nhan electron xay ra:
Fe - 3e ->-Fe'" Z n - 2e - > Zn"*
0 , l - > 0 , 3 0 , 1 2 ^ 0 , 2 4
= > I n ^ (cho) = 0,3 + 0,24 = 0,54 (mol)
N " ' + 3e - > N ^ - * ^ " '
• 0,45 0,15 V . • ; ) , )
V i 0,45 < 0,54 => con c6 san phdm khir la NH4NO3 (dung dich):
+ 8e ^>
( 0 , 5 4 - 0 , 4 5 ) - > 0,09/8 . ^ , a;/
H6n hop mu6'i thu duoc g6m:
0,1 mol F e ( N 0 3)3; 0,12 mol Zn(N03)2 va 0,09/8 mol NH4NO3.
Cty TNHH MTV DWH Khanx Viet
4 Z n + 1 0 H " + 2NO3 4 Z n + l O H " + N O ;
0 O Q 8 0
Vay m = 0,1 . 242 + 0,12 . 189 + ^ ^ ^ ^ ^ = 4 7 , 7 8 ( g ) p d p an dung la A .
cau 3: dung dich HNO3 loang, d u thu duoc dung dich A va h6n hop k h i gom N j va N j O . Thdm N a O H d u vao dung dich A , tha'y c6 khi mui khai thoat ra. Vife't phuong t n n h hoa hoc ciia t^t ca cac phan ling xay ra dudi dang phuong trinh ion riit gon.
Hu&ng ddn gidi
PTHH: 5Zn + 12H" + 2 N 0 J 5Zn-" + N ^ f + 6H2O 4Zn^" + N j O t + 5H2O . 3 — 4 Z n - " + N H; + 3H20 Dung dich A c6 cac ion Zn-\4 , H* va N O J .
Them N a O H d u vao dung dich A : ' H " + O H ~ ^ H, 0
'<>'•)•
N H; + O H " ^ N H 3 T + H20 (mui khai) Zn-" + 2 0 H ~ ^ Z n ( O H ) 4
Zn(0H)2 + 2 0 H - - > Z n Of + 2H2O
cau 4: Cho 2,16 gam h6n hop g6m A l va M g tan ha't trong dung dich axit HNO3 loang, dun nong nhe tao ra dung dich A va 448 m l (do d 354,9K va 988 m m H g ) h6n hop k h i B g6m 2 k h i kh6ng m^u, kh6ng d6i mau trong khdng k h i . T i khd'i ciia B so vdi o x i bang 0,716 l&n ti kh6'i ciia CO2 so v6i nito. L ^ m khan A m6t each c^n than thu duoc chat ran D , nung D dd'n kh6'i luong kh6ng d6i thu dirge 3,84 gam chat ran E. Tinh khd'i luong D va thanh phdn phdn tram kh6'i luong m6i k i m loai trong h6n hop ban diu.
Huong ddn gidi Theo gia thie't thi B chiia N , va N j O
" N J O + n N 2 =0,448.(988/760)/(0,082.354,9) = 0,02 nN20-44 + .28 = 0,02.32.0,716.44/28
=> So m o l electron nhan d^ tao ra 2 khi nay la: 0,01. (10 + 8) = 0,18 m o l ( I )
=> D C O A 1( N 0 3) 3 , M g( N 0 3 ) , c6 thd' c6 NH4NO3.
NH4NO3 - ^ N 2 0 T + 2H20 2NH4NO3 -> N2T + O, T + 4H20t 4 A 1( N 0 3)3 -> 2AI2O3 + I2NO2T + 3 0 . 1 2 M g( N 0 3)2 - > 2 M g O + 4NO2T + O2T
=> E chi C O AI2O3 va M g O . - i . , ., r-c c^l w'in)] <$>l^^' Ta C O •
n N 2 O= 0 , 0 1 11^2=0,01
K t thu^t mm gidi tihanh BT Hod hgc, t^p 2-Cu Thanh Toan Goi X , y l^n lirot la s6' mol cua A l va M g ta c6 h6:
27x + 24y = 2,16 : 1 0 2 . | + 40y = 3,84 -
=> X = n^i = 0,04 mol v^ y = n^g = 0,045 mol
=> s6' mol electron cho = 0,04.3 + 0,045.2 = 0,21 mol (II) + Tir ( I , II) suy ra phai c6 NH4NO3.
