Theo chiều từ trái sang, từ trên xuống nhé

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

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

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

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

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

\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

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

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

$C_2H_4 + H_2O \xrightarrow{t^o,H^+} C_2H_5OH$
$C_2H_5OH + CH_3COOH \buildrel{{H_2SO_4}}\over\rightleftharpoons CH_3COOC_2H_5 + H_2O$
$CH_3COOC_2H_5 + H_2O \buildrel{{H_2SO_4}}\over\rightleftharpoons CH_3COOH + C_2H_5OH$

$2CH_3COOH + Mg \to (CH_3COO)_2Mg + H_2$
$(CH_3COO)_2Mg + Ca(OH)_2 \to (CH_3COO)_2Ca + Mg(OH)_2$

$(CH_3COO)_2Ca + K_2CO_3 \to 2CH_3COOK + CaCO_3$

$C_6H_{12}O_6 \xrightarrow{t^o,men\ rượu} 2CO_2 + 2C_2H_5OH$

$C_2H_5OH + O_2 \xrightarrow{men\ giấm} CH_3COOH + H_2O$

$C_2H_5OH + CH_3COOH \buildrel{{H_2SO_4}}\over\rightleftharpoons CH_3COOC_2H_5 + H_2O$
$CH_3COOC_2H_5 + H_2O \buildrel{{H_2SO_4}}\over\rightleftharpoons CH_3COOH + C_2H_5OH$

$2CH_3COOH + CuO \to (CH_3COO)_2Cu + H_2O$

$(CH_3COO)_2Cu + NaOH \to 2CH_3COONa + Cu(OH)_2$

$2Na + 2H_2O \to 2NaOH + H_2$
$2NaOH + SO_2 \to Na_2SO_3 + H_2O$
$Na_2SO_3 + H_2SO_4 \to Na_2SO_4 + SO_2 + H_2O$
$Na_2SO_4 + BaCl_2 \to 2NaCl + BaSO_4$
$2NaCl + 2H_2O \xrightarrow{dpdd,cmn} 2NaOH + H_2 + Cl_2$
$NaOH + HCl \to NaCl + H_2O$
$NaCl + AgNO_3 \to AgCl + NaNO_3$

\(\left(1\right)Na+2H_2O\rightarrow2NaOH+H_2\\ \left(2\right)2NaOH+SO_2\rightarrow Na_2SO_3+H_2O\\ \left(3\right)Na_2SO_3+H_2SO_4\rightarrow Na_2SO_4+H_2O+SO_2\uparrow\\ \left(4\right)Na_2SO_4+BaCl_2\rightarrow2NaCl+BaSO_4\downarrow\\ \left(5\right)2NaCl+2H_2O\rightarrow^{\left(đpdd,cmm\right)}2NaOH+Cl_2\uparrow+H_2\uparrow\\ \left(6\right)NaOH+HCl\rightarrow NaCl+H_2O\\ \left(7\right)NaCl+AgNO_3\rightarrow NaNO_3+AgCl\downarrow\)

\(4Na+O_2\underrightarrow{^{^{t^0}}}2Na_2O\)

\(Na_2O+H_2O\rightarrow2NaOH\)

\(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)

\(Na_2SO_4+BaCl_2\rightarrow2NaCl+BaSO_4\)

\(2SO_2 + O_2 \xrightarrow{t^o,xt} 2SO_3\\ SO_3 + H_2O \to H_2SO_4\\ Na_2O + H_2SO_4 \to Na_2SO_4 + H_2O\\ Na_2SO_4 + BaCl_2 \to BaSO_4 + 2NaCl\)

2SO2+ O2 --> 2SO3
SO3+H2O--> H2S04
2Na+ H2SO4-->Na2SO4+ H2
mik cân bằng rồi còn BaS04 mik chịu XD

a)

$FeS + 2HCl \to FeCl_2 + H_2$
$2H_2S + SO_2 \to 3S + 2H_2O$

$S + O_2 \xrightarrow{t^o} SO_2$
$2SO_2 + O_2 \xrightarrow{t^o,xt} 2SO_3$
$SO_3 + H_2O \to H_2SO_4$
$Ba + H_2SO_4 \to BaSO_4 + H_2$
b)

$SO_2 + 2H_2S \to 3S + 2H_2O$
$Fe + S \xrightarrow{t^o} FeS$
$FeS + 2HCl \to FeCl_2 + H_2S$
$H_2S + \dfrac{3}{2} O_2 \xrightarrow{t^o} SO_2 + H_2O$

$SO_2 + 2NaOH \to Na_2SO_3 + H_2O$
$Na_2SO_3 + H_2SO_4 \to Na_2SO_4 + SO_2 + H_2O$

C2H4 + H2O => (140^oC,H2SO4đ) C2H5OH

C2H5OH + CH3COOH => (pứ hai chiều, t^o,H2SO4) CH3COOC2H5 + H2O

CH3COOC2H5 + NaOH => (t^o) CH3COONa + C2H5OH

C2H4 + H2O -axit-> C2H5OH

C2H5OH + CH3COOH <-H2SO4đ,to-> CH3COOC2H5 + H2O

CH3COOC2H5 + NaOH -to-> CH3COONa + C2H5OH

\(a)2Ca+O_2\xrightarrow[]{t^0}2CaO\\ CaO+H_2O\rightarrow Ca\left(OH\right)_2\\ Ca\left(OH\right)_2+2HCl\rightarrow CaCl_2+2H_2O\\ CaCl_2+2AgNO_3\rightarrow Ca\left(NO_3\right)_2+2AgCl\)

\(b)S+O_2\rightarrow SO_2\\ 2SO_2+O_2\xrightarrow[V_2O_4]{t^0}2SO_3\\ SO_3+Na_2O\rightarrow NaSO_4\\ Na_2SO_4+BaCl_2\rightarrow2NaCl+BaSO_4\)