To solve the problem of how much calcium phosphate can be produced from 379.4 grams of calcium chloride when reacting with an excess of sodium phosphate, we'll follow these steps:
1. Write the balanced chemical equation for the reaction.
2. Calculate the moles of calcium chloride.
3. Determine the moles of calcium phosphate produced using stoichiometry.
4. Calculate the mass of calcium phosphate produced.
### Step 1: Balanced Chemical Equation
The balanced chemical equation for the reaction between calcium chloride (CaCl₂) and sodium phosphate (Na₃PO₄) is:
3 CaCl₂ + 2 Na₃PO₄ → Ca₃(PO₄)₂ + 6 NaCl
This equation tells us that 3 moles of CaCl₂ react to produce 1 mole of Ca₃(PO₄)₂.
### Step 2: Determine Moles of Calcium Chloride
The molar mass of calcium chloride (CaCl₂) is 110.98 g/mol. We have 379.4 grams of CaCl₂.
Moles of CaCl₂ = (Mass of CaCl₂) / (Molar Mass of CaCl₂)
= 379.4 g / 110.98 g/mol
= 3.4186 mol
### Step 3: Determine Moles of Calcium Phosphate
From the balanced equation, 3 moles of CaCl₂ produce 1 mole of Ca₃(PO₄)₂.
Moles of Ca₃(PO₄)₂ = (Moles of CaCl₂) / 3
= 3.4186 mol / 3
= 1.1395 mol
### Step 4: Calculate Mass of Calcium Phosphate
The molar mass of calcium phosphate (Ca₃(PO₄)₂) is 310.18 g/mol.
Mass of Ca₃(PO₄)₂ = (Moles of Ca₃(PO₄)₂) × (Molar Mass of Ca₃(PO₄)₂)
= 1.1395 mol × 310.18 g/mol
= 353.5 g
So, the mass of calcium phosphate that can be produced is approximately 353.5 grams. Therefore, the correct answer is:
A. 353.5 g
