The balanced chemical equation for the reaction given is:
[tex] 6Li (s) + N_{2}(g) --> 2Li_{3}N (s) [/tex]
Moles of [tex] N_{2} = 33.6524 g * \frac{1 mol}{28 g} = 1.20187 mol N_{2} [/tex]
Moles of Li = [tex] 58.7032 g Li * \frac{1 mol Li}{6.94 g Li} = 8.4587 mol Li [/tex]
Determining the limiting reagent: The reactant that gives least amount of the product will be the limiting reactant.
Mass of [tex] Li_{3}N from N_{2} = 1.20187 mol N_{2}*\frac{2 mol Li_{3}N}{1 mol N_{2}}*\frac{34.83 g Li_{3}N}{1 mol Li_{3}N} = 83.7223 g Li_{3}N [/tex]
Mass of [tex] Li_{3}N from Li = [/tex][tex] 8.4587 mol Li *\frac{2 mol Li_{3}N}{6 mol Li} *\frac{34.83 g Li_{3}N}{1mol Li_{3}N} = 98.2055 g Li_{3}N [/tex]
[tex] N_{2} [/tex] will be the limiting reactant and the mass of lithium nitride produced will be dependent on the mass of limiting reactant.
Therefore, mass of lithium nitride produced = 83.7223 g
