la Na2COị
NâCO, + H^SO, (/) -> Na^SO, + CÔ t + H^O
- Hop chát nao phan ling, tao ket tua trang va c6 khi bay ra la CaCO,: CaCO, + H^SO,(/) ^ CaSO, + CÔ T +H,0 CaCO, + H^SO,(/) ^ CaSO, + CÔ T +H,0
1.16. Dap an diing la D.
La'y tii m6i lo ra m6t it mu6'i 6i lam thi nghiem. Cho dung dich HCl phan ling vdi tiimg mu6'i: Cho dung dich HCl phan ling vdi tiimg mu6'i: - Mu6'i nao c6 khi thoat ra la BaCO,:
BaCO, + 2HC1 -> BaCl^ + CÔ T + H^O
- Lafy dung dich BaClz vilra tao thanh d phan ling tr6n lam thu6'c t h i dé phan biet BaClj va C U S O 4 : phan biet BaClj va C U S O 4 :
Cho dung dich BaCl2 tac dung vdi hai mu6'i: Mu6'i nao cho kfít tua trSng laCuS04: laCuS04:
CUSO4 + BaClj -> BaSÔ i + CuClj
Mu6'i con lai la BaClj.
1.17. Dap an diing la Ạ So d6 chuyen hoa: So d6 chuyen hoa:
Cu(OHX - ^ C u O >Cu >Cufe _ t N 2 2 H^ C u ( O H ) , .
1.18. Dap an diing la D.
Kh6ng dung axit H2SO4 dac de lam kho khi N H 3 dupe, vi xay ra phan ling:
2 N H 3 + H,SO, - >(NH4) ,S O 4 • • • ' • •
1.19. Dap an diing la C.
Phuong trinh hoa hoc: '*
H2SO,+2NaOH->Na2SO,+2H20 (1)
Theo (1): So mol NaOH = 2 s6' mol H2SO4 = 2.0,2.0,1 = 0,04 (mol)
- > niN^oH = 0,04.40 = 1,6 (gam).
100
Khdi lupng dung dich NaOH 10% = 1,6. = 16 (gam).
10 .! '
1.20. Dap an diing la B.
Co cac phuong trinh hoa hoc: f» ;. j 4, ;
C u ( 0 H) 2 —^ C u O + HjO
2Fe(OH)3 — F e ^ O j + SH^O M g C O j — ^ M g O + CO^T M g C O j — ^ M g O + CO^T
1.21. Dap an diing la C. 1
Ba chat d day nay deu dg bi nhiet phan, tao thknh oxit. ^
1.22. Dap an dung la Ạ
Khoi lupng chát tan: m^i = 25 (gam). (adsbygoogle = window.adsbygoogle || []).push({});
Khd'i lupng dung dich A: m^^ = m,^ + mj„ = 25 + 100 = 125 (gam). , -> N6ng do C% cua dung dich A: •.. .:;)-if aaf'!: gii ^|ịCKTS.i -> N6ng do C% cua dung dich A: •.. .:;)-if aaf'!: gii ^|ịCKTS.i
. C%-—.100% = 20%. .;;xn:ifcrHl,.iij;ti;,/-i.<iq
125
Th^ tich dung dich A: X,, = ^ = - ^ = 109,36 (ml).
1.23. Dap an diing la D. Cac phuong trinh hoa hpc: Cac phuong trinh hoa hpc:
B. CuO + 2HC1 -> CuClj + HjO
CUCI2 + 2 N a O H -> Cu(0H)2 + 2 N a C l
C. Cu + 2 H j S O , (d.n) -> CuSO, + SÔ t + 2 H 2 O CUSO4 + 2 N a O H ^ Cu(0H)2 + 2 N a C l CUSO4 + 2 N a O H ^ Cu(0H)2 + 2 N a C l 1.24. Dap an diing la C.
20
S6' mol NaOH = — - 0,5 (mol). 40 40
Phuang trinh hoa hoc:
CuCl^ + 2NaOH ^ CuCOH)^ + 2NaCl (1) Theo (1): S6' mol NaOH = 2 s6' mol CUCI2 = 2.0,2 = 0,4 (mol)
Du NaOH, s6' mol Cu(OH)2 = 0,2 (mol). Kh6'i lugng Cu(OH)2 = 0,2.98.80% = 15,68 (gam). 1.25. Dap an dung la B.
