BAI TAP VAN D g N G
B a i 1: Hoa tan 3,28 gam h5n hgp muoi M g C l 2 va Cu(N03)2 vao n\ldc dirge
dung dich Ạ Nhiing vao dung dich A mot t h a n h sSt. Sau mot khoang thdi gian lay t h a n h sSt ra can lai thay tang t h e m 0,8 gam. Co can dung dich sau phan ufng t h u dirge m gam muoi khan. Gia t r i m 1^
Ạ 4,24 gam. B. 2,48 gam. C. 4,13 gam. D. 1,49 gam.
HU&ng dan giai
Ap dung dinh luat bao toan khoi lirgng: Sau mot khoang thdi gian do tSng ' khoi lirgng cua thanh Fe bSng do giam khoi lirgng eua dung dich muoị
Do do: m = 3,28 - 0,8 = 2,48 gam. Chon dap an B.
B a i 2: {CD - Khoi A 2007): Cho 5,76 gam axit hufu ca X dan chdc, mach
ho tac dung het v d i CaCOa t h u dirge 7,28 gam muoi cua axit hiJu cỌ
Cong thiJc cau tao t h u ggn cua X la
Ạ C H 2 = C H - C O O H . B. C H 3 C O O H .
HUdng dan giai
D a t CTTQ cua axit hOru ccf X don chufc la RCOOH.
2 R C 0 0 H + CaCOs > (RC00)2Ca + COzt + H 2 O
Cuf 2 mol axit p h a n lifng tao muoi thi khoi lirgng tang (40 - 2) = 38 gam.
X mol axit < (7,28 - 5,76) = 1,52 gam.
(40 la Ca vao - 2 la H ra 2 R C 0 0 H > (RC00)2Ca)
1^52^ ^ ^ ^ ^ 5 ^ ^ ^2 ^ R = 27
38 0,08 ^ A x i t X: C H 2 = C H - C O O H . ^ A x i t X: C H 2 = C H - C O O H .
Chon dap an Ạ
B a i 3: Hoa tan hoan toan 23,8 gam hon hop mot muoi cacbonat cua kim loai hoa tri (I) va mot muoi cacbonat ciia k i m loai hoa tri (II) bSng dung
dich H C l thay thoat ra 4,48 lit k h i CO2 (dktc). Co can dung dich thu
diicfc sau phan umg thi khoi lugng muoi k h a n thu diicfc la bao nhieủ
Ạ 26,0 gam. B. 28,0 gam. C. 26,8 gam. D. 28,6 gam.
HUcfng dan gidi
Cur 1 mol muoi cacbonat tao t h a n h 1 mol muoi clorua cho nen k h o i
luong muoi k h a n t a n g (71 - 60) = 11 gam, ma J^CÔ ~ " m u t f l cacbonat = 0,2 mol.
Suy ra k h o i Itfcfng muói k h a n t a n g sau p h a n ufng l a 0,2.11 = 2,2 gam. Vay tong k h o i liTOng muoi k h a n thu difdc la 23,8 + 2,2 = 26 gam.
Chon dap an Ạ
B a i 4: Cho 3,0 gam mot axit no, don chiJc A ta c dung vCfa du v d i dung dich N a O H . Co c a n dung dich sau p h a n iJng thu dtfoc 4,1 gam muoi
k h a n . CTPT cua A la (adsbygoogle = window.adsbygoogle || []).push({});
Ạ H C O O H B. C3H7COOH C. CH3COOH D. C2H5COOH.
HU&ng dan gidi
Cuf 1 mol axit don chufc tao t h a n h 1 mol muoi th i k h o i li/cJng tang (23 - 1) = 22 gam, ma theo dau bai khoi liTOng muoi tang (4,1 - 3) = 1,^ gam nen so mol axit la
1 1 3
naxit = = 0,05 mol. -> Maxit = — — = 60 gam. D a t CTTQ cua axit no, don chuTc A la C„H2n+ i C 0 0 H nen t a c6:
14n + 46 = 60 n = 1.
Vay CTPT cua A la CH3COOH.
Chon dap an C.
p a i 5: Ngam mot vat b&ng dong c6 khoi lUdng 15 gam trong 340 gam dung dich AgNOa 6%. Sau mot thcJi gian lay vat r a thay khoi luong AgNOa dich AgNOa 6%. Sau mot thcJi gian lay vat r a thay khoi luong AgNOa trong dung dich giam 25% Khoi liiOng cua vat sau phan ufng la
Ạ 3,24 gam. B. 2,28 gam. C. 17,28 gam. D. 24,12 gam.
