Question

What mass of copper(II) chloride is required to prepare sodium chloride from a `3.65*g` mass of sodium metal?

Answer 1

Approx. `4*g` of `"copper salt"` are required.......

Explanation:

We have a molar quantity of `(3.65*g)/(58.44*g*mol^-1)=0.0625*mol` with respect to `NaCl`.

There is thus a molar quantity of `0.0625*mol` with respect to the chloride ion? Agreed.

And thus we need a `1/2*"equivalent"` of `"cupric chloride"`, because by composition, each mole of `CuCl_2` supplies 2 moles of chloride ion...........

So thus we need `1/2xx0.0625*molxx134.45*g*mol^-1`

`=4.20*g` `CuCl_2` to supply the chloride ions.