Tir do d l dang ti'nh duoc ket qua sau: '
. Dg6m: AKNO,), (8,52 gam); MgCNOa)^ (6,66 gam); ' NH,N03 (0,3 gam) = 15,48 gam. , , - Hon hop ban d^u c6 50% luong m6i kim loai.
Cau 5: Hoa tan he't 31,89 gam h6n hop A g6m 2 kim loai A l va M g trong luong dir dung dich HNO3 loang, thu duoc 10,08 lit (dktc) h6n hop khi X (g6m NO va N 2 O ) va dung djch Y. Ti kh6'i hoi ciia X so v<5ri khi hidro la 59/3. C6 can dung dich Y thu duoc 220,11 gam mu6'i khan. Tinh ph^n tram kh6'i luong m6i kim loai trong h6n hop A.
- - - rs i. ' C i' . V' v : Hu&ngddngidi ' + Dat s6 mol ciia NO va N 2 O Mn luot la a va b, ta c6:
a..b = i ^ = 0,45 22,4
59 O S
30a + 44b = —.2.0,45 = 17,7
a = 0,15 b = 0,3
IV' - ( 7 •
Dat s6' mol cua A l va M g Mn luot la x va y, ta c6: 27x + 24y = 31,89 (1) Khi cho h6n hop A tac dung vdi dung dich HNO3:
A l x
> A l ' ^ +
X
+ 3e H
0,45 3e 3x . N^^
0,15
Mg y 2,4
Mg^^ + 2e 2y N^' y
+ 4e 0,6 Ne'u san p h ^ khiJr chi c6 NO va N^O thi:
m^„,i = 31,89 + 62. (0,45 + 2,4) = 208,59 gam < 220,11 gam: V6 If
= > C O mu6'i NH4NO3 tao thanh trong dung djch Y N*' + 8e N-^
8z
. Ta c6: 3x + 2y = 0,45 + 2,4 + 8z hay 3x + 2y - 8z = 2,85 (2)
Matkhac:213x+148y + 80z = 2 2 0 , l l (3) Giai he (1), (2), (3) ta duoc: x = 0,47; y = 0,8; z = 0,02
mil •
Vay: %A1 = 0,47.27.100%
31,89
% M g = 100% - 39,79 = 60,21 % .
= 39,79%
raw 6' ^ ^^"^ ^'^ ® ^ ^ trong dung dich pjfjOj, sau khi phan ling ket thiic thu duoc dung dich B va 1,12 lit h6n hop khi jsjO va NjO C O s6' mol bang nhau. C6 can dung dich B thu duoc 31,75 gam muO'i.
Tinh th^ tich dung dich HNO3 0,5M t6'i thi^u d^ hoa tan hoan toan A . Hu&ng ddn gidi
S6 mol h6n hop khi = 0,05 mol; s6' mol m6i khi = 0,025 mol.
M g - a Z n -
• Mg^"" + 2e 2a .Zn2* + 2 e
A l b
•Al^++3e
2c
3b N*^ + 3e->.NO 0,025 0,075 2N +5
0,05
N2O N*^ + 8e N H ;
+ 8e
0,2 X 8x
Ta c6: 2a + 3b + 2c = 0,275 + 8x
31,75 = 7,5 + 62.(0,275 + 8x) + 80x.-> x = 0,0125
S6' mol HNO3 tham gia phan iJng = s6' mol HNO3 tao khi + s6 mol HNO3 tao mu6'i = 0,025 + 0,05 + 0,275 + 8. 0,0125 = 0,475 (mol)
''HNO3
0,475 0,5
= 0,95(1).
cau 7: H6a tan hoan toan 30,0 gam h6n hop X g6m M g , A l , Zn trong dung dich HNO3, sau khi phan ling ket thuc thu duoc dung djch Y va h6n hop khi g6m 0,1 mol NjO va 0,1 mol NO. C6 can c^n than dung dich sau phan ihig thu duoc 127 gam mu6'i. Tinh s6' mol HNO3 t6'i thi^u cin de tham gia cac phan ling tren.
