balance the reaction of potassium hydroxide and iron(ii) chloride to give iron(ii) hydroxide and potassium…

Explanation:

Step1: Check element - by - element balance

For the reaction between potassium hydroxide ($KOH$) and iron(II) chloride ($FeCl_2$) to form iron(II) hydroxide ($Fe(OH)_2$) and potassium chloride ($KCl$), we need to ensure that the number of each type of atom is the same on both sides of the chemical equation. The unbalanced equation is $KOH + FeCl_2\rightarrow Fe(OH)_2+KCl$. On the left - hand side: 1 $K$, 1 $O$, 1 $H$, 1 $Fe$, 2 $Cl$. On the right - hand side: 1 $K$, 2 $O$, 2 $H$, 1 $Fe$, 1 $Cl$.

Step2: Balance the chlorine atoms

To balance the chlorine atoms, since there are 2 chlorine atoms in $FeCl_2$, we need 2 moles of $KCl$. So the equation becomes $KOH + FeCl_2\rightarrow Fe(OH)_2 + 2KCl$. Now, on the right - hand side, we have 2 $K$ atoms.

Step3: Balance the potassium and hydroxide groups

To balance the potassium and hydroxide groups, we need 2 moles of $KOH$. The balanced equation is $2KOH+FeCl_2\rightarrow Fe(OH)_2 + 2KCl$.

Answer:

$2KOH + FeCl_2\rightarrow Fe(OH)_2+2KCl$ (the third option)