a, \(C_2H_6+Cl_2\xrightarrow[1:1]{as}C_2H_5Cl+HCl\)

b, \(nCH_2=CH-CH_3\underrightarrow{t^o,xt,p}\left(-CH_2-CH\left(CH_3\right)-\right)_n\)

c, \(CH_2=CH-CH_3+H_2\underrightarrow{t^o,Ni}CH_3CH_2CH_3\)

d, \(CH_3C\equiv CH+AgNO_3+NH_3\rightarrow CH_3C\equiv CAg_{\downarrow}+NH_4NO_3\)

e, \(CH_3-CH_2-CH_3+Cl_2\underrightarrow{as}CH_3-CHCl-CH_3+HCl\)

f, \(CH_2=CH-CH_3+H_2O\underrightarrow{t^o,xt}CH_3-CH\left(OH\right)-CH_3\)

g, \(CH\equiv CH+H_2O\underrightarrow{t^o,xt}CH_3-CHO\)

h, \(CH_2=CH-CH_3+HCl\underrightarrow{t^o,xt}CH_3-CHCl-CH_3\)

26)

a) \(C_2H_6+Cl_2\xrightarrow[]{a/s}C_2H_5Cl+HCl\)

b) \(nCH_2=CH-CH_3\xrightarrow[]{t^o,p,xt}\left(-CH_2-CH\left(CH_3\right)-\right)_n\)

c) \(CH_2=CH-CH_3+H_2\xrightarrow[]{Ni,t^o}CH_3-CH_2-CH_3\)

d) \(CH_3-C\equiv CH+AgNO_3+NH_3\rightarrow CH_3-C\equiv CAg\downarrow+NH_4NO_3\)

e) \(CH_3-CH_2-CH_3+Cl_2\xrightarrow[]{a/s}CH_3-CH_2-CH_2Cl+HCl\)

f) \(CH_2=CH-CH_3+H_2O\xrightarrow[]{t^o,xt}CH_3-CH\left(OH\right)-CH_3\)

g) \(CH\equiv CH+H_2O\xrightarrow[]{Hg^{2+},t^o}CH_3CHO\)

h) \(CH_2=CH-CH_3+HCl\xrightarrow[]{t^o}CH_3-CHCl-CH_3\)

I)

1) 

\(2CH_4\xrightarrow[lln]{1500^oC}C_2H_2+3H_2\\ 2CH\equiv CH\xrightarrow[]{t^o,p,xt}CH\equiv C-CH=CH_2\\ CH\equiv C-CH=CH_2+H_2\xrightarrow[]{Pd/PbCO_3,t^o}CH_2=CH-CH=CH_2\\ CH_2=CH-CH=CH_2+2H_2\xrightarrow[]{Ni,t^o}CH_3-CH_2-CH_2-CH_3\\ CH_3-CH_2-CH_2-CH_3\xrightarrow[]{cracking}CH_2=CH-CH_3+CH_4\)

2)

\(CH_3COONa+NaOH\xrightarrow[]{CaO,t^o}CH_4+Na_2CO_3\\ 2CH_4\xrightarrow[lln]{1500^oC}C_2H_2+3H_2\\ CH\equiv CH+H_2\xrightarrow[]{Pd/PbCO_3,t^o}CH_2=CH_2\\ CH_2=CH_2+Br_2\rightarrow CH_2Br-CH_2Br\)

3)

\(2CH_4\xrightarrow[]{1500^oC}C_2H_2+3H_2\\ 2CH\equiv CH\xrightarrow[]{t^o,p,xt}CH\equiv C-CH=CH_2\\ CH\equiv C-CH=CH_2+H_2\xrightarrow[]{Pd/PbCO_3,t^o}CH_2=CH-CH=CH_2\\ nCH_2=CH-CH=CH_2\xrightarrow[]{t^o,p,xt}\left(-CH_2-CH=CH-CH_2-\right)_n\)

4) 

\(C_4H_8+H_2\xrightarrow[]{Ni,t^o}C_4H_{10}\\ C_4H_{10}\xrightarrow[]{cracking}CH_4+C_3H_6\\ 2CH_4\xrightarrow[lln]{1500^oC}C_2H_2+3H_2\\ CH\equiv CH\xrightarrow[]{t^o,p,xt}CH\equiv C-CH=CH_2\)

II)

1)

\(CH\equiv C-CH_2-CH_3+2Br_2\rightarrow CHBr_2-CBr_2-CH_2-CH_3\\ CH_3-C\equiv C-CH_3+2Br_2\rightarrow CH_3-CBr_2-CBr_2-CH_3\\ CH\equiv C-CH_2-CH_3+AgNO_3+NH_3\rightarrow CAg\equiv C-CH_2-CH_3\downarrow+NH_4NO_3\)

2)

