What volume of 0.181 M #Na_3PO_4# solution is necessary to completely react with 91.0 mL of 0.107 M #CuCl_2#?

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Answer 1 · 2016-03-10T22:24:59

#35.9"ml"#

Start with the balanced equation:

#3CuCl_(2(aq))+2Na_3PO_(4(aq))rarrCu_3(PO_4)_(2(s))+6NaCl(aq)#

This tells us that 3 moles of #CuCl_(2)# react with 2 moles #Na_3PO_4#.

#:.# 1 mole #CuCl_2# will react with 2/3 moles #Na_3PO_4#

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

#c=n/v#

#:.n=cxxv#

#:. nCuCl_2=0.107xx91.0/1000=9.737xx10^(-3)#

I divided by 1000 to convert #"ml"# to #"L"#

#:.nNa_3PO_4=9.737xx10^(-3)xx2/3=6.491xx10^(-3)#

#v=n/c=(6.491xx10^(-3))/(0.181)=35.86xx10^(-3)"L"#

#:.v=35.86"ml"#