Fe(OH)2+ H2SO4=FeSO4+H2O

The reaction you've provided is a neutralization reaction between iron(II) hydroxide \((\text{Fe(OH)}_2)\) and sulfuric acid \((\text{H}_2\text{SO}_4)\) that produces iron(II) sulfate \((\text{FeSO}_4)\) and water \((\text{H}_2\text{O})\).

To ensure that the chemical equation is balanced, let's write it out:

1. Reactants: Fe(OH)₂ + H₂SO₄
2. Products: FeSO₄ + H₂O

Now, let's check the balancing:

• On the left side:

  • 1 Fe
  • 2 OH
  • 2 H from H₂SO₄
  • 1 SO₄

• On the right side:

  • 1 Fe
  • 1 SO₄
  • 2 H from 1 H₂O

The total number of atoms for each element is the same on both sides.

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

So, the final balanced equation should be:

\[ \text{Fe(OH)}_2 + \text{H}_2\text{SO}_4 \rightarrow \text{FeSO}_4 + 2\text{H}_2\text{O} \]

In this case, you would have 1 Fe, 1 SO₄, and 4 H atoms on both sides, confirming the reaction is balanced.