3m 3 —m _ ^ , ^
— =: + 0,075 m = 2,52 (g) 56 8 56 8
BAI TAP AP DUNG
Bai 1. Dot chay hoan loan 6,2g photpho trong oxi dif. Cho san pham tao thanh tac dung
vira du vdi dung dich NaOH 32% tao ra muoi NaaHPO*.
a) Viet cac phUdng trinh hoa hoc.
b) Tinh khoi luTcJng dung dich NaOH da dung.
c) Tinh nong dp phan trim cua muoi trong dung dich thu difPc.
Gidi
a) 4P + 5O2 2P2O5
0,2 mol 0,1 mol
P2Ọ, + 4 N a O H —'->• 2Na2HP04 + H2O
0,1 mol 0,4 mol 0,2 mol
Hp = = 0,2 (mol). 31
b) Khoi lurpng chat tan NaOH : m,, = 0,4.40 = 16 (g) Khoi lirpng dung dich NaOH : mjd = ^^'^^^^ = 50 (g). Khoi lirpng dung dich NaOH : mjd = ^^'^^^^ = 50 (g).
32%
c) Khoi liTdng chat tan Na2HP04 : m,., = 0,2.142 = 28,4 (g)
Khói lifPng dung dich sau phan uTng :
niđ = mp^o, + nijj NaOH = 0.1 • 142 + 50 = 64,2 (g)
28 4
Nong dp % muoi tao thanh : C% = —^.100% = 44,24%. 64,2 64,2
Bai 2, Hoa tan 8,65g hon hdp X gpm Al va Zn vap 200ml dung djch HNO3 thi tbu diTPc
3,36 lit bpn hPp khi A gpm NO va N2O (dktc). Ti khpi cua A sp vdi heli la 8,665. De trung boa hét axit trong dung dich thu dtfPc sau phan ilng da diing bet 400ml dung trung boa hét axit trong dung dich thu dtfPc sau phan ilng da diing bet 400ml dung
dich KOH 0,5M.
a) Tinh % theo the tich moi khi trong Ạ b) Tinh khoi liTdng moi kim loai trong X. b) Tinh khoi liTdng moi kim loai trong X.
100
c) Tinh nong dp mol dung dich HNỌ, ban daụ
Gidi D&l a la so mol NO, b la so mol N2Ọ D&l a la so mol NO, b la so mol N2Ọ
a + b= ^ = 0,15(mol) (1) Ta CO Ta CO 22,4 MA = 8,665 x 4 = 34,66 30a + 44b - . = 34,66 (2) a + b o 30a + 44b = 34,66(a + b) => 9 , 3 4 b = 4.66a r=> a = 2b
The a = 2b vao (1) => b = 0,05 mol => a = 0,1 mol 0 1 22 4 0 1 22 4
Thanh phan % the tich cac khi: %VN() = ' " '
3,36 => %V,^o =33.33%. => %V,^o =33.33%.
b) Al + 4 H N O , > A1(N0.,)3 + NO + 2 H 2 O
8Al(NO,h + 3N2O + I5H2O
> 3Zn(NO,)2 + 2 N 0 + 4H2O
->4Zn(NO,)2 + N20 + 5H20 8A1 + 30HNO, -
3Zn + 8HNO,-4Zn+ lOHNỌ,- 4Zn+ lOHNỌ,-
PhiTPng trinh cho nhan electron : + 5 +2 + 5 +2 N +3e > N (NO) 0,3 0.1 +5 ti 2N +8e > 2N (N2O) 0.4 0.05
Tdng .SP mol electron nhan : 0.3 + 0,4 = 0,7 (mol)
.100% = 66,67% (I) (I) (2) (3) (4) 0 Al — X mol 0 Zn - -> Al + 3e 3x mol +2 -» Zn + 2e y mol 2x mol
Tdng só mol electron cho : 3x + 2y
f3x + 2y = 0,7 Ta CO hO phiTdng trinh : Ta CO hO phiTdng trinh :
x = 0,2
y = 0,05 (mol) 27x + 65y = 8,65 27x + 65y = 8,65
Khoi liTdng moi kim loai trong X: = 0,2.27 = 5.4 (g); iii/,, ^ 3,25 (g). c) HNO, + KOH > KNO, + H 2 O c) HNO, + KOH > KNO, + H 2 O
0,2 mol 0,2 mol
HKOH = 0,5 X 0,4 = 0,2 (mol)
Theo cac phiTdng trinh (1), (2), (3), (4):
Phan dgng vk phuong p h i p g i ^ i H6a hpc 1 1 VP ca - D5 Xuan Hung
"HNOaptr = 4nNo + lOnn^o = 4.0,1 +10.0,05 = 0,9 (mol)
= > n H N 0 3 b a n d 4 u = 0,9 + 0.2 = 1,1 (mol)
Nong do mol HNO3 ban dau : C M = = = 5,5 (M).