Cac phuong trinh hoa hoc:
FeCl^ + 2NaOH Fe(0H)2 i + 2NaCl AlCl, + 3NaOH -> Al(OH), i + 3NaCl Loc lay ket tua Fe(0H)2 va AKOH), rui nung:
4Fe(OH)2 + O2 — ^ 2 F e 2 0 3 + 4 H 2 O
2Al(OH)3 — ^ A I 2 O 3 + 3H2O
Chat ran thu dugc g6m F e j O j va A I 2 O 3 .
1.26. C6ng thiic dung la B.
Dat CTPT cua hidroxit la X ( 0 H ) 3
2X(OH)3 — ^ X 2 O 3 + 3H2Ọ
^ 2X 52,94 Theo de bai: = —-—
2X + 3.16 100
200X = 105,88X + 2.541,12-> X = 27.
Nguyen t6' X la A l , CTPT cua hidroxit la A1(0H)3.
1.27. Dap an dung la C. Phuong trinh hoa hoc:
CaClj + K 2 S O , ^ CaSO, i + 2KC1 1.28. Dap an diing la D.
Cho bot dong vao, Cu day bac ra va tao thanh Cu(N03)2: Cu + 2AgN03 -> 2Ag 4- + Cu(N03 \
Loc bo Ag, thu duoc dung dich Cu(N03)2 tinh khiet. 1.29. Dap an diing la B.
Dat CTPT cua oxit la X20„ (n la hoa tri ciia X). Phuong trinh hoa hoc: (adsbygoogle = window.adsbygoogle || []).push({});
X20„ +2nHCl ^ 2XC1„ + nH^O (2X + 16n)gam 2(X + 35,5n)gam 56 gam 11,1 gam 2X + 16n 2(X + 35,5n) ^ —> = > A = zUn. 56 11,1 Khi n = 1 X = 20: loaị n = 2 - > X = 4 0 - > X l a C a . ' n = 3 -> X = 60: loaị
Vay kim loai X la Ca ^ oxit la CaỌ 1.30. Dap an dung la Ạ
Cac phuong trinh hoa hoc: - ^ ^fo -
C a C O , — ^ C a O + COjT . (1)
CaO + H 2 0 ^ C a ( O H ) , (2)
Theo (1) va (2): So mol Că0H)2 = s6' mol CaC03 = — = 0,1 (mol) 100
^C^(CăOH),) = - ^ = 0,125M. 1.31. Dap an dung la Ạ
S6'mol NaOH = 80. — . — = 0,5 (mol) 100 40
Goi X , y la s6' mol NaHCO, va NâCOy
Cac phuong trinh hoa hoc:
CO, + NaOH -> NaHC03 ' ^ ^''^ ^'^'^ ^^^^ '
X mol X mol x mol
CO2 +2NaOH^Na2C03 + H20 (2) r w -
y m o l 2ymol ymol
Theo di bai:
^NaHco, • "Na,co, 2:3 = X : y
^ x = | y (I)
"Theo (1) va (2): SÓ mol NaOH = x + 2y = 0,5 (II) Giai he phuong trinh (I), (II) ta dupe: '
X = 0,125 (mol), y = 0,1875 (mol)
So mol C O 2 = X + y = 0,125 + 0,1875 = 0,3125 (mol).
IỊ BAI TAP TU LUAN
1.32. Chat ran con lai sau phan iJng la Fẹ TO khd'i luong oxit sat (Fe,Oy) va
kh6'i luong Fe, ta suy ra khd'i luong oxi c6 trong oxit. Tir do tim dugc ti 16 s6' moj; nf^:n„ = x : y . , , ..^
Phuong trinh hoa hoc:
Fe,0^ + yCO ^ xFe + yCOj
m o = n i p , „ - m p ^ - 3 2 - 2 2 , 4 = 9,6(gam)
Tijr phuong trinh phan ling ta CO so d6: ' '
Fe,0^ -> xFe + yO ' 56x 16y
22,4g 9,6g : 56x 16y X 2 ^ _
224 " "96" y " 3 ^"^ ' ^^^Ov (adsbygoogle = window.adsbygoogle || []).push({});
1.33. a) Cac phuong trinh hoa hoc:
Zn + Cu(NO02->Zn(NO,)2+Cu>l (1) r . J Zn + Pb(NO,),->Zn(N03),+Pb>l (2)
b) Goi X la so mol Zn da tham gia timg phan ting.
