Khi CO can dung dich M thu dircKc khoi lufdng muoi khan bang:
Ạ l l . l g B. 5,55g C. 16,5g D. 22,2g
HiC&ng dan gidi
1 12
n^o = —— = 0,05mol CO, 22,4
Ap dung cong thufc: rtir^^ii clorua = nihh muoi cacbonat + n^ọ^ .(71 - 60) = 5 + 0,05.11 = 5,55 gam = 5 + 0,05.11 = 5,55 gam
Chon dap an B.
Bai 3: Cho 42 gam hon hap muoi MgCOg, CuCOs, ZnCOg tac dung vdi
dung dich H2SO4 loang, thu diTcrc 0,25 mol CO2, dung dich A va chat ran B. Co can dung dich A, thu diTdc 38,1 gam muoi khan. Dem nung ran B. Co can dung dich A, thu diTdc 38,1 gam muoi khan. Dem nung liiong chat rSn B tren cho den khoi liiong khong doi thi thu duac 0,12 mol CO2 va con lai cac chat rSn B'. Khoi lUcfng cua B va B' 1^:
Ạ 10,36 gam; 5,08 gam B. 12,90 gam; 7,62 gam C. 15, 63 gam; 10,35 gam D. 16,50 gam; 11,22 gam C. 15, 63 gam; 10,35 gam D. 16,50 gam; 11,22 gam
Hitdng dan gidi
Dat h5n hop muoi la A CO3
A CO3+ H2SO4 ^ A SO4+ H2O + CO2T
0,25 0,25 0,25mol
mrSnB = nihon hop muoi + "^HjSÔ ~ " ^ C O j ~ ^ihO ^^muei khan
= 42 + 0,25.98 - 0,25.44 - 0,25.18 - 38,1 = 12,9 gam
mrinB- = 12,9 - (0,12 .44) = 7,62 gam
Chon dap an B.
Dang 7. KIM LOAI SAT B| 0X1 HOA THANH HON HPP
(Fe, FeO, FegOạ Fe304)
Dang nay c6 rat nhieu each giai khac nhau nhitog qua dai nen chi dUa ra 2 each giai saụ ra 2 each giai saụ
Cdch 1: Vi Fe bi oxi hoa thanh cac exit s^t c6 the con s^t dif (Fe,
FeO, FeaOs, Fe304)
Tom lai ta c6 the coi hon hop chi c6 sat va oxi nen ta c6 CTPT chung la FêOy . CTPT chung la FêOy .
FêOy - (3x - 2y)e xFê^ + yỐ Ta luon c6 he phiiong trinh sau Ta luon c6 he phiiong trinh sau
56x + 16y - m,5„
hop oxit sat
3x - 2y = mol san pham khuf.so e nhan
Cdch 2: Van dung cong thitc tinh nhanh
nipe = 0,7.mh6n hop oxit sit
+ (molsan phim kM-SO e nhan.5,6)
BAI T A P V A N DyNG
Bai 1: (De KA-2008): Cho 11,36 gam mot hon hap gom Fe, FeO, FegOs,
Fe304 phan ufng het vdi dung dich HNOg loang da, thu dtfac 1,344 lit NO (la san pham khuf duy nhat d dktc) va dung dich X. Co can dung NO (la san pham khuf duy nhat d dktc) va dung dich X. Co can dung dich X thu dufoc m gam muoi khan. Gia tri cua m lạ
Ạ 49,09 gam B. 34,36 gam C. 35,50 gam D. 38,72 gam.
Hii&ng dan gidi Cdch 1: Ap dung cong thitc Cdch 1: Ap dung cong thitc
mpe = 0,7.mh6n hap oxit s2it + (mol san pha'm khii-so 6 nhan.5,6) = 0,7.11,36 + ( i ^. 3 . 5 , 6 ) = 8,96 gam = 0,7.11,36 + ( i ^. 3 . 5 , 6 ) = 8,96 gam
n,,, =0,16mol 56 Fe Fe(N03)3 0,16 -> 0,16 mol Ta c6: m „ u 6 i = 0,16.242 = 38,72 gam Chon dap an D.