Bai 3. Cho 44g NaOH vao dung djch chtfa 39,2g axit H.,P04, c6 can dung dich.
Hoi nhOfng muói nao dUdc lao nen va khói li/dng bao nhicủ
(DH Y DU(/cTP.HCM) Gidi Gidi 44 39 2 Ta CO : nNaOH = = ' ("^"0; n^^^m^ = = ^^'4 (mol) TacolilO: 2 < - ! ^ ^ ^ = - ^ = 2,75<3 nH,ro4 0.4
=>Tao hai muoi Na2HP04 va NaiPOj.
H3PO4 + 2NaOH > Na;HP04 + 2H2O (1)
0,4 mol 0,8 mol 0,4 mol
NaOH + NạHPOa > Na,,P04 + H.O (2) 0,3 mol 0.3 mol 0,3 mol 0,3 mol 0.3 mol 0,3 mol
nNaOH<ip.r(:)= 1,1 - 0 , 8 = 0,3 (mol)
^ " N . ^ H P O, ọn i,i = 0.4 - 0,3 = 0,1 (mol)
Na2HP04 :0,1 mol Na3PO4:0,3 mol Na3PO4:0,3 mol
Khoi lifcJng moi muoi: mNâHPOj = 0,1.142 = 14,2 (g)
m N. , P 0 4 = 0,3.164 = 49,2 (g).
Bai 4. Trong mol binh kin dung tich 56 lit chita N2 va H2 thco li le the tich 1:4c} nhiet dc) 0"C vii 200 aim va mol it chát xuc Liẹ Nung ncMig binh mc)t thcii gian, nhiet dc) 0"C vii 200 aim va mol it chát xuc Liẹ Nung ncMig binh mc)t thcii gian, sau do diTa ve 0"C ihay ap suaft trong binh giiim 10% so vc'Ji ap suiít ban daụ a) Tinh hiOu suaft phan uTng dieu che NH3.
b) Neu lay ^ liTc^ng NH., iren ihi dieu che diMc bao nhieu lit dung dich HNO1
70% (d = 1,41 g/ml) hieu suat qua trinh dieu che la 85%..
c) NS'u lay ^ lifcfng NH3 ihu diTcJc d iren thi dieu che diWc bao nhiCni lit dung
dich NH3 25%) (d = 0,91 g/ml).
\m
Vay cac muoi tao lhanh
Gidi a) N2 + 3H2 ? a) N2 + 3H2 ? a mol 3a mol PV 200.56 nhhkhi - RT 0.082.273 2NH3 2a mol = 500 (mol) = lOOmol va n^, =400mol => Só mol hem hdp sau phiin iJng :
n = (100 - a) + (400 - 3a) + 2a = 500 - 2a
VI trong Cling dieu kicn, nhiet dc), the tich => ap suiíl ti le vc'Ji sc) mol ta cc) :
"hh Jill _ PjJu p..„ 500 'hh sau 200 500-2a 180 => a = 25 (mol) 25
=> Hieu suál phan ỉng di6u ché NH,: H% = j ^ - 1 0 0 % = 25%. b) Lay - liMng NH,: n^H, = - ^ = ămol) b) Lay - liMng NH,: n^H, = - ^ = ămol)
Sa 60 dieu che HNO3 liT NH.,
NH3 > NO > NO2 > HNO, a mol a mol a mol a mol a mol a mol a mol a mol
Khoi \Mng HNO3 v(3i hieu suál qua Irinh 85% :
m UNO, = ạ—.63 = 25.—.63 = 1338,75 (g) 100 100
Khoi luTdng dung dich HNO,:
n,.,.ioo%jm7...ia».^,^
C % 70%
The tich dung djch HNO3: 1912,5 1.41 1.41
1
= 1356,38 (ml) = 1,35638 (lil). c) Khoi lirc^ng NH3 (vc^i - liTcJiig NH3): m^H, =25.17 = 425(g) c) Khoi lirc^ng NH3 (vc^i - liTcJiig NH3): m^H, =25.17 = 425(g)
Khoi lirctng dung dich NH3:mjj = ^^-^-'00% ^ j^y^, . The lich dung dich NH3: V = 25% . The lich dung dich NH3: V = 25%
d 0,91
Bai 5. De dieu che 5 tan axil nitric mmg dc) 60%, can dung bao nhicu Ian
amoniac? Biel rang sifhao hut amoniac Irong qua irinh siin xual la 3,89{.
