Dua vao san phdm chay la CO2 va H^O ta biét chát X chac chan phai c6
cacbon va hidrọ Con oxi c6 the c6 trong X hoac cua kh6ng khị"
4.6. Dap an diing la A .
1 mol Q H g chay cho 6 mol COj.
4.7. Dap an diing la D.
K h i dua h6n hop CH4 va CI2 ra ngoai anh sang, xay ra phan ling: CH, + CI2 ^ 5 _ > C H , C 1 + HCl
K h i HCl lam do giáy quy tím udt.
K h i HCl tan trong nudfc tao thanh axit H C l , axit nay phan ling voi da voi, tao thanh khi CO2 (tao bot khi).
4.8. Cau CO n6i dung sai la C.
Benzen kh6ng phan ling vdi dung dich nude brom. Benzen chi phan ling thé voi brom long c6 b6t Fe xiic tac.
4.9. Nhan xet sai la C.
CH2 = C H - CH3 chay cho san phim la khi CO2 va hoi H2Ọ
4.10. Dap an diing la B.
Dan khi axetilen c6 iSn khi CO2 qua dung dich NaOH, CO2 phan ling vdfi NaOH va bi g i i i lai:
CO2 + 2NaOH ^ NâCO, + H^O K h i thoat ra la axetilen tinh khi6't.
4.11. Dap an diing la A . So 66 chuy^n hoa: So 66 chuy^n hoa:
CaCO, CaO _J252!£_>CaC, ) C^H^.
4.12. Dap an diing la D. Cac phuong trinh hoa hoc: Cac phuong trinh hoa hoc:
Q H « + C 1 , Fe, t"
•>C«H,C1 + HC1 C , H , + B ^ - ^ C , H , B i i
4.13. Dap an diing la C. Phuong trinh hoa hoc: Phuong trinh hoa hoc:
CH, + 3C1, CHCI3 + 3HC1
4.14. Dap an diing la B.
C2H2 va CfiHfi vira tham gia phan irng thé, vtra tham gia phan ling c6ng: CH = CH + 2 A g N 0 , > AgC = CAg i + 2NH,NO,
CH = CH + 2B^ ) CHBr^ - CHBr, Q H« + 3 C l 2 - ^ Q H , C 1 ,
C,H,C1, + Bi^ >C,H,Br + HBr
4.15. Bien phap sai la A . ^ , Neu cung cap du khong khi hay oxi thi se mát nhiet lugng lam nong Neu cung cap du khong khi hay oxi thi se mát nhiet lugng lam nong kh6ng khi hay oxi du, do do sir chay khong c6 hiSu suát caọ
4.16. C6ng thiic diing la C.
S6'mol H , 0 = — = 0,3(mol). ' ' Tir 4 dap an ta dat CTPT cua X la C,H . Ta c6: , : , . 4
M x = 1 2 x + y = 30 Phuong trinh phan iJng chay:
(D 1 mol 1 mol 3 x + y . 4 j O, •xCO, + ^ H , 0 ' 2 ' ^ m o l 30 mol y 3 0 _ ^ 3 ~0,3 2 0,3 mol •y = 6 (II)
Giai he phuong trinh g6m 2 phuong trinh (I) va (II) duoc x = 2.
4.17. Dap an diing la A .
S6' mol C H , - — - 0,5 (mol). ' 16
Phucfng trinh hoa hoc:
C H , + ^ CÔ + 2 H , 0 (1) Theo (1): S6' mol C O 2 = s6' mol C H 4 = 0,5 (mol) Theo (1): S6' mol C O 2 = s6' mol C H 4 = 0,5 (mol)
mc„, = 0,5.44 = 22 (gam)
S6' mol H 2 O = 2s6' mol C H 4 = 2.0,5 = 1 (mol) ^ m „ ^ o = 1 . 1 8 = 18(gam).
4.18. Dap an diing B.
S6' mol etilen = — - 0,25 (mol). 22,4
Phuomg trinh hoa hoc:
C,H, + Bt,^C,li,Bt, (1)
Theo (1): SÓ mol Bfj = s6' mol C2H4 = 0,25 (mol)
mp, = 0,25.160 = 40 (gam). 4.19. Dap an diing la C .
Cac phuong trinh hoa hoc:
C H = C H + 2Bii ^ C H B ^ - C H B r , (1)
CH4 + Bi^: Kh6ng c6 phan img (2)
C H 4 kh6ng phan ling vdi Bij va c6 th^ tich la 1,12 (lit). ^ The tich C2H2 la 2,24 - 1,12 = 1,12 (lit).
Theo (1): S6' mol Bfz = 2 s6' mol Q H j = 2. - i ^ = 0,1 (mol)
->mB,^ =0,1.160 = 16 (gam). 4.20. Dap an dung la D.