\(CH\equiv CH+2AgNO_3+2NH_3\rightarrow CAg\equiv CAg\downarrow+2NH_4NO_3\\ CH_2=CH_2+Br_2\rightarrow CH_2Br-CH_2Br\)

III)

1) \(n_C=n_{CO_2}=\dfrac{1,344}{22,4}=0,06\left(mol\right)\)

\(\Rightarrow n_H=\dfrac{0,88-0,06.12}{1}=0,16\left(mol\right)\Rightarrow n_{H_2O}=\dfrac{1}{2}n_{H_2O}=0,08\left(mol\right)\\ \Rightarrow n_{ankan}=n_{H_2O}-n_{CO_2}=0,02\left(mol\right)\\ \Rightarrow M_{ankan}=\dfrac{0,88}{0,02}=44\left(g/mol\right)\)

Đặt CTHH của ankan là C_nH_2n+2

=> 14n + 2 = 44 => n = 3

Vậy X là C₃H₈ \(\left(CTCT:CH_3-CH_2-CH_3:propan\right)\)

2) \(n_{H_2O}=\dfrac{1,62}{18}=0,09\left(mol\right)\Rightarrow n_H=2n_{H_2O}=0,18\left(mol\right)\)

\(\Rightarrow n_{CO_2}=n_C=\dfrac{1,62-0,18}{12}=0,12\left(mol\right)\\ \Rightarrow n_{ankin}=n_{CO_2}-n_{H_2O}=0,03\left(mol\right)\\ \Rightarrow nM_{ankin}=\dfrac{1,62}{0,03}=54\left(g/mol\right)\)

Đặt CTHH của ankin là C_nH_2n-2

=> 14n - 2 = 54 => n = 4

Vậy X là C₄H₆

CTCT: 

\(CH\equiv C-CH_2-CH_3:\) but-1-in

\(CH_3-C\equiv C-CH_3:\) but-2-in

3)

Sửa đề: 1,17 -> 11,7

\(\left\{{}\begin{matrix}n_{H_2O}=\dfrac{11,7}{18}=0,65\left(mol\right)\\n_{CO_2}=\dfrac{17,6}{44}=0,4\left(mol\right)\end{matrix}\right.\Rightarrow n_{H_2O}>n_{CO_2}\Rightarrow\) hh thuốc dãy đồng đẳng ankan

Ta có: \(n_{hh}=n_{H_2O}-n_{CO_2}=0,25\left(mol\right)\)

Theo BTNT: \(\left\{{}\begin{matrix}n_C=n_{CO_2}=0,4\left(mol\right)\\n_H=2n_{H_2O}=1,3\left(mol\right)\end{matrix}\right.\Rightarrow m_{hh}=0,4.12+1,3=6,1\left(g\right)\)

\(\Rightarrow M_{hh}=\dfrac{6,1}{0,25}=24,4\left(g/mol\right)\)

Đặt CT chung của hh là C_nH_2n+2

=> 14n + 2 = 24,4 => n = 1,6 

=> Hai hiđrocacbon là CH₄ và C₂H₆

Đổi 10kg = 10000g

Ta có: \(n_{CH_3COOH\left(LT\right)}=\dfrac{10000.5\%}{92\%}=\dfrac{12500}{23}\left(mol\right)\)

PTHH:
\(C_2H_5OH+O_2\xrightarrow[]{\text{men giấm}}CH_3COOH+H_2O\)

\(\dfrac{12500}{23}\)<---------------------\(\dfrac{12500}{23}\)

\(\Rightarrow m_{C_2H_5OH}=\dfrac{12500}{23}.46=25000\left(g\right)=25\left(kg\right)\)

LT là gì vậy ạ

1)

a) \(2H_2+O_2\xrightarrow[]{t^o}2H_2O\)

b) \(2KClO_3\xrightarrow[]{MnO_2,t^o}2KCl+3O_2\)

c) \(4H_2+Fe_3O_4\xrightarrow[]{t^o}3Fe+4H_2O\)

d) \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

2)

a) \(3Fe+2O_2\xrightarrow[]{t^o}Fe_3O_4\)

b) \(2KClO_3\xrightarrow[]{MnO_2,t^o}2KCl+3O_2\)

c) \(3H_2+Fe_2O_3\xrightarrow[]{t^o}2Fe+3H_2O\)

d) \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

3)

a) \(n_{Zn}=\dfrac{26}{65}=0,4\left(mol\right)\)

PTHH: Zn + 2HCl ---> ZnCl₂ + H₂ 

           0,4-->0,8---------------->0,4

b) \(V_{H_2}=0,4.22,4=8,96\left(l\right)\)

c) \(m_{HCl}=0,8.36,5=29,2\left(g\right)\)

PTHH:
\(C_2H_4+3O_2\xrightarrow[]{t^o}2CO_2+2H_2O\\ C_4H_4+5O_2\xrightarrow[]{t^o}4CO_2+2H_2O\)

