What volume of 0.205 M K3PO4 solution is necessary to completely react with 134 mL of 0.0102 M NiCl2?

the balanced equation for the above reaction is as follows;

3NiCl₂ + 2K₃PO₄ - - - > Ni₃ (PO₄) ₂ + 6KCl

stoichiometry of NiCl₂ to K₃PO₄ is 3:2

the number of NiCl₂ moles reacted - 0.0102 mol/L x 0.134 L = 0.00137 mol

according to molar ratio

if 3 mol of NiCl₂ reacts with 2 mol of K₃PO₄

then 0.00137 mol of NiCl₂ reacts with - 2/3 x 0.00137 = 0.000911 mol of K₃PO₄

molarity of given K₃PO₄ solution - 0.205 M

there are 0.205 mol in 1000 mL

therefore volume of 0.000911 mol - 0.000911 mol / 0.205 mol/L = 4.44 mL

volume of K₃PO₄ needed is 4.44 mL