1. Cu➜ CuO➜CuSO4➜CuCl2➜Cu(NO3)2➜Cu(OH)2➜Cu 2.C➜CO2➜CO➜CO2➜NaHCO3➜Na2CO3➜CaCO3➜CO2 3. Cl2➜HCl➜Cl2➜NaCl➜Cl2➜NaClO➜NaCl 4. H2O➜H2➜H2O➜HClO➜Cl2➜KClO➜KCl

0 bình luận về “1. Cu➜ CuO➜CuSO4➜CuCl2➜Cu(NO3)2➜Cu(OH)2➜Cu 2.C➜CO2➜CO➜CO2➜NaHCO3➜Na2CO3➜CaCO3➜CO2 3. Cl2➜HCl➜Cl2➜NaCl➜Cl2➜NaClO➜NaCl 4. H2O➜H2➜H2O➜HClO➜Cl2➜KClO➜KCl”

1. Đáp án:

    

   Giải thích các bước giải:

    1)

   $2Cu+O2—>2CuO$

   $CuO+H2SO4—>CuSO4+H2O$

   $CuSO4+BaCl2–>CuCl2+BaSO4$

   $CuCl2+2AgNO3->2AgCl+Cu(NO3)2$

   $Cu(NO3)2+2NaOH—>Cu(OH)2+2NaNO3$

   $Cu(OH)2–>CuO+H2O$

   2

   $C+O2–>CO2$

   $CO2+C–>2CO$

   $2CO+O2–>2CO2$

   $CO2+NaOH—>NaHCO3$

   $2NaHCO3–>Na2CO3+CO2+H2O$

   $Na2CO3+CaCl2–>2NaCl+CaCO3$

   $CaCO3–>CaO+CO2$

   3)

   $Cl2+2H2—>2HCl$

   $2HCl–>2H2+Cl2$

   $Cl2+2Na—>2NaCl$

   $2NaCl–>2Na+Cl2$

   $Cl2+2NaOH—>NaClO+NaCl+H2O$

   $NaClO+2HCl–>NaCl+Cl2+H2O$

   4)

   $2H2O–>2H2+O2$

   $2H2+O2–>2H2O$

   $2H2O+Cl2–>2HClO+2HCl$

   $2HClO+2HCl–>Cl2+2H2O$

   $Cl2+2KOH—>KCl+KClO$

   $KClO+2HCl–>2KCl+H2O$

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2. 1/ 

   $2Cu+ O_2 \buildrel{{t^o}}\over\to 2CuO$

   $CuO+ H_2SO_4 \to CuSO_4+ H_2O$

   $CuSO_4+ BaCl_2 \to CuCl_2+ BaSO_4$

   $CuCl_2+ 2AgNO_3 \to Cu(NO_3)_2+ 2AgCl$

   $Cu(NO_3)_2+ 2NaOH \to Cu(OH)_2+ 2NaNO_3$

   $Cu(OH)_2 \buildrel{{t^o}}\over\to CuO+ H_2O$ 

   2/

   $C+ O_2 \buildrel{{t^o}}\over\to CO_2$

   $CO_2+ C \buildrel{{t^o}}\over\to 2CO$

   $2CO+ O_2 \buildrel{{t^o}}\over\to 2CO_2$

   $CO_2+ NaOH \to NaHCO_3$

   $NaHCO_3+ NaOH \to Na_2CO_3+ H_2O$

   $Na_2CO_3+ CaCl_2 \to CaCO_3+ 2NaCl$

   $CaCO_3+ 2HCl \to CaCl_2+ CO_2+ H_2O$ 

   3/

   $Cl_2+ H_2 \buildrel{{as}}\over\to 2HCl$

   $MnO_2+ 4HCl \to MnCl_2+ Cl_2+ 2H_2O$

   $Cl_2+ 2Na \to 2NaCl$ 

   $2NaCl \buildrel{{đpnc}}\over\to 2Na+ Cl_2$

   $Cl_2+ 2NaOH \to NaCl+ NaClO+ H_2O$

   $NaClO+ 2HCl \to NaCl+ Cl_2+ H_2O$ 

   4/

   $2Na+ 2H_2O \to 2NaOH+ H_2$

   $2H_2+ O_2 \buildrel{{t^o}}\over\to 2H_2O$ 

   $Cl_2+ H_2O \rightleftharpoons HCl+ HClO$

   $HClO+ HCl \to Cl_2+ H_2O$ 

   $Cl_2+ 2KOH \to KCl+ KClO+ H_2O$

   $3KClO \to 2KCl+ KClO_3$

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