Ten ciia kim loai va the' tich dung dich HNO3 da phan dug la
A. D6ng;61,5ml t v i / C. Thuy ngan; 125,6 ml ô B. Chi; 65,1 ml . D. Sat; 82,3 ml
Hu&ng ddn gidi
O n M la kim loai hoa trj I I dn tim d6ng thoi cung la kh6'i lugmg mol nguyfin tir
ciia no. Phan ling: ' > - • i M + 4HNO3 M(N03)2 + 2N02t + 2H2O
J (mau mau do) , v
8 96
S6' mol khi NO. = - — = 0,40 (mol). Theo phuong trinh hoa hoc:
• 22,4
n 40 0 40 4
= ^ = 0,20 mol va nHN03 =^~^= 0,80 (mol)
2 2 12 8
KheJi lirong mol nguyen tii ciia kim loai M : M = = 64,0 (g/mol).
=>kimloailaCu(d6ng). <•
Kh6'i lircmg HNO3 tham gia phan ling = 0,80.63 = 50,40 (gam) 50,40.100
Kh6'i luong dung djch HNO3 60,0% da dung = — — - — = 84 (gam) ou,u
210
Th^ tfch dung dich HNO3 c^n dung la V = — = -^1^ = 61,5 (ml) D 1,365
DdpdnA. . . . , ^ l f i. R * J ; : . • . . ^ . m ^ . • . ^ • .
C&u 7: H6n hop X g6m Fe va kim loai M hoa tri 3. Chia 38,6 gam X th^nh 2 phSn bang nhau. PhSn 1 cho tan hoan to^n trong dung dich HNO3 loang du thu duoc cic san ph^m khii chi c6 NO, NjO (h6n hop Y) vdi t6ng th^ tfch 6,72 1ft, ti kh6'i ciia Y so vdi la 17,8. P h ^ 2 cho vao dung djch ki^m sau m6t then gian tha'y lucmg H , thoat ra vuot qud 6,72 1ft. Bie't cac khf do 6 di^u kien tieu chudn.
a) Xac dinh ten kim loai M va % khd'i luong ciia kim loai trong X.
b) Tfnh kh6'i luong HNO3 da phan ling.
Hu&ng ddn gidi
^ Do HNO3 du nen Fe se tao mu6'i Fe^^ => Coi Fe v^ M c6 c6ng thiic chung M .
i=> ny = 0,3 mol > •,
I Kh<5i luong trung binh ciia Y: 35,6 g/mol. ' ' ' '' '
I H6n hop Y la 0,3 mol; a la s6' mol
> 30a + (0,3 - a ) 4 4 = 35,6 => a = 0,18 mol [ => Ti le mol NO/N2O = 3/2
j Phuong trinh hoa hoc ciia phdn 1:
25M + 96HNO3 —!—>25M(N03)3 + + ^ ^ 2 ^ + 48H2O ( l ) 0.18.25
= 0,5 m o l .
X tac dung voi ki^m c6 khf thoat ra nen M se phan ling.
Phuong trinh hoa hoc ciia phin 2:
+ 3H2O + OH" M(OH)J"+3/2H2 (2) 4!".
| > 2.0,3/3 = 0,2 >0,3 mol
>i X la s6 mol ciia M=>s6 mol Fe: 0,5 - x mol
>Mx + ( 0 , 5 - x ) 5 6 = 1 9 , 3 ^ M = ^ ^ ^ ^ ^ ^ vdi 0 . 2 < x < 0 , 5 8,7
5 6 - M
•X = 0,3 mol
•0,2< 8.7
5 6 - M < 0.5 => 12,5 < M < 38,6 => Chi c6 A l .
f&y %m^i = ^ 2 1 1 0 0 % = 41,97%; %niF^ =58,03%
Theo (1) nHNOj = ^ ^ f ^ = 1.92 (mol)
=> Khefi luong l i N O j phan ling = 63.1,92= 120,96 gam.
