The balanced chemical equation for the reaction is:
3NiCl2(aq) + 2Na3PO4(aq) → Ni3(PO4)2(s) + 6NaCl(aq)
From the equation, we can see that 3 moles of NiCl2 react with 1 mole of Ni3(PO4)2.
We can use the following steps to calculate the volume of 0.244 M NiCl2 solution required to produce 65.9 g of Ni3(PO4)2 precipitate:
1. Calculate the moles of Ni3(PO4)2 precipitate produced:
molar mass of Ni3(PO4)2 = 380.61 g/mol
moles of Ni3(PO4)2 = mass / molar mass = 65.9 g / 380.61 g/mol = 0.173 moles
2. Calculate the moles of NiCl2 required to produce 0.173 moles of Ni3(PO4)2:
moles of NiCl2 = (1/3) x moles of Ni3(PO4)2 = (1/3) x 0.173 moles = 0.058 moles
3. Calculate the volume of 0.244 M NiCl2 solution containing 0.058 moles of NiCl2:
molarity = moles of solute / volume of solution in liters
volume of solution in liters = moles of solute / molarity
volume of 0.244 M NiCl2 solution = 0.058 moles / 0.244 mol/L = 0.238 L
Therefore, the volume of 0.244 M NiCl2 solution required to produce 65.9 g of Ni3(PO4)2 precipitate is 0.238 L.