To balance the reaction of zinc (II) hydroxide (Zn(OH)₂) with phosphoric acid (H₃PO₄) to produce zinc (II) phosphate (Zn₃(PO₄)₂) and water (H₂O), we begin by writing the unbalanced equation:
\[ \text{Zn(OH)}_2 + \text{H}_3\text{PO}_4 \rightarrow \text{Zn}_3(\text{PO}_4)_2 + \text{H}_2\text{O} \]
To balance this reaction, we can follow these steps:
- Count the number of each atom in the reactants and products.
- Adjust coefficients to balance the atoms one type at a time.
Step 1: Write the unbalanced equation
\[ \text{___ Zn(OH)}2 + \text{___ H}3\text{PO}4 \rightarrow \text{_ Zn}3(\text{PO}4)2 + \text{__ H}_2\text{O} \]
Step 2: Balance Zinc (Zn)
From the product Zn₃(PO₄)₂, we see that we need 3 zinc atoms:
- Therefore, we need 3 Zn(OH)₂:
\[ \text{3 Zn(OH)}_2 \]
Step 3: Balance Phosphate (PO₄)
From the product Zn₃(PO₄)₂, we see we need 2 phosphate groups (PO₄). Since each H₃PO₄ contains one phosphate, we need 2 H₃PO₄:
- Therefore, we have:
\[ \text{2 H}_3\text{PO}_4 \]
Step 4: Balance Water (H₂O)
To find the water produced, we can use the total number of hydrogens:
From the reactants:
- Each Zn(OH)₂ has 2 hydrogens and we have 3, contributing 6 H.
- Each H₃PO₄ has 3 hydrogens and we need 2, contributing 6 H. So total H in reactants = 6 + 6 = 12 H
From the products:
- Each H₂O has 2 hydrogens. To balance 12 H, we need:
\[ \text{6 H}_2\text{O} \]
Step 5: Combine everything into the balanced equation
Putting it all together, we have the balanced equation:
\[ \text{3 Zn(OH)}_2 + \text{2 H}_3\text{PO}_4 \rightarrow \text{1 Zn}_3(\text{PO}_4)_2 + \text{6 H}_2\text{O} \]
Final Balanced Equation:
\[ 3 \text{ Zn(OH)}_2 + 2 \text{ H}_3\text{PO}_4 \rightarrow 1 \text{ Zn}_3(\text{PO}_4)_2 + 6 \text{ H}_2\text{O} \]
So the coefficients are:
- 3, 2, 1, 6 for Zn(OH)₂, H₃PO₄, Zn₃(PO₄)₂, and H₂O, respectively.
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