2 1 1
C&u 8: Chia 16,68 gam h6n hop X g6m Fe va kim loai R (hoa tri kh6ng d6i) thanh ba phdn bang nhau.
PhSn 1 cho vao dung dich HCl du, sau phan ling xay ra hoan toan chi thu dugc dung djch va 3,136 lit Hj.
Phdn 2 cho vao dung dich HNO3 loang, du sau khi phan ihig xay ra hoan toan thu duoc 2,688 lit khi NO (san ph£m khir duy nha't). Cac th^ tich khi do of dktc.
a. Xac dinh kim loai R va tinh thanh phSn phin tram kh6'i luong m6i kim loaj trong h6n hop ban dSu.
b. Cho phSn 3 vao V lit dung djch CUSO4 I M , sau khi phan ung xay ra hoan toan thu diroc 8,64 gam cha't r l n . Tinh V.
Hu&ng ddn gidi
a. M6i phSn c6 kh6'i 5,56 gam; goi trong m6i phSn c6 chiia x mol Fe va y mol kim loai R.
Fe + 2HCl->FeCl2 + H2 ( l ) , , x (mol) -> x ' '•'* ' mnM&&
R + n H C l - > R C l „ + | H 2 (2) n
nH2 =0,14(mol)=c>x + | . y = 0,14 ( l )
Phdn 2:
, Fe + 4HNO3 -> Fe(N03)3 + NO + 2H2O (3)
'. • ' I. X ^ . •• , „ • X -
3 R + 4 n H N 0 3 ^ 3 R ( N 0 3 ) ^ + n N O + 2 n H 2 0 (4)
n N o - 0 , 1 2 ( m o l ) = ^ x + | . y = 0,12 ( l l )
0 12 Giai hf phuong trinh (I), (H) ta c6: x = 0,08(mol);y = Theo kh6'i luong h6n hop ban dSu ta c6: 56x + M R .y = 5,56
O M R ==9.n;n lahoatri. ; f v S!^
n = l= > M R - 9 ( l o a i ) "
n = 2= > M R- 1 8 ( l o a i )
n = 3 ^ M R =27=> K i m l o a i R l a A l ; n = 3.
X = 0,08(mol);y = 0,04(mol)
=:>%mFe =80,67%;%mAi =19,33%.
b.EhloJ:
Sau phan utig khd'i luong chat ran tang = 8,64-5,56 = 3,08 gam
Khi A l phan ung hfi't, khd'i luong chat rSn tang .,
= 64.0,06 - 27.0,04 = 2,76 gam < 3,08 gam=>Al phan ling he't.
Khi Fe phan ling het, khd'i luong chat ran tang
= 2,76 + 0,08(64-56) = 3,4 gam >3,08 gam ^ Fe phan img chua hd't.
Vay AI phan utig hd't, Fe phan ling m6t phSn, goi s6' mol Fe phan ling la z (mol) 2A1 + 3CUSO4 ^ AI2 (SO4 \ 3Cu (5)
0,04^0,06 0,06(mol) K I ' Fe + CuS04->FeS04+Cu (6) .^00 4 z(mol)->z ^ z , a ' ' ..^ ^
=^ 2,76 + (64 - 56) .z = 3,08 => z = 0,04 (mol)
=>Z"cuS04 Phaniing =.0,06 + 0,04 = 0,1 (mol) ^ " f ' ' ,
=> V = 0,1 lit = 100 ml.
Q u 9: Hoa tan hoan toan 1,62 gam nh6m vao 280 ml dung dich HNO3 I M duoc dung dich A va khi NO (san ph^m khij duy nha't). Mat khac, cho 7,35 gam hai kim loai kiem thudc hai chu ki lifin tie'p vao 500 ml dung djch HCl, duoc dung djch B va 2,8 lit khi H . (dktc). Khi trdn dung djch A vao dung djch B tha'y tao thanh 1,56 gam ket tiia.
