Phương trình hóa học:

(1) 2Na + Cl₂ → 2NaCl

(2) Cl₂ + H₂ → 2HCl

(3) 3HCl + Fe(OH)₃ → FeCl₃ + 3H₂O

(4) FeCl₃ + Cu → CuCl₂ + FeCl₂

\(H_2+\dfrac{1}{2}O_2\xrightarrow[]{t^o}H_2O\)

\(H_2O+SO_3\rightarrow H_2SO_4\)

\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\uparrow\)

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

\(Cu+\dfrac{1}{2}O_2\xrightarrow[]{t^o}CuO\)

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

\(\dfrac{1}{2}O_2+Cu\xrightarrow[]{t^o}CuO\)

\(CuO+CO\xrightarrow[]{t^o}Cu+CO_2\)

\(2KMnO_4\xrightarrow[]{t^o}K_2MnO_4+MnO_2+O_2\uparrow\)

\(Zn+\dfrac{1}{2}O_2\xrightarrow[]{t^o}ZnO\)

\(ZnO+CO\xrightarrow[]{t^o}Zn+CO_2\)

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

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

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

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

 C₂H₄ → C₂H₅OH → CH₃COOH → CH₃COOC₂H₅ → C₂H₅OH

(1)   C₂H₄ + H₂O \(\underrightarrow{axit}\) C₂H₅OH

(2)  C₂H₅OH  + O₂  \(\xrightarrow[25^0-30^0C]{mengiam}\) CH₃COOH + H₂O

(3)   CH₃COOH + C₂H₅OH → CH₃COOC₂H₅ + H₂O

(4) CH₃COOC₂H₅ + NaOH \(\underrightarrow{t^0}\) CH₃COONa + C₂H₅OH

\(\left(-C_6H_{10}O_5-\right)_n+nH_2O\rightarrow nC_6H_{12}O_6\\ C_6H_{12}O_6\underrightarrow{\text{men rượu}}2C_2H_5OH+2CO_2\\ C_2H_5OH+O_2\underrightarrow{\text{men giấm}}CH_3COOH+H_2O\\ CH_3COOH+C_2H_5OH\xrightarrow[t^o]{H_2SO_{4\left(đ\right)}}CH_3COOC_2H_5+H_2O\)

1)nH2O+(C6H10O5)n→nC6H12O6
(2)C6H12O6lm→2C2H5OH+2CO2
(3)C2H5OH+O2to,xt→CH3COOH+H2O
(4)C2H5OH+HCOOH→H2O+HCOOC2H5
 

\(\left(-C_6H_{10}O_5-\right)n+nH_2O\underrightarrow{t^o}nC_6H_{12}O_6\)

\(C_6H_{12}O_6\underrightarrow{men.rượu}2C_2H_5OH+2CO_2\uparrow\)

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

\(CH_3COOH+C_2H_5OH\underrightarrow{t^o}CH_3COOC_2H_5+H_2O\)

a) 

\(Mg+\dfrac{1}{2}O_2\xrightarrow[]{t^o}MgO\)

\(MgO+2HCl\rightarrow MgCl_2+H_2O\)

\(MgCl_2+K_2CO_3\rightarrow2KCl+MgCO_3\downarrow\)

\(MgCO_3+2HNO_3\rightarrow Mg\left(NO_3\right)_2+H_2O+CO_2\uparrow\)

\(Mg\left(NO_3\right)_2+2KOH\rightarrow2KNO_3+Mg\left(OH\right)_2\downarrow\)

a) 2Mg + O2 ----------to---------> 2MgO

MgO + 2HCl -----------> MgCl2 + H2O

MgCl2 + K2CO3 ---------> MgCO3 + 2KCl

MgCO3 + 2HNO3 --------> Mg(NO3)2 + H2O + CO2

Mg(NO3)2 + 2KOH ----------> Mg(OH)2 + 2KNO3

b) 

(1) \(4FeS_2+11O_2\xrightarrow[]{t^o}2Fe_2O_3+8SO_2\uparrow\)

(2) \(SO_2+Br_2+2H_2O\rightarrow H_2SO_4+2HBr\)

(3) \(2H_2SO_{4\left(đ\right)}+S\rightarrow3SO_2+2H_2O\)

(4) \(SO_2+BaO\rightarrow BaSO_3\)

(5) \(BaSO_3+SO_2+H_2O\rightarrow Ba\left(HSO_3\right)_2\)

(6) \(Ba\left(HSO_3\right)_2+2HCl\rightarrow BaCl_2+2SO_2+2H_2O\)

(7) \(SO_2+\dfrac{1}{2}O_2\xrightarrow[V_2O_5]{t^o}SO_3\)

(8) \(SO_3+H_2O\rightarrow H_2SO_4\)

\(CH_2=CH_2+H_2O\rightarrow\left(H_2SO_4,t^o\right)CH_3CH_2OH\\ CH_3CH_2OH+CH_3COOH⇌\left(H^+,t^o\right)CH_3COOC_2H_5+H_2O\)

a)

C_nH_2n+2 + (1,5n + 0,5)O₂ → nCO₂ + (n + 1)H₂O

1          → 1,5n + 0,5               n          n + 1

Theo đề bài: n_{khí trước pứ} = n_{khí sau pứ} → 1 + 1,5n + 0,5 = n + n + 1 → n = 1

→ A: CH₄

b)

R là chất dẻo nên R có thể là: PE, PVC, PA, PS → X có chứa liên kết đôi