Zn + 2 H C l ^ Z n C l 2 + H21
CuO + 2HC1 -> CuClj + HjO ^ . Cu + HCl: kh6ng phan irng. < • • .
Lgc, thu dugc Cụ y.huT
- Cho dung dich NaOH du vao phSn nude lgc chiia ZnCl2 va CUCI2:
CuCl, + 2NaOH ^ Cu(0H)2 i + 2NaCl
ZnCl, + 2NaOH ^ Zn(OH)2 ^ + 2NaCl - •
Zn(0H)2 + 2NaOH du -> Na2Zn02 + 2H2O ^,
- Loc lay két tiia Cu(0H)2. Nung Cu(0H)2 thu h6i CuO:
Cu(0H)2 — ! ^ C u O + H2O t . Sue khi CO2 vao dung dich Na2Zn02:
NâZnO, + 2C0, + 2H2O -> Zn(0H)2 i + 2NaHC03.
lu h6i Zn tCr Zn(OH)2 theo so đ sau: . . _ . ; Zn(OH), > ZnCl, Zn.
5 . 3 1 . a) Cac phuong trinh hoa hgc:
, Mg + 2 H C l ^ M g C l , + H 2 1 (1)
4 48
b) Theo(l): n^ ^ = n H ^ = - ^ - 0 , 2 ( n i o l ) .
n„c„n =2nH^ =0,4(mol). ^ m ^ ^ = 0 , 2 . 2 4 = 4,8(gam).
^ M g o = 8,8 - 4,8 = 4 (gam) hay 0,1 (mol). c) Theo (2): n„c„„ - 211^^0 - 0,2 (mol).
" H C I = " H C K I) + nHci(2) = 0'6 (mol). The tich dung dich HCI phai dung:
V = = 0,4 (lit) hay 400 (ml). 1,5
5.32. a) Dat CTPT cCia A la C,H Ọ ' x * y
Phiiong trinh d6't chay A:
y 1
x + - —
4 1)
Tinh s6' mol cac chát:
nccx = ^ - 0,35 (mol); n„^o = = 0,252 (mol).
Dua vao phuong trinh phan ung va s6' mol cac chát, tim duoc CTPT ci A la CjHfiỌ
b) V i A + Na ^ t nen A la rugu, phan tur c6 m6t nhom - O H .
v a y CTCT ciia A la C2H,OH.
5.33. - S6' mol cac khi: n = — = 0,25 (mol).
22,4
^ n ^ , =0,25.4,16% = 0,0104 (mol); H^H^ =0,25.89,6% = 0,224.
n^^H^ = 0,25.2,24% = 0,0056 (mol); n^ô = 0,01 (mol). - Cac phirong trinh phan ling chay:
CH, + 20j ^ CO2 + 2H2O (1)
C,H, + | o , -> 2CO, + SH^O (2)
S6' mol CO2 = 0,224 + 2.0,0056 + 0,01 = 0,2452 (mol). DSn khi CO2 vao dung dich NaOH xay ra phan ling:
2NaOH + CO, ^NâCOj+H^O (3)
NaOH + CÔ-»NaHC03 (4)
son 8
S6' mol NaOH = ^ ^ ^ ^ = 1 (mol) ^ S6' mol NaOH > 2 s6' mol CO,. 40.100
Nhir vay phan ung (4) khdng c6, sau phan ung trong dung dich chi c6 rtiud'i NajCO, vdri s6' mol = 0,2452 mol (bang sG' mol CO2).
S6' mol NaOH du = 1 - 2.0,2452 = 0,5096 (mol).
Khfíi lugng dung dich: m^^ =500 + 10,7888 + 8,3664 = 519,1552(gam).
vay C%(Na,CO,) = "'^^^^'^"^.100% = 5%. 519,1552 C%(NaOH du) = 100% = 3,93%. 519,1552 11,2 5.34. So mol O, = — ^ = 0,5 (mol). 22,4 a) Phuong trinh hoa hoc:
CH, + 2O2 COj + 2H2O
0,25 <-0,5 ^ 0 , 2 5 mol b) The tich cac khi:
VcH, =Vco, =0,225.22,4 = 5,6 (lit). 5.35. a) Tinh s6' mol cac chát:
22 13,5
''CO, = — = 0,5(mol); n„^o = — ^ = 0,75(mol). 44 18
Dat CTPT cua hidrocacbon la C H.,. Phuang trinh hoa hoc:
C . H , + 4 O,-^xCO, + ^ H , 0 4 O,-^xCO, + ^ H , 0 1 mol a mol -> x mol ^ m o l 0,5 mol 0,75 mol • -> y = 3x (I) i.,i,-,t(- 0,5 1,50 12x + y = 30 (II) v :
jiai he phuong trinh (I) va (II) dugc: X = 2, J ' = 6. b) Cong thiic phan tir cua A la: CH, - CH3.