Hitdng dan gidi
_ _ 3 4 0 ^ _
n A g N o , , p h . . , g . = 0 , 1 2 . — = 0,03 mol. 25 100
Cu + 2AgNO,3 > Cu(N03)2 + 2Agi 0,015 <- 0,03 > 0,03 mol
n i v a t sau phan ijfng — H l v a t ban dau + ^ l A g (bam) "~ m C u (tan)
= 15 + (108.0,03) - (64.0,015) = 17,28 gam. Chon dap a n C .
B a i 6: Cho dung dich AgNOs dir tac dung vdi dung dich hon hgtp c6 hoa
tan 6,25 gam hai muoi K C l va K B r t h u C^iac 10,39 gam hon hop AgCl
va AgBr. Hay xac d i n h so mol h6n hop daụ
Ạ 0,08 mol. B. 0,06 mol. C. 0,03 mol. D. 0,055 mol.
Hitdng dan gidi
CUT 1 mol muoi halogen tao t h a n h 1 mol ket tua
> k h o i li/(?ng tang: 108 - 39 = 69 gam; 0,06 mol < khoi li/ong tang: 10,39 - 6,25 = 4,14 gam. Vay tong so mol h o n hop dau la 0,06 mol.
Chon dap a n B.
B a i 7: N h i i n g mot t h a n h graphit diiOc phu mot Idp k i m loai hoa t r i (II) vao dung dich CUSO4 di/. Sau phan ufng k h o i lii(?ng cua t h a n h graphit giam di 0,24 gam. Cung t h a n h graphit nay neu difoc nhiing vao dung dich AgN03 t h i k h i phan ufng xong thay k h o i liiong t h a n h graphit tang len 0,52 gam. K i m loai hod t r i (II) la k i m loai nao sau daỷ Ạ Pb. B. Cd. C. A l . D. Sn.
Hii&ng dan gidi
Dat k i m loai hoa t r i (II) la M v d i so gam la x (gam). M + CUSO4 > MSO4 + Cu
Cuf M gam k i m loai tan r a t h i se c6 64 gam Cu bam vaọ Vay k h o i
li^Ong k i m loai giam (M - 64) gam;
Vay: x (gam) = ^''^'^•^ ^ ^^loi liiong k i m loai giam 0,24 gam. (adsbygoogle = window.adsbygoogle || []).push({});
M a t khac: M + 2AgN03 > M(N03)2 + 2Ag
Cuf M gam k i m loai tan r a t h i se c6 216 gam Ag bam vaọ Vay khS'i
luang k i m loai tSng (216 - M ) gam;
0 52 M
Vay: x (gam) = — — ' - — < k h o i luong k i m loai tSng 0,52 gam. 216
^ , 0,24.M 0,52.M - . - . o n - i • nÂ
Ta c6:— = — > M = 112 ( k i m loai Cd). M - 6 4 2 1 6 - M
Chon dap an B.
B a i 8: Co 1 l i t dung dich hon hop NazCOa 0,1 mol/1 vk (NH4)2C03 0,25
mol/1. Cho 43 gam h6n hgfp BaCh va CaCh vao dung dich dọ Sau k h i cac phan ufng ket thiic ta t h u diTcfc 39,7 gam ket tiia A va dung dich B. T i n h % k h o i luTOng cac chat trong Ạ
Ạ % m 3 , ^ ^ = 50%, %mc„co, = 50%-
B. %m^,co, = 50,38%, %m^^^ = 49,62%. C. %m^co, = 49,62%, %mc„co, = 50,38%. C. %m^co, = 49,62%, %mc„co, = 50,38%. D. Khong xac d i n h duoc.
Ht^&ng dan gidi
Trong dung dich: NazCOg > 2Nâ + COg^'
(NH4)2C03 > 2NH4" + COâ- BaClz > Bâ" + 2Cr
CaCla > Câ^ + 2Cr
Cac phan iJng: Bâ^ + COg^- > B a C 0 3 i (1)
Cá" + CO3'- > CaCOg^ (2) Theo (1) va (2) cuf 1 mol BaCl2, hoSc CaClz bien t h a n h BaCOg hoSc CaCOg t h i k h o i iLfOng muoi giam (71 - 60) = 11 gam. Do do tong so
43 — 39 7
mol hai muoi BaCOg va CaCOg bang: ' = 0,3 mol
Ma tong so mol COg^' = 0,1 + 0,25 = 0,35, dieu do chufng to du COg^'.