Cac phirong trinh hoa hoc:
CHj = CH2 + Bii ^ CHjBr - CH^Br (1)
CH4 + Br, : Kh6ng c6 phan ling (2)
Theo (1): S6' mol Br2 = s6' mol C^H^ = — = 0,025 (mol). 160
-> Wc,H, = 0,025.22,4 = 0,56 (lit) V c „ = 2,8-0,56 = 2,24 (lit). 4.21. Dap an dung m A .
Cac phuong trinh hoa hoc:
C H 4 + 2 0 2 ^ C 0 2 + HjO (1)
C2H4 + 3 0 , - > 2CO2 + 2 H , 0 (2)
Sau phan ling chay, s6' mol C O 2 la: 0,1 + 0,2 = 0,4 (mol).
CO, + CăOH)j du ^ CaCO, i + H^O (3)
Theo (3): S6 mol C a C O , = s6' mol C O 2 = 0,4 (mol) m^^co, 4 = 0,4.100 = 40 (gam). 4.22. C 6 n g thiic diing la C.
Dat C T P T cua hidrocacbon \h C . H ^ . Ta c6: m^^H^ = 12x + y = 0,875.32 = 28 cap nghiem duy nhát phu hop la: x = 2, y = 4. C T P T cua X la C2H4.
4.23. Dap an diing la B. Cac phirong trinh hoa hoc: Cac phirong trinh hoa hoc:
CaCj + 2 H 2 O- > Că0H)2 + CjHj (1)
C,U,+2BT,^C,U,Bt, (2)
Theo (1) va (2): S6'mol C a C 2 = - s6'mo] Br2 = - . — = 0,01 (mol) 2 2 160
^m^^c, =0,01.64 = 0,64 (gam). 4.24. Dap an diing la D.
Phucmg trinh hoa hoc:
C , H , + Bi^ y >C6H,Br + HBr
Theo (1): S6' mol C^Hfi = s6' mol C^H^Brj = 0,1 (mol). ^m^^n^ =0,1.78 = 7,8 (gam).
4.25. C 6 n g thiic dung la Ạ Phuong trinh hoa hoc: Phuong trinh hoa hoc:
(1) Theo P T H H : Theo P T H H : sTheo de bai: 1 lit 1 lit 4 J O2—!—J-xCÔM-^HjO x l i t 3 lit y i i t 2 4 lit Hị jfi r y X 2 X 3 3 4 y 8 [Vay C T P T cua X la C^HR.
4.26. Dap an dung la A . S6'mol cac chat: '* S6'mol cac chat: '*
8,8 44 18 = 0,2 (mol) -> mc = 0,2.12 = 2,4 (gam). = 0,3 (mol) ^ m ^ =0,6.1 = 0,6 (gam). - ^ m ^ . + m H = 2 , 4 + 0,6 = 3(gam) = m^.
Vay trong phan tir X khdng c6 oxị
4.27. C6ng thiic dung la C.
Dat CTPT cua hidrocacbon la C H„. mo
Phuang trinh hoa hoc:
4 Theo PTHH: Theo d6 bai: 1 l i t 1 l i t x + - 4 j lit - ^ x C O , + ^ H , 0 ' 2 ' hO lor 2 lit x + y ^ = 1 -> 4x + y = 1
Cap nghiem duy nhát phu hop la: x = 1, y = 4. Hidrocacbon Y la CH4.
4.28. Dap an dung la B. Cac phuang trinh hoa hoc: Cac phuang trinh hoa hoc:
CH4+2O, CO2+2H2O
CjHj + 2,50^ -> 2 C O 2 +H j O
Goi X va y la the tich CH4 va Q H j c6 trong 28 ml h6n hop. Ta c6: x + y = 28 (I)
T h e o ( l ) : V , , , „ = 2 V „ , =2x. . TTieo(2): V , , , , , =2,5V,^„^ =2,5y
= X,,„ + X, = 2x + 2,5y = 6,72 (H) Giai he phuong trinh (I) va (II) dugc: x = 5,6; y = 22,4 (ml).
•%v„, = ^ "'^ 28 "'^ 28 %Vc,„, =80%. .100% = 20%. (1) (2) 4.29. Dap an dung la A .
K h i dun nong c6 N i xiic tac, trong h6n hop xay ra phan ling:
N i. t "
0,2 mol 0,2 mol 0,2 mol + 2H, N i, t " ->C,H,
(1)
(2) 0,2 mol 0,4 mol 0,2 mol
Sau phan ung trong h6n hop c6:
CjHfi: n^^H, = 0,2 + 0,2 + 0,1 = 0,5 (mol).
