6.69 g.
Make sure the chemical equation is balanced:
[tex]6\; \text{Li}\; (s) + \text{N}_2 \; (g) \to 2\; \text{Li}_3\text{N}\; (s)[/tex]
Assuming that one mole of N₂ is consumed. How many grams of each reactant will that take? Refer to a periodic table for data on atomic mass.
4.00 grams of each reactant are available. How many moles of N₂ can they consume?
However, only 0.09606 moles of the reaction is possible, since Li would have ran out before all 4.00 grams of N₂ are used up.
Each mole of the reaction makes 2 moles of Li₃N. 0.09606 moles of the reaction will produce 0.09606 × 2 × (3 × 6.94 + 14.01) = 6.69 grams of Li₃N.