- Kh6'i luong Zn tham gia (1) la 65x. Kh6'i luong Cu tao thanh la 64x.
Kh6'i luong la Zn giam: 65x - 64x = 0,15 ^ x = 0,15 (mol). - Kh6'i luong Pb tao thanh la 207x.
Kh6'i luong la Zn thir hai tang: 207x - 65x (207-65).0,15 = 21,3 (gam). 1.34. a) Cac phuong trinh hoa hoc:
2A1 + 6HC1->2A1C1,+3H2 ^ 2Fe + 6HCl->2FeCl,+3H2 ^
b) S6 mol H, = ^ = 0,5 (mol).
Dat X va y la s6' mol A l va Fe c6 trong h6n hop, dat phuong trinh va
giai, tim dugc:
m = 0,2.27 = 5,4 (gam); mp, = 11,2 (gam).
c) Phuong phap tach Fe ra khoi h6n hcfp:
Ngam h6n hop trong dung dich NaOH dụ A l tan h€t vao dung dich, chi
c6n lai t»Ot sat. Loc bo dung dich, thu dugc sat. > ^' - - 1.35. Dat CTPT cua mu6'i sat la FeCl„, ta c6 phuong trinh hoa hoc:
FeCl„ + nAgNO, -> Fe(N03)„ + nAgCl i
S6 mol AgCl i = = 0,06 (mol).
143,5
Theo phuong'trinh phan ting, ta c6:
(56 + 35,5n) gam FeCl„ tham gia phan ung, sinh ra n mol AgCl 4^. Theo d^ bai, ta c6:
3,25 gam FeCl„ tham gia phan ling, sinh ra 0,06 mol AgCl i . . .. ^
Ta CO phuong trinh: 3,25n = 0,06(56 + 53,5n).
Giai phuong trinh dugc: n = 3. _^
V a y CTPT cua mu6'i sat clorua la FeCl,.
1.36. a) Dat CTPT cua oxit sat la FêỘ Sau phan ling, kh6'i lugng oxit sat giam 4,8 gam. Do chinh la kh6'i lugng ciia oxi trong oxit sat.
So do phan ling: FePy ->xFe + yO t i i ^ ' ) ! / -^ :db (56x + 16y)g ymol . I6g M. . o , 3( m o l ) . ^ ^ - ^ " ' - ' 16 56x + 16y y x ^ 2 -^^inm. v^.- ? ị , f 16 " 0,3 y 3 ' v a y CTPT cua oxit sat la Fe203.
b) Phuong trinh hoa hoc:
Fe203 + 3 C 0 ^ 2 F e + 3C02
S 6 m o l F e A = ^ = 0'l(mol). o-d Ha,, ^ " c o =0-1-3 = 0,3 (mol). - .
Toan bg khi CO2 bi giiJ lai trong binh dung dung dich NaOH dụ Do do kh6'i lugng binh tang: 0,3.44 = 13,2 (gam).
c) Theo phucmg trinh phan ling: (adsbygoogle = window.adsbygoogle || []).push({});
Khur hoan toan 0,1 mol Fe203 cfin 0,3 mol CỌ Th^ tich CO cdn dung (du 10%):
0,3.22.4.110^ 3^ ™ 100
1.37. a) Phuorng trinh hoa hoc:
Al + 3AgNO, ^ AKNO,), + 3Ag i
b) Theo phucmg trinh phan ung:
Khi CO 1 mol A l phan irng thi khdi lugmg la nh6m tang:
3 . 1 0 8 - 2 7 = 297 (gam).
Do 66, khi kh6'i lucmg la nhom tang 2,97 gam thi:
2 97
Da CO = 0,01 (mol) A l phan ling va sinh ra 0,03 mol Ag.
Kh6'i luomg A l da phan ung: 0,01.27 = 0,27 (gam). Kh6'i lugng Ag tao thanh: 0,03.108 = 3,24 (gam). c) Sau phan ling, trong 250 ml dung dich c6:
+ Mu6'i A K N O , ) , : n^,^^,,^^^ =n^, =0,01 (mol).