Hu&ng ddn gidi Dat s6' mol Mg, A l , Zn Mn luot la x, y, z mol.
.2+
' A P . ' ( V S ' X ) . '
2+
Cac ban phan iJng:
Cha'tkhix: M g - 2 e - > M g '
X 2x
A l - 3e -J y 3y Zn - 2e - > Z n z 2z
T6ng s6' mol electron chflt khijr nhucmg Ik: 2x + 3y + 2z
Cac mu6'i tao ra la Mg(N03)2: x mol, A1(N03)3: y mol, Zn(N03)2: z mol s6' mo g6c NO3 trong mu6'i = 2x + 3y + 2z
Gia sir san phdm khu HNO3 chi c6 N 2 O va NO thi tdng s6' mol electron chSft oxi hoa nhan la: 0,1.8+ 0,1.3= 1,1 mol
Phuong trinh bao toan electron: 2x + 3y + 2z = 1,1
165
—> s6mol g6c NO3" trong m u D i ^ 2x + 3y + 2z = 1,1 Vay khd'i lucmg mu6'i khan thu diroc la:
m„„i; = n i K L + m„i,„, = 30 + 62.1,1 = 98,2 g a m < 127 (theo b^i cho) Chihig to ngoai N2O va NO, san ph^m khu HNO3 con CO NH4NO3 Goi s6' mol NH4NO3 tao ra la a mol;
S6 mol electron ma chSit oxi hoa nhan la: n f:^^^) s/ii if,, i 1
0,1. 8 + 0,1. 3 + 8a = 1,1 + 8a '^,^1,. , ,ằ <
Phuong trinh bao toan electron: 2x + 3y + 2z = 1,1 + 8a
* ->s6'molg6'c N O; trong mu6'i MgCNOj), + A K N O J) , + Z n( N 0 3) 2
2x + 3y + 2z = 1,1 +8a
'• KhO'i luong mu6'i tao thanh = kh6i luong M g( N 0 3 ) 2 + A1(N03)3 + Z n( N 0 3 ) , + NH4NO3 = 30 + 62. (1,1 + 8a ) + 80a = 127
=> a = 0,05 mol v d .
Bao tokn nguydn t6' nito, ta c6:
S6' mol HNO3 cSn phan iJng = s6' mol NO3" trong mu6'i 3 kim loai + s6' mol N trong N2O, NO, NH4NO3 = 1,1 + 8. 0,05 + 0,1. 2 + 0,1 + 0,05. 2 = 1,9 mol.
C^u 8: Hoa tan hoan toan 25,3 gam h6n hop X gdm M g , A l , Zn blng dung dich H N O 3. Sau khi phan ting ka't thiic thu duoc dung dich Y va 4,48 lit (dktc) khf Z (gdm hai hop chat khi khOng mau) c6 khdi lucmg 7,4 gam. C6 can dung djch Y thu dugc 122,3 gam hdn hop mudi. Tinh so mol H N O 3 da tham gia phan ung.
Hu&ng ddn gidi rf Z khdng mau => kh6ng c6 NOj
Cac khi la hop chat => kh6ng c6 N ,
=> Hai hop chat khf la N^O va NO. .xt
n N 2 0 + "NO 4,48/22,4 rnN2o=0.1mol Theo d l ta cd:
44.n N2O + 3 0. n N o= 7 , 4 n^Q =0,1 mol
Hdn hop mudi gdm M g( N 0 3 ) 2, Z n( N 0 3 ) 2 , A1 (N 0 3 ) 3 N H 4 N O 3
mi
Goi sd mol ciia N H 4 N O 3 la x mol (x ^ O).
