Question

Chem help please!!!

Answer 1

The balanced chemical equation for the reaction given is:

$6Li (s) + N_{2}(g) --> 2Li_{3}N (s)$

Moles of $N_{2} = 33.6524 g * \frac{1 mol}{28 g} = 1.20187 mol N_{2}$

Moles of Li = $58.7032 g Li * \frac{1 mol Li}{6.94 g Li} = 8.4587 mol Li$

Determining the limiting reagent: The reactant that gives least amount of the product will be the limiting reactant.

Mass of $Li_{3}N from N_{2} = 1.20187 mol N_{2}*\frac{2 mol Li_{3}N}{1 mol N_{2}}*\frac{34.83 g Li_{3}N}{1 mol Li_{3}N} = 83.7223 g Li_{3}N$

Mass of $Li_{3}N from Li =$$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$

$N_{2}$ 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