To determine if the equation \( \text{Mg(OH)}_2 + 2 \text{HNO}_3 \rightarrow \text{Mg(NO}_3\text{)}_2 + 2 \text{H}_2\text{O} \) is balanced, we need to count the number of atoms of each element on both sides of the equation.
Left Side:
- Mg: 1
- O: 2 (from \(\text{Mg(OH)}_2\)) + 6 (from \(2 \text{HNO}_3\)) = 8
- H: 2 (from \(\text{Mg(OH)}_2\)) + 2 (from \(2 \text{HNO}_3\)) = 4
- N: 2 (from \(2 \text{HNO}_3\))
Right Side:
- Mg: 1
- O: 6 (from \(\text{Mg(NO}_3\text{)}_2\)) + 2 (from \(2 \text{H}_2\text{O}\)) = 8
- H: 4 (from \(2 \text{H}_2\text{O}\))
- N: 2 (from \(\text{Mg(NO}_3\text{)}_2\))
Now comparing both sides:
- Mg: 1 (left) = 1 (right)
- O: 8 (left) = 8 (right)
- H: 4 (left) = 4 (right)
- N: 2 (left) = 2 (right)
Since the number of each type of atom is the same on both sides of the equation, the equation is balanced.
So, the reaction \( \text{Mg(OH)}_2 + 2 \text{HNO}_3 \rightarrow \text{Mg(NO}_3\text{)}_2 + 2 \text{H}_2\text{O} \) is balanced.