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What volume of 0.194 MNa3PO4 solution is necessary to completely react with 85.5 mL of 0.109 MCuCl2 ?

Sagot :

Answer:

35.9 ml

Explanation:

Start with the balanced equation:

3CuCl2(aq)+2Na3PO4(aq)→Cu3(PO4)2(s)+6NaCl(aq)

This tells us that 3 moles of CuCI2 react with 2 moles Na3PO4-

∴  1 mole CuCl2 will react with 2/3 moles Na3PO4

We know that concentration = moles/volume i.e:

c= n/v

∴n=c×v

∴nCuCl2=0.107×91.01000=9.737×10−3

I divided by 1000 to convert ml to L

∴nNa3PO4=9.737×10−3×23=6.491×10−3

v=nc=6.491×10−30.181=35.86×10−3L

∴v=35.86ml

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