Phuong trinh hoa hoc d6't chay CH4:
C H 4 + 2 O 2 •CO2+2H2O (3)
"cH, =nco, =0,05 = x (III)
Giai he phuong trinh (I), (II), (III) duoc: x = 0,05, y = z = 0,1. Vay: % m c H ^ =12,9%,%mc^„^ =45,1%, %m^^^^ =42%. Vay: % m c H ^ =12,9%,%mc^„^ =45,1%, %m^^^^ =42%.
2.41. Tinh so mol cac chát: ^ * ' '
fill 39 2
n. = ^ = 0,3(mol); n^^ = ^ = 1,75(mol)(0,35molO,) 22,4 22,4
8,96 22,4 1,8
nco, = = 0,4 (mol); n„„ = y - = 0,1 (mol).
Dat CTPT cua hidrocacbon la C^H^. Cac phuong trinh phan ung chay: Cac phuong trinh phan ung chay:
2CO + 0 , 2CO i^ii .,J i^ii .,J a mol 0,5a 2 a mol C,H^ + (x + ^)02->xC03 2 ' (1) (2)
Trong do: a, b la s6' mol CO va C^H^ c6 trong h6n hop Ạ
Theo dau bai, ta c6: a + b = = 0,3 mol (3)
0,5a + (x + ^)b = n„ =0,35 (4)
4 '
a + xb = nc(, =0,4 (5)
l y . b = nH^„=0,l . .i^ - : (6)
Giai he phuong trinh (3, 4, 5 va 6) dugc: a = 0,2, b = 0,1, x = 2, y = - C6ng thiic phan til cua hidrocacbon la C2H2.
- Ph^n tram ihi tich cac chát trong h6n hop:
%V„, = — . 100% = 66,7%; %YCH, =33,4%.
0,3
2.42. - Tinh kh6'i luong cac nguyfin t6' c6 trong hop chát A:
m,, =^^.12 = 1,8(gam); m„ = ^ ^ ^ . 1 = 0,4(gam).
44 18 m^ =mc + mH +m() ->m„ = 3 - l , 8 - 0 , 4 = 0,8(gam). m^ =mc + mH +m() ->m„ = 3 - l , 8 - 0 , 4 = 0,8(gam).
Trong phan tir A, ngoai cacbon va hidro con c6 nguyen t6' oxị a) Dat CTPT cua A la C H O . Ta c6:
12x_ y _ 1 6 z _ 6 0 m^, m^^ m() 3
I 2 ^ . J L . 1 ^ . 2 0 - > x = 3,y = 8,z = l .
1,8 0,4 0,8 ^
C6ng thiic phan tur cua A la C^HsỌ b) C6ng thiic cau tao c6 th^ c6 cua A:
CH, - CH^ - CH^OH; CH, - CH(OH) - CH,; CH, - O - CH^ - CH,.
(1) (2) (3)
Chiromg 3
PHI KIM - S O LLTOC V E BANG TUAN HOAN C A C NGUYEN TO HOA HOC C A C NGUYEN TO HOA HOC
••': r',<
, T R A C NGHIEM K H A C H QUAN
3.1. Dap an diing la C.
CI2 va O2 kh6ng phan ung true tiép vdi nhaụ
3.2. Cong thiic dung la B.
Dat CTPT cua oxit X la C Ộ
Theo de bai: M^^ô = 12x + 16y = 28 (I) 16y _ 57,14
''j .liJ.
12x + 16y 100
Giai he phuomg trinh (I), (II) dugc x = 1, y = 1. -> CTPT cua A la CỌ
3.3. cau CO n6i dung sai la C.
Nude Gia-ven la dung dich h6n hop hai muCi'i NaCl va NaClỌ
3.4. Dap an dung la D.
Cac phuong trinh hoa hoc: ' '" ''^ >« t - ^ ^ n ^
MnOj + 4HC1 MnCl^ + CI, T + 2H,0 * '"^ n-JYt/sH
C a C O , — ^ C a O + COj t
JO) -hi
Na,SO, + 2HC1 ^ 2NaCl + SOj T + HjO ' [ 3.5. Dap an diing la B.