a. Xac djnh tan hai kim loai ki^m. *;
b. Tinh n6ng d6 moi/1 ciia dung djch HCl da diing. | £ Hu&ng ddn gidi
Phuong trinh hoa hoc:
A l + 4 H N 0 3 ^ A l ( N 0 3 ) 3 + N O + 2H20 (1)
2M + 2 H C 1 ^ 2 M C I + H2 (2) 2 M + 2 H 2 0 ^ 2 M O H + H2 (3) Ban dSu: s6 mol A l : 0,06 mol; s^ mol HNO3: 0,28 mol |; jj^J
Sau phan ihig HNO3 con du: 0,04 mol
Khi cho h6n hop hai kim loai ki^m vao dung dich H Q thi xay ra phan iJng (2) CO the CO phan ling (3):
^- Theo PTHH: s6' mol M = s6' mol H , = 0,25 mol => M = 29,4
^> hai kim loai ki^m thu6c hai chu ki lidn tie'p nen Na, K thoa man ( 2 3 < 2 9 , 4 < 3 9 ) g , , , , j i
Khi tr6n hai dung dich A B c6 kfi't tua tao ra chumg to ban d^u c6 phan vtng (3), ta CO phucmg trinh hoa hoc: , . . .
H N O 3 + MOH ^ M N O 3 + H 2 O (4)
Al(N03)2+3MOH-^-Al(OH)3+3MN03 (5)
S6 mol k6't tua: A l ( 0H) 3 = 0,02 mol nho Hon s6 mol A l( N O 3 ) 3. Nfin c6 2 kha nang:
+) Tru&nghapl: Al(N03)3 condirthi s6'mol MOH = 0,04 + 0,02.3 = 0,1 mol
=>S6molMphan umg (2) =0,25-0,1=0,15 mol
=^ S6 mol HQ = 0,15 mol =^ C M(H C I) = 0,3M. ,, ^ . ^
+) Tru&ng hap 2: MOH con du, A1(0H)3 tan trd lai m6t phSn: ' '
A l (OH)3 + MOH M A I O 2 + 2 H 2 O (6)
S6mol A1(0H)3 tan = 0,06 - 0,02 = 0,04 mol. ' - Tilf cdc phirong trinh (4, 5, 6) ta c6: ''" •* ' S6' mol MOH = 0,04+ 0,06.3+ 0,04 = 0,26 mol (loai, vi 1dm hom s6' mol M ban dSu). v ;
C a u 10: Khi ho^ tan cung m6t lirgmg kim loai R v^o dung djch HNO3 loang va vao
dung dich H2SO4 loang thi thu duoc khi NO va H2 c6 th^ tich bang nhau (do 6 cung di^u kien). Bie't kh6'i luong mu6'i nitrat thu duoc bang 159,21% kh6'i luong mu6i sunfat. Xac djnh kim loai R. ijj yiv A
Hu&ng ddn gidi
Goi n la hoa trj ciia kim loai R, a la s6' mol R da dung of m6i phan ling.
PTPlT: 3R + 4nHN03 -> 3R(N03)„ + nNO + 2nH20
a a an/3 'v ''-.'J fi'' 2R + H2SO4 ^ R2(S04)„ + nH,
a a/2 an/ 2 Theo bai ra: an/ 3 = an/ 2 => v6 If.
Didu nay chihig to R c6 hoa trj thay d6i. Khi R tac dung vdi HNO3 la ch&t oxi hoa manh th^ hidn hoa tri cao la m:
PTPlT: 3R + 4mHN03 3R(N03)„ + mNO + 2mH20 a a am/ 3
Ta c6 he: 3 2
a.(R + 62m) = i ^ . - . 2 R + 96n (2)
^ 100 2 ^ ^ Tiir (1) => m = 1,5n. Thay m vao (2):
=> R + 62. l,5n = 1,5921. (R + 48n) R + 93n = 1,5921. R + 76,42n
=> 0,592IR = 16,58n => R = 28n.
Xa CO bang:
n 1 2 3
R 28 56 84
Kl Lx)ai Fe Loai
Chi CO n = 2 (m = 3) ihig vdi R = 56 la thoa man. Vay kim loai R la Fe.