Ta cd cac qua trinh nhan electron:
lOH^ + 2 N 0 ; + 8e -> N 2 O + SHjO
1 0,1 0,5 (mol)
4 H ^ + NOJ + 3e NO + 2 H 2 O
0,4 0,1 0 , 2 (mol)
iOH^ + 2 N 0 ; + 8e ^ N H 4 N O 3 + 3 H 2 O
lOx , ,. X 3x (mol)
H I ' ằ••
^ n H N 0 3 =n^+ = l , 4 + 1 0 x ( m o l);nH2o= 0 . 7 + 3x(mol)
-jTieo phuong phdp bao toan khdi luong ta c6:
m i c i m i o a i + ' ^ H N O j = m„,„tfi + m , + m H^ o ''''''''''''''' 'T',^:^'^^
^ 25,3 + 63(1,4 + lOx) = 122,3+ 7,4 +18(0,7+ 3x) = > x = 0,05
=:>nHN03 = l + 0'4 +10.0,05 = 1,9 (mol).
9: Hoa tan hoan toan 12,42 gam A l bang m6t luong viifa du dung djch HNO3
loang, thu duoc dung dich X va 1,344 lit (d dktc) hdn hop khi Y gdm hai khf N.O va N , . T i khdi cua hdn hop khf Y so vdi khf H , la 18. Tfnh sd mol HN63 da dung. .
Hu&ng dart gidi ;;:u.i
M2khf = 18.2 = 36
Goi a, b l^n lirot la sd mol cua N^O va N j
" fa + b = 0,06
=>a = b = 0,03
[44a + 28b = 2,16 ' ^'
Al ^ Al^+ + 3e f+ 3 N O ; ) ( l )
0,46 mol 1,38 1,38
2 N O; + 8 e ^ N 2 O (2) 2 N O ; + 1 0 e^ N 2 (3)
0,06 0,24 0,03 0,06 0,3 0,03
Do sd mol electron Al nhudng > tdng sd mol electron HNO3 nhan d^ tao NjO va N, ntn sd mol electron du nay HNO3 se nhan de tao N H 4
Sd mol electron do HNO3 nhan tao NH^ = l , 3 8 - ( 0 , 2 4 + 0,3) = 0,84 mol
N O; + 8 e- ằ N H 4 (4)
0,105 0,84 0,105 ' ' ^
^ "HN03cSn "N03-(mutfi) + "N03-(oxi hoa)
= 1,38+ 0,105+ 0,06+ 0,06+ 0,105 = 1,71 (mol) Vay da dung 1,71 mol HNO3.
Dang 4: Hop chat tdc dung vdi axit nitric '
• U thuyet van dung va phuong phap giai:
Axit HN03tac dung duoc vdi nhieu hop chat (cd tfnh khu, tinh bazo,...) iTiidu: 3H2S + 2HN03(,) > 2S + 2N0 + 4H2O
3 F e O + IOHNO3,,) > 3Fe(N03)3 + NO + S H P 3Fe( O H ) . + 1OHNO3 > 3Fe(N03)3 + NO + 8H2O
167
- Axit HNO3 oxi ho^ cic hop chat sat (II) thknh hop chaft (FeCNO,),):
Fe (FeO, FeClj, FeS04, FeCOj, Fe304, Fe(OH)2) +2 ^ H N O - ,
Fe(N03)3
- C6n cic hop chat sat (III) nhir oxit (FcjOj), hidroxit (Fe(OH)3) tac dung voi axjj
HNO3 theo kie'u axit - bazo:
Fe^Oj + 6HNO3 -> 2Fe(NO,)3 + 3 H, 0
— > Fe(N03)3 + 3H2O.
Fe(OH)3 + 3HNO3
* Phirong phap giai nhanh cic bai tap dang n^y:
- Phirong phap bao toan electron. ^, .j , , . y, - Phucng phap qui ddi.
- Phuong phap bao toan dien tich. * '
- Phtfong phap bao toan nguyen to. '^^'^ • - Phuong ph^p sir dung phuong trinh ion thu gpn. j - i |Qt>5 itt?'