Gidi 4 N H 3 + 5O2 4 N O + •6H2O 4 N H 3 + 5O2 4 N O + •6H2O (mol) (mol) 21 21 2N0 + O2 > 2NO2 _ ( m o l ) -^(mol) 4 N O 2 + O2 + 2H2O > 4 H N O 3 (mol) (mol) 21 21 Khoi liTdng HNOj nguyen chat:
^ 60 ^ ^, 3.10' 10" , „
n^HNo, = 5 . = 3 tan; nuMn = = (mol)
HNO, HNO, 63 21
m^H, = — . 1 7 = 0,809.10'(g) = 0,809 tan
Vi trong qua Irinh san xuat hao hut 3,8% ncn khoi lifdng NH3 la : m = 0 , 8 0 9 . — =0,841 (tan). 96,2 m = 0 , 8 0 9 . — =0,841 (tan). 96,2
Bai 6. Nung 6,58g Cu(N03)2 trong binh kin, sau mot thcJi gian thu du"ac 4,96g
chát ran vii hon hdp khi X. Hap thu hoan toan hon hdp X viio nu^dc, difcrc 300ml dung djch Ỵ Viet phufttng trinh hoa hoc ciia cac phan iJug xiiy ra vii 300ml dung djch Ỵ Viet phufttng trinh hoa hoc ciia cac phan iJug xiiy ra vii linh pH ciia dung dich Ỵ
(Trick TSDH. Hat I)
Gidi
Taco: nc„(N03)2 = - ^ = 0,035 (mol)
Cu(N0,)2 CuO + 2NO2 + - O ,
2 ' a mol a mol 2a mol a mol a mol 2a mol
4 N O 2 + O2 + 2 H 2 O > 4 H N O 3
2a mol 2a mol
N O 2 O2 O2
Hon hdp X
Dung dich Y : HNO,
Cu(N0.,)2 dir: 0,035 - a 104 104
KHAJgG VBET
=> Khoi Ming chát ran : m = mc„o + m c u i N o , ) ; dir
o4,96 = 80a + 188(0,035 - a) => a = 0,015 (mol)
n H N O 3 = 2.0,015 = 0,03 (mol)
H N O 3 > H^ + NO3
0,03 0,03
=^[H1 = Y^ = 0,1M =>pH = -lg[Hl=-lgO,l = 1 =^pH= 1.
Bai 7. Hoa tan 2,095g hon hdp X gom Fc vii m()t kirn loai R (R c6 hoa trj khong doi) trong dung dich HCl diT, thu diTdc 0,784 lit khi H2. Ncu hoa tan ciing doi) trong dung dich HCl diT, thu diTdc 0,784 lit khi H2. Ncu hoa tan ciing lifcJng hon hdp X trcn trong dung djch H N O 3 diT thu du^dc 1,12 lit hon help khi gom NO va NO2 c6 ti khoi so vdi hidro lii 19,8. Xac dinh kim loai R.
Gidi
Goi X, y Ian liTdt lii so mol ciia NO, NO2.
Ta c6: X + y = 1,12 22,4 Mhh khf = 19,8.2 = 39,6 Mhh khf = 19,8.2 = 39,6 30x + 46y = 0,05 (mol) (1) x + y Giiii(l) vii (2) = 39,6 x = 0,02 y = 0,03 (2) (mol)
Phifdng trinh cho nhan electron khi cho hon hdp kim loai tac dung vdi H N O 3 : 0 0 Fc a mol R b mol Fc + 3c 3 a mol +11
R + ne (n la hoa trj ctia R) nb mol Tong so electron nhufdng : 3a + nb (mol) Tong so electron nhufdng : 3a + nb (mol)
+2 ^ N ^ N +5 N + 3e 0,06 mol 0,02 mol | , +5 i N + le +4 N f o;()3mol 0,03 mol
:, Tdng so mol electron thu : 0,06 + 0,03 = 0,09 (mol)
I => 3a + nb = 0,09 (I)
Phúdng trinh hoa hoc khi cho hon hilp kim loai tiic dimg v6i HCl d\i:
Phan dang va phuong phap giai H6a hqc 11 V6 co - D5 XuSn Hang Fe + 2HC1 a mol R +nHCl b mol nb -^FeClr+H. a mol n R C l n + - H2 nb mol 0,784 = 0,035 22,4 Mat khdc ta c6: 56a + bR = 2,095 fa = 0,02 (II) (III) Giiii (I), (II) va (III) ^ b = 0,015 n = 2
R = 65 Vay kim loai R la Zn. Vay kim loai R la Zn.
Bai 8. Mot hon hdp A gom hai khi N2 va H2 theo ti le mol 1 : 3. Tao phan iJng giffa N2 va H2 cho ra N H 3 . Sau phan iJng thii diTOc hon hdp khi B. Ti khoi hdi giffa N2 va H2 cho ra N H 3 . Sau phan iJng thii diTOc hon hdp khi B. Ti khoi hdi cua A doi vdi B la dA/B = 0,6.
a) Tinh hieu suat cua phan i t n g l o n g hdp N H 3 .
b) Cho hon hdp khi B qua niTdc thi con lai hon hdp khi C. Tinh ti khoi hdi cua A
doi vdti C.
(Trich TS DHQG, dot I)
Hon hdp A Nj :H, :3x mol X mol N, + 3H2 N, + 3H2 Giai • nA = 4x (mol)