- Zn la nguyen t6'd c 62 electron hoa tri nhom IIB.
2Al+ 6H2S04d n^ Al2(S04)3 +3S02 t+ 6H20 (4)
Sd'mol SO2 = X + ^y = ^ = 0,4 (II)
Giai he phucfng trinh (I) vk (II) duoc x = 0,1 , y = 0,2.
vay h6n hop c6 0,2 mol Cu \k 0,4 mol Al.
7.60. Ddp in dung Ik C.
XCln + nAgN03 ^ nAgOi + X(N03)n
0,1 mol n.0,1 mol
43,05
= 0,3 mol-> 0,ln = 0,3 ^ n = 3.
Kh6i luong phan tir cua X(N03)3 = 24,2 0,1 = 242 -> Mv = 56 -> X = Fẹ
Vay mu6'i la FeCls . 7.61. Dap an dung la D.
Phircmg trinh phan ihig :
2R + 2nH2S04d,n R2(S04)n +nS02t + 2nH20
Ta c6 ti Id : = —14,93 0,4 ^ R = i8,667n. Chi c6 n = 3 va R =56 la phu hop. R la Fẹ
7.62. Dap s6'diing la Ạ
Phircmg trinh phan img: . . ' Fe304 +4C0 3Fe + 4 C 0 2 t (1)
Tru6c he't xac dinh CO du, vi neu CO he't, khi trong binh chi c6 CO2, ti
44
kh6'i se la — = 1,57 khac v6i d^ bai (1,457). 28 Du6i day 1^ m6t cich giai:
Goi x la s6' mol CO da phan irng, ta c6 :
0,5.28
1 04
Theo (1): nFe304 = - r-co = ^ = 0,1 mol ^ a = 0,1 .232 = 23,2 gam. 7.63. DdpsfídunglaC.
Dua vao dinh lu|n BTKL ta c6 :
KL(FeO + Fe203) + KL(CO) = KL(Fe) + KL( CO2 ) CO2 + CăOH)2 CaCOj i + HjO
m + 0,05.28 = 5,5 + 0,05.4 -> m = 6,3 gam. 7.64. C6ng thiic diing la C. 7.64. C6ng thiic diing la C.
234
Phirang trinh phan img hoa tan ;
2 FêOy +(6x-2y) H2SO4 d.n-» x Fe2(S04 )3 +(3x-2y) SO2 t+(6x-2y) H 2 O
2 24 120
nso, = =0.1 mol; nFe,(so4h = — =0.3 mol.
22,4 400 (adsbygoogle = window.adsbygoogle || []).push({});
Theo phurorng phuong trinh phan img :
3x-2y ^ ^ y 0.1 X 3 ^ ^ > - = - Fe304 0,3 V 4 ^ * 7.05. Dap an diing la B.
Dira vao 5 hien tu-ong dac liimg : '
1. CuO + 2HCI CuCh + H2O : dung dich mau xanh.
2. Mn02 + 4HC1 MnCl2 + C l 2 t + 2 H 2 O : khi m^u vang luc, mili hiẹ
3. Ag20+2HC1 ->2Aga>l' + H 2 O : ket tua trdng. 4. CuS + HCl kh6ng phan ling.
5. FeS + 2HC1 -> FeCl2 + H2S T : mui trumg th6'ị
7.66. Ddp an diing ^ Ạ
I Giai theo phuong phap cSn bang electron : Cu-2e = Cu2+ iSOj = Zn-2e = Zn 3,136 2+ S^^+6e = S« S^"+2e = S^+ = 0,14 mol; n^ = 0,64 = 0,02 mol. 22,4 " ' 32
Goi X va y la s6 mol Cu va Zn c6 trong 12,9 gam h6n hcrp. Tac6 : 64x + 65y = 12,9(1)
S6 mol electron cho = s6 mol electron nhSn :
2x + 2y = 0,14.2+ 0,02.6(11) Giai he (I) va (II), ta c6 : x = y = 0,1.
0,1.64
vay : % m c u =
12,9 .100% =49,61%; %mzn= 50,39%.
7 . 6 7 . Ddp s6 dung la B.
VicondirFe, HNO3 het va do d6 rci:N03)3 bikhiJrthanh Fe(N03)2 vdi
112 — 28
s6' mol bang s6 mol Fe da phan ling va bang — — =0,15. 56
Do d6 khd'i lircmg mu6'i Fe(N03 )2 = 0,15 .180 = 27 gam.
7 . 6 8 . Dap s6'diing la D.
Cdc phucfng trinh phan ihig :
Fe + 4 H N 0 3 Fe(N03)3 + N O t + 2H2O (1)
Wk Fe + 2Fe(N03)3 -)> Fe(N03)2 ( 2 ) ' Theo(l):nFe(i) = "NO = =0,15 mol.
^ nFe(2) = ^ -0,15 =0,05mol.
Theo ( 2 ) : npe(N03)2 = 3- nFe(2) = 0-05.3 = 0,15 mol.
-> nFe(N03)3 conlai = 0,15 -0,05.2 = 0,05 mol.
Tdng kh6'i lugmg mu6i trong X bang : 0,15.180 + 0,05.242 = 39,1 gam.
2 3 6 \ Chuang 8 PHRN 8I€T MOT S O CHRT V O C O CHURN D O D U N G DICH 8 . 1 . Dap an diing la C. 8 . 2 . Dap an dung la B. 8 . 3 . Dap an dung la Ạ
Lay tir m6i lo ra mot it dung dich de lam thi nghiem : Cho dung dich NaOH vao timg dung dich thi nghifim : (adsbygoogle = window.adsbygoogle || []).push({});