Dang 8: Bdi tap ve photpho va hop chat cua photpho U thuyet van dung va phuofng phap giai:
Photpho la phi kim tucmg d6'i boat ddng. Photpho trang boat d6ng hoa hoc manh hem photpho do. Trong cac hc>p chat, photpho c6 %6 oxi hoa -3, +3 va +5. Do do, khi tham gia phan ting hoa hoc photpho th^ hidn ti'nh oxi hoa hoac tinh khijf.
Photpho the' hifin tinh oxi hoa khi tac dung vdi m6t s6' kim loai boat ddng tao ra photphua kim loai. . u. a •„,, :.n:i .iiV^:
Thidu: 2 P + 3Ca ^ Cajp' canxi phophua
Photpho th^ hidn tinh khix khi tac dung vdi cic phi kim boat ddng nhu oxi, halogen, liru huynh,... va cac hop chat c6 tinh oxi hoa manh khac.
Photpho chay duoc trong khdng khi khi d6't ndng:
thie'uoxi: 4P + 30, +3 dix oxi: 4P + 5 0 . 0
-> 2P2O3 diphotpho trioxit -> 2P2O5 diphotpho pentaoxit +5
Hiotpho tac dung d^ dang vdi khi clo khi dd't nong: Ah thie'u clo:
du clo:
0
2P + 302
0
2P + 5C1.
+3
2PCI3 photpho triclorua
+5
-> 2PCI5 photpho pentaclorua
2 r
Axit photphoric la axit ba nSc, cd d6 manh trung binh, c6 ta't ca nhOng tinh cha't chung ciia axit. Khi tac dung vdi dung djch ki^m, tuy theo lucmg chSt t&c dung ma axit photphoric tao ra mu6'i axit, hoac mu6'i trung boa, hoac hop cac mu6i dd.
D a t k = - 5 ^ i ^ . n e ' u :
" H 3 P O 4
k < 1: Tao ra KH2PO4, H3PO4 du.
k = l : TaoraKH2P04.
1 < k < 2: Tao ra KH2PO4 va K2HPO4.
k = 2: Tao ra K2HPO4.
2 < k < 3: Tao ra K2HPO4 va K3PO4.
k = 3: Tao ra K3PO4.
k > 3 : Tao ra K3PO4, K O H du.
Bi^u d i l n tr&n true s6':
K H 2 P 0 4 KH,P04 K2HPO4 K3PO4 H3PO4 du
1
K2HPO4
f 2
K3PO4
3 KOHdu K H 2 P 0 4
2. C a c thi du minh hoa:
K2HPO4 K 3 P O 4
Thi du 1: Oxi hoa hoan toan 3,1 gam photpho trong khi oxi du. Cho toan b6 san ph^m vao 200 m l dung dich NaOH IM de'n khi phan ting xay ra hoan toan, thu duoc dung djch X. Kh6'i luong mu6'i trong X la
A. 16,4 gam B. 14,2 gam C. 12,0 gam D. 11,1 gam (Trich de thi tuyen sink DH khoi A nam 2013)
J * v :
Hu&ng ddn gidi
Theo bai ra: np = 3,1/31 = 0,1 (mol); nN,0H = 0,2.1 = 0,2 (mol) PTHH: 4P + 5 0 2^ 2 P A n ,
0,1 0,05 (mol) ' * ' P2O5 + 3H2O - > 2H3PO4
I f e i €lM 0,05 ^ 0,1 (mol) i l N a O H _ = 2l^ = 2 =^ Xay ra phan ling:
nH3P04
H 3 P O 4 + 2NaOH -> Na2HP04 + 2H2O 0,1 ^ 0,2 0,1 (mol)
Vay kh6'i luomg mu6'i (Na2HP04): m = 0,1. 142 = 14,2 (gam).
Dap an dung la B.
Thidu 2: Cho 1,42 gam PjO, tac dung hoan toan vdd 50 ml dung djch KOH IM, thu duoc dung djch X. C6 can dung djch X thu duoc chat ran khan gdm A. KH2PO4 va K2HPO4. B. K2HPO4 va K3PO4.