Sue khi CI2 vao dung djch NaOH (loang, ngu6i) xay ra phan dug:
CI2 + 2NaOH -> NaCl + NaClO + HjO Clo bi loai bo, cho nudfc Gia-ven.
3.6. Dap an dung la C. - - -
Dot h6n hop: N2 khong chay, CO ^ COj. :
Cho san pham qua nudrc vOi, CO2 bi giu lai do tao thanh CaCO,; r6i qua ^aSÔ dac, hoi H2O bi giu laị CuC'i cung duoc N2 nguyen chát.
3.7. Dap an diing la D. ' ''^ Cacbon the hiSn tinh khii khi phan ung vdri oxi: Cacbon the hiSn tinh khii khi phan ung vdri oxi:
Cacbon thi hitn tinh oxi hoa khi phan ung vdri hidro: C + 2H, >CH, C + 2H, >CH,
3.8. Dap an dung la Ạ
Phucfng phap A la phuong phap thich hcfp nhát. Cac phucfng phap khae
(B, C, D) d^u CO the dieu che dugc clo nhung kh6ng phii hop vdi phong thi
nghifim (cAn diiu che luong nho khi clo dl lam thi n g h i 6 m ) . 3.9. Phan ling B kh6ng xay rạ 3.9. Phan ling B kh6ng xay rạ
S i 0 2 kh6ng phan iJng vdi axit HCl.
3.10. Dap an dung la D.
Phuong trinh phan ling:
2C + FejO, — 5 ^ 3Fe + 2 C O 21
C da la'y oxi cua FêÔ no la chat khuf.
3.11. Cau sai la C.
Axit khCng an mon thuy tinh nfin đng lo thuy tinh dung dung dich axit
ra't tot. Chi kh6ng dung lo thuy tinh dung kiem, vi kiem an mon thuy tinh.
(SiOj + 2NaOH -> NâSOj + H^O)
3.12. Dap an dung la C.
Nguyen t6' X c6 3 electron => X 6 chu ki 3.
Nguyen t6' X c6 5 electron d Icrp ngoai ciing ^ X cf nhom VẠ
Nguyen t6' X c6 dien tich hat nhan 15+ =^ X c6 s6' thii tu cr nguyen t6' la 15. Vay X la nguydn t6' photpho (P). Vay X la nguydn t6' photpho (P).
3.13. Dap an dung la D. 3.14. Dap an dung la Ạ 3.14. Dap an dung la Ạ
Goi kim loai hoa tri II la X, Ta c6: X + C l 2 - > X C l 2 X + C l 2 - > X C l 2 Xgam (X +35,5.2) gam 12,8 gam 27 gam X _X + 35,5.2 ~^12,8" 27 •im 27X = 12,8(X + 71)->X = 64^KimloaiXlad6ng. 3.15. Dap an dung la B.
Phuong trinh hoa hoc: H^SO, + 2NaHC03 NâSO, + 2 C O 21 + 2H,0 (1)
980
Theo (1): S6' mol C O 2 = 2 so mol H2SO4 = 2 . — = 20 (mol).
Thé tich khi CO2 la: V„,^ = 20.22,4.90% = 403,2 (lit).
3,16. C6ng thiic dung la C.
Dat CTPT cua oxit la R^Ộ Ta c6:
-'^•'2. = 0.7272
^•R + 16y 100 .
Theo d^ bai: „^ = x.R + 16y = 44
^ — = 0,7272->y = 2 ? ^; 44
x.R = 44-16.2 = 12->R = —
x Khix= 1 ^ R = 12(cacbon) Khix= 1 ^ R = 12(cacbon)
X = 2 ^ R = 6 (loai) Vay CTPT cua oxit la COj. Vay CTPT cua oxit la COj. 3.17. Dap an diing la B.
