Answer
$37.1 \space mL$ of $0.102 \space M \space CuCl_2$
Work Step by Step
1. Determine the conversion factors.
- This $Na_3PO_4$ solution gives:
$\frac{0.175 \space mole \space Na_3PO_4}{1 \space L \space (soln. \space Na_3PO_4)}$ and $\frac{1 \space L \space (soln. \space Na_3PO_4)}{0.175 \space mole \space Na_3PO_4}$
- This $CuCl_2$ solution gives:
$\frac{0.102 \space mole \space CuCl_2}{1 \space L \space (soln. \space CuCl_2)}$ and $\frac{1 \space L \space (soln. \space CuCl_2)}{0.102 \space mole \space CuCl_2}$
- According to the balanced equation:
$\frac{2 \space moles \space Na_3PO_4}{3 \space moles \space CuCl_2}$ and $\frac{3 \space moles \space CuCl_2}{2 \space moles \space Na_3PO_4}$
2. Calculate the needed volume of $0.175 \space M$ $Na_3PO_4$.
$95.4 \space mL \space (soln. CuCl_2) \times \frac{0.102 \space mole \space CuCl_2}{1 \space L \space (soln. \space CuCl_2)} \times \frac{2 \space moles \space Na_3PO_4}{3 \space moles \space CuCl_2} \times \times \frac{1 \space L \space (soln. \space Na_3PO_4)}{0.175 \space mole \space Na_3PO_4} = 37.1 \space mL (soln. \space Na_3PO_4)$
