Using the following equation, calculate the volume (in mL) of 0.660 M Na3PO4 that will react with 38.0 mL of 0.395 M MgCl2.
3 MgCl2(aq) + 2 Na3PO4(aq) → Mg3(PO4)2(s) + 6 NaCl2(aq)
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