Ta có: \(n_{CO_2}=4n_{hh}=0,4\left(mol\right)\)

          \(n_{hh}=\dfrac{n_{CO_2}-n_{H_2O}}{2}\Rightarrow n_{H_2O}=0,4-0,1.2=0,2\left(mol\right)\)

Vậy \(\left\{{}\begin{matrix}m_{CO_2}=0,4.44=17,6\left(g\right)\\m_{H_2O}=0,2.18=3,6\left(g\right)\end{matrix}\right.\)

a) Sửa đề: `3,7285 -> 3,7185` và `2,075 -> 20,75`

\(n_{H_2}=\dfrac{3,7185}{24,79}=0,15\left(mol\right)\)

Đặt CT chung của 2 kim loại kiềm là R

\(2R+2H_2O\rightarrow2ROH+H_2\)

 0,3<-------------0,3<-----0,15

\(ROH+HCl\rightarrow RCl+H_2O\)

  0,3------------->0,3

\(\Rightarrow M_{RCl}=\dfrac{20,75}{0,3}=\dfrac{415}{6}\left(g/mol\right)\\ \Rightarrow M_R=\dfrac{415}{6}-35,5=\dfrac{101}{3}\left(g/mol\right)\)

`=>` 2 kim loại là Na và K

b) Đặt \(\left\{{}\begin{matrix}n_{Na}=a\left(mol\right)\\n_K=b\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a+b=0,3\\58,5+74,5=20,75\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,2\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Na}=0,1.23=2,3\left(g\right)\\m_K=0,2.39=7,8\left(g\right)\end{matrix}\right.\)

a, Ta có: \(n_{H_2}=\dfrac{3,7185}{24,79}=0,15\left(mol\right)\)

Giả sử 2 KL cần tìm là A.

PT: \(A+2HCl\rightarrow ACl_2+H_2\)

Theo PT: \(n_A=n_{H_2}=0,15\left(mol\right)\)

\(\Rightarrow M_A=\dfrac{4,4}{0,15}=29,33\left(g/mol\right)\)

Mà: 2 KL nằm ở 2 chu kì kế tiếp.

→ Mg và Ca.

b, Ta có: 24n_Mg + 40n_Ca = 4,4 (1)

BT e, có: 2n_Mg + 2_Ca = 0,15.2 (2)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{Mg}=0,1\left(mol\right)\\n_{Ca}=0,05\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Mg}=0,1.24=2,4\left(g\right)\\m_{Ca}=0,05.40=2\left(g\right)\end{matrix}\right.\)

- Trích mẫu thử, đánh STT

- Dễ dàng nhận biết:
+ Chất rắn: K, MgO, K₂O, P₂O₅ (nhóm A)
+ Chất lỏng: KOH, H₂O (nhóm B)

- Cho quỳ tím vào các mẫu thử nhóm B, nếu thấy:

+ QT hóa xanh: KOH

+ QT không đổi màu: H₂O

- Hòa tan các mẫu thử A vào nước có pha sẵn quỳ tím, nếu thấy:
+ Tan, có khí không màu, không mùi thoát ra, dd thu được có màu xanh: K

`2K + 2H_2O -> 2KOH + H_2`

+ Tan, dd thu được có màu xanh: K₂O

`K_2O + H_2O -> 2KOH`

+ Tan, dd thu được có màu đỏ: P₂O₅

`P_2O_5 + 3H_2O -> 2H_3PO_4`

+ Không tan: MgO

Sửa đề: 5,6 (g) → 5,4 (g)

a, PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}=0,3\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\)

b, Ta có: \(n_{CuO}=\dfrac{40}{80}=0,5\left(mol\right)\)

PT: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)

Xét tỉ lệ:\(\dfrac{0,5}{1}>\dfrac{0,2}{1}\), ta được CuO dư.

Theo PT: \(n_{Cu}=n_{CuO\left(pư\right)}=n_{H_2}=0,2\left(mol\right)\)

⇒ n_{CuO (dư)} = 0,5 - 0,2 = 0,3 (mol)

⇒ m _{chất rắn} = m_{CuO (dư)} + m_Cu = 0,3.80 + 0,2.64 = 36,8 (g)

a) Sửa đề: Al -> Fe

\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

          0,1-------------------------->0,1

\(\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)

b) \(n_{CuO}=\dfrac{40}{80}=0,5\left(mol\right)\)

PTHH: \(CuO+H_2\xrightarrow[]{t^o}Cu+H_2O\)

Xét tỉ lệ: \(0,5>0,1\Rightarrow CuO\) dư

Theo PT: \(n_{CuO\left(p\text{ư}\right)}=n_{Cu}=n_{H_2}=0,1\left(mol\right)\)

\(\Rightarrow m=0,4.80+0,1.64=38,4\left(g\right)\)