C. K3PO4 va KOH. D . H3PO4 va KH2PO4.
(Trich de tuyen sinh Cao dang khoi A) Hu&ng ddn gidi
Theo bai ra: np^Oj =0,01(mol); n^ o H =0,05 (mol) P205 + 3H20^2H3P04
0,01 0,02 V i 2<
nH3P04 0'02
= 2,5<3
Tao ra 2 mu6'i la K 2 H P O 4 va K 3 P O 4 H3PO4 + 2 K O H K 2 H P O 4 + 2H2O 0,02 0,04 -> 0,02
K 2 H P O 4 + K O H ^ K 3 P O 4 + H2O
0,01 ^ 0 , 0 1 0,01 m^P4pMV'-4nM:
Pap dung la B.
'jfiidu3: Thanh phSn chinh cua quSng photphorit la
A.Ca3(P04)2. B. N H 4 H 2 P O 4 . C. Ca(H2P04)2. D . CaHP04.
(Trich De thi tuyen sinh DH - CD khoi B) Hu&ng ddn gidi
Thanh phSn chinh cua quang photphorit la Ca3(P04)2 (canxi photphat).
Pap an diing la A.
fhidu 4: Cho 0,1 mol P2O5 vao dung djch chiia 0,35 mol KOH. Dung djch thu duoc CO cac chat
A. K 3 P O 4 , K 2 H P O 4 . B. K 2 H P O 4 , K H 2 P O 4 . C. K 3 P O 4 , K O H . , , * D . H 3 P O 4 , K H 2 P O 4 .
(Trich De thi tuyen sinh DH - CD khoi B) Cac PTPir (CO thi CO):
Hu&ng ddn gidi
P2O5 + 3H2O — > 2H3PO4
0,1 (mol) 0,2 (mol) ' Vi 1 < J 1 K O H _ ^ 2:2^ ^ 2 tao ra K H 2 P O 4 va K , H P 0 4 : *>ôằi
nH3P04 0-2
."•Vj
H3PO4+ K O H > K H 2 P 0 4 + H 2 0 feaw 1 H3PO4 + 2 K O H ). K 2 H P O 4 + 2H2O
Dap an dung la B. <^
Thidu 5: Cho 100 ml dung djch KOH 1,5M vao 200 ml dung djch H 3 P O 4 0,5M.
thu duoc dung djch X. C6 can dung djch X, thu duoc h6n hop g6m cac chat la
A. K 3 P O 4 va KOH. B. KH2PO4 va H 3 P O 4 . C. KH2PO4 va K2HPO4. D. KH2PO4 va K 3 P O 4 .
. (Trich De thi tuyen sinh Dai hoc khoi B) Hu&ng ddn gidi
Theo bai ra: n^oH = 0,1. 1,5 = 0,15 (mol); nHjpo^ = 0,2. 0,5 = 0,1 (mol).
V. i < - i l K O H _ = 0J5 ^ j ^ ^ ^ 2 n e n :
nH3P04 0,1
KOH + H 3 P O 4 > KH2PO4 + H 2 O KOH + K H 2 P O 4 > K 2 H P O 4 + H2O Vay thu duoc h6n hop KH2PO4 va K2HPO4.
V ''i .A
^ E>ap an dung la C.
Jb.< 217
3. Cac bai tap t u luyen:
Cku 1: Thuy phan hoan to^n hgrp chat 20,625 gam P Q , thu diroc dung dich X chiig h6n hcrp 2 axit H3PO3 va HCl. Th^ tich dung dich NaOH 2M d^ trung ho^ dung dich X la
A. 450 ml B. 225 ml C. 750 ml D. 375 ml
:.v^Ar'':^ f-^^ :A':,;V'I' Hu&ngddngidi Theobaira: npcij = = 0,15 (mol)
PTPlT: P C I 3 + 3 H 2 O >H3P03+3HC1 _ Axit photphorof ''''''' '':'''\}'^^
0,15 > 0 , 1 5 ^ 0,45 (mol) ':[}Y]n/^'^4:f):
HCl + NaOH ^NaCl + H j O
' 0,45-> 0,45 (mol) ^ H3PO3 + 2NaOH > Na2HP03 + 2H2O
0,15 -> 0,3 (mol) Mud'i trung hoa
=> "NaOH =0-45+0,3 = 0,75(mol)
W H = ^ = 0.375 (1ft) = 375 (ml) ^
^'•'^ • . ' 1.'-)
Dap An dung la D.
