B a i 4. Cho V lit dung dich A chufa dong th6i FeCl3 I M va Fe2(S04)3 0,5M
tac dung vdfi dung dich Na2C03 c6 du, phan ufng ket thiic thay khoi
lirong dung dich sau phan iJng giam 69,2 gam so vdi tong khoi li/ong cua cac dung dich ban daụ Gia t r i cua V la: cua cac dung dich ban daụ Gia t r i cua V la:
Ạ 0,2 lit. B. 0,24 lit. C. 0,237 lit. D.0,336 lit.
Bai 5. Cho luong khi CO di qua 16 gam oxit sat nguyen chat duac nung
nong trong mot cai ong. IChi phan ufng thiic hien hoan toan va ket
thiic, thay khoi liTOng ong giam 4,8 gam. Xac dinh cong thiJc va ten oxit sat dem dung. oxit sat dem dung.
Bai 6. Dung CO de khuf 40 gam oxit Fe203 thu difcfc 33,92 gam chat rSn B gom Fe203, FeO va Fẹ Cho - B tac dung vdi H2SO4 loang dir, thu B gom Fe203, FeO va Fẹ Cho - B tac dung vdi H2SO4 loang dir, thu
2
dacfc 2,24 lit khi H2 (dktc).
Xac dinh thanh phan theo so mol chat ran B, the tich khi CO (dktc)
toi thieu de c6 duoc ket qua naỵ
Bai 7. Nhiing mot thanh sat nSng 12,2 gam vao 200 ml dung dich CUSO4
0. 5M. Sau mot thdi gian lay thanh kim loai ra, c6 can dung dich diroc 15,52 gam chat ran khan. 15,52 gam chat ran khan.
a) Viet phLfOng trinh phan ling xay ra, tim khoi Itfcfng tCrng chat c6 trong 15,52 gam chat r^n khan. trong 15,52 gam chat r^n khan.
b) Tinh khoi lucfng thanh kim loai sau phan ling. Hoa tan hoan toan thanh kim loai nay trong dung dich HNO3 dSc nong, du thu dtfOc thanh kim loai nay trong dung dich HNO3 dSc nong, du thu dtfOc
khi NO2 duy nhat, the tich V lit (do d 27,3 °C, 0,55 atm). Viet cac phuong trinh phan ufng xay rạ Tinh V. phuong trinh phan ufng xay rạ Tinh V.
Bai 8. Ngam mot thanh dong c6 khoi lirong 140,8 gam vao dung dich
AgN03 sau mot thdi gian lay thanh dong dem can lai thay nSng 171,2
gam. Tinh thanh phan khoi liTcyng cua thanh dong sau phan iJng.
Dap an
1. B 2. D. 3. B. 4. Ạ 5. FegOs.
6. Vco = 8,512 l i t ; %nFe = 46,51% ; %nFeo = 37,21% ;%nF,^o, = 16,28%.
7. a) 6,4 gam CUSO4 va 9,12 gam FeS04.
b) m K L = 12,68 gam; V^ô = 26,88 lit.
VIIỊ PHlJONG P H A P GIA T R j T R U N G BJNH
Nguyen tSc cua p h L f O n g phap n h u sau: K h o i Itfcfng phan tuf trung binh ( K L P T T B ) ( k i hieu M ) cung n h i / k h o i l i T O n g nguyen tuf t r u n g binh ( K L N T T B ) c h i n h l a k h o i lifong cua m o t m o l h 5 n hop, nen no diioc t i n h theo cong thufc:
^ _ t6ng khoi luong h6n hap (tinh theo gam) long so mol cac chat trong h6n hop
_ M,n, + M , n , + M3n3 + ••• _ X ^ j ^ j ( D
n ,+ n 2 + n3 + ... J n ^
trong d o Mi, M a , - l a K L P T (hoSc K L N T ) cua cac chat trong h o n hop;
ni, n2,... la so mol ttfong u f n g cua cac chat. Cong thufc (1) CO the viet t h a n h :
M = M , . ^ + M ^ . ^ + M 3 . ^ + ...
Z^. 2. " i
M = M,x, + M2X2 + M3X3 + ... (2)
trong d o xi, X 2 , . . . l a % s o m o l ti/ong u f n g (cung chinh l a % k h o i luong) cua cac chat. DSc biet doi v d i chat k h i t h i xi, X2, ... cung chinh la % the tich nen cong thufc (2) c6 the viet t h a n h :
- M,V,+M,V,-hM3V3^-..._lM.V,
^ - v , . v , . V 3 . . . . ' - Zv.
trong d o V i , V2,... l a the t i c h cua cac chat k h i . N e u h o n hop chi c6 2
chat t h i cac cong thufc (1), (2), (3) tuong ufng t r d t h a n h (1'), (2'), ( 3 ' )
nhif sau:
n
trong d o n la tdng s o só mol cua cac chat trong hon hop,
M = M i X i + M 2( l- X i ) (2')
trong do con so 1 iJng vdi 100% va
^ ^ M,V.+M,(V-V.) ^g,j
trong do V i l a the tich k h i thdf n h a t va V la tong the tich h o n hap.