Cdch 2: H 5 n hdp chi c6 s^t va oxi nen ta c6 the coi la FexOy ap dung
cong thufc
•56x + 16y = m,.„ y,,^,,;,,^,
<
3x - 2y = mol san pham khuf.so e nhan 56x + 16y = 11,36 fx = 0,16
3 x - 2 y = 0,06.3 ' ^ | y = 0,15
Ta CO scf do hop thufc: Fe Fe(N03)3
0,16 0,16mol
m ^ u o i = 0,16.242 = 38,72 gam Chon dap an D.
- B a i 2: (DH KA 2007): Nung m gam hot s^t trong oxi t h u dtfoc 3 gam
hon hop chat r ^ n X. Hoa tan het X trong dung dich HNO3 dU thay thoat ra 0,56 l i t (dktc) k h i NO duy nhat. Gia t r i cua m?
A . 2 , 5 2 g B.2,62g C.2,22g D.2,32 g
Hii&ng dan gidi Cdch 1: Van dung cong thiic
mpe = 0,7.m/,6-„ hap oxit sdt + ( m o l s d n p h d m khit-so c nhdn.5,6)
= 0,7 .3 + ( 0,56 . 3 .5,6) = 2,52 gam 22,4
Cdch 2: Ap dung cong thiic
'56x + 16y = m,,„
3x - 2y = mol san pham khijf.so e n h a n 56x + 16y = 3 fx = 0,045
3x - 2y = 0,025.3 [ y = 0,03
m p e = 0,045.56 = 2,52 gam Chon dap an Ạ
Ta c6:
B a i 3: Nung n6ng 16,8 gam bot sSt trong khong k h i t h u duoc m gam h6n hop X gom bon chat r ^ n . Hoa tan het m gam X bang H 2 S O 4 dSc nong diT thoat ra 5,6 l i t S O 2 (dktc). Gia t r i cua m:
A . 2 2 g B . 2 6 g C.20g D.24 g
Hitdng dan gidi
V g " dung cong thiic
m p e = 0,7 . vcihSn hgp oxit sdt + (moljdn pArfm * A i ? - s o 6 nhdn.5,6)
16,8-(0,25.2.5,6)
=> mx = — ^-^ — = 20 gam 0,7
Chon dap an C.
B a i 4: De m gam bot sSt ngoai khong k h i , sau mot t h d i gian se chuyen
th&nh hon hop B gom 4 chat rAn c6 k h o i Itfong 12 gam. Cho hon hap
B phan iJng het v d i dung dich HNO3 du thay thoat r a 2,24 l i t N O (dktc). T i n h m va k h o i iuong HNO3 da phan iJng?
I Ạ10,08 g va 34,02 g B. 10,8 g va 34,02 g C.10,8 g va 40,32 g D. 10,08 g va 40,32 g.
Hiidng đn gidi Vdn dung cong thiic
mpe = 0,7.mA<5„ hifp out sdt + (molsdn phdm khu-sd e nhdn.5,6)
9 9 4 .
= 0,7.12 + ( . 3 . 5 , 6 ) = 10,08 gam 22,4
Fe Fe(N03)3
0,18 0,18mol
Theo d i n h luat bao toan nguyen tuf N
Ta c6: n N / H N o , = ^^N,Feim,), + H N / N O = 3-0.18 + 0,1 = 0,64 mol
= > " ^ H N O a = 0,64.63 = 40,32 gam Chon dap an D.
B a i 5: Hoa tan het m gam hon hop X gom Cu va hai oxit s^t can vCfa du 500ml dung dich H C l 1,2M. Co can dung dich sau phan urng t h u daoc 38,74 gam hon hop h a i muoi khan, m nhan gia trỉ
Ạ 22,24 B. 20,72 C. 23,36 D. 27,04
Hiiofng dan giai
Hai muoi k h a n d day la CuCl2 va FeCl2 .
