AI2(504)3 + 6Na0H ^ 2A1(0H)3 i + m&^SO,
(keo trang) A1(0H)3 + NaOH ^ NaAlÔ + 2H2O A1(0H)3 + NaOH ^ NaAlÔ + 2H2O 2.14. Dap an diing la C.
Phuang trinh hoa hoc: 2M + SH^SÔ ^ (SO4 )3 + SH^ t (1) 2 2 2 24 2 2 2 24
Theo(l): S6'moI M = -s6'molH, = - . - ^ = 0,067(mol). 3 ' 3 22,4 3 ' 3 22,4
Kh6'i lugng mol cua kim loai M la: - i ^ = 27-> M la nh6m (Al). - i ^ = 27-> M la nh6m (Al). 0,067
2.15. Dap an dung la D. Phuang trinh hoa hoc: Phuang trinh hoa hoc:
Zn +CUSO4->ZnS04+Cu (1)
X mol X mol
Goi X la s6' mol Zn da phan ung.
Theo (1): S6' mol Cu tao thanh va bam vao thanh Zn la x. Ta c6: 65x - 64x = 50 - 49,82 = 0,18 (gam) . r •: , Ta c6: 65x - 64x = 50 - 49,82 = 0,18 (gam) . r •: ,
- > x = 0,18 (mol).
Vay khdi luang Zn da phan ling la: 0,18.65 = 11,7 (gam).
2.16. Dap an dung la B. Phuang trinh hoa hoc: Phuang trinh hoa hoc:
Zn + FeS04 ZnS04 + Fe (1)
FeS04 da phan ung het, Fe duoc tao thanh bam vao thanh Zn.
Theo (1): S6' mol Zn phan ling = s6' mol Fe tao thanh = s6' mol FeS04
^ - mp^^ p = 2,25 = 65.0,Ix - 56.0,Ix ^ x - 2 , 5 M . ^ x - 2 , 5 M .
2.17. Dap an diing la C. Phuang trinh hoa hoc: Phuang trinh hoa hoc:
2M + 2nHC1^2MCl„+nH21 (1) 200 7 3 200 7 3
mol HCl = = 0,4 (mol) , .
S6' mol cac chát trong phuang trinh phan ung Ik:
0 4 0 4 0 4 > • ' = ^ ( m o l ) ; n^e, =—(mol);nH = - ^ - 0 , 2 ( m o l ) . = ^ ( m o l ) ; n^e, =—(mol);nH = - ^ - 0 , 2 ( m o l ) .
n " n 2 ' i '
0 4 0 4M
Khdi luang mudi MC1„ = — ( M + 35,5n) = +14,2
n n
H I = (a+ 14,2) (gam). Khdi lugfng dung dich sau phan ung: Khdi lugfng dung dich sau phan ung:
m^j = 200 + a - 0,2.2 -199,6 + a Ndng đ % cia dung dich mud'i: Ndng đ % cia dung dich mud'i:
C% = ±tMil.ioo% = 11,96% 199,6 +a 199,6 +a
Giai ra dugc a = 11 (gam). 11 l l . n 11 l l . n
OA ~ 0,4
n
Khi n = 1 ^ M = 27,5: loaị
n = 2 ^ M = 55 (Mn). ''' n = 3 ^ M = 82,5: loaị " n = 3 ^ M = 82,5: loaị " vay kim loai M la Mn (mangan). " *'
2.18. Dap an dung la B. Phirong trinh hoa hoc: Phirong trinh hoa hoc:
2A1 + SH^SO, ^ Al^ (SO,), + 3H2 T
Theo (1): S6' mol = s6 mol H 2 S O 4 = 0,2.1 = 0,2 (mol). -> =0,2.22,4 = 4,48(lit) -> =0,2.22,4 = 4,48(lit)
Hieu suát cua phan iJng: H = -—.100% = 75,0%. 4,48 4,48
2.19. Dap an diing la Ạ Phuong trinh hoa hoc: Phuong trinh hoa hoc:
2A1 + 2 N a O H + 2 H 2 O2 N a A 1 0 3 + 3 H 21
Fe +NaOH :kh6ngc6 phan ling
Theo (1): S6' mol A l = s6' mol NaOH = 0,2.1 = 0,2 (mol). m^, =0,2.27 = 5,4 (gam) m^, =0,2.27 = 5,4 (gam)
%m^, = ^ . 1 0 0 % = 33, 7 5 %->%mp, =66,25%. 16 16
2.20. Dap an diing la D.
Kh6'i lugng thanh sat tang la: 50. = 2 (gam).
Phuong trinh hoa hoc:
Fe + CuSO,->FeSO,+Cui
Goi X la s6' mol Cu thoat ra da bam vao thanh sat. Theo (1): 64x - 56x = 2 ^ X = 0,25 (mol). Theo (1): 64x - 56x = 2 ^ X = 0,25 (mol). Khái luomg Cu thoat ra la: 0,25.64 = 16 (gam).