Goi X , y la so mol BaCOg va CaCOg trong A ta c6:
X + y = 0,3 ' l 9 7 x + 100y = 39,7 = > X = 0,1 mol; y = 0,2 mol. n i l Q7 T h a n h phan cua A: % m 3 , c o 3 = ' .100 = 49,62%; ' 39,7 % " i c a c o , = - 49,62 = 50,38%. Chon dap an C.
a i 9: Hoa t a n hoan toan 104,25 gam hon hop X gom NaCl v^ N a l vao Vnuf<Jc dugc dung dich Ạ Sue k h i CI2 du vao dung dich Ạ K e t thuc t h i
nghiem, c6 can dung dich t h u daoc 58,5 gam muoi k h a n . K h o i laong
NaCl C O trong h6n hop X la
Ạ 29,25 gam. B. 58,5 gam. C. 17,55 gam. D. 23,4 gam.
Hiiceng dan gidi
K h i CI2 dii chi khijf difcfc muoi N a l theo phifofng t r i n h 2 N a I + CI2 > 2NaCl + I2
CiJ 1 mol N a l tao t h a n h 1 mol N a C l > K h o i lUdng muoi giam 127 - (adsbygoogle = window.adsbygoogle || []).push({});
35,5 = 91,5 gam.
Vay: 0,5 mol < Khoi lugrng muoi giam 104,25 - 58,5 = 45,75 gam.
=> mNai = 150.0,5 = 75 gam mNaCi = 104,25 - 75 = 29,25 gam. Chon dap an Ạ
B a i 10: Nhiing mot t h a n h k e m va mot t h a n h s^t vao cung mot dung dich C U S O 4 . Sau mot t h 6 i gian lay hai t h a n h k i m loai r a thay trong dung dich c6n l a i c6 nong do mol ZnS04 bang 2,5 Ian nong do mol
FeS04. M a t khac, k h o i liTcfng dung dich giam 2,2 gam. K h o i lifcfng
dong bam len t h a n h k e m va bam len t h a n h ski Ian liiOt la
Ạ 12,8 gam; 32 gam. B. 64 gam; 25,6 gam. C. 32 gam; 12,8 gam. D. 25,6 gam; 64 gam.
Hii&ng dan gidi
Yi trong cung dung dich c6n l a i (cung the tich) nen:
[ZnS04] = 2,5 [FeS04]
^ "znso, =2,5np^sô
Zn + C U S O 4 > ZnS04 + C u i (1)
2,5x <- 2,5x < 2,5x mol
Fe + CuS04 >FeS04 + Cu^ (2)
X <- x < X - > X mol
TCr (1), (2) n h a n dirac do giam k h o i li/ang cua dung dich la
nicu (b^m) - nizn (tan) " T^Fe (tan)
^ 2,2 = 64.(2,5x + x) - 65.2,5x - 56x
= > X = 0,4 mol.
Vay: m c u ( W m l e n t h a n h kgra) = 64.2,5.0,4 = 64 gam;
m c u ( b d m l e n t h a n h s ^ t ) = 64.0,4 = 25,6 gam. Chon dap an B.
^ a i 1 1 : N h u n g t h a n h kem vao dung dich chiJa 8,32 gam CdS04. Sau k h i khuf hoan toan ion Cd^^ k h o i luong t h a n h kem tSng 2,35% so v d i ban
daụ H o i k h o i \\iCng t h a n h k e m ban daụ
HiCcfng đn giai
Goi k h o i luong t h a n h k e m ban dau la a gam t h i k h o i lugng tan^
^ 2,35a
t h e m la — gam.
100
Zn + CdS04 > ZnS04 + C d
65 ^ 1 mol > 112, tang (112 - 65) = 47 gam (adsbygoogle = window.adsbygoogle || []).push({});
^ (= 0,04 mol) > gam 208 100 1 47 T a CO t i le: = -^^-^ >• a = 80 gam. 0,04 2,35a 100 Chon dap an C .
B a i 12: N h i i n g t h a n h k i m loai M hoa t r i 2 vao dung dich CUSO4, sau
mot t h d i gian lay t h a n h k i m loai ra thay k h o i luong giam 0,05%. Mat khac nhiing t h a n h k i m loai t r e n vao dung dich Pb(N03)2, sau mot then
gian thay k h o i liiOng t a n g 7,1%. X a c d i n h M , biet r a n g so mol CuSO, va Pb(N03)2 t h a m gia d 2 triTdng hgfp nhir nhaụ
Ạ A l . B . Zn. C . Mg. D. F e .
Hiic/ng đn giai
Goi m la k h o i liiong t h a n h k i m loai, M la nguyen tilf k h o i cua kim loai, X la só mol muoi phan lifng.