U^:n^, = 0 , 9 - ( 0 , 2 + 0,4) = 0,3 (mol). Doi vori chat khi, t i 16 % mol cung la t i le % the tich.
0,3 Do do: %V H ^ =
0,3 + 0,5 .100% = 37,5%. =6,25%. =6,25%.
4.30. Dap an diing la D . Cac phuong trinh hoa hoc: Cac phuong trinh hoa hoc:
CH4 + Bi^: khong c6 phan ling C^H.+B^-^C^H^Br,
Theo (2): S6' mol C2H4 = s6' mol Br2 = 0,1.1 = 0,1 (mol)
-^Vc.H, =0,1.22,4 = 2,24 (lit). 2 24 %W = ^ . 1 0 0 % = 33,33%. %Vc„^ =66,67%. (1) (2) n C , H „ 0 ,
IỊ BAI T A P 11/ LUAN
4.31. a) Cac phuong trinh hoa hoc: ( - C , H , „ 0 , - ) „ + n H , 0 - ( - C , H , „ 0 , - ) „ + n H , 0 -
n Q H „ 0 , ;ZTc >2nC,H,OH + 2nCO, t • b) Ta CO so d6 di^u che rugu tir tinh bot:
( - C , H , „ 0 , - ) „ - > n C , H „ 0 ,
S6 mol rucm thu duoc:
81 100.1000 (1) (1) (2) 100 162n -.2n.75% = 750(mol). r So lit ruou 46" = 750.46.10 0,8.0,46 1-3 = 93,75 (lit).
4.32. Phuang trinh phan ling xa phong hoa:
(C.jHj.COOX^qH, + 3NaOH ^ 3C,5H3,COONa + C3H,(OH)3
- Kh6'i lucmg chat beo tham gia phan ung: M = 200.(100 - 19,4)% = 161,2 (kg). - Theo phucfng trinh phan ung:
806 kg chát beo na'u dugc 834 kg xa phong. 161,2 kg chat beo na'u dugc m kg xa phong. 161,2 kg chat beo na'u dugc m kg xa phong.
834.161,2 , - ^ ^ ^ — ^ — = 166,8 (kg). - ^ ^ ^ — ^ — = 166,8 (kg).
Nhung vi qua trinh san xuát xa phong hao hut 15% ntn kh6'i lircmg xa
phong thu dugc la: 166,8.85% = 141,78 (kg).
4.33. a) Phucfng trinh phan ung dot chay metan:
CH4 + 2 O 2 ^ C O 2 + 2 H 2 O
b) Tmh th^ tich (dktc) cua CH4 dem dÓt va C O 2 tao thanh: - S6' mol O, = = 0,5 (mol).
22,4
Theo phuong trinh phan ung:
S6' mol CH, = ^s6' mol = ^ = 0,25 (mol).
S6' mol CO2 = |s6' mol = 0,25 (mol).
Tinh th^ tich CH4 va C O 2 : V^H^ = 0,25.22,4 = 5,6 lit = •
4.34. - Dat X la s6' mol khi thien nhien dem d6't.
^ S6' mol CH4 la 0,96x, s6 mol C O 2 = s6' mol N 2 = 0,02x. a) Cac phuang trinh hoa hoc:
CH4 + 2 O 2 ^ C O 2 + 2 H 2 O (1)
0,96x mol 0,96x mol
CO, + CăOH), ^ C a C O , ^ + H , 0 (2)
1 mol lOOg 0,8 mol <- 80g
Sau khi d6't chay hoan toan khi thien nhiSn, s6' mol C O 2 la:
Hcô =0,96x + 0,02x = 0,8 (mol)->x = 0,82 (mol). Thé tich khi dem d6't chay: V = 0,82.22,4 = 18,4 (lit).
b) So mol NaOH = 0,6.2 = 1,2 (mol).
Khi hap thu khi CO2 bang dung dich NaOH c6 thé xay ra cac phan ung:
CO, + 2NaOH -> NâCO, + H^O (3)
CO, + NaOH ^ NaHCO, (4) Ggi a la s6' mol CO2 tham gia phan ung (3).
b la s6' mol CO2 tham gia phan ung (4). Theo phuong trinh phan ung ta c6:
'CO, = a + b = 0,8
'NaOH = 2a + b = l,2
Giai he phuong trinh dugc: a = 0,4, b = 0,4.
Trong dung dich sau phan ung c6 hai mu6'i: Na2C03 va NaHCO,.
m N a , a o , = 0,4.106 = 42,4 (gam).
mNaHco, =0,4.84 = 33,6 (gam).
4.35. Cac phuong trinh hoa hoc:
CH = CH + H , ^ C H , =CH2 (1) (2)
I V I V I V C H ^ C H + 2H2-).CH3-CH3