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Sagot :
To determine how many grams of calcium phosphate can be made from reacting 17.373 grams of lithium phosphate, we start by analyzing the given balanced chemical equation:
[tex]\[ 2 \text{Li}_3\text{PO}_4(aq) + 3 \text{CaCl}_2(aq) \rightarrow 6 \text{LiCl}(aq) + \text{Ca}_3(\text{PO}_4)_2(s) \][/tex]
This balanced equation tells us that 2 moles of lithium phosphate (Li₃PO₄) react with 3 moles of calcium chloride (CaCl₂) to form 1 mole of calcium phosphate (Ca₃(PO₄)₂) and 6 moles of lithium chloride (LiCl).
### Step-by-Step Solution:
1. Determine the molar mass of lithium phosphate (Li₃PO₄):
- Lithium (Li): 6.94 g/mol
- Phosphorus (P): 30.97 g/mol
- Oxygen (O): 16.00 g/mol
The molar mass of Li₃PO₄:
[tex]\[ 3 \times 6.94 + 30.97 + 4 \times 16.00 = 20.82 + 30.97 + 64.00 = 115.79 \text{ g/mol} \][/tex]
2. Calculate the number of moles of lithium phosphate from its given mass:
[tex]\[ \text{Mass of Li}_3\text{PO}_4 = 17.373 \text{ grams} \][/tex]
[tex]\[ \text{Number of moles of Li}_3\text{PO}_4 = \frac{\text{Mass}}{\text{Molar mass}} = \frac{17.373}{115.79} \approx 0.150 \][/tex]
3. Use the stoichiometric ratio from the balanced equation to find the number of moles of calcium phosphate produced:
According to the balanced equation, 2 moles of Li₃PO₄ produce 1 mole of Ca₃(PO₄)₂. Therefore, the number of moles of Ca₃(PO₄)₂ will be half the number of moles of Li₃PO₄:
[tex]\[ \text{Number of moles of Ca}_3(\text{PO}_4)_2 = \frac{0.150}{2} = 0.075 \][/tex]
4. Determine the molar mass of calcium phosphate (Ca₃(PO₄)₂):
- Calcium (Ca): 40.08 g/mol
- Phosphorus (P): 30.97 g/mol
- Oxygen (O): 16.00 g/mol
The molar mass of Ca₃(PO₄)₂:
[tex]\[ 3 \times 40.08 + 2 \times (30.97 + 4 \times 16.00) = 3 \times 40.08 + 2 \times (30.97 + 64.00) = 120.24 + 2 \times 94.97 = 120.24 + 189.94 = 310.18 \text{ g/mol} \][/tex]
5. Calculate the mass of calcium phosphate formed:
[tex]\[ \text{Mass of Ca}_3(\text{PO}_4)_2 = \text{Number of moles} \times \text{Molar mass} = 0.075 \times 310.18 \approx 23.27 \text{ grams} \][/tex]
### Conclusion:
From reacting 17.373 grams of lithium phosphate, we can produce approximately 23.27 grams of calcium phosphate.
[tex]\[ 2 \text{Li}_3\text{PO}_4(aq) + 3 \text{CaCl}_2(aq) \rightarrow 6 \text{LiCl}(aq) + \text{Ca}_3(\text{PO}_4)_2(s) \][/tex]
This balanced equation tells us that 2 moles of lithium phosphate (Li₃PO₄) react with 3 moles of calcium chloride (CaCl₂) to form 1 mole of calcium phosphate (Ca₃(PO₄)₂) and 6 moles of lithium chloride (LiCl).
### Step-by-Step Solution:
1. Determine the molar mass of lithium phosphate (Li₃PO₄):
- Lithium (Li): 6.94 g/mol
- Phosphorus (P): 30.97 g/mol
- Oxygen (O): 16.00 g/mol
The molar mass of Li₃PO₄:
[tex]\[ 3 \times 6.94 + 30.97 + 4 \times 16.00 = 20.82 + 30.97 + 64.00 = 115.79 \text{ g/mol} \][/tex]
2. Calculate the number of moles of lithium phosphate from its given mass:
[tex]\[ \text{Mass of Li}_3\text{PO}_4 = 17.373 \text{ grams} \][/tex]
[tex]\[ \text{Number of moles of Li}_3\text{PO}_4 = \frac{\text{Mass}}{\text{Molar mass}} = \frac{17.373}{115.79} \approx 0.150 \][/tex]
3. Use the stoichiometric ratio from the balanced equation to find the number of moles of calcium phosphate produced:
According to the balanced equation, 2 moles of Li₃PO₄ produce 1 mole of Ca₃(PO₄)₂. Therefore, the number of moles of Ca₃(PO₄)₂ will be half the number of moles of Li₃PO₄:
[tex]\[ \text{Number of moles of Ca}_3(\text{PO}_4)_2 = \frac{0.150}{2} = 0.075 \][/tex]
4. Determine the molar mass of calcium phosphate (Ca₃(PO₄)₂):
- Calcium (Ca): 40.08 g/mol
- Phosphorus (P): 30.97 g/mol
- Oxygen (O): 16.00 g/mol
The molar mass of Ca₃(PO₄)₂:
[tex]\[ 3 \times 40.08 + 2 \times (30.97 + 4 \times 16.00) = 3 \times 40.08 + 2 \times (30.97 + 64.00) = 120.24 + 2 \times 94.97 = 120.24 + 189.94 = 310.18 \text{ g/mol} \][/tex]
5. Calculate the mass of calcium phosphate formed:
[tex]\[ \text{Mass of Ca}_3(\text{PO}_4)_2 = \text{Number of moles} \times \text{Molar mass} = 0.075 \times 310.18 \approx 23.27 \text{ grams} \][/tex]
### Conclusion:
From reacting 17.373 grams of lithium phosphate, we can produce approximately 23.27 grams of calcium phosphate.
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