posted by drwls .
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.