To balance the equation:
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Start with the unbalanced equation: \[ \text{CaO} + \text{P}_2\text{O}_5 \rightarrow \text{Ca}_3(\text{PO}_4)_2 \]
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Count the number of each type of atom on both sides of the equation:
- For \(\text{Ca}_3(\text{PO}_4)_2\):
- Calcium (Ca): 3
- Phosphorus (P): 2
- Oxygen (O): 8 (4 from PO4 * 2)
- For \(\text{Ca}_3(\text{PO}_4)_2\):
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We need 3 calcium (Ca) from CaO, which leads to: \[ 3 \text{CaO} \] This gives us 3 calcium.
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Now we consider phosphorus (P). We have 2 phosphorus in \(\text{Ca}_3(\text{PO}_4)_2\) and need one \(\text{P}_2\text{O}_5\): \[ 1 \text{P}_2\text{O}_5 \]
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Now, let's check the balance of oxygen. We have:
- From \(3 \text{CaO}\): \(3 \times 1 = 3\) oxygen
- From \(1 \text{P}_2\text{O}_5\): \(5\) oxygen
- Total oxygen on the left: \(3 + 5 = 8\)
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Now consider the right side:
- From \(\text{Ca}_3(\text{PO}_4)_2\): \(8\) oxygen
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Each side now has:
- Ca: 3
- P: 2
- O: 8
Thus, the balanced equation is: \[ 3 \text{CaO} + 1 \text{P}_2\text{O}_5 \rightarrow 1 \text{Ca}_3(\text{PO}_4)_2 \]
The coefficient in front of \(\text{Ca}_3(\text{PO}_4)_2\) is 1.