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222-9+11+12:2*14+14 = ? ( )

Hóa học Lớp 9: a/ Al2O3 => Al => Al(NO)3 => Al(OH)3 => NaAlO2 b/ Fe –> FeCl3 –> Fe(NO3)3 –> Fe(OH)3 –> Fe2O3 c/

Hóa học Lớp 9: a/ Al2O3 => Al => Al(NO)3 => Al(OH)3 => NaAlO2
b/ Fe –> FeCl3 –> Fe(NO3)3 –> Fe(OH)3 –> Fe2O3
c/ FeCl2=>Fe(NO3)2=> Fe(OH)2=> Fe2O3=> Fe2(SO4)3
d/ Fe => FeCl2 => Fe(OH)2 => Fe(OH)3 => Fe2(SO4)3, giúp em giải bài hóa này ạ, em cảm ơn thầy cô và các bạn nhiều.

Comments ( 1 )

  1. a)
    – 2Al_2O_3$\xrightarrow[criolit]{đpnc}$4Al+3O_2
    – Al+3AgNO_3→Al(NO_3)_3+3Ag
    – Al(NO_3)_3+3NaOH→Al(OH)_3+3NaNO_3
    – Al(OH)_3+NaOH→NaAlO_2+2H_2O
    b)
    – 2Fe+3Cl_2$\xrightarrow[]{t^o}$2AlCl_3
    – FeCl_3+3NaNO_3→Fe(NO_3)_3+3NaCl
    – Fe(NO_3)_3+3NaOH→Fe(OH)_3+3NaNO_3
    – 2Fe(OH)_3$\xrightarrow[]{t^o}$Fe_2O_3+3H_2O
    – Fe_2O_3+3H_2SO_4→Fe_2(SO_4)_3+3H_2O
    c)
    – FeCl_2+2NaNO_3→Fe(NO_3)_2+2NaCl
    – Fe(NO_3)_2+2NaOH→Fe(OH)_2+2NaNO_3
    – 4Fe(OH)_2+O_2$\xrightarrow[]{t^o}$2Fe_2O_3+4H_2O
    – Fe_2O_3+3H_2SO_4→Fe_2(SO_4)_3+3H_2O
    d)
    – Fe+2HCl→FeCl_2+H_2
    – FeCl_2+2NaOH→Fe(OH)_2+2NaCl
    – 4Fe(OH)_2+O_2+2H_2O→4Fe(OH)_3
    – 2Fe(OH)_3+3H_2SO_4→Fe_2(SO_4)_3+6H_2O
     

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222-9+11+12:2*14+14 = ? ( )

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