Since we have sufficient sodium fluoride (NaF), the magnesium (Mg) will be our limiting reactant, and we will use its coefficient in the balanced equation to determine our molar ratio between the product of interest which, in this case, is sodium (Na).
We are given the mass of magnesium, 23.56 grams, which we must convert to moles by dividing the mass by the molar mass of magnesium: (23.56 g)/(24.305 g/mol) = 0.9693 mol Mg.
According to the balanced equation given, the molar ratio between sodium and magnesium is 2:1. That is, for every 2 moles of Na that are produced, one mole of Mg is consumed or reacted. The upshot is that the number of moles of sodium produced will be double the number of moles of Mg reacted. Since we have 0.9693 moles of Mg available to react, we will obtain (0.9693 mol Mg)(2 mol Na/1 mol Mg) = 1.939 mol Na.
Finally, we multiply our moles of sodium by the molar mass of sodium to get the mass: (1.939 mol Na)(22.99 g/mol) = 44.57 grams Na.
So, 44.57 grams of sodium will be produced.
The answer is given to four significant figures.
