Answer:
22.2 g of MgF₂
Explanation:
We star from the reaction:
NaF → sodium fluoride
Mg → Magnesium
2NaF + Mg → MgF₂ + 2Na
2 moles of sodium fluoride react to 1 mol of Mg in order to produce 1 mol of magnesium fluoride and 2 moles of sodium.
We convert mass of the reactant to moles:
30 g . 1mol / 41.98 g = 0.715 mol
We assume that Mg is in excess
As ratio is 2:1, if we have 0.715 moles of NaF we may produce the half of moles, of MgF₂
0.715 mol /2 = 0.357 moles of MgF₂
We convert moles to mass: 0.357 mol . 62.30g /mol = 22.2 g