Tii cong thufc t i n h K L P T T B t a suy r a cac cong thufc t i n h K L N T T B .
Vdi cac cong thuTc:
C . H ^ O , ; n , m o l
C^.HỵO,.; n^mol
Ta c6: - Nguyen tijf cacbon t r u n g binh: x =
n, + + . . . - Nguyen tuf hidro t r u n g binh: y -. - _ y i H , + y ^ n ^ +
n , H: n2 + ...
BAI TAP AP D U N G
B a i 1 : (Khoi A - TSDH nam 2007):
Cho 4,48 l i t hon hop X (d dktc) gom 2 hidrocacbon mach h d l o i iii tii
qua binh chufa 1,4 l i t dung dich Br2 0,5M. Sau k h i phan iJng hoan toan, so mol Br2 giam d i m o t niJfa va k h o i li/ong binh tang t h e m 6,7 gam. Cong thufc phan tuf cua 2 hidrocacbon la
Ạ C2H2 va C4H6. B. C2H2 va C4H8.
C. C3H4 va C4H8. D . C2H2 va C3H8.
Hii&ng dan gidi
4 48
"hhxhhx = ' = 0,2 mol 22,4
nBr.,ba„diu = 1,4x0,5 = 0,7 mol 0 7
^^Br,p«i,g = = 0>35 mol. (giam mot nufa la k h o i li/Ong phan ufng) Khoi lifOng binh Br2 tang 6,7 gam la so gam ciia hidrocabon khong nọ Dat CTTB ciia h a i hidrocacbon mach h d la C^H2s^2-2s so lien ket 7t trung binh).
Phuong t r i n h phan iJng: C-H,-^,_,- + aBr, > <^T^^2T,.i-2^h^
0,2 mol -> 0,35 m o l 0 ^5 6 7
^ _ ^ ^ ^ _^ i 4 n + 2 - 2 a = - ^ ^ i i = 2,5.
0,2 0,2 Do h a i hidrocacbon mach h d phan ufng hoan toan v d i dung dich Br2
nen chiing deu la hidrocacbon khong nọ Vay h a i hidrocacbon do la
C2H2 va C4H8. Chon dap a n B.
B a i 2: Hoa tan hoan toan 2,84 gam hon hop h a i muoi cacbonat cua hai
hoan bang dung dich H C l ta t h u dirgrc dung dich X va 672 m l CO2 (a dktc).
1. Hay xac d i n h ten cac k i m loaị
Ạ Be, M g . B. M g , Cạ C. Ca, Bạ D. Ca, Sr. 2. Co can dung dich X t h i t h u duoc bao nhieu gam muoi khan?
Ạ 2 gam. B. 2,54 gam. C. 3,17 gam. D. 2,95 gam.
Hiidng dan gidi
1. Goi A la k i m loai dai dien cho 2 k i m loai can t i m Cac phtfong t r i n h phan ilng la
ACO3 + 2HC1 > ACI2 + H2O + C02t
0,03mol <- 0,03mol n^o = ^ ^ - 0 , 0 3 m o l .
''''' 22,4
Vay K L P T T B cua cac muoi cacbonat la
M = ^ = 94,67 va MA,B = 94,67 - 60 = 34,67
0,03
V i thuoc 2 chu ky lien tiep nen hai k i m loai do 1^ M g ( M = 24) va Ca ( M = 40).
Chon dap an B.
2. KI^PTTB cua cac muoi clorua:
M „ „ « c , c ™ . = 34,67+ 71 = 105,67.
Hin^uoi clorua = 105,67 . 0,03 = 3,17 gam. Chon dap an C.
B a i 3: H5n hop k h i SO2 va O2 c6 t i k h o i so vdi CH4 bang 3. Can them bao nhieu l i t O2 vao 20 l i t hon hap k h i do de cho t i k h o i so v d i CH4
giam di - , tufc bkng 2,5. Cdc h6n hop k h i d cung dieu k i e n nhi$t do 6
va ap suat.
Ạ 10 l i t . B. 20 l i t . C. 30 l i t . D. 40 l i t .
Hiidng dan gidi
Cdch 1:
Goi X la % the tich cua SO2 trong hon hop ban dau, ta c6:
M = 16.3 = 48 = 64.x + 32(1 - x) ^ x = 0,5
Vay: moi k h i chiem 50%. NhiT vay trong 20 l i t , moi k h i chiem 10 l i t . Goi V la so l i t O2 can t h e m vao, ta c6:
M ' = 2,5.16 = 40 = ^ ^ , i ^ ± 3 2 a 0 l V )
2 0 + V
Giai ra CO V = 20 h t .
