## Word equation

** iron(II) cation + hydroxide anion →→ iron(II) hydroxide**

## Input interpretation

** F****e****2+**** ****+ O****H****-**** → Fe(OH****)****3**

** **** iron(II) cation hydroxide anion iron(II) hydroxide**

## Balanced equation

Balance the chemical equation algebraically:

** F****e****2+**** ****+ O****H****-**** → Fe(OH****)****3**

Add stoichiometric coefficients, c1, to the reactants and products:

**c****1****F****e****2+**** ****+ ****c****2**** O****H****-**** → ****c****3**** Fe(OH****)****3**

Set the number of atoms and the charges in the reactants equal to the number of atoms and the charges in the products for **Fe, O** and** H**:

**Fe: ****c****1****= ****c****3**

**O: ****c****2**** = 2 ****c****3**

** H: ****c****2****= 2****c****3**

**Charges: 2 ****c****1****- ****c****2****=0**

Since the coefficients are relative quantities and underdetermined, choose a coefficient to set arbitrarily. To keep the coefficients small, the arbitrary value is ordinarily one. For instance, set **c****1** = 1 and solve the system of equations for the remaining coefficients:

**c****1**** = 1 **

**c****2****=2 **

**c****3**** = 1**

Substitute the coefficients into the chemical reaction to obtain the balanced equation:

**Answer:**

** F****e****2+**** ****+ O****H****-**** → Fe(OH****)****3**