Chu V. -CrCTcuaaxitphotphoraHjPOj: > . W^^^^^^^ f i : 0 ' ' , ; f H - O - P - O - H
H3PO3 mdiaxit: H 3 P O 3 + HjPO^ :,"
H2P03?:iH^+HPỐ
Do vay H3PO3 chi tac dung vdi NaOH theo ti Id toi da
Ndn mu6'i NajHPOj tuy con nguydn t6' hidro trong g6c axit nhimg 1^ mu6i trung hoa.
Cau 2: Cho m gam PjO, vho 300 ml dung djch NaOH 2 M thi dung djch sau phan ling chi chiia 2 mu6'i NaH2P04, Na2HP04 c6 n6ng d6 mol bang nhau. Gia trj cua mia
A. 14,2 . V i B.28,4 C.21,3 D.71,0
^ O Hu&ngddngidi ,^.,,^^„, Theo bai ra: nNaOH = 0-3.2 = 0,6(mol) , j , ,,
Vi: Vi 2 mu6'i NaH2P04 Na2HP04 c6 n6ng d6 mol bang nhau {thd' tich cung blng phau) => 2 mu6'i c6 s6' mol bang nhau (x mol)
Yi nguyfin t6'natri bao toan n6n:
X + 2x = 0,6 => 3x = 0,6 => X = 0,2 (mol) 1 ; ; ; Vi nguydn t6'photpho bao toan ndn: ,>i
X + X - 0,2 + 0,2 = ^ m = 71.0.4 = 28,4 (gam)
£)ap an dung la B. f X i v j i j
y: Cac phuong trinh phan ihig xay ra: • P2O5 + 3H2O -> 2H3PO4 . • , H3PO4 + NaOH ^ NaH2P04 + H2O * - ' ' ' '
H3PO4 + 2NaOH Na2HP04 + 2H2O v^fp cau 3: Cho 100 ml dung djch H3PO4 I M tac dung vdi 21,875 ml dung dich NaOH
25% (d = 1,28 gam/ml) sau do dem pha loang bang nirdc ca't thu duoc 250 ml dung dich X. Hoi trong X c6 nhSng hop chat nao ciia photpho va n6ng d6 mol Ih bao nhidu (bo qua su thuy phan ciia cac mu6i)?
A.Na3PO4 0,4M B. NaH2PO4 0, l M vaNa2HPO4 0.3M C. NaH2P04 0,4M • D. Na2HP04 0.1M va Na3P04 0.3M.
Hu&ng dan gidi
Theo bai ra: n^^po^ = 0,1.1 = 0,1 (mol); ' ; ^, 21.875.1,28.25 .
"NaOH = = 0,175(mol) , 100.40 ' V ;
( 1
NaOH + H3PO4 > NaH2P04 + HjO ' 0,1 < - 0,1 - > 0.1 (mol)
C6n: 0,075 0 ' ^ * 0.1 (mol) ^ NaH2P04 + NaOH > Na2HP04 + H2O
0,075 < - 0,075 0,075 ^ C6n: 0,025 0 0,075 (mol) "/^ ' ;
Vay trong X c6 NaH2P04 va Na2HP04
C 0 , 0 2 5 ^ , , , . ^ 0,075
'NaH2P04 = " 5 ; ^ = ° ' ' ^ ^ CN, 3 „PO4 = - ^ = 0 , 3 M I / ' " '
Dap an dung la B. • ^1 v • '
^^u 4: D6't chay hoan toan 6,2 g photpho trong oxi du. Cho san phim tao thanh tac dung vCta du vdi dung djch NaOH 32% tao ra mu6'i Na2HP04.