Chon dap an B.
Cdch 2:
Hon hop k h i ban dau coi nhiT k h i thuf nhat (20 l i t c6 M = 16x3 = 48), con O2 t h e m vao coi nhif k h i thur hai, ta c6 phiTOng t r i n h :
M = 2.5.16 = 40 = i M ^ ± ^ ,
2 0 + V Rut ra V = 20 l i t . Rut ra V = 20 l i t .
Chon dap an B.
B a i 4: Co V l i t k h i A gom H2 va hai olefin la dong dang lien tiep, trong do H2 chiem 60% ve the tich. D a n h6n hc*p A qua bot N i nung ,n6ng
I dugtc h8n hcfp k h i B. Dot chay hoan toan k h i B di/dc 19,8 gam CO2 va 13,5 gam H2Ọ Cong thiJc cua hai olefin la
Ạ C2H4 va C3H6. B. CaHe va C4H8. C . C4H8 va C5H10. D. C5H10 va CgHiz.
HUofng dan gidi
D a t CTTB cua hai olefin la C^H^s.
(j cung dieu kien nhiet do va ap suat t h i the tich ty le v d i só moi k h i .
H 6 n h o p k h i A c 6 : = M . 2
n „ 0,6 3
Ap dung dinh luat bao toan khói luong va dinh luat bao toan nguyen
tuf Dot chay h5n hop k h i B cung chinh la dot chay hon hop k h i Ạ Ta c6: C . H ^ R + — > i i CO2 + i i H2O (1)
2
Theo phuong t r i n h (1) ta c6: n^ô = n„^o = 0>45 mol. 0,45 , =^ "c„n,„ = mol.
Tong: n„^,, = ^ = 0,75 mol => n„^o,^.,, = 0,75 - 0,45 = 0,3 mol => n,,^ = 0,3 mol.
T a c 6 : ^ = - M 5 . 2 ^ _ ^
n,,^ 0,3.n 3
=> H a i olefin dong d^ng lien tiep la C2H4 va CsHẹ Chon dap an B.
B a i 5: Tach nUdc hoan toan til hon hop X gom 2 ancol A va B ta duoc
hon hop Y gom cac olefin. Neu dot chay hoan toan X t h i thu dtfdc 1,76 gam C O 2 . K h i dot chay hoan toan Y t h i tong khoi lucfng H 2 O vk CO,
tao ra la
Ạ 2,94 gam. B. 2,48 gam. C. 1,76 gam. D. 2,76 gam.
HU&ng dan gidi
Hon hop X gom hai ancol A va B tach n\l6c dtroc olefin (Y) -> hai
ancol la ruou no, don chiJc.
Dat CTTB cua hai ancol A, B la C-II^j^^OH ta c6 cac phiicfng trinh phan iJng sau:
C„H2„,,0H + — > nCÔ + (n + DH^O 2 C„H,„,.OH C , H , , + H 2 O (Y) C„H,„ + ^O, nCO, + h H , 0 Nhan xet:
- K h i dot chay X va dot chay Y cumg d i o so mol C O 2 n h u nhaụ
- Dot chay Y(olefin hay anken) t h i n^ô = M^^Q .
Vay dot chay Y cho tong
(mcô + m„^o) = 0,04.(44 + 18) = 2,48 gam. Chon dao an B.
•
B a i 6: Trong tu n h i e n , dong (Cu) t o n t a i dudi hai dang dong vi " C u va *,Cu . K L N T (xap xi khoi liiOng trung binh) cua Cu la 63,55. T i n h % ve khoi luong cua m6i loai dong v i .
Ạ '^^Cu: 27,5% ; "^^Cu: 72,5%. B. ^^Cu: 70% ; ^^Cu: 30%. C. '^^Cu: 72,5% ; '^^Cu: 27,5%. D. ^^Cu: 30% ; ^^Cu: 70%.
HiCcfng dan gidi
Goi X la % cua dong v i ^^Cu ta c6 phucrng t r i n h : M = 63,55 = 65.x + 63(1 - x) :^ x = 0,275
Vay: dong v i ^^Cu chiem 27,5% va dong v i ^^Cti chiem 72,5%. Chon dap an C.
B a i 7: Dot chay hoan toan a gam h6n hop hai rUcfu no, đn chuTc lien tiep trong day dong d i n g thu di/oc 3,584 lit C O 2 d dktc va 3,96 gam H 2 O . T i n h a va xac dinh CTPT cua cac rirgụ
Ạ 3,32 gam; C H 3 O H va C 2 H 5 O H . B. 4,32 gam; C 2 H 5 O H va C 3 H 7 O H .