Answer to Question #95058 in General Chemistry for Zabiba

Question #95058
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)
1
Expert's answer
2019-09-23T03:47:14-0400

3\ mole\ MgCl_2\ react\ completely\

with 2 mole Na3P04with\ 2\ mole\ Na_3P0_4

Moles of MgCl2 taken=.395×0.038=0.01501 molesMoles\ of\ MgCl_2\ taken=.395\times0.038=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\ ml


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