Question

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`?

Answer 1

`35.9"ml"`

Explanation:

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"`