Answer
37.1 mL of 0.175 M $Na_3PO_4$ are necessary to react with 95.4 mL of $0.102 $ M $CuCl_2$
Work Step by Step
1. Use the molarities and other information as conversion factors:
$$ 95.4 \space mL \space CuCl_2 \times \frac{1 \space L}{1000 \space mL} \times \frac{ 0.102 \space mol \space CuCl_2 }{1 \space L \space CuCl_2 } \times \frac{ 2 \space mol \space Na_3PO_4 }{ 3 \space mol \space CuCl_2 } \times \frac{1 \space L}{ 0.175 \space mol \space Na_3PO_4 } \times \frac{1000 \space mL}{1 \space L} = 37.1 \space mL $$
