To balance the reaction between zinc(II) hydroxide \((\text{Zn(OH)}_2)\) and phosphoric acid \((\text{H}_3\text{PO}_4)\) that produces zinc(II) phosphate \((\text{Zn}_3(\text{PO}_4)_2)\) and water \((\text{H}_2\text{O})\), we start by writing the unbalanced equation:
\[ \text{Zn(OH)}_2 + \text{H}_3\text{PO}_4 \rightarrow \text{Zn}_3(\text{PO}_4)_2 + \text{H}_2\text{O} \]
Next, we need to identify the number of each type of atom on both sides to ensure balanced coefficients.
- Zinc (Zn): In \(\text{Zn}_3(\text{PO}_4)_2\) there are 3 Zn.
- Phosphate (PO₄): In \(\text{Zn}_3(\text{PO}_4)_2\) there are 2 \(\text{PO}_4\), so we need 2 moles of \(\text{H}_3\text{PO}_4\).
- Water (H₂O): Each \(\text{Zn(OH)}_2\) provides 2 OH⁻ ions.
After some trial and error, we find the coefficients to balance the reaction:
The balanced equation will be:
\[ 3 , \text{Zn(OH)}_2 + 2 , \text{H}_3\text{PO}_4 \rightarrow 1 , \text{Zn}_3(\text{PO}_4)_2 + 6 , \text{H}_2\text{O} \]
Thus, the corresponding coefficients are:
- 3 for \(\text{Zn(OH)}_2\)
- 2 for \(\text{H}_3\text{PO}_4\)
- 1 for \(\text{Zn}_3(\text{PO}_4)_2\)
- 6 for \(\text{H}_2\text{O}\)
The complete balanced equation can be stated as:
\[ 3 , \text{Zn(OH)}_2 + 2 , \text{H}_3\text{PO}_4 \rightarrow 1 , \text{Zn}_3(\text{PO}_4)_2 + 6 , \text{H}_2\text{O} \]
So the answers to your question about the coefficients in order are: 3, 2, 1, 6.