45.7 g of P4o10 is reacted with an excess of calcium oxide. what mass of calcium phosphate is produced. ?
The correct answer and explanation is :
To solve this problem, let’s start by analyzing the chemical reaction and then work through the calculations step by step.
Step 1: Write the balanced chemical equation
The reaction between phosphorus pentoxide (P₄O₁₀) and calcium oxide (CaO) produces calcium phosphate (Ca₃(PO₄)₂). The balanced equation for the reaction is:
[
P_4O_{10} (s) + 6CaO (s) \rightarrow 2Ca_3(PO_4)_2 (s)
]
Step 2: Calculate the molar mass of P₄O₁₀
The molar mass of phosphorus pentoxide (P₄O₁₀) can be calculated by adding the atomic masses of phosphorus and oxygen:
- Atomic mass of phosphorus (P) = 30.97 g/mol
- Atomic mass of oxygen (O) = 16.00 g/mol
The molar mass of P₄O₁₀ is:
[
M_{P_4O_{10}} = 4 \times 30.97 \, \text{g/mol} + 10 \times 16.00 \, \text{g/mol}
]
[
M_{P_4O_{10}} = 123.88 \, \text{g/mol} + 160.00 \, \text{g/mol} = 283.88 \, \text{g/mol}
]
Step 3: Calculate the moles of P₄O₁₀
You are given 45.7 grams of P₄O₁₀. To find the number of moles, use the formula:
[
\text{moles of P₄O₁₀} = \frac{\text{mass of P₄O₁₀}}{\text{molar mass of P₄O₁₀}}
]
[
\text{moles of P₄O₁₀} = \frac{45.7 \, \text{g}}{283.88 \, \text{g/mol}} = 0.1616 \, \text{mol}
]
Step 4: Use stoichiometry to find the moles of Ca₃(PO₄)₂
From the balanced equation, we know that 1 mole of P₄O₁₀ reacts to produce 2 moles of Ca₃(PO₄)₂. Therefore, the moles of Ca₃(PO₄)₂ produced are:
[
\text{moles of Ca₃(PO₄)₂} = 2 \times \text{moles of P₄O₁₀}
]
[
\text{moles of Ca₃(PO₄)₂} = 2 \times 0.1616 \, \text{mol} = 0.3232 \, \text{mol}
]
Step 5: Calculate the mass of Ca₃(PO₄)₂
Now, calculate the molar mass of calcium phosphate (Ca₃(PO₄)₂):
- Atomic mass of calcium (Ca) = 40.08 g/mol
- Atomic mass of phosphorus (P) = 30.97 g/mol
- Atomic mass of oxygen (O) = 16.00 g/mol
The molar mass of Ca₃(PO₄)₂ is:
[
M_{\text{Ca}3(\text{PO}_4)_2} = 3 \times 40.08 \, \text{g/mol} + 2 \times (2 \times 30.97 \, \text{g/mol} + 8 \times 16.00 \, \text{g/mol}) ] [ M{\text{Ca}3(\text{PO}_4)_2} = 120.24 \, \text{g/mol} + 2 \times (61.94 \, \text{g/mol} + 128.00 \, \text{g/mol}) ] [ M{\text{Ca}_3(\text{PO}_4)_2} = 120.24 \, \text{g/mol} + 2 \times 189.94 \, \text{g/mol} = 120.24 \, \text{g/mol} + 379.88 \, \text{g/mol} = 500.12 \, \text{g/mol}
]
Now, use the number of moles of Ca₃(PO₄)₂ to calculate the mass:
[
\text{mass of Ca₃(PO₄)₂} = \text{moles of Ca₃(PO₄)₂} \times M_{\text{Ca}_3(\text{PO}_4)_2}
]
[
\text{mass of Ca₃(PO₄)₂} = 0.3232 \, \text{mol} \times 500.12 \, \text{g/mol}
]
[
\text{mass of Ca₃(PO₄)₂} = 161.22 \, \text{g}
]
Final Answer:
The mass of calcium phosphate (Ca₃(PO₄)₂) produced is 161.22 grams.
Explanation:
This calculation involves determining the amount of calcium phosphate produced from a given amount of phosphorus pentoxide (P₄O₁₀) in excess calcium oxide. We first used stoichiometric relationships from the balanced chemical equation, converting the given mass of P₄O₁₀ into moles and using the molar ratio between P₄O₁₀ and Ca₃(PO₄)₂ to determine the moles of calcium phosphate formed. Finally, we converted the moles of Ca₃(PO₄)₂ into mass using its molar mass. This approach ensures accurate calculation of the product’s mass in a chemical reaction.