Cac phuong trmh hoa hoc:
NajCOj + 2HC1 ^ 2NaCl + CÔ T + H2O (1)
CaCO, + 2HC1 CaCl^ + CÔ T + H^O (2) Ggi X va y la s6' mol Na2C03 va CaCO, c6 trong 31,2 gam h6n hcrp. Ggi X va y la s6' mol Na2C03 va CaCO, c6 trong 31,2 gam h6n hcrp.
Taco: 106x + lOOy = 31,2 (I) Theo(l) va(2): Theo(l) va(2):
S6'mol HCl = 2 so mol Na2C03 + 2 s6'mol CaCOj.
n„a =2x + 2y = 0,4.1,5 = 0,6 (mol) (II) ^^.^ Giai he phuomg trinh (I) va (II) dugc: X = 0,2; y = 0,1. Giai he phuomg trinh (I) va (II) dugc: X = 0,2; y = 0,1.
Thé tich CO2 thu dugc la:
Vco, = (0,2 + 0,1).22,4 = 6,27 (lit). 3.18. 1. Dap an dung la Ạ 3.18. 1. Dap an dung la Ạ
1 12 So mol kim loai M: a = — M So mol kim loai M: a = — M Phuong trinh hoa hoc:
2M +xCl2->2MCl, ' * .1 ' * .1 1 ! O'/d'i a mol a mol 1.12 _ 3,25 M ~M + 35,5.x 1 19 3 75 a = iilf = — >2,13M = 39,76x
Theo (1): So mol Clj = - so mol Fe = = 0,03 (mol). Khi X = 1 va 2 d6u khong phu hop. Khi X = 1 va 2 d6u khong phu hop.
x = 3 ^ M = 56-> kim loai M la sat.
2. Dap an diing la B. ; t
Cac phuong trinh hoa hoc:
2Fe + 3Cl2 ^2FeCl, (1) •' I » • ' . 4HC1 + M n 0 2 — ^ M n C l 2+ C l j+ 2 H 2 0 (2) 3 . ,^ 3.1,12 - so mol Fe = 2 2.56 Theo (2): So mol HCl = 4 s6 mol CI2 = 4.0,03 = 0,12 (mol).
- V , , . , , , = ^ = 0,02(lft). • - ^ ^ ^ ^ . O ^ . ^ l^ l . . . .
3.19. Dap an dung la C. 4 48 4 48
86'mol HCl = ^ = 0,2 (mol). • . ' v 22,4 22,4
Kh6'i lugng chat tan HCl = 0,2.36,5 = 7,3 (gam). Kh6'i lugng dung dich HCl = 7,3 + 42,7 = 50 (gam). Kh6'i lugng dung dich HCl = 7,3 + 42,7 = 50 (gam).
C%(HC1) = ^.100% = 14,6%. Nong do mol cua dung dich: Nong do mol cua dung dich:
C^ (HCl) = - = = 4,7 (M). V 0,0427
3.20. 1. Dap an dung la B. Phuong trinh hoa hgc: Phuong trinh hoa hgc:
CO, + Că0H)2 ^ CaCO, i + Wfi (1)
Mu6'i duy nhát khOng tan la CaCO,.
Theo (1): So mol Că0H)2 = so mol CaCO, = : j ^ = 0,1 (mol). -> N6ng do mol ciia dung dich CăOH)2 la: -> N6ng do mol ciia dung dich CăOH)2 la:
C,(CăOH),) = ^ = l(M).
2. Dap an diing la Ạ
^ Theo (1): So mol COj = so mol CaCO, = 0,1 (mol). V„,, =0,1.22,4 = 2,24 (lit). V„,, =0,1.22,4 = 2,24 (lit).
