Cau 2. Dap an diing la D.
HSu hét cac kim loai phan ling duoc \di halogen, axit va oxị /
Cau 3. Dap an. dung la C. Cac phuomg trinh hoa hoc:
Fe + 2 H C l ^ F e C l 2 + H 2 T (1)
, . Cu + HCl: Khdng xay ra phan ung (2) Theo (1): S6' mol Fe = s6' mol = ^ — = 0,15 (mol).
->mp^-0,15.56 = 8,4 (gam). Thanh phSn % kh6'i luong cuạm6i kim loai:
M
10
%mc„ = 100% - 84% = 16%.
%mp, =-^.100% = 84%.
2. Tir L U A N
a) Cac phuong trinh hoa hoc:
Fe + 2HC1 -> FeClj + t (1) FeO + 2 H C l - ^ F e C l , + H 2 0 (2)
b) Thanh phSn % cac chat trong h6n hop ban dSu: Theo (1): S6' mol Fe = s6' mol H , = = 0,05 (mol)
' 22,4 mp, =0,05.56 = 2,8 (gam). mp^„ = 1 0 - 2 , 8 = 7,2 (gam). % m p , = 100% = 28%; %mp^.o = 72%. De so 1.5 1. T R A C N G H I E M K H A C H Q U A N
C a u 1. Cau CO n6i dung sai la C.
Phi kim tac dung vdi oxi tao thanh oxit axit.
cau 2. Dap an dung la B.
Phucfng trinh hoa hoc:
+ a, 2HC1 (khi hidro clorua).
C a u 3. Dap an dung la C.
Phuong trinh hoa hoc: Q,,; \ r Clj + 2KOH ^ KCl + KCIO + H^O u; \
San phdm phan ling la KCl, KCIO, HjO va KOH dụ
C ^ u 4. Dap an dung la Ạ
Phuomg trinh hoa hoc: :. .< .
CÔ + CăOH), ^ CaCO, i + H^O f>> - -
N^O + CăOH)2 ^ khSng c6 phan ling
Khi thoat ra la NjO, c6 th^ tich 1,12 lit (d dktc).
V i Mco, = MN,O = 44 nen % V = %m = %mol
1.12 4,48 C a u 5. Dap an diing la D. ^ %VN,O = - ^ . 1 0 0 % = 25%. u^H 2 5 J Theo d^bai: — ^ = ^ ^ = ^ mxH, X + 2 100 -> X = 32 (S)
Nguyen t6' X la luu huynh.
2. T l / LUAN
a) Phuong trinh hoa hoc:
= 0,0588.
C + 0 , - ^ C O , V ( 1 )
2C + 0 , - ^ 2 C O (2) ^
CuO + C O — ^ C u + CÔ (3)
Nhu vay sau phan ung (do du CuO) toan h6 cacbon da chuyen thanh C O 2 .
b) Dan khi C O 2 vao nude voi trong du, xay ra phan ung:
Ca(0H)2 + CO2 ^ CaCO, i + H^O ^, > ( 4 )
Theo (1), (2), (3) va (4): S6' mol cac chát: 7 2
" c a a ) , = " c o , = He = = 0,6 (mol). Kh6'i luorng kdt tua a = m^^co = 0,6.100 = 60 (gam).
De so 1.6
1. TRAC N G H I E M K H A C H Q U A N
C a u 1. Dap an diing la B. C a u 2. Dap an dung la C.
C a u 3. Dap an dung la C.
Cac phuong trinh hoa hoc:
^« ; CO + FeO Fe + CÔ t 3CO + FêO, 2Fe + SCOj t
(1) (2) (2)
Theo (1) va (2): CO da láy oxi cua cac oxit, tao ra C O 2 . S6' mo] nguyen tur trong cac oxit bang s6' mol CO va bang 0,2 mol.
Vay khoi luong oxi trong cac oxit la 0,2.16 = 3,2 (gam). Khd'i lugng Fe thu dirge sau phan ling la:
mp, =17,6-3,2 = 14,4 (gam).
C a u 4. Dap an dung la C.
Phuong trinh hoa hoc:
M C O , + 2 H C l^ M C l 2 + C02 t + H^O (1)
Theo (1): 1 mol mudi cacbonat -> 1 mol mudi clorua, tao ra: 1 mol khi CO2 va khoi lugng mudi tang:
( M + 7 l ) - ( M + 60) = ll(gam)
Theo de bai: Khdi lugng mudi tang = 5,1 - 4 = 1,1 (gam)
- > CO 0,1 mol C O 2 bay rạ
Vco, =0,1.22,4 = 2,24 (lit).
2 . T L / L U A N
Khi hap thu COj vao nude vdi trong, cd (hi xay ra cac phan dug:
CO2 + CăOH)2 -> CaCO, i + Bfi (1)
Neu du C O 2 thi xay ra phan ling:
CO2 + CaCOj + H^O ^ Ca(HC03 \)
Sd mol CO, = ^ = 0.1 (mol)
0
De sd 1.7
^ TRAC NGHIEM KHACH QUAN
C a u 1. Dap an dung la C.
Dfin hdn hop qua dung dich brom, etilen va axetilen phan irng vdi dung (jich brom va bi giu laị Metan khdng phan irng, thoat ra d dang tinh khiét.
C a u 2. Dap an dung la D.
Nhung y ntu sai:
4. Metan rát it trong nudẹ
7, 8. Metan khdng tham gia phan ling cdng va thé.
f # c d u 3 .
