hoa Iri lhap nhat va bao nhicu lit khi NO tao ra (dktc)?
Gidi
a) Dill cong ihuTc muoi nilrat kim loai X lii X(NO.0n-
2X(N03)., X2O,, + 2nN02 + ^ 0 3
9,4
'x(No,>„-j^_^g2n " x , o „ = •
2X + 16n Theo phifdng irinh : n,„,„-H = 2n.,xii Theo phifdng irinh : n,„,„-H = 2n.,xii
9,4 ^ 4 = 2.-
X + 62n 2X + 16n
=> X = 32n => n = 2 X = 64
Vay X la Cụ
0,05 mol 0,1 mol 0,025 mol
Cu(N03)2 CuO + 2NO2 +
4 n c u o = — = 0,05 (mol) n c u o = — = 0,05 (mol) 80 PV 0,984.0,5 n„ = — = • • = 0,02 (mol) RT 0,082(27 + 273)
So mol hon hdp khi sau khi nung :
nhh = + nN02 + " o , = 0.02 + 0,1 + 0,025 = 0,145 (mol) V i trong Cling dieu kien nhict do, ap suat, la co : .
^ = i l ^ P, = Pj. = 0 , 9 8 4 . ^ ^ = 7,134 (atm).
P, n , n , 0,02
b) So mol khi ihu diTdc (vdi ^ liTdng khi)
n^o, = 0,01 (mol); nô = 0,0025 (mol)
4NO2 + O2 + 2H2O > 4 H N 0 3 0,01 0,0025 0,01 (mol) => "HNO, = 0 . 0 1 mol HNO, > + NO3- 0,01 mbl 0,01 mol =>1H1= ^ = 0,()4(M) pH = - l g [ H l = -lgO,04 = l,4 V a y p H = 1,4. 3CỤ0 + 14HN0, > 6Cu(N03)2 + 2 N 0 + 7H2O 1.0,0, 0,0, M l 14 7 m c „ ^ o = ^ - 1 4 4 = (),3(g); V,o = ^ .22,4 = 0,032 (Hi). ^ 14 7
Bai 10. Hoa lan 0,368g hon hdp gom A l va Zn dung viTa dii 25 lit dung dich
HNO3 CO pH = 3. Sau phan rfng la chi ihu diTdc ba muoị Tinh thanh phan % thco khoi liTdng moi kim loai irong hon hdp.
Gidi
pH = 3 =>IH^] = 10-^M
i n s
=> [HNO3] = l O - ' M HHNO, =25.10"-' =0,025 (mol) V i Ihu dtfdc ba mu6'i n6n san pham lao thanh la NH4NO3.
4Zn + IOHNO3 > 4Zn(N03)2 + NH4NO3 + 3H2O
a mol 2,5a mol
8A1 + 3OHNO3 > 8A1(N03)3 + 3NH4NO3 + 9H2O
u . 30b , b mol mol
8 Ta c6 h^ phiTdng trinh :
65a + 27b = 0,368 Thanh phtin % theo khoi liTdng :
2,5a+ — = 0,025 fa = 0,004
8 (mol) b = 0,004
%mz„ = - ^ ^ ^ ^ ^ . 1 0 0 % = 70,65% => %mM = 29.35%. 0,368
B ^ i 11. Cho hon hdp A gom FeCOj va FeSj. A tac dung vdi dung dTch axit
HNO3 63% (khoi lifdng rieng 1,44 g/ml) theo cac phan tfng sau :
FeC03 + HNO3 > mu6'i X + CO2 + NO2 + H2O (1) FeSz + HNO3 > muoi X + H2SO4 + NO2 + H2O (2) difdc hon hdp khi B va dung dich C. T i khoi ciia B doi vdi oxi bkng 1,425. De
phan tfng vilfa het vdi cac chS't trong dung dich C can diing 540ml dung dich Bă0H)2 0,2M. Loc hfy ket tua dem nung den khoi liTdng khong doi dufdc
7,568g chát r^n (BaS04 coi nhU" khong bi nhiet phan, cac phan tfug xay ra
hôn loan).
a) X Ih mu6l gỉ Hoan thanh c^c phiTdng tnnh phdn tfng (1) va (2). b) Tinh khS'i liTdng tijfng chat trong hon hdp Ạ
c) Xac dinh the tich dung djch HNO3 da diing (gia thiet HNO3 khong b| bay hdi trong qua trinh phan uTng).