- C^c cdng thurc de giai nhanh bai tap dang n^y: f\
+ C6ng thii'c tinh kh6'i luong mu6'i thu diroc khi cho h6n hop sit cdc oxit sat tac dung vdd HNO3 dir, giai phong khi NO:
242 ' '• '*
(m
Kin hi?p + 24.nNo)
m 80
C6ng thiic tinh kh6'i lirong mu6'i thu duoc khi hok tan hfi't h6n hap g6m Fe, FeO, F e A . Fe304 bang H N O 3 dac, nong du giai ph6ng khi NO2:
242 , o x •
80
C6ng thiic tinh khd'i luong sat da dung ban <i^u, bie't oxi hoa luong sat n^y bang oxi duoc h6n hop ran X. Ho^ tan hd't ran X trong HNO3 dac, nong, du tao san ph^m khix NO, duy nha't: ^ •
56 , „ , ' r :m
2. Cac thi du minh hoa:
Thidu 1: Cho 18,4 gam h6n hop X g6m Cu2S,CuS,FeS2 va FeS tac dung het vdri HNO3 (dac ndng, du) thu duoc V lit khf chi c6 NO2 (o dktc, san phim khu duy nha't) va dung djch Y. Cho toan b6 Y vao m6t luong du dung dich BaCl2 , thu duoc 46,6 gam ke't tiia; con khi cho toan b6 Y tac dung v6i dung dich NH3 du thu duoc 10,7 gam ket tiia. Gia tri ciia V 1^
A. 11.2. B. 38,08. C. 16,8. D. 24,64.
(Trich de thi tuyen sink DH khoi A)
168
'l 4. :i:H0):;3?JC''
Hu&ng ddn giai
PhUffng phdp giai nhanh hai nay Id phuong phdp quy ddi nguyen tu (vi c6 su lap lai car nguyen to trong cdc hap chat). Ngodi ra kit hap dinh lugt hdo toan nguyen tdvd hdo toan electron de lap cdc phuang trinh dqi sd'.
K6't tiia dSu m BaS04: n^^^^ = ^ = 0,2 (mol) Kd'ttuasau laFe(OH)3: Hp^^Qj^^^^ =1^ = 0,1 (mol) Quy d6i h6n hop X thanh x mol Fe; y mol Cu z mol S i
=s>56x + 64y + 32z = 18,4 ( l )
Mat khac: z = n ^ = nBaso4 =0,2 (bao toan nguyfin t6' luu huynh) Ta lai c6: x = np^ = Hp^^Q^^^ = 0,1 (bao toan nguyfin t6' sat)
TO(l)=>yJMz^5*iz22*2=o., , C - i ^'
+3 2+ +^
T6ngs6'mol electron do F e F e ; Cu ^ Cu ; S-> S \k \. 1 v r He (cho) = 0,1.3+ 0,1.2+ 0,2.6= 1,7 (mol) , v/i-
+5 +4
=> T6ng s6' mol electron do N+ le - ằ N 1^ 1,7 (mol) S>''
=>nN02 =1.7(mol)=>VN02 =1,7. 22,4 = 38,08(1) Dap an diing la B.
Thi du 2: D6t 5,6 gam Fe trong kh6ng khi, thu duoc h6n hop cha't ran X. Cho toan b6 X tac dung vdi dung dich HNO3 loang (du), thu duoc khf NO (san ph^m khu duy nha't) va dung dich chiJa m gam mu6'i. Gia tri ciia m la
