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QUESTION

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

"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.

1. All of Group 1 metals are soluble.
2. All hydroxides are insoluble except those of Group 1 metals and "Ba"^(2+), "Ca"^(2+), and "Sr"^(2+).

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