2.21. Dap an dung la B.
Phuong trinh d'ltn phan nong chay: 2A1,03 )4Al + 30, 2A1,03 )4Al + 30, C + 02(dii)-^C0j
w Theo (1): S6' mol 0^=~ s6 mol Al^O, = 0,0375 (mol).
Theo (2): S6' mol C = s6' mol O2 = 0,0375 (mol). ^ m ^ =0,0375.12 = 0,45 (gam). ^ m ^ =0,0375.12 = 0,45 (gam).
2.22. Dap an diing la C. S6' mol cac chát: S6' mol cac chát:
n„^s =0,2.1 = 0,2 (mol);
"NaOH =0'3-l = 0,3(mol). _o
Cac phuong trinh hoa hoc: •
H^S + NaOH-> NaHS+ H,0 (1) 0,2 mol 0,2 mol 0,2 mol 0,2 mol 0,2 mol 0,2 mol
Sau phan img (1), NaOH con du 0,3 - 0,2 = 0,1 (mol). Do do xay ra phan ling: Do do xay ra phan ling:
NaHS +NaOH^Na2S + H20 (2)'
0,1 mol <- 0,1 mol 0,1 mol ; r Nhu vay dung djch thu dugc c6 0,1 mol NaHS va 0,1 mol NajS. ' ' Nhu vay dung djch thu dugc c6 0,1 mol NaHS va 0,1 mol NajS. ' '
2.23. Dap an diing la Ạ '
Cac phuong trinh hoa hgc: r i ' » ; r r Y T V-
MgC03+2HCl->MgCl2+C02 t + H,0 (1)
CO,+CăOH)j ^ C a C O , i + H,0 (2)
•20
Theo (1) va (2): S6' mol MgCO, = s6' mol CaCO, = — = 0,2 (mol).
-> m^gco, = X = 0,2.84 = 16,8 (gam).
2.24. Dap an diing la B. ,
2.25. Dap an diing la D.
2.26. Dap an diing la C. '5''
Phuong trinh hoa hgc hoa tan X: , 2X + aH2SO, ^X2( S O , ) , + a H j (a la hoa tri ciia X) ( D 2X + aH2SO, ^X2( S O , ) , + a H j (a la hoa tri ciia X) ( D T h e o ( l ) : S 6 ' m o l X = - = s 6 m o l H , = - . ^ = — (mol).
a a 22,4 a
a = 2 ^ s6' mol X = 0,075 -> M„ = = 18: loaị 0,075 a = 3 -^s6'molX = 0,05->M„ = ^ ^ = 27(A1). 0,05 X l k A l( n h 6 m ) . 2.27. Dap an dung la Ạ Phuong trinh hoa hoc:
- CuO + C O — ^ C u + COj t (1) Theo (1): 80 gam CuO -> 64 gam Cu, kh6'i luong giam 16 gam.
m gam CuO -> Cu, kh6'i luong giam 20 - 16,8 - 3,2 (gam). 80.3,2 ,
-> m = — — - = 16 (gam). 16
% CuO bi phan huy la —.100% = 80%. 20
2.28. Dap an diing la Ạ Phuong trinh hoa hoc:
Fe + 2FeCl3 ^ 3 F e C l 2 (1)
Theo (1): S6' mol FeClj = 2 s6' mol Fe = 2.^^"^'^ = 0,05 (mol) 56
- > x = ^ - 0 , l ( m o l / / ) . 0,5
2.29. C6ng thiic diing la D. Phuong trinh hoa hoc:
X^O, + x H 2 S O, ^ X , ( S O , ) ,+ X H 2 O (1) 1 mol 98x gam 1 mol
Khd'i lugfng dung dich axit = = 980.x (gam). N6ng d6 % ciia muoi sunfat:
C%^12,9%=(^^^96).100% 2X + 16x + 980x - > x = 18,66x.
Nghiem duy nhát phu hop la x = 3 ^ X = 56 (Fe).
CTPT cua oxit la Fe203.
2.30. Dap an diing la C.
De di^u che Na kim loai chi c6 the diing phuong phap difn phan mu6'i jvfaCl nong chay hoac NaOH nong chaỵ
2.31- Dap an diing la D. Cac phucfng trinh hoa hoc: Cac phucfng trinh hoa hoc:
BaCO, + 2 H C l ^ B a C l , +CO, T + SH^O y,^^, HCl + NaOH-^NaCI + HjO (2);r. " S6' mol cac chát:
3 94
"Bacx,, = ^ = 0,02 (mol); n„e, = 0,5.0,4 = 0,2 (mol).
Theo (1): Di hoa tan 0,02 mol BaCO, dung hét 0,04 mol HCl.
S6' mol HCl du la: 0,2 - 0,04 = 0,16 (mol).
Theo (2): nf,^^^^ = nHa,„ = 0,16 (mol).