CjHfi la hidrocacbon no c6 tinh chat tucfng tu metan: tham gia phan ung thé voi CI2, phan ling chaỵ..
8 96
5.36. - S6' mol h6n horp: n^h = —— = 0,4 (mol). 22,4 22,4
a) Cho h6n hop vao binh dung dung dich nu6c brom, xay ra cac phan ung: CH, = CH2 + Bii ^ CH^Br-CH^Br (1)
Goi X, y la s6' mol C2H4 va C2H2 trong h6m hop.
Kh6'i lugfng binh nu6c brom tang chinh la khoi luong cua 8,961ft h6n hop. rx + y = 0,4
Theo Mu bai ta c6: < ' , , < .
[28x + 26y = l l
T h e o ( l ) v a ( 2 ) : ' , . y. ^ i / . ^ ) . ,
Giai he phircfng trinh duac: X = 0,3, y = 0,1.
%v =M.ioo% = 75%; %Vc,„ =25%. w .
b) Phucfng trinh hoa hoc: ; , s > , „ > , ,
q H , + ^ 2CO2 + 2H2O .' - (3)
0,3 mol 3.0,3 2.0,3 ''>'"-'•
' ' • + - O j ^ 2CO2 + H^O (4)
0,1 0,25 0,2
^ H o , = 3.0,3 + 0,25 = 1,15 (mol)-^V„, =1,15.22,4 = 25,76 (lit).
Hco, = 2.0,3 + 0,2 = 0,8 (mol) ^ V^^^ = 0,8.22,4 = 17,92 (lit).
5.37. a) S6' mol H , = = 0,1 (mol). 22,4 22,4
Phucfng trinh hoa hoc:
e ' . ^ 2CH3 - CH2OH + 2Na -> 2CH3 - CH^ONa + t (1)
CH3 - O - CH, + Na: khong c6 phan ling. (2)
Theo (1): S6' mol C2H,0H = 2 so mol H2 = 0,2 (mol)
mc,H,oH = 0'2.46 = 9,2 (gam).
, . %ni,,.H,oH =||.100% = 40%; %m^,„,^„ =60%.
b) So mol H , 0 = — = 1,2 (mol); S6' mol H , = — = 0,05 (mol).
^ 1 8 22,4 Phuomg trinh hoa hoc:
CjH^OH + -> 2C0, + 3H2O (3)
., (CH, )2 O + 3O2 ^ 2CO2 + 3H2O (4) 2C2H,OH + 2Nâ2CjH,ONa + H21 (5)
Theo (5): n^^^^^^^^ =2.nH, =0,05.2 = 0,1 (mol).
Theo(3) vă4): n^^,^ =^(l,2-0,1.3) = 0,3(mol).
Phan III
GI6I THIEU MOT SO DE KIEM TRA
A . DE KIEM TRA
Ị Bi K I ^ M T R A 15 PHUT ; ^
... , D e s o l . l :ii-iyh^ -i
1. TRAC NGHIEM KHACH QUAN (5 DIEM) ^ '^^^
. Cau 1 (1,5 diem) ^
Nhimg chat khi nao sau day lam đi mau giáy quy tim im thanh dỏ
Ạ S02vaC0. B. S02vaC02. C. C02vaN0. D. COvaN20s. C. C02vaN0. D. COvaN20s.
' Cau 2 (1,5 diém)
Chi dung thu6'c thir la BăOH)2 c6 th^ nhan biét tijmg dung dich trong day chat nao sau daỷ
Ạ H^SỘFeClj.NaCl. B. H2SO,,NaOH,Na2C03.
C. BaCl2,HCl,NaOH. D. HCl.CaClj.KOH.
Cau 3 (2,0 diem)
Hoa tan 50 gam CaCO, vao dung dich HCl dụ Hieu suát phan ling la 85%. Thé tich khi thu ducfc (cf dktc) la
Ạ 11,2 lit. B. 9,52 lit. ' " ^ , C. 22,4 lit. D. 13,18 lit. r,,^>
2. TLrLUAN(5 0IEM) ' ^ ^ ' ' " , . . iii^^^
Hoa tan hoan toan 11,0 gam h6n hcfp b6t Fe va A l bang dung dich
H2SO41M viia du, thu dirge 8,96 lit khi H2 (6 dktc).
a) Viet cac phucfng trinh hoa hgc. ' b) Tinh khd'i lugng m6i kim loai c6 trong h6n hop daụ
D e s d l . 2 C NGHIEM KHACH QUAN (5 DIEM) Cau 1(1,5 diem)
Nhiing cap chát nao sau day phan dug duac vdri dung dich HCl va dung ^ich H2SO4 loang?
Ạ Cu va ZnỌ B. Zn va CuỌ