1) 

3Fe + 2O2 -to-> Fe3O4 

C + O2 -to-> CO2 

4P + 5O2 -to-> 2P2O5

4Al + 3O2 -to-> 2Al2O3 

2Cu + O2 -to-> 2CuO 

2Zn + O2 -to-> 2ZnO 

S + O2 -to-> SO2 

4Na + O2 -to-> 2Na2O 

2Ba + O2 -to-> 2BaO 

CH4 + 2O2 -to-> CO2 + 2H2O 

C2H5OH + 3O2 -to-> 2CO2 + 3H2O 

NO + 1/2O2 -to-> NO2 

CO + 1/2O2 -to-> CO2 

b) 

2KClO3 -to-> 2KCl + 3O2 

2KMnO4 -to-> K2MnO4 + MnO2 + O2 

\(a)\\ 3Fe + 2O_2 \xrightarrow{t^o} Fe_3O_4\\ C + O_2 \xrightarrow{t^o} CO_2\\ 4P + 5O_2 \xrightarrow{t^o} 2P_2O_5\\ 4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3\\ 2Cu + O_2 \xrightarrow{t^o} 2CuO\\ 2Zn + O_2 \xrightarrow{t^o} 2ZnO\\ S + O_2 \xrightarrow{t^o} SO_2\\ 4Na + O_2 \xrightarrow{t^o} 2Na_2O\\ 2Ba + O_2 \xrightarrow{t^o} 2BaO\)

\(CH_4 + 2O_2 \xrightarrow{t^o} CO_2 + 2H_2O\\ C_2H_5OH + 3O_2 \xrightarrow{t^o} 2CO_2 + 3H_2O\\ 2NO + O_2 \to 2NO_2\\ 2CO + O_2 \xrightarrow{t^o} 2CO_2\\ b)\\ 2KClO_3 \xrightarrow{t^o} 2KCl + 3O_2\\ 2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2\)

\(C_2H_5OH\underrightarrow{^{^{H^+,t^0}}}C_2H_4+H_2O\)

\(2C_2H_4+O_2\underrightarrow{^{^{Pd,t^0}}}2CH_3CHO\)

\(CH_3CHO+O_2\underrightarrow{^{^{t^0,Mn^{2+}}}}CH_3COOH\)

\(CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O\)

a) CH4+Cl-----to------>CH3Cl+HCl( đây là phản ứng thế)

C6H6+Br2----to-->C6H5Br+HBr( phản ứng thế)

b) C6H6+3H2------to->C6H12( phản ứng hóa hợp)

C2H4+H2--to->C2H6 ( phản ứng hóa hợp)

C2H2+H2--to-->C2H4(phản ứng hóa hợp)

c) CH4+2O2---To->CO2+2H2O

C2H4+3O2------to------To->2CO2+2H2O

C2H2+5/2O2--to->2CO2+H2O

C6H6+12/5O2--to->6CO2+3H2O

-->đây đều là phản ứng thế

a. \(CH_4+Cl_2\underrightarrow{^{as}}CH_3Cl+HCl\)

\(C_6H_6+Br_2\underrightarrow{^{xt,to}}C_6H_5Br+HBr\)

b. \(C_6H_6+3H_2\rightarrow C_6H_{12}\)

\(C_2H_4+H_2\underrightarrow{^{Ni,to}}C_2H_6\)

\(C_2H_2+2H_2\underrightarrow{^{Ni,to}}C_2H_6\)

c. \(CH_4+2O_2\rightarrow CO_2+2H_2O\)

\(C_2H_4+3O_2\rightarrow2CO_2+2H_2O\)

\(2C_2H_2+5O_2\rightarrow4CO_2+2H_2O\)

\(2C_6H_6+15O_2\rightarrow12CO_2+6H_2O\)

Theo chiều từ trái sang, từ trên xuống nhé

\(C_2H_2+H_2\underrightarrow{t^o,Pd/PbCO_3}C_2H_4\)

\(C_2H_4+H_2O\underrightarrow{H^+,t^o}C_2H_5OH\)

\(C_2H_5OH+O_2\underrightarrow{men.giấm}CH_3COOH+H_2O\)

\(CH_3COOH+C_2H_5OH\underrightarrow{H_2SO_{4\left(đ\right)},t^o}CH_3COOC_2H_5+H_2O\)

\(CH_3COOC_2H_5+NaOH\underrightarrow{t^o}CH_3COONa+C_2H_5OH\)

\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

\(2CH_3COOH+MgO\rightarrow\left(CH_3COO\right)_2Mg+H_2O\)

\(C_2H_2+H_2\xrightarrow[t^o]{Pd}C_2H_4\\ C_2H_4+H_2O\underrightarrow{axit}C_2H_5OH\\ C_2H_5OH+O_2\underrightarrow{men.giấm}CH_3COOH\\ CH_3COOH+C_2H_5OH\xrightarrow[t^o]{H_2SO_4đặc}CH_3COOC_2H_5+H_2O\\ CH_3COOC_2H_5+NaOH\rightarrow CH_3COONa+C_2H_5OH\\ C_2H_4+Br_2\rightarrow C_2H_4Br_2\\ 2CH_3COOH+Mg\rightarrow\left(CH_3COO\right)_2Mg+H_2\)

2 H 2 S  + 3 O 2   → t ° 2S O 2  + 2 H 2 O

2 H 2 S  + S O 2  → 3S + 2 H 2 O

H 2 S  + 4 Cl 2  + 4 H 2 O →  H 2 SO 4  + 8HCl

Trong các phản ứng trên,  H 2 S  thể hiện tính khử.

3Fe+2O2 -t--> Fe3O4
 2Na + Cl2 ---> 2NaCl 
2K + S --> K2S
4P+ 5O2-t-> 2P2O5 

Fe + O2=> Fe3O4

2Na + Cl2 => 2NaCl

4P + 5O2 → 2P2O5

\(CH_3OH+C_2H_5OH\underrightarrow{H_2SO_{4\left(đ\right)},140^oC}CH_3-O-C_2H_5+H_2O\)

\(CH_3OH+CH_3OH\underrightarrow{H_2SO_{4\left(đ\right)},140^oC}CH_3-O-CH_3+H_2O\)

\(C_2H_5OH+C_2H_5OH\underrightarrow{H_2SO_{4\left(đ\right)},140^oC}C_2H_5-O-C_2H_5+H_2O\)

\(n_{H_2}=\dfrac{V_{H_2}}{22,4}=\dfrac{2,4}{22,4}\approx0,11\left(mol\right)\)

\(n_{O_2}=\dfrac{V_{O_2}}{22,4}=\dfrac{1,6}{22,4}\approx0,07\)

\(2H_2+O_2\rightarrow2H_2O\)

2 mol-1mol---2 mol

Ta có: \(\dfrac{n_{H_2}}{2}=\dfrac{0,11}{2}\)

\(\dfrac{n_{O_2}}{1}=\dfrac{0,07}{1}\)

\(\Rightarrow\dfrac{n_{H_2}}{2}< \dfrac{n_{O_2}}{1}\)

Vậy \(O_2\) dư

Số mol O₂ dư:

\(n_{O_2}=\dfrac{0,07.1}{2}=0,035\left(mol\right)\)

Khối lượng O₂ dư

\(m_{O_2}=0,035.32=1,12\left(g\right)\)

Khối lượng nước thu được:

\(n_{H_2O}=\dfrac{0,07.2}{2}=0,07\left(mol\right)\)

\(\Rightarrow m_{H_2O}=n_{H_2O}.M_{H_2O}=0,07.18=1,26\left(g\right)\)