First of all, you need to know how many g of Mg actually reacted in the first reaction. You can determine this by determining the limiting reactant.

The reaction is
Mg + F2 -> MgF2

16.2 g of Mg is 16.2/24.32 = 0.666 moles
25.3 g of F2 is 25.3/38.0 = 0.666 moles

So equal numbers of moles are consumed. There is no limiting reactant in this case. I should have realized this wehen the problems stated that the Mg reacts exactly with 25.3 g.

10.5 g is 64.8% of 16.2. When you have excess fluorine, you will get 64.8% as much product as you did in the first (complete) reaction. That would be 0.648 x 0.666 = 0.432 moles of MgF2. Each mole of MgF2 has a mass of 62.32 g. That means you make 26.9 g of MgF2.

I quicker way to get the answer is:
MgF2 formed = (10.5/16.2)(16.2 + 25.3), which is 64.8% of the total mass of reactants in the first (complete) reaction.

To find the amount of MgF2 formed in the second reaction, where 10.5 g of Mg is used, you can use the concept of percent yield.

First, calculate the percent yield by dividing the mass of the actual yield (10.5 g) by the theoretical yield (which you determined in the complete reaction) and multiplying by 100:

Percent yield = (10.5 g / (16.2 g + 25.3 g)) * 100

Next, multiply the percent yield by the theoretical yield of the complete reaction, which you found to be 0.666 moles or 26.9 g of MgF2:

MgF2 formed = (Percent yield / 100) * 26.9 g

Simplifying the expression, you can also calculate it directly:

MgF2 formed = (10.5 / 16.2) * (16.2 + 25.3)

So, the amount of MgF2 formed in the second reaction is 26.9 g (or you can calculate it as 0.648 * (16.2 + 25.3) = 26.9 g)