Vaa NaOH = = 0,32 (lit) hay 320 (ml).
2.32. Dap an diing la B.
Cac phuong trinh hoa hoc: '^HÓffD
AlCL + 3NaOH -> AKOH), i + 3NaCl (1)
0,1 mol 0,3 mol 0,1 mol
AI(OH)3 +NaOHdu->NaA102+2H20 (2)
0,02 mol <- 0,02 mol i
S6 mol AKOH), i thu dupe la: 0,1 - 0,02 = 0,08 (mol).
Khoi luong két tiia AKOH), thu dupe la: 0,08.78 = 6,24 (gam).
IỊ BAI TAP T L / L U A N
2.33. - X la khi clo (CI2).
- Tinh s6' mol cac chát: n^,„o, = — = 0,8 (mol). • 87
"NaOH =0.5-4 = 2 (mol). - Cac phuong trinh hoa hpc:
MnOj + 4HC1 ^ MnCl^ + 2H2O + CL, t (1) 0,8 mol 0,8 mol
CIj + 2 N a O H ^ N a C l + NaC10 + H20 (2) 0,8 2.0,8 0,8 0,8 mol
.C^(NaCl) = C^(NaC10) = ^ = l , 6 M .
C^(NaOH) = ^ - ^ = 0 , 8 M .
tinh n6ng d6 phSn tram (C%) cua cac chát trong dung dich A , ta tinh khd'i lucmg dung dich A va kh6'i lucmg cac chat NaCl, NaClO va NaOH dụ Két qua:
C%(NaCl) =6,86%. C%(NaC10) -8,74%. C%(NaOHdu) = 2,35%.
2.34. a) Cac phucmg trinh hoa hoc:
C + O2 ^ C O j (1)
2C + O2 ^ 2 C 0 (2)
CuO + C O ^ C u + COj (3)
N h u vay toan b6 cacbon da chuy^n thanh khi COj. b) DSn khi CO2 vao nu6c v6i trong, xay ra phan ling:
CăOH)j + CO2 ^ CaCO, i + H^O (4) Theo (1), (2), (3) va (4): n^^co, ^n^o, ^ n ^ = ^ = 0,6(mol).
m^co, - 0,6.100 = 60 gam := ạ
2.35. Dat CTPT cua mu6'i cacbonat ngam nudrc: xNa2C03. yH20 K h i dun nong, thu dugc mu6'i khan: K h i dun nong, thu dugc mu6'i khan:
xNâCOj.yH^O—^xNâCO, + yH^O t
mH,„ = 3,1 - 2,65 = 0,45 (gam).
106 18
v a y c6ng thiic phan tCr ciia mu6'i cacbonat ngam nude: Na2C03.H20.
2.36. a) Cac phuomg trinh hoa hoc:
2CuO + C — ^ 2 C u + C 0 21 (1)
2PbO + C — 5 ^ 2 P b + C 0 21 . (2)
, COj+CăOH)2 ^ C a C O j + H^O (3)
Theo (1), (2), (3): n^ô ^n^^co, = ^ = 0,055 (mol). ^
1 uu
Dat X va y la s6' mol CuO va PbO c6 trong 10,23 gam h6n hop. Ta c6: '80x + 223y = 10,23
' 0 , 5 x + 0,5y = 0,055 ,5.--,. ?
Giai he phucmg trinh dugc: x = 0,1; y = 0,1. ' - Thanh phdn phSn tram theo kh6'i lucmg ciia cac oxit:
^""icuo = ? ? ^ . 100% = 78,2%. 10,23
2 2 3 0 ni
% m , _ = ' .100% = 21,8%. 10,23
2.37. a) Dat CTPT cda A la C^H^N,. Theo dSu bai ta c6 t i 16 kh6'i lucmg:
12x 53,33 ^ x = 2. 45 100 15,55 45 100 14z 31,12 •z = l . 45 100 v a y CTPT ciia A la C 2 H 7 N .
b) Cong thtic cáu tao c6 th^ c6 ciia A :
CH, - CH^ - H N j ; CH, - HN - CH,
(1) (2)
2.38. S6' mol khi va oxi: n^ = 0 ^ , , = 6,72 22,4 22,4
= 0,3 (mol). a) Dat X va y la s6' mol CO va CH4 c6 trong h6n hop A . Phuomg trinh hoa hoc d6't chay h6n hop A :
2CO + O, X 2 0,5x + 2 0 , ^4 ' y 2y
Theo dSu bai ta c6:
2C0, X mol COJ+2H2O 6'^ A jhG (1) (2) x + y =0,3 0,5x + 2y = 0,3
Giai he phuong trinh dugc: x = 0,2, y = 0,1. Kh6'i lugng cac chát trong h6n hgp A :
- Phfin tram thé tich: = ^ . 1 0 0 % = 66,7%; %^cu, -33,3%.