2 94
%V^„ = ^ . 1 0 0 % = 20%. 11,2
Ị: .21. Dap an diing la D. Phuong trinh hoa hgc: Phuong trinh hoa hgc:
4P + 50j—^2P20, SÓ mol = = 0,3 (mol). SÓ mol = = 0,3 (mol). S6' mol P2O, = ^ = 0,1 (mol).
li,0 (1)
Theo(l): n„^(phaniing) = -sd mol P^O, =-.0,1 = 0,25 (mol) 2 2 2 2
Oxi du = 0,3 - 0,25 = 0,05 (mol). Tính khd'i lugng P theo P2O,:
n,, =2n,,„^ =2.0,1 = 0,2 (mol). ->m|, =0,2.31 =6,2 (gam). ->m|, =0,2.31 =6,2 (gam).
3.22. Dap an diing la C. Cac phuong trinh hoa hgc: Cac phuong trinh hoa hgc:
CaCO, — ^ CaO + CO, t MgCO, — ^ MgO + CO, t MgCO, — ^ MgO + CO, t
Ggi X va y la s6' mol CaCO, va MgCO,. Ta c6:
lOOx + 84y = 14,2 (I) Theo(l)vă2): Theo(l)vă2):
86' mol CO2 = s6' mol CaCO, + so mol MgCO, = ^ = 0,15 (mol). 22) 4 22) 4
=x + y = 0,15 (II)
Giai he phuong trinh (I) va (II) dugc: x = 0,1; y = 0,05.
Thanh phSn % khoi lugng cac chát trong h6n hgp: • 0,1.100 0,1.100
(1) (2)^^ (2)^^
%m, 'CaCO, 14,2 -.100% = 70,42%. %m^^c(>, = 100% - 70,42% = 29,58%. %m^^c(>, = 100% - 70,42% = 29,58%.
3.23. Cong thiic diing la B.
Ggi CTPT cua oxit sat la FêỘ Phuong trinh hoa hgc: Phuong trinh hoa hgc:
Fe,Oy + 2yHCl ^ xFeCl,^,, + yH^O
Fe,Oy+yCÔxFe + yC02
S6'molFe = ^ = 0,15(mol). 56
(1) (2) (2)
S6' mol HCl = 3.0,15 = 0,45 (mol).
S6' mol O = - s6 mol HCl = — = 0,225 (mol). 2 2 Ta CO t l le: _ 0,15 _ 2
0,225 3
CTPT cua oxit sat la FêOy
3.24. Dap an dung la Ạ Cac phucmg trinh hoa hoc:
FêO, + 3CO 2Fe + 3 C O 21 (1)
FêO, + 3H2 — ^ 2Fe + 3Ufi (2)
T h e o ( l ) v a (2):
Cir 6.22,4 lit h6n hop khi CO va H2 phan ling, thu dugc 4.56 gam Fẹ vay 268,8 lit h6n hgp khi CO va H2 phan ling, thu dugc x gam Fẹ
268,8.4.56 ,
- > X = • = 448 (gam).
6.22,4 3.25. 1. Dap an diing la Ạ
Cac phuong trinh hoa hoc:
NaHCO, + NaOH -> NâCO, + H^O (1) NajCO, + NaOH: Khong xay ra phan dug (2) Theo (1): S6' mol NaHCO, = s6' mol NaOH = 0,2.1 = 0,2 (mol).
- > m N a H c o , = 0.2.84 -16,8 (gam)
% m , , , o , ~ - 1 0 0 % = 44,21%.
%m,,^eo, =55,79%. 2. Dap an diing la D.
Sau phan ting, trong dung dich chi c6 muoi NajCOj. S6' mol Na2C03 = 0,2 + ' = 0,4 (mol)
1 Uo
' ' ^ C J N a , C O , ) ^ ^ = 2,0(M).