M + CUSO4 > MSO4 + C u i
M (gam) -> 1 mol > 64 gam, giam (M - 64) gam.
0,05.m
X mol > giam gam.
0,05.m
= ^ x = - ^ l 0 0 - (1) M - 6 4
M + Pb(N03)2 -> M(N03)2 + P b i
M (gam) -> 1 mol > 207, tang (207 - M) gam 7,l.m 7,l.m 100 , 7 , l . m X mol > t a n g gam 7,l.m ^ X = ^ 0 0 - (2) 2 0 7 - M 0,05.m 7,l.m T C ( l ) v a ( 2 . t a c 6 : ^ = ^ ,3, TCf (3) giai r a M = 65. V a y k i m loai M la k e m . C h o n dao a n B.
B a i 13: Cho 3,78 gam hot A l p h a n urng vCra du vdi dung dich muoi XCI3
tao t h a n h dung dich Ỵ K h o i ItTofng chat tan trong dung dich Y g i a m
4,06 gam so vdi dung dich XCI3. X a c d i n h cong thufc cua muoi XCI3. Ạ FeCl3. B. AICI3. C. C r C l s . D. K h o n g x a c dinh.
HU&ng đn gidi
G o i A la nguyen tijf k h o i cua k i m loai X .
A l + XCI3 — ^ AICI3 + X
o no
= (0,14 mol) -> 0,14 0,14 mol
T a c6: (A + 35,5.3).0,14 - (133,5.0,14) = 4,06
Giai ra difOc: A = 56.
Vay k i m loai X l a F e va muoi F e C l s .
Chon dap an Ạ
B a i 14: Nung 100 gam hon hop gom Na2C03 va NaHCOg cho den, k h i khoi
iLfOng hon hop khong doi diTOc 69 gam chat rSn . Xac dinh phan t r a m
khoi \\igng cua m5i chat tuong ijfng trong hon hop ban daụ (adsbygoogle = window.adsbygoogle || []).push({});
Ạ 15,4% va 84,6%. B. 2 2 , 4 % va 77,6%. C. 16% va 8 4 % . D. 2 4 % va 7 6 % .
Hii&ng đn gidi
Chi CO NaHCOg bi phan buỵ D S t x l a so gam N a H C O g . 2 N a H C 0 3 Na2C03 + C O a t + H2O
Cur nung 168 gam > kho i li/Ong giam: 44 + 18 = 62 gam
x > k h o i ItfOng g i a m : 100 - 69 = 31 gam m . 1 6 8 62 m . 1 6 8 62 l a co: > x = 84 gam. X 0 1 Vay N a H C 0 3 c h i e m 8 4 % va Na2C03 c h i e m 16%. C h o n dap an C.
B a i 15: H o a tan 3,28 gam h 6 n hop muoi CuCl2 va Cu(N03)2 vao nildc difOc dung dich Ạ Nhiing M g vao dung dich A cho den k h i m a t m a u x a n h cua dung dich. L a y t h a n h M g ra c a n lai thay t^ng t h e m 0,8 gam. Co can dung dich sau p h a n ufng thu dugc m gam muoi k h a n . T i n h m ?
HuCdng dan giai
Ta c6:
mtang = mcu - mMg phan tog = nip^2, - nij^^a, = 3,28 - (nig^,^;, + m^^^, j = 0,8 ^ m = 3,28 - 0,8 = 2,48 gam.
Chon dap an B.
B A I T A P V A N DgNG
Bai 1 . Cho 115 gam hon hgfp gom ACO3, B2CO3, R2CO3 tac dung het vdi dung dich HCl thay thoat ra 22,4 Ht CO2 (dktc). Khoi Itrgng muoi clorua tao ra trong dung dich la
Ạ 142 gam. B. 126 gam. C. 141 gam. D. 132 gam.
B a i 2. Ngam mot la sMt trong dung dich C U S O 4 . Neu biet khoi luong dong bam tren la sSt la 9,6 gam thi khoi lifcfng la sSt sau ngam tang dong bam tren la sSt la 9,6 gam thi khoi lifcfng la sSt sau ngam tang
them bao nhieu gam so vdx ban daủ
Ạ 5,6 gam. B. 2,8 gam. C. 2,4 gam. D. 1,2 gam.
B a i 3. Cho hai thanh sSt c6 khoi lifdng bang nhaụ
- Thanh 1 nhiing vao dung dich c6 chula a mol AgNOạ - Thanh 2 nhiing vao dung dich c6 chiJa a mol Cu(N03)2;