Question
Asked Oct 10, 2019
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Consider the following precipitation reaction: 2Na3PO4(aq)+3CuCl2(aq)→ Cu3(PO4)2(s)+6NaCl(aq)

What volume of 0.188 M Na3PO4 solution is necessary to completely react with 86.7 mL of 0.109 M CuCl2?

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Expert Answer

Step 1

The precipitation reaction is given below.

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Step 2

Given data:

Molarity of Na3PO4 is 0.188 M. Volume of CuCl2 is 86.7 ...

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Actual moles of CuCl, = Molarity x Volume 1L =0.109 mol x86.7 mL x- L 1000 mL 9.4503 x103 mol. Hence the actual moles of Na3PO4 can be calculated as given below 2 -x9.4503 x 10mol = 6.3002x10 mol. 3 Actual moles of Na,PO

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