Question

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)

Answer 1

3\ mole\ MgCl_2\ react\ completely\

$with\ 2\ mole\ Na_3P0_4$

$Moles\ of\ MgCl_2\ taken=.395\times0.038=0.01501\ moles$

$Volume\ of\ Na_3PO_4\ required=\frac{0.01501}{0.660}\times\frac{2}{3}=15.1\ ml$