Answer
2.15 mol $NiCl_{2}$.
Work Step by Step
Balanced chemical reaction:
3 $NiCl_{2}$(aq) + 2 $Na_{3}PO_{4}$(aq) ${\longrightarrow}$ $Ni_{3}PO_{4}$ (s) + 6 NaCl (aq)
3 moles $NiCl_{2}$ give 1 mol $Ni_{3}PO_{4}$. To find out the number of moles of $NiCl_{2}$ that would produce 0.715 moles $Ni_{3}PO_{4}$, the conversion factor should be: $\frac{3_{mole (NiCl_{2})}}{1_{mol (Ni_{3}PO_{4})}}$.
So the number of moles of $NiCl_{2}$ needed to produce 0.715 moles $Ni_{3}PO_{4}$, is:
0.715 mol $Ni_{3}PO_{4}$ x $\frac{3_{mole (NiCl_{2})}}{1_{mol (Ni_{3}PO_{4})}}$ = 2.145 mol $NiCl_{2}$ or 2.15 mol $NiCl_{2}$.
