Final answer:
To completely react with 95.4 mL of 0.102 M CuCl2, approximately 37.2 mL of 0.175 M Na3PO4 solution is necessary.
Step-by-step explanation:
The given reaction is 2 Na3PO4(aq) + 3 CuCl2(aq) → Cu3(PO4)2(s) + 6 NaCl(aq). To determine the volume of 0.175 M Na3PO4 solution necessary to completely react with 95.4 mL of 0.102 M CuCl2, we can use the concept of stoichiometry. First, we need to determine the moles of CuCl2 present in the given volume:
Moles of CuCl2 = (Volume of CuCl2 solution in L) x (Molarity of CuCl2)
Moles of CuCl2 = (95.4 mL ÷ 1000 mL/L) x (0.102 mol/L)
Moles of CuCl2 = 0.00977 mol
Based on the balanced equation, the molar ratio between CuCl2 and Na3PO4 is 3:2. Therefore, the moles of Na3PO4 required can be calculated as:
Moles of Na3PO4 = (Moles of CuCl2) x (2 mol Na3PO4 ÷ 3 mol CuCl2)
Moles of Na3PO4 = 0.00977 mol x (2 ÷ 3)
Moles of Na3PO4 = 0.00651 mol
Finally, we can calculate the volume of the Na3PO4 solution using the molarity:
Volume of Na3PO4 solution = (Moles of Na3PO4) ÷ (Molarity of Na3PO4)
Volume of Na3PO4 solution = 0.00651 mol ÷ 0.175 mol/L
Volume of Na3PO4 solution = 0.0372 L = 37.2 mL