Fe(OH)2+ H2SO4=FeSO4+H2O

The given equation represents a chemical reaction between iron(II) hydroxide (Fe(OH)₂) and sulfuric acid (H₂SO₄). To balance this reaction properly, let's go through the steps:

1. Identify Reactants and Products:

   • Reactants: Fe(OH)₂ and H₂SO₄
   • Products: FeSO₄ and H₂O

2. Write the Unbalanced Equation: \[ \text{Fe(OH)}_2 + \text{H}_2\text{SO}_4 \rightarrow \text{FeSO}_4 + \text{H}_2\text{O} \]

3. Balance the Equation:

   • On the left side, we have 1 iron (Fe), 2 hydroxide ions (OH), and 1 sulfate ion (SO₄).
   • On the right side, we have 1 iron in FeSO₄ and 2 hydrogen (H) from H₂O and 1 sulfate from FeSO₄.

Now, balance the hydroxide to produce water:

• One Fe(OH)₂ will produce:
  • 1 FeSO₄ (from Fe)
  • 2 H₂O (to get 2 OH from 2 H)

The balanced equation will be: \[ \text{Fe(OH)}_2 + \text{H}_2\text{SO}_4 \rightarrow \text{FeSO}_4 + 2\text{H}_2\text{O} \]

So the final balanced equation for the reaction is: \[ \text{Fe(OH)}_2 + \text{H}_2\text{SO}_4 \rightarrow \text{FeSO}_4 + 2\text{H}_2\text{O} \]

This shows that 1 mole of iron(II) hydroxide reacts with 1 mole of sulfuric acid to produce 1 mole of iron(II) sulfate and 2 moles of water.