Gidi
FeCOa + 4HNO3 > Fe(N03)3 + CO2 + NO2 + 2H2b
a 4a a a a
FeS2 + I8HNO3 > Fc(N03)3 + 2H2SO4 + 15N02 + 7H2O
b 18b b 2b 15b a) X la muoi Fe(N03)3.
b) Hon hdp khi B gom C O 2: a mol
NO2 :(a + 15b) mol
dêoj =1.425 M B = 1,425x32 = 45,6
PhSn dgng phuong phap giai H6a hgc 11 V6 ca - E)5 XuSn Hung _ 44a + 46(a + 15b) MB - Dung djch C -=45,6 =>b = 0,2a n BăOH)2 a + a + 15b
FeCNO,), :(a + b) mol
H 2 S O 4 : 2b mol
HNO3 dii
= 0.54.0,2 = 0.108(mol)
Phan tfng : H2SO4 + BaCOH).
2b 2b 2HNO3 + BăOH)2 > BăN0,)2 + 2H:0
> BaS04i + 2H:0 2b 2b 0,04 2Fe(N03)3 + 3BăOH)2 (a + b) 2Fe(OH)3 (a + b) | ( a + b) 0,02 2Fe(OH)3i + 3Ba(N03)2 (a + b) (3) (4) (5) • Fe203 + 3H2O a + b FcjO, + m BaS04 a + b .160 + 233.2h => 80(a + b) + 466b = 7.568 => 80a + 546b = 7,568 The b = 0,2a => a = 0,04 mol => b = 0.008 mol
roFcco, = 116.0,04 = 4,64 (g); m^.^^ =120.0,008 = 0,96 (g). c) mol Bă0H)2 d phan tfng (3), (5): c) mol Bă0H)2 d phan tfng (3), (5):
nB.<oH„ (3,. (5, = | ( a + b) + 2b = 0.88 mol => nB,(OH)j (4, =0,108-0.088 = 0,02 mol => nB,(OH)j (4, =0,108-0.088 = 0,02 mol
S6' mol HNO3 ban dau : n„No, = 4a + 18b + 0,04 = 0,344 (mol) => m„No, =0,344.63 = 21,672 (g) => m„No, =0,344.63 = 21,672 (g)
Khoi lirdng dung djch HNOj : 21,672.100% 21,672.100%
= 34.4(g)r:> X
Bai 12. Cho hon hcJp gom 6,72 gam Mg vk 0,8 gam MgO Idc dung het vdi liTdng
dir dung dich HNO3. Sau khi cac phan tfng xay ra hoan loan, thu diTcfc 0,896
lit mot khi X (dktc) v^ dung djch Ỵ Lam bay hdi dung dich Y thu di/dc 46
gam muoi khan. Tim khi X.
A.NO2 B.N2O C.NO D.N2
no I
m
d" 34,4 = 23,89 (ml).
Giai
Ta c6: ny^^= 0,28 mol ; n^^^a= 0,02 mol
HNO3 dung dÚr^ Mg va MgO het M g - M g - 0,28 MgO- 0,02 HNO, HNÔ •MgCNOj)^ 0,28 -^MgCNOj), 0,02 => E"Mg,No,. = 0>28 + 0,02 = 0,3 mol =^ m^^,,,,,., = 44,4 (g) Ta thay: m M ; , , N o , ) , = 44,4 < m„,ur;i .ho = 46 (g)
=> Trong 46 (g) muoi khan c6 muoi NH4NO3
Va m N H. N o . , = 46 - 44,4 = 1,6 g =^ n^H^NOj = 0.^2 mol Ta c6; n^hi = = 0,04 mol Ta c6; n^hi = = 0,04 mol Mg - 2 e - 0,28 0,56 +5 N + nc 22,4 Mg 2+ 2 N + ge 0,16 0,02 NH4NO3 X 0,04n 0,04
Ap diing dinh luat bao lojln electron la c6: 0,56 = 0,16 + 0,04n n = 10 , So mol electron trao doi = 10 khi X la N2 => Dap an D. , So mol electron trao doi = 10 khi X la N2 => Dap an D.
