Question

The reaction of zinc (II) hydroxide with phosphoric acid produces zinc (II) phosphate and water.

___ Zn(OH)2+____ H2PO4 Arrow right ___ Zn3(PO4)2 + _____H2O

Balance the reaction by completing the chemical equation with the molar coefficients.

Answer 1

To balance the reaction of zinc (II) hydroxide with phosphoric acid to produce zinc (II) phosphate and water, follow these steps:

1. Write down the unbalanced equation:
\[
\text{Zn(OH)}_2 + \text{H}_3\text{PO}_4 \rightarrow \text{Zn}_3(\text{PO}_4)_2 + \text{H}_2\text{O}
\]

2. Identify the compounds:
- Zinc hydroxide: Zn(OH)₂
- Phosphoric acid: H₃PO₄
- Zinc phosphate: Zn₃(PO₄)₂
- Water: H₂O

3. Count the number of atoms for each element on both sides and adjust coefficients to balance them.

4. Zinc (Zn): On the right, there are 3 Zn in Zn₃(PO₄)₂, so we need 3 Zn(OH)₂ on the left:
\[
3 \text{Zn(OH)}_2
\]

5. Phosphate (PO₄): We have 2 PO₄ on the right (from Zn₃(PO₄)₂), so we need 2 H₃PO₄ on the left:
\[
2 \text{H}_3\text{PO}_4
\]

6. Count the hydrogen atoms after adding 2 H₃PO₄. Each H₃PO₄ has 3 H, so in 2 H₃PO₄ we have 6 H atoms. Each Zn(OH)₂ contributes 2 H, so we will need 3 Zn(OH)₂.
7. The total water produced will be given by the remaining hydrogen after combining with OH from Zn(OH)₂:
- H in Zn(OH)₂ comes from two molecules and combines with the remaining:
\[
2 \text{H}_3\text{PO}_4 \rightarrow 3 \text{Zn(OH)}_2
\]

Putting it all together, we get:
\[
3 \text{Zn(OH)}_2 + 2 \text{H}_3\text{PO}_4 \rightarrow 1 \text{Zn}_3(\text{PO}_4)_2 + 6 \text{H}_2\text{O}
\]

Thus, the balanced equation is:
\[
3 \text{Zn(OH)}_2 + 2 \text{H}_3\text{PO}_4 \rightarrow 1 \text{Zn}_3(\text{PO}_4)_2 + 6 \text{H}_2\text{O}
\]

By inserting the coefficients, it looks as follows:
\[
\boxed{3} \, \text{Zn(OH)}_2 + \boxed{2} \, \text{H}_3\text{PO}_4 \rightarrow \boxed{1} \, \text{Zn}_3(\text{PO}_4)_2 + \boxed{6} \, \text{H}_2\text{O}
\]