+ Mu6'i AgNOj du: n^^^^^ j„ = (0,24.0,25) - 0,03 = 0,03 (mol).
Dodo: C^(Al(NO3)3) = ^ = 0,04M.
C^(AgNO3du) = ^ = 0,12M.
1.38. a) Cac phuofng trinh hoa hoc:
Fe203 + 3C0 ^ 2Fe + 3CO2 t (1) Fe + HjSO, ->FeSO,+HjT (2)
Sef mol = = 0,1 (mol) = np„
22,4
Theo (1) va (2): n^^^^ =lnp„ =0,05 (mol).
mFe,o, = 0,05.160 = 8 (gam),
^^^p^o =—.100% = 80%.
c) Mu6'n CO 1 tan gang cSn 0,96 tán sat hoac
CO khd'i luong FcjO, nay c^n: 160.0,96.100
= 1,1714 (tan) quang hematit. 112.80
1,39. a) Cac phucmg trinh hoa hoc:
Mg + 2 H C l- > M g C l 2 + H 21 (1) 2A1 + 6HC1 ^ 2A1C1, + t (2) Cu + HCl: kh6ng c6 phan irng
b) Khdi lucmg bot A l va b6t Mg trong h6n hop:
n ^ A i + m M g = 1 0 - l = 9 ( g a m ) .
tanPêOj. (adsbygoogle = window.adsbygoogle || []).push({});
mol H 2 = ^ ^ ^ = 0,45 (mol). 22,4
Dat X va y la s6' mol A l va M g trong h6n hop.
•27x + 24y = 9 Ta c6: <^ 3
- x + y = 0,45
I2
Giai he phuong trinh ta dugc: x = 0,2; y = 0,15 (••. >.ẠX\
Trong 10 gam h6n hap c6: ,
xvif^^ =lgam->%mQ, =10%. . iieh ' u . K^^*; VS
m^, = 0,2.27 = 5,4gam ^ = 54%. •„ '
m^j = 0,15.24 = 3,6 gam -> %mMg = 36%. •
1.40. a) Ngam cac la kim loai vao dung dich CUSO4:
- Ngam la Cu: Khdng c6 phan ling. Khdi lugng la Cu khdng d6ị - Ngam la Zn: Xay ra phan ung:
Zn + CUSO4 ZnS04 + Cu Theo phucfng trinh phan ling:
1 mol Zn (= 65g) tan ra thi c6 1 mol Cu (= 64g) tao thanh. Do do, khdi lugng la kem giam (< m).
- Ngam la Fe: Xay ra phan ling:
1 mol Fe (= 56g) tan ra thi c6 1 mol Cu (= 64g) tao thanh. Do do, khdi lugng la Fe tang (> m).
b) Ngam cac la k i m loai trfin trong dung djch A g N O , :
Kh6'i luong cac la k i m loai d6u tang v i kh6'i lugng m o l ciia cac k i m loai
d^u nho hcfn kh6'i lugng m o l cua bac. r > i i 4 r > v j
22 4 7 "SQ
1.41. a) Kh6'i lugng m o l phan tir k h i A : = ' ' ' = 34 (gam).
S6' m o l k h i A dem d6'r chay: n^ = | ^ - 0,1 (mol). ' ' ' " ' ^ ' ^ ^
S6' mol cac san pham d6't chay A :
2 2 4 1 8
n.so, = r r — = 0 , l ( m o l ) ; n^^o = Y - = 0 , l ( m o l ) . , }
N h u vay: 1 mol A — ^ 1 mol SÔ va 1 mol H^Ọ
Trong phan tir A c6 1 nguyfin tuf S va 2 nguydn tir H .
M A = 3 4 = 3 2 + 2.
C5ng thiic phan tijr cua A la HjS. b) Phuong trinh phan ung d6't chay H j :
2 H 2 S + 3 O 2 - > 2SO2 + 2 H 2 O (adsbygoogle = window.adsbygoogle || []).push({});
c) Theo phuong trinh phan ting: V^^ = |^H,.S = " '^'^ (l^^)-
Chiromg 2 KIM L O A I KIM L O A I