Question

The reaction of zinc (II) hydroxide with phosphoric acid produces zinc (II) phosphate and water.
Balance the reaction by completing the chemical equation with the molar coefficients.
Zn(OH)2 + H3PO4 → Zn3(PO4)2 + H2O
2
3
4
5
6

Answer 1

To balance the reaction of zinc (II) hydroxide with phosphoric acid, we start with the unbalanced equation:

\[ \text{Zn(OH)}_2 + \text{H}_3\text{PO}_4 \rightarrow \text{Zn}_3(\text{PO}_4)_2 + \text{H}_2\text{O} \]

First, let's identify how many of each type of atom we have on both sides:

- **On the left:**
- Zinc (Zn): 1 (from Zn(OH)₂)
- Oxygen (O): 2 (from Zn(OH)₂) + 4 (from H₃PO₄) = 6
- Hydrogen (H): 2 (from Zn(OH)₂) + 3 (from H₃PO₄) = 5
- Phosphorus (P): 1 (from H₃PO₄)

- **On the right:**
- Zinc (Zn): 3 (from Zn₃(PO₄)₂)
- Oxygen (O): 8 (from Zn₃(PO₄)₂) + 1 (from H₂O) = 9
- Hydrogen (H): 2 (from H₂O)
- Phosphorus (P): 2 (from Zn₃(PO₄)₂)

To balance the equation, we can adjust the coefficients. Here are the steps:

1. **Balance zinc (Zn)**: Since there are 3 Zn on the product side, place a coefficient of 3 in front of Zn(OH)₂.
2. **Balance phosphorus (P)**: There are 2 phosphate ions (PO₄³⁻) in zinc (II) phosphate, so we need to have 2 H₃PO₄. Place a coefficient of 2 in front of H₃PO₄.
3. **Balance water (H₂O)**: After adding those coefficients, count the remaining hydrogen atoms and make sure the resulting water molecules balance.

The balanced equation is:

\[
3 \text{Zn(OH)}_2 + 2 \text{H}_3\text{PO}_4 \rightarrow \text{Zn}_3(\text{PO}_4)_2 + 6 \text{H}_2\text{O}
\]

So the complete balanced chemical equation is:

\[
\boxed{3} \text{Zn(OH)}_2 + \boxed{2} \text{H}_3\text{PO}_4 \rightarrow \text{Zn}_3(\text{PO}_4)_2 + \boxed{6} \text{H}_2\text{O}
\]

Thus, the molar coefficients are 3, 2, 1, and 6 for Zn(OH)₂, H₃PO₄, Zn₃(PO₄)₂, and H₂O respectively.