Answer:
No precipitate is formed.
Explanation:
Hello,
In this case, given the dissociation reaction of magnesium fluoride:

And the undergoing chemical reaction:

We need to compute the yielded moles of magnesium fluoride, but first we need to identify the limiting reactant for which we compute the available moles of magnesium chloride:

Next, the moles of magnesium chloride consumed by the sodium fluoride:

Thus, less moles are consumed by the NaF, for which the moles of formed magnesium fluoride are:

Next, since the magnesium fluoride to magnesium and fluoride ions is in a 1:1 and 1:2 molar ratio, the concentrations of such ions are:
![[Mg^{2+}]=\frac{3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =3.75x10^{-4}M](https://tex.z-dn.net/?f=%5BMg%5E%7B2%2B%7D%5D%3D%5Cfrac%7B3x10%5E%7B-4%7DmolMg%5E%7B%2B2%7D%7D%7B%280.3%2B0.5%29L%7D%20%3D3.75x10%5E%7B-4%7DM)
![[F^-]=\frac{2*3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =7.5x10^{-4}M](https://tex.z-dn.net/?f=%5BF%5E-%5D%3D%5Cfrac%7B2%2A3x10%5E%7B-4%7DmolMg%5E%7B%2B2%7D%7D%7B%280.3%2B0.5%29L%7D%20%3D7.5x10%5E%7B-4%7DM)
Thereby, the reaction quotient is:

In such a way, since Q<Ksp we say that the ions tend to be formed, so no precipitate is formed.
Regards.