3\ mole\ MgCl_2\ react\ completely\
with 2 mole Na3P04with\ 2\ mole\ Na_3P0_4with 2 mole Na3P04
Moles of MgCl2 taken=.395×0.038=0.01501 molesMoles\ of\ MgCl_2\ taken=.395\times0.038=0.01501\ molesMoles of MgCl2 taken=.395×0.038=0.01501 moles
Volume of Na3PO4 required=0.015010.660×23=15.1 mlVolume\ of\ Na_3PO_4\ required=\frac{0.01501}{0.660}\times\frac{2}{3}=15.1\ mlVolume of Na3PO4 required=0.6600.01501×32=15.1 ml
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