Ta c6: 2H^ + Ố > H2O ( H C l ^ H^ + C D 0,6 0,3mol
B T K L : mcu va Fe = m^.ai - m^,. = 38,74 - 0,6.35,5 = 17,44 gam ^ mx = mcuvi Fe + mo(ox.t) = 17,44 +0,3.16 = 22,24 gam
Chon đp an Ạ
DANG MUOI NHQM TAG DUNG VOI BAZQ
Lf THUYET
A I C I 3 + 3 N a O H A l ( 0 H ) 3 i + 3NaCl
A1(0H)3 + N a O H N a A l O j + 2H2O PhiiOng t r i n h ion:
Al^^ + 3 0 H - ^ A1(0H)3
A1(0H)3 + 0 H - AlO" + 2H2O
Chu y: A1(0H)3 khong tan dUdc trong dung dich NH3, trong axit cacbonic - CO2 day duoc goc aluminat ra k h o i muoi
NaA102 + CO2 + 2H2O ^ A l ( 0 H ) 3 i + NaHC03
CO2 khong hoa tan diioc A1(0H)3 nen phan ling dCrng l a i d ket
tua keo trSng
- Neu suf dung axit m a n h day t h i tao ket tua keo t r ^ n g sau do tan ra
NaAlOa + H C l + H2O A l ( 0 H ) 3 i + NaCl A1(0H)3 + 3HC1 ^ A I C I 3 + 3H2O
Cach t i n h nhanh cho loai bai tap baza tac dung v d i muoi
nhom (Al^*).
T H I : n ^ „ - „ i „ =3.n^„o„,,
T H 2 : n ^ H - _ = 4 . n ^ . - - " A U O H , ,
* Cach t i n h n h a n h cho l o a i b a i t|ip axit tac d u n g vofi NaAlOa (AlÔa)
T H 2 : n „ . _ ^ 4 . n ^ , ^ ^ . - 3 n ^ , o H , 3 BAI TAP VAN DUNG
B a i 1: Cho 3,42gam A l 2 ( S 0 4 ) 3 tac dung vdi 25 m l dung dich N a O H tao r a di/oc 0,78 gam ket tuạ T i n h nong do mol cua N a O H da diing.
Ạ 1,2M hoSc 2,8M B. 0,4M hoac 1,6M C. 1,6M hoac 2,8M D. 0,4M hoSc 1,2M.
Hit&ng dan giai Vg.n dung cong thUc
3 42 " A M S O, , , = ^ = 0 , 0 1 m o l ; n ^ 3 . = 2.0,01 - 0,02mol 342 =>C.. 0,025 T H I : n^„. = 3.n.,,oH, =>CM = ^ ^ ^ ^ = 1,2M T H 2: n„„. = 4 . n , , 3 . - n ^ , , o H ) CM = ^ ' " ' " ^ " ' " ^ = 2 , 8 M
Al'' AKOlUj Mf^iOH Q Q25 Chon dap an Ạ
B a i 2: Cho 150 m l dung dich K O H 1,2M tac dung vdi 100 m l dung dich
A I C I 3 nong do x mol/1, t h u difOc dung dich Y va 4,68 gam két tuạ Loai bo k e t tua, t h e m tiep 175 m l dung dich K O H 1,2M vao Y, t h u difoc 2,34 gam ket tuạ Gia t r i cua x la
Ạ 1,2 M B. 0,8 M C. 0,9 M D. 1,0 M .
Hiicfng dan giai
T N I : H K O H = 0,18 mol; n^^oH), = 0.06 mol
T N 2 : H K O H = 0,21 mol; H A K O H ) . = 0,03 mol Al^" + 3 0 H - > A1(0H)3 a + 0,03 3(a + 0,03) a + 0,03 mol Al(OH)3 + O H " > A l O - + 2H2O a a mol a + 3(a + 0,03) = 0,21 => a = 0,03 0,06 + 0.03 + 0,03 , 04 Chon dap an Ạ
B a i 3: Trong mot coc dUng 200ml dung dich AICI3 2M. Rot vao coc 200ml dung dich NaOH c6 nong do a mol/lit, ta duac mot ket tua; 200ml dung dich NaOH c6 nong do a mol/lit, ta duac mot ket tua; Dem say kho va nung den khoi Itfgrng khong đi diioc 5,lg chat r^n. Tinh ạ
Ạ 1,2M hay 3,5M B. I M hay 0,5M C. 1,5M hay 7,5M D. 1,5M hay 0,5M. C. 1,5M hay 7,5M D. 1,5M hay 0,5M.
