Considering the following equation and that Fluorine is in excess to allow all the molecular nitrogen to react
N₂ + 3F₂ → 2NF₃
We will use the molar mass and reaction stoichiometry to determine the grams of NF₃ produced, using the following calculation:
[tex]10.9\text{ g N}_2\frac{1\text{ mol N}_2}{14.0067\text{ g N}_2}\frac{2\text{ mol NF}_3}{1\text{ mol N}_2}=1.5564\text{ mol NF}_3[/tex]