To balance the reaction of zinc (II) hydroxide with phosphoric acid to produce zinc (II) phosphate and water, follow these steps:
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Write down the unbalanced equation: \[ \text{Zn(OH)}_2 + \text{H}_3\text{PO}_4 \rightarrow \text{Zn}_3(\text{PO}_4)_2 + \text{H}_2\text{O} \]
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Identify the compounds:
- Zinc hydroxide: Zn(OH)₂
- Phosphoric acid: H₃PO₄
- Zinc phosphate: Zn₃(PO₄)₂
- Water: H₂O
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Count the number of atoms for each element on both sides and adjust coefficients to balance them.
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Zinc (Zn): On the right, there are 3 Zn in Zn₃(PO₄)₂, so we need 3 Zn(OH)₂ on the left: \[ 3 \text{Zn(OH)}_2 \]
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Phosphate (PO₄): We have 2 PO₄ on the right (from Zn₃(PO₄)₂), so we need 2 H₃PO₄ on the left: \[ 2 \text{H}_3\text{PO}_4 \]
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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.
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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: \[ \boxed{3} , \text{Zn(OH)}_2 + \boxed{2} , \text{H}_3\text{PO}_4 \rightarrow \boxed{1} , \text{Zn}_3(\text{PO}_4)_2 + \boxed{6} , \text{H}_2\text{O} \]
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