Hiidng dan giai Vqin dying cong thitc Vqin dying cong thitc
nẠc.3 = ^M^' - 0' 2.2 ^ 0,4mol n^,^03 = ^ = ^' 0^"^°^
pt: A l ' " +30H- -> A1(0H)3; 2A1(0H)3 AI2O3
0,1 0,05
Chon dap C.
B a i 4: (DHKA- 2007)
Tron dung dich chiia a mol AICI3 vdi dung dich chufa b mol NaOH. De thu daoc ket tua thi can cd t i le thu daoc ket tua thi can cd t i le
Ạ a: b = 1: 4. B. a: b < 1: 4. C. a: b = 1: 5. D. a: b > 1: 4. C. a: b = 1: 5. D. a: b > 1: 4.
Hiidng dan giai
Trpn a mol AICI3 vdi b mol NaOH de thu dtfac ket tua thi
Al=*^ + 30H- > A1(0H)3^ A1(0H)3 + OH- > AlO- + 2H2O
Al'" + 40H- > A10-+2H,0
a 4 mol
De ket tua tan hoan toan thi > 4 ^ - > 4. Vay de c6 ket tua thi - < 4 => a : b > 1 : 4. Vay de c6 ket tua thi - < 4 => a : b > 1 : 4.
a Chon dap ^n D. Chon dap ^n D.
B a i 5: Cho 200ml dung dich AICI3 1,5M tdc dung vdi V lit dung dich
NaOH 0,5M, Itfong ket tua thu diTcfc la 15,6 gam. Tinh gia t r i Idn nhat
cua V.
Ạ 2 lit B. 2,5 lit C. 1,5 lit D. 1 lit.
Hii&ng dan giai
15 6
nAi(0H)3 = = 0,2mol n^ic, =0,2.1,5 = 0,3 mol
nAici, > nAuoH)3 chi xay ra trirdng hop 2 Tacd: _ = 4 . n ^ , 3 . - n A „ 0 H ) 3
nxaOH = no„- = 4.0,3 - 0,2 = 1 mol VN«OH T T ^ = 2 lit. Chon dap an Ạ
U,5
B a i 6: Cho V lit dung dich NaOH 0,3M vao 200 ml dung dich A l 2( S 0 4 ) 3
0,2M thu dtroc mot ket tua tr^ng keọ Nung ket tua nay den khoi
lugng lUOng khong đi thi diigfc l,02g r^n. Tinh the tich dung dich
NaOH da dung.
Ạ 0,4 lit hoSc 1,2 lit B. 0,2 lit hôc 1,2 lit C. 0,2 lit hoac 1 lit D. 0,4 lit hoSc 1 lit. C. 0,2 lit hoac 1 lit D. 0,4 lit hoSc 1 lit.
HUcfng dan giai
nA.,,so,,3= 0,2.0,2 = 0,04mol 1,02 ^ ,
n., n = — — = 0,01mol
n^,3. =2.0,04 = 0,08mol '"^"^ 102 pt: A P " + 30H- ^ A1(0H)3; 2A1(0H)3 ^ AI2O3 pt: A P " + 30H- ^ A1(0H)3; 2A1(0H)3 ^ AI2O3
0,06 0,02 0,02 0,01
T H I : n„„. . = 3.n^„o„, => V = ^ - 0 , 2 l i t
"^^2: n „ H , n « x = 4 - n ^ , - - n A u o n , 3
nNaOH = noir = ^-^'^^ ' 0'02 = 0,3 mol