Consider the precipitation reaction: 2 Na3PO4(aq) + 3 CuCl2(aq) → Cu3(PO4)2(s) + 6 NaCl(aq). What volume of 0.175 M Na3PO4 solution is necessary to completely react with 95.4 mL of 0.102 M CuCl2?

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1. Identify the balanced chemical equation: $2{\text{Na}}_3{\text{PO}}_4(aq)+3{\text{CuCl}}_2(aq)\to {\text{Cu}}_3({\text{PO}}_4{)}_2(s)+6\text{NaCl}(aq)$.
2. Determine the moles of ${\text{CuCl}}_2$ using its concentration and volume: $\text{moles of }{\text{CuCl}}_2=0.102\text{ M}\times 0.0954\text{ L}$.
3. Use the stoichiometry of the reaction to find the moles of ${\text{Na}}_3{\text{PO}}_4$ needed. According to the balanced equation, 2 moles of ${\text{Na}}_3{\text{PO}}_4$ react with 3 moles of ${\text{CuCl}}_2$.
4. Calculate the moles of ${\text{Na}}_3{\text{PO}}_4$ required using the ratio from the balanced equation: $\text{moles of }{\text{Na}}_3{\text{PO}}_4=\frac{2}{3}\times \text{moles of }{\text{CuCl}}_2$.
5. Determine the volume of $0.175\text{ M }{\text{Na}}_3{\text{PO}}_4$ solution needed using its concentration: $\text{Volume}=\frac{\text{moles of }{\text{Na}}_3{\text{PO}}_4}{0.175\text{ M}}$.