Bai 13. HNO3 (vira du), thu dugrc dung dich X (chi chura hai muoi sunfai) va khi duy nhal NỌ Gid Irj cua a 1^ duy nhal NỌ Gid Irj cua a 1^
Gidi
T a c6 sa d6 : 2FeS2 * ^ > Fe2(S04),
0.12 0,06 CuzS >2CuS04 CuzS >2CuS04
a 2a
Ap dung dinh luat bao tôn nguyen to S: 0.12.2 + a = 0,06.3 + 2a -> a = 0,06 mol 0.12.2 + a = 0,06.3 + 2a -> a = 0,06 mol
Bai 14. Cho 61,2 gam hon hgrp X g6m Cu v^ Fe304 tdc dung vdi dung djch HNO3
loang, dun ndng va khuS'y deụ Sau khi cac phan ung xay ra hoan loan, thu difcJc 3,36 lit khi NO (san pham khOr duy nhat. a (dktc), dung djch Y va c6n lai difcJc 3,36 lit khi NO (san pham khOr duy nhat. a (dktc), dung djch Y va c6n lai 2,4 gam kim lôị Co cgn dung djch Y, thu difdc m gam mu6'i khan. Tinh m.
^ Gidi
Sau phdn iJng con 2,4 gam kim loai chirng to Cu du
'han d^ng v^i phuong ph^p giSi H6a hgc 11 V6 cd - P i Xufln Hung
=> kh6i lugng Cu va Fe304 phan urng = 61,2 - 2,4 = 58,8g.
3Cu + 8 H N O 3 3Cu(N03)2 + 2 N O + 2H2O ( 1 )
a a 2a/3
3Fe304 + 2 8 H N O 3 9Fe( N 0 3 ) 3 + NO + 14H2O ( 2 )
b 3b b/3
Cu + 2Fe( N 0 3 ) 3 Cu(N03)2 + 2Fe(N03)2 (3)
1,5b 3b 1,5b 3b
Gpi n c u pit(i) = a m o l ; so mol Fe304 ban dau = b mol. Ta c6 he <=> 64a + 328b = 58,8 2a + b = 0,45 <=> a = 0,15 b = 0,15 64(a+ 1,5b) + 232b = 58,8 2a/3 +b/3 = 0,15 Vay: m„^M= 188.(a + 1,5b) + 3b. 180 = 188.0,375 + 180.0,45 = 151,5g
Jai 15. Cho 1,82 gam hon hdp bpt X gom Cu va Ag (ti 16 so mol tifdng tfug 4 : 1) v^o 30 m l dung dich gom H2SO4 0,5M va H N O 3 2 M , sau khi cac phan ij-ng xay ra hôn toan, thu diTdc a mol khi NO (san pham khuf duy nhát cua N*^).
Trpn a mol NO tren vdi 0,1 mol O2 thu dUdc hon hdp khi Ỵ Cho toan bp Y
tic dung vdi H2O, thu dúdc 150 ml dung dich c6 pH = z. Tinh z.
Gidi
Gpi nAg = X => ncu = 4x => 108.x + 64.4x = 1,82 x = 0,005 mol Tac6: HH^SO^ = ÓÔ-'^mol
" H N O, = 0,06mol
n +=0,09 mol H
n _ = 0,06 mol
— . 1 NO3
Gia suf Cu va Ag phan tfng hét
=> He „h„.-,„g = 0.005.4.2+ 0,005.1 =0,045 mol
Bdn phSn tfug: 4H'^ + N O 3 ' + 3e ^ NO + 2 H 2 0
0,06 0,015 0,045 0,015
n T T + ji, = 0,03 ; n^,^_ = 0,045 => dieu nid siidiin^.
n NO3 V a y : U N O =0,015 mol 2 N 0 + O2 - > 2NO2 0,015 0,0075 0,015 2 N O 2 + - O 2 + H 2 0 - > 2 H N O 3 2 0,015 0,015 [ H N O 3 ] = = O.IM => [H*] = O.iM 15 • p H = 1 B A I T A P T R A C N G H I $ M
CSu 1. Cho sd do bien hoa sau:
NH3 +C0, ->A, ->A,
+ H 2 S O 4 loang -> A3(khi)
+ NaOH-> A 4 (khi) Vay A i , A2, A 3 va A 4 Ian luTdt 1^:
Ạ (NH2)2CO, (NH4)2C03, CO2 va NH3
B. (NH2)2CO, (NH4)2C03, CO va NH3
C. N H 2 C O , (NH4)2C03, CO va NH3