> i * 0,2
3.26. Dap an diing la B.
Phuong trinh hoa hoc xay ra khi d6't h6n hgp:
Fe + S — ^ F e S (1)
S6moIFe = — = 0,l(mol). 56
S6 mol S = — = 0,05 (mol). 32
Theo (1): Fedu = 0,1 - 0,05 = 0,05 (mol). Chát ran A gdm: 0,05 mol Fe va 0,05 mol FeS. H6n hgp A tac dung vdi dung dich HCl:
Fe + 2 H C l^ F e C l 2 + H 2 t FeS + 2HC1 ^ FeCl^ + H^S T
H6n hgp khi B g6m H2 va H2S.
Theo (2) va (3):
S6' mol HCl = 2 s6' mol Fe du + 2 s6' mol FeS. = 2.0,05 + 2.0,05 = 0,2 (mol) 0,2 (2) (3) ' d d H C l l M 1 = 0,2 (lit). B A I T A P T i ; L U A N
3.27. a) Cac phuong trinh hoa hoc san xuat axit H2SO4 tiir quang pirit (FeSj):
4FeS2 +1IO2 —^ 2 F e, 0 , + SSÔ (1)
0 , + 2 S 0 2 ^ - > 2 S 0 3 (2)
SO3 + H2O -> H2SO4 (3)
T h e o ( l ) , ( 2 ) v a ( 3 ) :
- Khd'i lugng dung dich axit H2SO4 98% thu dugc tCr 1kg quang pirit la:
1^.2.98.1^ = 1666,67(gam).
120 98
|Hieu suát qua trinh san xuat chi dat 90% ntn thuc té thu dugc: = 1666,67.90% = 1500 (gam).
|b) S6' mol axit H2SO4 nguyen chat thu dugc: • - 1500 98
. „ o f , . = 15 (mol).
"^""^ 3,75 100
[Thé tich dung dich H^SÔ 3,75M: V^^,^^ 15
Cdch pha che: Láy c6'c thuy tinh c6 dung tich Ion hon 4 lit c6 vach chi^
d6 (ghi the tich). DSu tien cho vao c6'c 3 lit H p , r6i cho tir tir dung dich axit
H2SO4 98% viJra san xuát duac vao c6'c, khuáy deụ Sau khi da cho het axit vâ coc, them HjO vao c6'c d^ dat t6i the tich 4 lit.
3.28. Dat c6ng thiic phan tir cua sat oxit la Fê^. Cac phucfng trinh hoa hoc: Cac phucfng trinh hoa hoc:
. Fe,0^ + yH, — ^ x F e + y H , 0 (1) (56x + 16y) g y mol 1,6 g ^ ^ = 0,03 (mol) • ^0,03(56x + 16y)-l,6y (I) >• Fe + 2HC1 ^ F e C l ^ + H ^ t (2) n 448 0,02 mol <- = 0.02 (mol). 22,4
Theo (1) va (2): S6' mol Fe = x = 0,02 (mol).
Thay gia tri cua x = 0,02 vao phuong trinh (I), tim dugc y = 0,03. x 0,02 2
Lapti \t: ~ = —— = -.
• y 0,03 3 -> CTPT cua oxit sat la FcjO,.
3.29. - Tong s6' mol NaOH la 0,02. - S6' mol AICI3 la 0,006. - S6' mol AICI3 la 0,006. Cac phirong trinh hoa hoc:
AICI3 + 3NaOH -> A1(0H)3 i + 3NaCl (1)
Al(OH), + NaOH -> NaAlÔ + 2H2O (2)
Cac phucfng trinh phan ung:
So mol NaOH d pu (1) = 3 s6' mol AlCl, = 3.0,006 = 0,018. S6' mol NaAlOj = s6' mol NaOH b pu (2) = 0,02 - 0,018 = 0,002. Sau phan ling (1) va (2), trong dung dich c6 mud'i NaCl va NaAlOj.
CM(NaCl) - = 1,286 (M).
0 009
C^(NaA10, ) = = 0,0143 (M).
> 0,14
3,30. a) Phucfng trinh hoa hoc:
HNO3 + N a O H N a N O , + H2O
b) Chat tan c6 th^ c6 trong dung dich D: