What is the net ionic equation of the reaction of FeCl_2 with NaOH?

    0

    ##Fe^(2+) (aq) + 2OH^(-)(aq)## ##Fe(OH)_2(s) ##
    Step 1. Write the word equation for the reaction.
    ##FeCl_2## reacts with ##NaOH## to produce ##Fe(OH)_2## and ##NaCl##
    Step 2. Write the molecular equation for the reaction.
    You use the Solubility Rules to determine if any of the products is a solid.
    This tells you that ##Fe(OH)_2## is a solid and ##NaCl## is soluble. The molecular equation is then
    ##FeCl_2(aq)+ 2NaOH(aq) Fe(OH)_2(s) + 2 NaCl(aq)##
    Step 3. Write the ionic equation.
    Iron(II) chloride when dissolved produces ##Fe^(2+)## ions and ##Cl^-## ions.
    Sodium hydroxide produces ##Na^+## ions and ##OH^-## ions.
    You do not write solids as ions.
    The ionic equation is
    ##Fe^(2+)(aq) + 2Cl^(-)(aq) +2Na^(+)(aq) + 2OH^(-)(aq) Fe(OH)_2(s) + 2Na^(+)(aq) + 2Cl^(-)(aq)##
    Step 4. Cancel items that appear on each side of the equation (##2Cl^-## and ##2Na^+##) to get the net ionic equation.
    ##Fe^(2+)(aq) + color(red)(cancel(color(black)(2Cl^(-)(aq))) + color(red)(cancel(color(black)(2Na^(+)(aq))))) + 2OH^(-)(aq) Fe(OH)_2(s) + color(red)(cancel(color(black)(2Na^(+)(aq)))) + color(red)(cancel(color(black)(2Cl^(-)(aq)))##
    The net ionic equation is
    ##Fe^(2+)(aq) + 2OH^(-)(aq) Fe(OH)_2(s) ##
    And there you have it.
    Here’s another example this one using ##FeCl_3 + NaOH##:
    video from: Noel Pauller

                                                                                                                                      Order Now