A. 18,0. B.22,4. C. 15,6. D. 24,2.
..! (Trich de thi tuyen sinh DH khoi B) Hu&ng ddn giai
S6'mol sat: np^ =0,1 (mol) * So d6: F e > FcjO,,Fe304,FeO,Fe > Fe(NO3 )3
Bao toan nguydn tb'sat: np,(p,) = n^^^^^^^^^^^^ =^"Fe(N03)3 = " F e =0,1 (mol) Vay m =0,1.242 = 24,2(g) -4ô.fằiiHi;0(-
£5ap an diing la D. • . ;
169
Thi du 3: Hoa tan hoan toan h6n hop g6m FeSj 0,24 m o l CujS v^o dung dich HNO3 vCra du thu duoc dung dich X (chi chiia hai mu6'i sunfat) va V l i t khi N O duy nha't. Gia t r i cua V (dktc) la
A . 34,048 B. 35,84 C. 31,36 D . 25,088 (Trich de thi du bi dqi hoc) Huomg dan gidi
Sor d6 phan ling: , ^< ^,\i>^-lA'. ằ•
2FeS2 - > F e 2 ( 5 0 4 ) 3 + S O ^ . , 1 . * fcw'^ / "
0,24 0 , 1 2 ( m o l ) •^\\ <l . . f , v ô r > . i'l C u 2 S^ C u S 0 4 + Cu^^ Cu^^ + SO^- ^ CUSO4 'A •
X x( m o l ) X - > X / .}f<n$i
= ^ x= 0 , 1 2 ( m o l ) •W'*^'-*^^;.- ••y t ,f Qua trinh nhuomg, nhan electron:
+2 - 1 + 3 + 6 +1 -2 +2 +ô
FeS2 - > Fe + 2 S +15e ' " CU2S ^ 2 C u + S + lOe
0,24 ^ 3 , 6 ( m o l ) 0,12 l , 2 ( m o l ) Tdng s6' m o l electron cho: 3,6 + 1,2 = 4 , 8 ( m o l )
+ 5 + 2 , , N + 3 e ^ N ( N O ) ; ' ' '
4 , 8 - > l , 6 ( m o l ) ,
= > V N O = 1'6.22,4 = 35,84(1)
Dap an dung la B . _ ^ Thi du 4: Cho 11,36 gam h6n hgfp g6m Fe, FeO, Fe^Oj va Fe304 phan ling het
vdi dung djch HNO3 loang (du), thu duoc 1,344 l i t k h i N O (san ph£m khir duy nha't, a dktc) va dung dich X . C6 can dung djch X thu duoc m gam mu6'i khan. Gia t r i ciia m la
A . 38,72. B. 49,09. C. 35,50. D . 34,36.
(Trich De thi tuyen sink DH - CD khoi A) C v : . ; . Hudtigddtigidi f:'vho''\
, 1 344
1 T h e o b a i r a : = = 0,06(mol)
22,4 *•
Ta CO the hlnh dung theo so d6 nhu sau: ' "•• !•'<•' '••0. 5,''' '
170
0 +3 Fe - 3e > Fe
X 3x(mol) X
Fe > Fe(N03)3
I ON
X O + ft i
o >^
t +
/No
3 o Fe, F e O , Fe304, Fe^Oj
16
= > 2 4 x = l l , 3 6 - 5 6 x + 1,44 ,
= : > 8 0 x = 1 2 , 8 z ^ x = 0 , 1 6 ( m o l ) Vay n p , ( N 0 3 ) 3 = n F e = x = 0,16 (mol)
^ "iFe(N03)3 = 0.16. 242 = 38,72 (g) Dap an dung la A .
1 i - ~
r/ijrfM 5: Nung m gam b6t sSt trong o x i , thu dirgfc 3 gam h6n hop chat ran X . Hoa tan het h6n hop X trong dung djch HNO3 (du), thoat ra 0,56 l i t ( d dktc) NO (la san ph^m k h u duy nha't). Gia t r i ciia m la
A- 2,62. B. 2,32. ' C. 2,52. D . 2,22.
(Trich De thi tuyen sinh DH - CD khoi B) Huong ddn gidi
Theo bai ra: nNo = 0,56 22,4
- 0 , 0 2 5 ( m o l ) Ta c6 so dd:
Fe
0
Fe - 3 e
+3 Fe
00
3 0 0 +
3m 56
Fe(N03)3 (mol)
t
FeO, Fe304, Fe^Oj, Fe (du) o' O
171
Tac6: ^ = ^ + 3.0.025 56 8
=> 3m = 2 1 - 7 m + 4,2=:> 10m = 25,2 =>m = 2,52.
Dap an dung la C.
Thl du 6: Hoa tan hoan toan h6n hcfp gdm 0,12 mol FeSj va a mol CujS vao axit
H N O 3 (vira du), thu duoc dung dich X (chi chiia hai mu6'i sunfat) va khi duy nha't NO. Gia tri ciia a Ik - <
A. 0,04. B. 0,075. C.0,12. D. 0,06.
(TrichDelhituyen sinkDH - CD khoJA) Hu&ng ddn gidi
Tac6sod6: • , , : J
2FeS, + Cu.S MTH^O > Fe,(S04)3 + CUSO4
- N O - H 2 O
0,12(mol) a(mol) ...^ '
: ^ a = ^ ^ ^ - 0 , 0 6 ( m o l ) >?:Ji=-xte:
Dap an diing la D.