1. Dap an dung la B.
Khi dun ndng va cd Ni xiic tac, trong hdn hgp da xay ra cac phan ling:
22,4
= 0,075 < n „ , <2.n^ = 2.0,075 = 0,15 (mol).
' C a ( O H ) , — ^ " C O , ^ • ^ • " C a ( ( ) H ) ,
Do vay, theo (1) va (2): San pham sau phan ung gdm: CaCO, CăHCO02.'
Ni,l°
0 , 1 mol 0 , 1 mol 0 , 1 mol
C2H2 + 2 H , - ^
(1) (2) (2)
0,1 mol 0,2 mol 0,1 mol
Sau phan ung, trong hdn hgp cd: 0,1 + 0,1 = 0,2 (mol) C-^Uf^.
va 0 , 5 - ( 0 , 1 + 0 , 2 ) = 0,2 (mol) H2.
2. Dap an dung la C. Cac phirong trinh hoa hoc:
CjH, + 3,50^ 0,2 mol 0,7 mol ->2C02+3H20 2H, + O, •2H2O ^2 ' ^2 0,2 mol 0,1 mol
Sd mol O2 cfo đt chay hdn hgp X la: 0,7 + 0,1 = 0,8 (mol). - ^ V ^ =5.0,8.22,4 = 89,6 (lit).
2- T l / LUAN
a) Phucfng trinh hoa hoc:
C^H, + Bi^ -> C , H , B ^ 3,76 (3) ^ (4) (1) LL b ) T h e o ( l ) : n,^,, =n^^ =n,^„^„,^ = - ^ = 0,02(mol). Khdi lugng brom tham gia phan ling la:
c) The tích cua Q H , : V^^H, = 0,02.22,4 = 0,448 (h't).
%V^„ = - 5 ^ . 1 0 0 % = 13,3%.
% V c H^=86,7%.
, Desdl.S "
1. T R A C NGHIEM KHACH QUAN C a u 1. Tinh chat neu sai la C. C a u 1. Tinh chat neu sai la C. C a u 2. Tinh chat nSu sai la D . C a u 3. Dap an dung la D .
Tinh s6' m o l cac chat va kh6'i lugng cac nguyfen t6':
= ^ = 0,4 (mol) ^ m^ = 0,4.12 = 4,8 (gam). 10 8
n„,o = — ^ = 0,6 ( m o l ) - > m H =1,2.1 = 1,2 (gam).
" 1 8
Kh6'i luong cua h6n hop hidrocacbon la: m = 4,8+ 1,2 = 6,0 (gam).
C a u 4. Dap an dung la B.
Phuofng trinh hoa hoc:
C , H, + 3 0 2 — ! % 2 C 0 2 + 2 H 2 0 (1)
Theo (1): The tich = 3 the tich C2H4 = 3.5,6 = 16,8 (lit) b dktc.
- ^ V ^ =5.16,8 = 84 (lit).
2. T l / LUAN
Phuong trinh hoa hoc:
C a C, + 2 H 2 0^ C a ( O H ) , + C 2 H 2 t , • . (1)
a) T h e o ( l ) : nêc, =nc,„^ . 0 , 2 3 ( m o l ) .
Hieu suát phan irng 95%, ham lircmg canxi cacbua 90% ntn:
S6 m o l CaC, cSn lay la: Ó^^-'QO-^QQ = o,269 (mol).
^ 95.90
->mc,c^ =0,269.64 = 17,216 (gam).
b) Theo (1): So m o l Că0H)2 = so m o l CjH^ = 0,23 (mol). ^ m ^ o H ) , = 0,23.74 = 17,02 (gam). ^ m ^ o H ) , = 0,23.74 = 17,02 (gam).
D e s 6 1.9 ^ T R A C NGHIEM KHACH QUAN
C ^ u 1. cau sai la C. .in
Chat beo khong tan trong nudfc.
C a u 2. Dap an diing la C. ^cr^
Chat A la ruou etylic, tac dung dupe vdi Nạ
Chat B la benzen, kh6ng phan ling vdi Na, N a O H va N a j C O , .
Chat C la axit axetic, phan ung dupe vdi Na, N a O H va N a j C O , .
C a u 3. Dap an dung la C.
Cac phuong trinh hoa hoc:
( Q H , „ 0 , ) „ + n H 3 0 ^ 1 1^ n Q H . , 0 , (1)
Glucozo
n Q H „ 0 « > 2 n C , H , O H + 2 n C 0 , t (2)
Khd'i lupng tinh b6t tinh khiet la: i
80 •
1000. = 800(kg) 100 ^
Theo (1) va (2): S6' m o l C^H^OH = 2 s6' m o l tinh bpt (CgH.oOsX 800
= 2 . — = 9,8765(kmol). ' ^'^^ M ^ ^ ' ^ "
162
Hieu suát qua trinh chi dat 75% ntn kh6'i lupng rupu thu dupe la:
75 "100
mK^n. = 9 , 8 7 6 5 . 4 6 . ^ = 340,74 (kg). Z.TL/LUAN
- Phuong trinh hoa hpc o x i hoa glucozo:
Q H „ 0 , + A g , 0 - i ^ ^ Q H „ 0 , +2Ag>t (1) 39
Theo (1): S6' m o l A g = 2 s6' m o l glucozo = 2. (mol).
180 ,
.'I
39 75
- > m . = 2 . .108. = 35,l(gam). ,. ,.•
180 100 ' ' A . - . '