3. Vife't phuong trinh hoa hoc ciia cac phan umg xay ra.
l ' . Tinh kh6'i luong dung djch NaOH da dung.
c. Tinh ndng d6 phin tram ciia mu6i trong dung djch thu diroc sau phan itng.
Hudng ddn gidi
a. PTHH: 4P + 2P2O5 (1)
P2O5 + 4NaOH > 2Na2HP04 + H j O (2)
b. S6'mol Ptham gia phan ling: = 0,2(mol)
(ft
(I H " .
^ ôr > ' i n
. Tir (1, 2) ta CO so d6 phan umg: , , , , ,
2P > P 2 O 3 — > 2Na2HP04
• 2nriol 1 mol 4 mol 2 mol • ^ ,r 0,2mol -)• 0,1 mol -> 0,4mol -> 02, mol
S6' gam dung dich NaOH da dung: ^^^-^^^^^=50 (gam) c) Dua vao (3) tinh duoc 0,2 mol Na2HP04.
Khd'i lucmg dung dich Na2HP04
mddNa2HP04 = 14,2 + 50=64,2(gam) N6ng d6 phSn tram ciia mu6'i Na2HP04:
0,2.142.100% _
^^''ddNa2HP04 4 4 , Z / o .
Cau 5: De thu ducrc mu6'i photphat trung hoa, cin la'y bao nhieu ml dung dich NaOH 1,00M cho tac dung \6i 50,0 ml dung dich H3PO4 0,50M?
Hu&ng ddn gidi
S6'mol H3P04 = 0,05.0,5 = 0,025 (mol) , y PTHH: H3PO4 + 3 N a O H > ^a^-j^O^ + SHjO
0,025 mol-)- 0,075 mol , \-,
=> So mol NaOH cin bang 0,075 mol ,^
The tich dung djch NaOH I M : ' ' ^ f
VddNaOH = ^ = 0,075(1) = 75(ml).
Cau 6: M6t loai quang photphat c6 chiia 35% Ca3(P04)2. Hay tinh ham luong phan tram P2O5 c6 trong quang tren.
! Hu&ng ddn gidi Ham luofng phSn tram P2O5 c6 trong loai quang trdn:
, . ^ ^ ^ . 3 5 ^ = 16,03%.
' ' ' ' ' M,,3(po4)^ 310
Cau 7: Cho 6,00 g P2O5 vao 25,0 ml dung dich H3PO4 6,00% (D = 1,03 g/ml). Tinh n6ng d6 ph^n tram cua H3PO4 trong dung dich tao thanh.
Hu&ng ddn gidi PTHH: P2O5 + 3 H 2 O > 2 H 3 P O 4
1 mol 2 mol 6 " 2.6^ ,^
- — 1 (mol)
142 142
Khd'i luongdungdjch sau phan ling: 25.1,03+ 6 = 31,75 (g) , v
Kh6'i luong H,P04 sau khi them PjO,: ,1 25.1,03.6 6 ^
100 71 '
N6ng do cua dung dich H3PO4 thu duoc: C%H^po^ = ^'^^'^^"^^ = 30,94%.
