What volume of 0.205 M K3PO4 solution is necessary to completely react with 130 mL of 0.0122 M NiCl2?

The equation for the reaction :
2K₃PO₄(aq) + 3NiCl₂(aq) → Ni₃(PO₄)₂(s) + 6KCl(aq)
Mole ratio K₃PO₄ : NiCl₂ = 2 : 3

No. of milli-moles of NiCl₂ reacted = (0.0122 mmol/mL) × (130 mL) = 1.586 mmol
No. of milli-moles of K₃PO₄ needed = (1.586 mmol) × (2/3) = 1.057 mmol
Volume of K₃PO₄ = (1.057 mmol) / (0.205 mmol/mL) = 5.16 mL