Thi du 7 : Cho 61,2 gam h6n hop X g6m Cu vk Fe304 tac dung voi dung dich
H N O 3 loang, dun nong va khua'y d^u. Sau khi cac phan ihig xay ra hoan toan, thu duoc 3,36 lit khi NO (san ph^m khit duy nha't, 0 dktc), dung dich Y va con lai 2,4 gam kim loai. C6 can dung dich Y, thu duoc m gam muoi khan.
Gia trj ciia m la
A. 108,9. B. 151,5. C. 137,1. D. 97,5.
(Trich De thi tuyen sink Dai hoc khd'i B) Hu&ng ddn gidi
3 66
Theobaira: n N o = - — =0,15 (mol) ' V
i . 22,4
Goi x, y \in lirot la s6' mol Fe304 va Cu phan iJng.
Theo bai ra, ta c6: 232x + 64y + 2,4 = 61,2 : ^ 232x + 64y = 58,8 (1) VI Cu (kim loai) con du nen trong dung dich Y c6 hai loai mu6'i \k Cu(N03)i Fe(N03)2. Do do c6 cac qua trinh cho - nhan electron xay ra la:
Cu - 2e > Cu-* ^®
2 '
y 2y y 3x 3. — X 3x *
+5 +2
3e > N 0,45 0,15 (mol)
1 7 2
Cty TNHH MTV DWH Khang Vift
r frhi
Vi ^ n e c c h ) = Z" e ( n h ) ^^n- 2y = 2x + 0,45
2) ta c6: x = 0,15; y = 0,375. ' '
Vfty: nl = mcu(N03)2 " ^ F e f N O j j j •>
= 188. y + 180. 3x '
= 188.0,375+ 180.3.0,15 = 151,5 (gam).
p ^ p d n ^ u n g j a B .
(2)
Thidu 8: Nung 2,23 gam h6n hop X g6m cac kim loai Fe, A l , Zn, M g trong oxi, sau m6t then gian thu dugc 2,71 gam hOn hop Y. Hoa tan hoan tokn Y vao dung djch H N O 3 (dir), thu diroc 0,672 lit khi NO (san phim khir duy nha't, a dktc). S6' mol H N O 3 da phan iJng la
A. 0,12 B.0,14 C.0,16 D. 0,18
(Trich dethi tuyen sinh Dai hgckhoi B) Hu&ng ddn gidi
Theobaira: n^o = 0,03(mol) ''
* Sad6: X + O j ^ Y 2,23g 2,7 Kg)
= > m o 2 =2,71-2,23 = 0,48(g) =>no =0,48/16 = 0,03(mol)
- 2
Sod6: O + 2e O 0,03 - > 0,06 (mol)
,;;yp,i li^ffb ,!i&y
N + 3e - > N 0,03 <-0,09 <-0,03 •
=^L"e ( K L cho) = 0,06 + 0,09 = 0,15(mol)
I I
Vi electron va ion NOJ mang didn tich bang nhau ntn
" N O J "'^''^('"^O = " H N O 3 (mfiitmJmg)
" H N 0 3 (bikhir) = n N o =0,03(mol) ' • ' - ^ : i *''4 r ' ^ U : ; - ' /
% t6ng s6'mol H N O 3 cdn diing: H H N O J = 0,03 + 0,15 = 0,18(mol).
- . ^ ^ a n dung la D.
^ht du 9: Hoa tan het 0,03 mol m6t oxit sat c6 c6ng thiic Fe,0, vao dung dich
H N O 3 loang du thu duoc 0,01 mol mot oxit nita c6 c6ng thitc N,0, (san ph^m
^hxi duy nha't). Md'i quan h6 gifra x, y, z, t la
A . 2 7 x + 1 8 y = 5 z - 2 t B. 9x - 8y = 5z - 2t C ^ x - 2y = 5z - 2t D. 9x - 6y = 5z - 2t^