Cau 8: D6't chay hoan toan 6,2 g photpho trong oxi la'y du. Cho san ph^m tao thanh tac dung vori 150,0 ml dung dich NaOH 2,0 M . Sau phan irng, trong dung dich thu duoc cac mu6'i:
A. NaH2P04 va Na2HP04 B. Na2HPb4 va Na3P04 C. NaH2P04 va Na3P04 D. Na3P04
Hu&ng ddn gidi Cac phan iJng c6 the' xay ra:
4P + 5O2 —!-^2P205 ( I ) .0
P2O5+ 2NaOH + H2O > 2NaH2P04 (2) ^5*
P205 + 4 N a O H )• 2Na2HP04 + H2O (3)
P2O5 + 6 N a O H > 2Na3P04 + 3H2O (4)
rac6: np = ^ = 0,20 (mol); n^^oH = ^^"'""^'^ = 0,30 (mol)
31 1000 I
San ph^m tao thanh khi dd't photpho la P2O5 Theo (1), s6' mol P2O5 = ^ . U p = ^ = 0,10 (mol)
Dat T = "NaOH ^0'30 ^ 3 ^ 2 ^; „ np^os 0,10 1
Ti le mol nam trong khoang 2 va 4, do do theo cac phan ling (2) va (3) trong dung djch thu duoc c6 hai mu6'i duoc tao thanh la NaH,P04 va Na.HP04.
OapanA.
CAU 9: D6't chay a gam photpho do trong kh6ng khi la'y dir, r6i hoa tan hoan toan san ph^m thu duoc vao 500,0 ml dung djch H3PO4 85,00% (D = 1,700 g/ml).
Sau khi hoa tan san ph^m, n6ng d6 cua dung djch H3PO4 xac djnh duoc la 92,60 %. Tinh gia trj ciia a.
Hu&ng ddn gidi Photpho chay trong kh6ng khi du theo phan dug:
4P + 5O2 — ^ 2 P A <1) 4.31,0 g 2.142,0 g '*
P2O5 tan trong nude tao thanh H3PO4 theo phan ung:
P A + 3H2O 2H3PO4 ' (1)
142,0 g 2. 98,0 g ' ^' ^ T h e o ( l ) : 4 . 31,0 g P tao ra 2. 142,0 gPjO, , 5* *
=> a g P tao ra = 2,29a (g) P A
4.31,0 -' . I ' , 1) . Theo cac phan img (1) va (2):
- - f ^ 4. 31,0 (g) P tao ra 4. 98,0 (g) H3PO4 ' - '
4 9 8 0 a
a g P tao ^ ^ ^ ^ ^ = 3,16a (g) H,P04
Khd'i luong H3PO4 c6 trong 500,0 ml dung dich 85,00%:
500,0.1,700.85,00 ^ , ,
N...^ [^5
Kh6'i lucmg H3PO4 sau khi da hoa tan PjO, = (722,5 + 3,16a) (g)
Kh6'i luong ciia dung dich H3PO4 sau khi da hoa tan P^O, : , ' ^ 500,0. 1,700 + 2,29a = 850,0 + 2,29a (g)
Theo bai ra khi hoa tan san ph^m, nong d6 H3PO4 la 92,60% ndn:
2 ? M ± ^ , 0 0 % =92,60%. ^ v.. 850,0+2,29a , ,
Giai ra ta duoc a = 62,11 g photpho. .
Cau 10: Cho 62,0 g canxi photphat tac dung voi 49,0 g dung djch axit sunfuric 64,0%. Lam bay hoi dung dich thu duoc d6'n can kh6 thi duoc m6t h6n hop ran Xac dinh thanh phSn khd'i luong m6i cha't trong h6n hop rdn, bie't rSng c^c phan ling d^u xay ra v6\u suSit 100%. *;ô
Hu&ng ddn gidi Cac phan ung c6 th^ x^y ra:
Ca3(P04), + H. S O 42 C a H P 0 4 + CaS04 (1) Ca3(P04)2 + 2H2SO4 Ca(H2P04)2 + 2CaS04 (2) Ca3(P04)2 + 3H2SO4 ^ 2H3PO4 + 3CaS04 (3) S 6 ' m o l C a 3 ( P 0 4 ) , =- ^ ^ = 0 , 2 ( m o l )
310,0 ^ ' ' Hi til'
49 0 64 0 '
S6' mol H2SO4 = ' ' = 0,32 (mol)
100.98,0 y
V i ti le s6' mol H2SO4 va Ca3(P04)2la: 1 < = 1,60 < 2 0,20
jsfen chi x^y ra phan iJng (1) va (2) tao ra hai mudi CaHI>04 va Ca(H2P04)2.