To balance the equation \( \text{Ca(OH)}_2 + \text{H}_3\text{PO}_4 \rightarrow \text{Ca}_3(\text{PO}_4)_2 + \text{H}_2\text{O} \), we will look at the number of each type of atom on both sides of the equation.
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Reactants:
- Calcium (Ca): 1 (in \( \text{Ca(OH)}_2 \))
- Oxygen (O): 2 (from \( \text{Ca(OH)}_2 \)) + 4 (from \( \text{H}_3\text{PO}_4 \)) = 6
- Hydrogen (H): 2 (from \( \text{Ca(OH)}_2 \)) + 3 (from \( \text{H}_3\text{PO}_4 \)) = 5
- Phosphorus (P): 1 (in \( \text{H}_3\text{PO}_4 \))
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Products:
- Calcium: 3 (from \( \text{Ca}_3(\text{PO}_4)_2 \))
- Oxygen: 8 (from \( \text{Ca}_3(\text{PO}_4)_2 \)) + 1 (from \( \text{H}_2\text{O} \)) = 9
- Hydrogen: 2 (from \( \text{H}_2\text{O} \))
- Phosphorus: 2 (in \( \text{Ca}_3(\text{PO}_4)_2 \))
To balance this equation, we need to equalize the number of each atom on each side.
By inspection (and testing the different options provided), the option that seems to make the most sense is:
3Ca(OH)2 + 2H3PO4 → Ca3(PO4)2 + 6H2O
Let’s verify:
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Reactants:
- Ca: \( 3 \) (from \( 3 \times \text{Ca(OH)}_2 \))
- H: \( 6 \) (from \( 3 \times \text{Ca(OH)}_2 \)) + \( 6 \) (from \( 2 \times \text{H}_3\text{PO}_4 \)) = \( 12 \)
- O: \( 6 \) (from \( 3 \times \text{Ca(OH)}_2 \)) + \( 8 \) (from \( 2 \times \text{H}_3\text{PO}_4 \)) = \( 14 \)
- P: \( 2 \) (from \( 2 \times \text{H}_3\text{PO}_4 \))
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Products:
- Ca: \( 3 \) (from \( \text{Ca}_3(\text{PO}_4)_2 \))
- H: \( 6 \) (from \( 6 \times \text{H}_2\text{O} \))
- O: \( 8 \) (from \( \text{Ca}_3(\text{PO}_4)_2 \)) + \( 6 \) (from \( 6 \times \text{H}_2\text{O} \)) = \( 14 \)
- P: \( 2 \) (in \( \text{Ca}_3(\text{PO}_4)_2 \))
Now both sides balance perfectly with:
3Ca(OH)2 + 2H3PO4 → Ca3(PO4)2 + 6H2O
Thus, the proper way to balance the equation is:
3Ca(OH)2 + 2H3PO4 → Ca3(PO4)2 + 6H2O
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