**Problem Statement:**
Determine the volume of 0.15 M Na3PO4 solution that will react with 250 mL of 0.15 M CaCl2 solution in the given chemical reaction: 2Na3PO4 (aq) + 3CaCl2 (aq) → Ca3(PO4)2 (aq) + 6NaCl (aq)

**Solution:**

To determine the volume of Na3PO4 solution that will react with the given volume of CaCl2 solution, we need to follow these steps:

1. Determine the moles of CaCl2 present in 250 mL of the 0.15 M solution.
2. Use the stoichiometry of the balanced chemical equation to determine the moles of Na3PO4 required to react with the moles of CaCl2.
3. Finally, calculate the volume of 0.15 M Na3PO4 solution required using the molarity and moles of Na3PO4.

**Step 1: Determine the moles of CaCl2**
Given:
Volume of CaCl2 solution = 250 mL
Molarity of CaCl2 solution = 0.15 M

Using the formula: moles = volume (L) × molarity (mol/L)

moles of CaCl2 = (250 mL ÷ 1000) L × 0.15 mol/L
= 0.0375 mol

**Step 2: Use stoichiometry to determine moles of Na3PO4**
From the balanced chemical equation, we can see that the stoichiometric ratio between CaCl2 and Na3PO4 is 3:2. This means that 3 moles of CaCl2 react with 2 moles of Na3PO4.

Using the ratio: moles of Na3PO4 = (moles of CaCl2 × 2) ÷ 3

moles of Na3PO4 = (0.0375 mol × 2) ÷ 3
= 0.025 mol

**Step 3: Calculate the volume of Na3PO4 solution**
Given:
Molarity of Na3PO4 solution = 0.15 M

Using the formula: moles = volume (L) × molarity (mol/L)

0.025 mol = volume (L) × 0.15 mol/L

volume (L) = 0.025 mol ÷ 0.15 mol/L
= 0.1667 L

Finally, convert the volume from liters to milliliters:

volume (mL) = 0.1667 L × 1000 mL/L
= 166.7 mL

Therefore, approximately 166.7 mL of 0.15 M Na3PO4 solution will react with 250 mL of 0.15 M CaCl2 solution in the given chemical reaction.

I think ans is 375ml